Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A)

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.2(A) Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A)

I. Integrate the following functions w.r.t. x:

Question 1.
\(\frac{(\log x)^{n}}{x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q1

Question 2.
\(\frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^{2}}}\)
Solution:
Let I = \(\int \frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^{2}}} d x\)
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q2

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A)

Question 3.
\(\frac{1+x}{x \cdot \sin (x+\log x)}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q3

Question 4.
\(\frac{x \cdot \sec ^{2}\left(x^{2}\right)}{\sqrt{\tan ^{3}\left(x^{2}\right)}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q4

Question 5.
\(\frac{e^{3 x}}{e^{3 x}+1}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q5
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q5.1

Question 6.
\(\frac{\left(x^{2}+2\right)}{\left(x^{2}+1\right)} \cdot a^{x+\tan ^{-1} x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q6

Question 7.
\(\frac{e^{x} \cdot \log \left(\sin e^{x}\right)}{\tan \left(e^{x}\right)}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q7

Question 8.
\(\frac{e^{2 x}+1}{e^{2 x}-1}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q8
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q8.1

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A)

Question 9.
sin4x . cos3x
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q9

Question 10.
\(\frac{1}{4 x+5 x^{-11}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q10

Question 11.
x9 . sec2(x10)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q11

Question 12.
\(e^{3 \log x} \cdot\left(x^{4}+1\right)^{-1}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q12

Question 13.
\(\frac{\sqrt{\tan x}}{\sin x \cdot \cos x}\)
Solution:
Let I = \(\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x\)
Dividing numerator and denominator by cos2x, we get
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q13

Question 14.
\(\frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q14
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q14.1

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A)

Question 15.
\(\frac{2 \sin x \cos x}{3 \cos ^{2} x+4 \sin ^{2} x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q15

Question 16.
\(\frac{1}{\sqrt{x}+\sqrt{x^{3}}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q16
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q16.1

Question 17.
\(\frac{10 x^{9}+10^{x} \cdot \log 10}{10^{x}+x^{10}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q17

Question 18.
\(\frac{x^{n-1}}{\sqrt{1+4 x^{n}}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q18

Question 19.
(2x + 1) \(\sqrt{x+2}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q19

Question 20.
\(x^{5} \sqrt{a^{2}+x^{2}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q20

Question 21.
\((5-3 x)(2-3 x)^{-\frac{1}{2}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q21

Question 22.
\(\frac{7+4 x+5 x^{2}}{(2 x+3)^{\frac{3}{2}}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q22
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q22.1

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A)

Question 23.
\(\frac{x^{2}}{\sqrt{9-x^{6}}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q23
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q23.1

Question 24.
\(\frac{1}{x\left(x^{3}-1\right)}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q24
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q24.1
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q24.2

Question 25.
\(\frac{1}{x \cdot \log x \cdot \log (\log x)}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) I Q25

II. Integrate the following functions w.r.t x:

Question 1.
\(\frac{\cos 3 x-\cos 4 x}{\sin 3 x+\sin 4 x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q1

Question 2.
\(\frac{\cos x}{\sin (x-a)}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q2

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A)

Question 3.
\(\frac{\sin (x-a)}{\cos (x+b)}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q3

Question 4.
\(\frac{1}{\sin x \cdot \cos x+2 \cos ^{2} x}\)
Solution:
Let I = \(\int \frac{1}{\sin x \cdot \cos x+2 \cos ^{2} x} d x\)
Dividing numerator and denominator of cos2x, we get
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q4

Question 5.
\(\frac{\sin x+2 \cos x}{3 \sin x+4 \cos x}\)
Solution:
Let I = \(\int \frac{\sin x+2 \cos x}{3 \sin x+4 \cos x} d x\)
Put, Numerator = A (Denominator) + B [\(\frac{d}{d x}\) (Denominator)]
∴ sin x+ 2 cos x = A(3 sin x + 4 cos x) + B [\(\frac{d}{d x}\) (3 sin x + 4 cos x)]
= A(3 sin x + 4 cos x) + B (3 cos x – 4 sin x)
∴ sin x + 2 cos x = (3A – 4B) sin x + (4A + 3B) cos x
Equating the coefficients of sin x and cos x on both the sides, we get
3A – 4B = 1 …… (1)
and 4A + 3B = 2 …… (2)
Multiplying equation (1) by 3 and equation (2) by 4, we get
9A – 12B = 3
16A + 12B = 8
On adding, we get
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q5

Question 6.
\(\frac{1}{2+3 \tan x}\)
Solution:
Let I = \(\int \frac{1}{2+3 \tan x} d x\)
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q6
Numerator = A (Denominator) + B [\(\frac{d}{d x}\) (Denominator)]
∴ cos x = A(2 cos x + 3 sin x) + B [\(\frac{d}{d x}\) (2 cos x + 3 sin x)]
= A (2 cos x + 3 sin x) + B (-2 sin x + 3 cos x)
∴ cos x = (2A + 3B) cos x + (3A – 2B) sin x
Equating the coefficients of cosx and sinx on both the sides, we get
2A + 3B = 1 …… (1)
and 3A – 2B = 0 ……. (2)
Multiplying equation (1) by 2 and equation (2) by 3, we get
4A + 6B = 2
9A – 6B = 0
On adding, we get
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q6.1

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A)

Question 7.
\(\frac{4 e^{x}-25}{2 e^{x}-5}\)
Solution:
Let I = \(\int \frac{4 e^{x}-25}{2 e^{x}-5} d x\)
Put, Numerator = A (Denominator) + B [\(\frac{d}{d x}\) (Denominator)]
∴ 4ex – 25 = A(2ex – 5) + B[\(\frac{d}{d x}\) (2ex – 5)]
= A(2ex – 5) + B(2ex – 0)
∴ 4ex – 25 = (2A + 2B) ex – 5A
Equating the coefficient of ex and constant on both sides, we get
2A + 2B = 4 …….(1)
and 5A = 25
∴ A = 5
from (1), 2(5) + 2B = 4
∴ 2B = -6
∴ B = -3
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q7

Question 8.
\(\frac{20+12 e^{x}}{3 e^{x}+4}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q8
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q8.1

Question 9.
\(\frac{3 e^{2 x}+5}{4 e^{2 x}-5}\)
Solution:
Let I = \(\int \frac{3 e^{2 x}+5}{4 e^{2 x}-5} d x\)
Put, Numerator = A (Denominator) + B [\(\frac{d}{d x}\) (Denominator)]
∴ 3e2x + 5 = A(4e2x – 5) + B [\(\frac{d}{d x}\) (4e2x – 5)]
= A(4e2x – 5) + B(4 . e2x × 2 – 0)
∴ 3e2x + 5 = (4A + 8B) e2x – 5A
Equating the coefficient of e2x and constant on both sides, we get
4A + 8B = 3 …….. (1)
and -5A = 5
∴ A = -1
∴ from (1), 4(-1) + 8B = 3
∴ 8B = 7
∴ B = \(\frac{7}{8}\)
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q9

Question 10.
cos8 x . cot x
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q10

Question 11.
tan5x
Solution:
Let I = ∫ tan5x dx
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q11

Question 12.
cos7x
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q12

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A)

Question 13.
tan 3x tan 2x tan x
Solution:
Let I = ∫ tan 3x tan 2x tan x dx
Consider tan 3x = tan (2x + x) = \(\frac{\tan 2 x+\tan x}{1-\tan 2 x \tan x}\)
tan 3x (1 – tan 2x tan x) = tan 2x + tan x
tan 3x – tan 3x tan 2x tan x = tan 2x + tan x
tan 3x – tan 2x – tan x = tan 3x tan 2x tan x
I = ∫(tan 3x – tan 2x – tan x) dx
= ∫tan3x dx – ∫tan 2x dx – ∫tan x dx
= \(\frac{1}{3}\) log | sec 3x| – \(\frac{1}{2}\) log |sec 2x| – log |sec x| + c.

Question 14.
sin5x cos8x
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q14

Question 15.
\(3^{\cos ^{2} x \cdot} \sin 2 x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q15

Question 16.
\(\frac{\sin 6 x}{\sin 10 x \sin 4 x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q16

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A)

Question 17.
\(\frac{\sin x \cos ^{3} x}{1+\cos ^{2} x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q17
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(A) II Q17.1

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

I. Choose the correct option from the given alternatives:

Question 1.
If the function f(x) = ax3 + bx2 + 11x – 6 satisfies conditions of Rolle’s theorem in [1, 3] and f'(2 + \(\frac{1}{\sqrt{3}}\)) = 0, then values of a and b are respectively.
(a) 1, -6
(b) -2, 1
(c) -1, -6
(d) -1, 6
Answer:
(a) 1, -6

Hint: f(x) = ax3 + bx2 + 11x – 6 satisfies the conditions of Rolle’s theorem in [1, 3]
∴ f(1) = f(3)
a(1)3 + b(1)2 + 11(1) – 6 = a(3)3 + b(3)2 + 11(3) – 6
a + b + 11 = 27a + 9b + 33
26a + 8b = -22
13a + 4b = -11
Only a = 1, b = -6 satisfy this equation.

Question 2.
If f(x) = \(\frac{x^{2}-1}{x^{2}+1}\), for every real x, then the minimum value of f is
(a) 1
(b) 0
(c) -1
(d) 2
Answer:
(c) -1

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

Question 3.
A ladder 5 m in length is resting against a vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of 1.5 m/sec. The length of the higher point of the ladder when the foot of the ladder is 4.0 m away from the wall decreases at the rate of
(a) 1
(b) 2
(c) 2.5
(d) 3
Answer:
(b) 2

Question 4.
Let f(x) and g(x) be differentiable for 0 < x < 1 such that f(0) = 0, g(0) = 0, f(1) = 6. Let there exist a real number c in (0, 1) such that f'(c) = 2g'(c), then the value of g(1) must be
(a) 1
(b) 3
(c) 2.5
(d) -1
Answer:
(b) 3

Hint: f(x) and g(x) both satisfies the conditions of LMVT in (0, 1).
∴ f'(c) = \(\frac{f(1)-f(0)}{1-0}=\frac{6-0}{1}=6\)
and g'(c) = \(\frac{g(1)-g(0)}{1-0}=\frac{g(1)-0}{1}\) = g(1)
But f'(c) = 2g'(c)
6 = 2g(1)
∴ g(1) = 3

Question 5.
Let f(x) = x3 – 6x2 + 9x + 18, then f(x) is strictly decreasing in
(a) (-∞, 1)
(b) [3, ∞)
(c) (-∞, 1] ∪ [3, ∞)
(d) (1, 3)
Answer:
(d) (1, 3)

Question 6.
If x = -1 and x = 2 are the extreme points of y = α log x + βx2 + x, then
(a) α = -6, β = \(\frac{1}{2}\)
(b) α = -6, β = \(\frac{-1}{2}\)
(c) α = 2, β = \(\frac{-1}{2}\)
(d) α = 2, β = \(\frac{1}{2}\)
Answer:
(c) α = 2, β = \(\frac{-1}{2}\)

Hint: y = α log x + βx2 + x
∴ \(\frac{d y}{d x}=\frac{\alpha}{x}+\beta \times 2 x+1=\frac{\alpha}{x}+2 \beta x+1\)
f(x) has extreme values at x = -1 and x = 2
∴ f'(-1) = 0 and f(2) = 0
α + 2β = 1
and \(\frac{\alpha}{2}\) + 4β = -1
By solving these two equations, we get
α = 2, β = \(\frac{-1}{2}\)

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

Question 7.
The normal to the curve x2 + 2xy – 3y2 = 0 at (1, 1)
(a) meets the curve again in the second quadrant
(b) does not meet the curve again
(c) meets the curve again in the third quadrant
(d) meets the curve again in the fourth quadrant
Answer:
(d) meets the curve again in fourth quadrant

Hint: x2 + 2xy – 3y2 = 0
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 I Q7
= slope of the tangent at (1, 1)
∴ equation of the tangent at (1, 1) is -1
∴ equation of the normal is
y – 1= -1 (x – 1) = -x + 1
∴ x + y = 2
∴ y = 2 – x
Substituting y = 2 – x in x2 + 2xy – 3y2 = 0, we get
x2 + 2x(2 – x) – 3 (2 – x)2 = 0
⇒ x2 + 4x – 2x2 – 3(4 – 4x + x2) = 0
⇒ x2 – 4x + 3 = 0
⇒ (x – 1)(x – 3) = 0
⇒ x = 1, x = 3
When x = 1, y = 2 – 1 = 1
When x = 3, y = 2 – 3 = -1
∴ the normal at (1, 1) meets the curve at (3, -1) which is in the fourth quadrant.

Question 8.
The equation of the tangent to the curve y = 1 – \(e^{\frac{x}{2}}\) at the point of intersection with Y-axis is
(a) x + 2y = 0
(b) 2x + y = 0
(c) x – y = 2
(d) x + y = 2
Answer:
(a) x + 2y = 0
Hint: The point of intersection of the curve with the Y-axis is the origin (0, 0).

Question 9.
If the tangent at (1, 1) on y2 = x(2 – x)2 meets the curve again at P, then P is
(a) (4, 4)
(b) (-1, 2)
(c) (3, 6)
(d) \(\left(\frac{9}{4}, \frac{3}{8}\right)\)
Answer:
(d) \(\left(\frac{9}{4}, \frac{3}{8}\right)\)
Hint: y2 = x(2 – x)2
= x(4 – 4x + x2)
= x3 – 4x2 + 4x
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 I Q9
= slope of the tangent at (1, 1)
∴ equation of the tangent at (1, 1) is
y – 1 = –\(\frac{1}{2}\) (x – 1)
∴ 2y – 2 = -x + 1
∴ x + 2y = 3
Only the coordinates \(\left(\frac{9}{4}, \frac{3}{8}\right)\) satisfy both the equations y2 = x(2 – x)2 and x + 2y = 3
∴ P is \(\left(\frac{9}{4}, \frac{3}{8}\right)\)

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

Question 10.
The approximate value of tan (44° 30′), given that 1° = 0.0175, is
(a) 0.8952
(b) 0.9528
(c) 0.9285
(d) 0.9825
Answer:
(d) 0.9825

II. Solve the following:

Question 1.
If the curves ax2 + by2 = 1 and a’x2 + b’y2 = 1, intersect orthogonally, then prove that \(\frac{1}{a}-\frac{1}{b}=\frac{1}{a^{\prime}}-\frac{1}{b^{\prime}}\)
Solution:
Let P(x1, y1) be the point of intersection of the curves.
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q1
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q1.1

Question 2.
Determine the area of the triangle formed by the tangent to the graph of the function y = 3 – x2 drawn at the point (1, 2) and the coordinate axes.
Solution:
y = 3 – x2
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q2
= slope of the tangent at (1, 2)
∴ equation of the tangent at (1, 2) is
y – 2= -2(x – 1)
⇒ y – 2= -2x + 2
⇒ 2x + y = 4
Let this tangent cuts the coordinate axes at A(a, 0) and B(0, b).
∴ 2a + 0 = 4 and 2(0) + b = 4
∴ a = 2 and b = 4
∴ area of required triangle = \(\frac{1}{2}\) × l(OA) × l(OB)
= \(\frac{1}{2}\) ab
= \(\frac{1}{2}\) (2)(4)
= 4 sq units.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

Question 3.
Find the equation of the tangent and normal drawn to the curve y4 – 4x4 – 6xy = 0 at the point M (1, 2).
Solution:
y4 – 4x4 – 6xy = 0
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q3
= slope of the tangent at (1, 2)
∴ the equation of normal at M (1, 2) is
y – 2 = \(\frac{14}{13}\) (x – 1)
∴ 13y – 26 = 14x – 14
∴ 14x – 13y + 12 = 0
The slope of normal at (1, 2)
\(=\frac{-1}{\left(\frac{d y}{d x}\right)_{\mathrm{at}(1,2)}}=\frac{-1}{\left(\frac{14}{13}\right)}=-\frac{13}{14}\)
∴ the equation of normal at M (1, 2) is
y – 2 = \(\frac{-13}{14}\) (x – 1)
14y – 28 = -13x + 13
13x + 14y – 41 = 0
Hence, the equations of tangent and normal are 14x – 13y + 12 = 0 and 13x + 14y – 41 = 0 respectively.

Question 4.
A water tank in the form of an inverted cone is being emptied at the rate of 2 cubic feet per second. The height of the cone is 8 feet and the radius is 4 feet. Find the rate of change of the water level when the depth is 6 feet.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q4
Let r be the radius of the base, h be the height and V be the volume of the water level at any time t.
Since, the height of the cone is 8 feet and the radius is 4 feet,
\(\frac{r}{h}=\frac{4}{8}=\frac{1}{2}\)
r = \(\frac{h}{2}\) ……..(1)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q4.1
Hence, the rate of change of water level is \(\left(\frac{2}{9 \pi}\right)\) ft/sec.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

Question 5.
Find all points on the ellipse 9x2 + 16y2 = 400, at which the y-coordinate is decreasing and the x-coordinate is increasing at the same rate.
Solution:
Let P(x1, y1) be the point on the ellipse 9x2 + 16y2 = 400 whose y-coordinate decreasing x-coordinate is increasing at the same rate.
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q5
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q5.1

Question 6.
Verify Rolle’s theorem for the function f(x) = \(\frac{2}{e^{x}+e^{-x}}\) on [-1, 1].
Solution:
The functions ex, e-x, and 2 are continuous and differentiable in their respective domains.
∴ f(x) = \(\frac{2}{e^{x}+e^{-x}}\) is continuous on [-1, 1] and differentiable on (-1, 1), because ex + e-x ≠ 0 for all x ∈ [-1, 1].
Now, f(-1) = \(\frac{2}{e^{-1}+e}=\frac{2}{e+e^{-1}}\) and f(1) = \(\frac{2}{e+e^{-1}}\)
∴ f(-1) = f(1)
Thus, the function f satisfies all the conditions of the Rolle’s theorem.
∴ there exist c ∈ (-1, 1) such that f'(c) = 0
Now, f(x) = \(\frac{2}{e^{x}+e^{-x}}\)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q6
Hence, Rolle’s theorem is verified.

Question 7.
The position of a particle is given by the function s(t) = 2t2 + 3t – 4. Find the time t = c in the interval 0 ≤ f ≤ 4 when the instantaneous velocity of the particle is equal to its average velocity in this interval.
Solution:
s(t) = 2t2 + 3t – 4
∴ s(0) = 2(0)2 + 3(0) – 4 = -4
and s(4) = 2(4)2 + 3(4) – 4 = 32 + 12 – 4 = 40
∴ average velocity = \(\frac{s(4)-s(0)}{4-0}\)
= \(\frac{40-(-4)}{4}\)
= 11
Also, instantaneous velocity = \(\frac{d s}{d t}\)
= \(\frac{d}{d t}\) (2t2 + 3t – 4)
= 2 × 2t + 3 × 1 – 0
= 4t + 3
∴ instantaneous velocity at t = c is \(\left(\frac{d s}{d t}\right)_{t=c}\) = 4c + 3
When instantaneous velocity at t = c equal to its average velocity, we get
4c + 3 = 11
4c = 8
∴ c = 2 ∈ [0, 4]
Hence, t = c = 2.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

Question 8.
Find the approximate value of the function f(x) = \(\sqrt{x^{2}+3 x}\) at x = 1.02.
Solution:
f(x) = \(\sqrt{x^{2}+3 x}\)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q8

Question 9.
Find the approximate value of cos-1(0.51), given π = 3.1416, \(\frac{2}{\sqrt{3}}\) = 1.1547.
Solution:
Let f(x) = cos-1 x
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q9
The formula for approximation is f(a + h)= f(a) + h . f'(a)
∴ cos-1 (0.51) = f(0.51)
= f(0.5 + 0.01)
= f(0.5) + (0.01) f'(0.5)
= \(\frac{\pi}{3}\) + 0.01 × (-1.1547)
= \(\frac{3.1416}{3}\) – 0.011547
= 1.0472 – 0.011547
= 1.035653
∴ cos-1 (0.51) = 1.035653.

Question 10.
Find the intervals on which the function y = xx, (x > 0) is increasing and decreasing.
Solution:
y = xx
∴ log y = log xx = x log x
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q10
y is increasing if \(\frac{d y}{d x}\) ≥ 0
i.e. if xx (1 + log x) ≥ 0
i.e. if 1 + log x ≥ 0 ……[∵ x > 0]
i.e. if log x ≥ -1
i.e. if log x ≥ -log e …….[∵ log e = 1]
i.e. if logx ≥ log \(\frac{1}{e}\)
i.e. if x ≥ \(\frac{1}{e}\)
∴ y is increasing in \(\left[\frac{1}{e^{\prime}}, \infty\right)\)
y is decreasing if \(\frac{d y}{d x}\) ≤ 0
i.e. if xx (1 + log x) ≤ 0
i.e. if 1 + log x ≤ 0 ……[∵ x > 0]
i.e. if log x ≤ -1
i.e. if log x ≤ -log e
i.e. if log x ≤ log \(\frac{1}{e}\)
i.e. if x ≤ \(\frac{1}{e}\) where x > 0
∴ y is decreasing is \(\left(0, \frac{1}{e}\right]\)
Hence, the given function is increasing in \(\left[\frac{1}{e^{\prime}}, \infty\right)\) and decreasing in \(\left(0, \frac{1}{e}\right]\)

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

Question 11.
Find the intervals on which the function f(x) = \(\frac{x}{\log x}\) is increasing and decreasing.
Solution:
f(x) = \(\frac{x}{\log x}\)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q11
f is increasing if f'(x) ≥ 0
i.e. if \(\frac{\log x-1}{(\log x)^{2}}\) ≥ 0
i.e. if log x – 1 ≥ 0 ……..[∵ (log x)2 > 0]
i.e. if log x ≥ 1
i.e. if log x ≥ log e ………[∵ log e = 1]
i.e. if x ≥ e
∴ f is increasing on [e, ∞)
f is decreasing if f'(x) ≤ 0
i.e. if \(\frac{\log x-1}{(\log x)^{2}}\) ≤ 0
i.e. if log x – 1 ≤ 0 ……..[∵ (log x)2 > 0]
i.e. if log x ≤ 1
i.e. if log x ≤ log e
i.e. if x ≤ e
Also, x > 0 and x ≠ 1 because f(x) = \(\frac{x}{\log x}\) is not defined at x = 1.
∴ f is decreasing in (0, e] – {1}
Hence, f is increasing in [e, ∞) and decreasing in (0, e] – {1}.

Question 12.
An open box with a square base is to be made out of the given quantity of sheet of area a2. Show that the maximum volume of the box is \(\frac{a^{3}}{6 \sqrt{3}}\).
Solution:
Let x be the side of square base and h be the height of the box.
Then x2 + 4xh = a2
∴ h = \(\frac{a^{2}-x^{2}}{4 x}\) …….(1)
Let V be the volume of the box.
Then V = x2h
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q12
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q12.1
Hence, the maximum volume of the box is \(\frac{a^{3}}{6 \sqrt{3}}\) cu units.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

Question 13.
Show that of all rectangles inscribed in a given circle, the square has the maximum area.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q13
Let ABCD be a rectangle inscribed in a circle of radius r.
Let AB = x and BC = y.
Then x2 + y2 = 4r2 …….(1)
Area of rectangle = xy
= \(x \sqrt{4 r^{2}-x^{2}}\) ……[By (1)]
Let f(x) = x2(4r2 – x2)
= 4r2x2 – x4
∴ f'(x) = \(\frac{d}{d x}\) (4r2x2 – x4)
= 4r2 × 2x – 4x3
= 8r2x – 4x3
and f”(x) = \(\frac{d}{d x}\) (8r2x – 4x3)
= 8r2 × 1 – 4 × 3x2
= 8r2 – 12x2
For maximum area, f'(x) = 0
⇒ 8r2x – 4x3 = 0
⇒ 4x3 = 8r2x
⇒ x2 = 2r2 ……..[∵ x ≠ 0]
⇒ x = √2r …..[x > 0]
and f”(√2r) = 8r2 – 12(√2r)2 = -16r2 < 0
∴ f(x) is maximum when x = √2r
If x = √2r, then from (1),
(√2r)2 + y2 = 4r2
⇒ y2 = 4r2 – 2r2 = 2r2
⇒ y = √2r ……[∵ y > 0]
⇒ x = y
∴ rectangle is a square.
Hence, amongst all rectangles inscribed in a circle, the square has maximum area.

Question 14.
Show that a closed right circular cylinder of a given surface area has maximum volume if its height equals the diameter of its base.
Solution:
Let r be the radius of the base, h be the height and V be the volume of the closed right circular cylinder, whose surface area is a2 sq units (which is given).
2πrh + 2πr2 = a2
⇒ 2πr(h + r) = a2
⇒ h = \(\frac{a^{2}}{2 \pi r}\) – r ……(1)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q14
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q14.1
Hence, the volume of the cylinder is maximum if its height is equal to the diameter of the base.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

Question 15.
A window is in the form of a rectangle surmounted by a semicircle. If the perimeter is 30 m, find the dimensions so that the greatest possible amount of light may be admitted.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q15
Let x be the length, y be the breadth of the rectangle and r be the radius of the semicircle.
Then perimeter of the window = x + 2y + πr, where x = 2r
This is given to be 30 m
⇒ 2r + 2y + πr = 30
⇒ 2y = 30 – (π + 2)r
⇒ y = 15 – \(\frac{(\pi+2) r}{2}\) ……..(1)
The greatest possible amount of light may be admitted if the area of the window is maximized.
Let A be the area of the window.
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q15.1
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q15.2
Hence, the required dimensions of the window are as follows:
Length of rectangle = \(\left(\frac{60}{\pi+4}\right)\) metres
breadth of rectangle = \(\left(\frac{30}{\pi+4}\right)\) metres
radius of the semicircle = \(\left(\frac{30}{\pi+4}\right)\) metres

Question 16.
Show that the height of a right circular cylinder of greatest volume that can be inscribed in a right circular cone is one-third of that of the cone.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q16
Given the right circular cone of fixed height h and semi-vertical angle a.
Let R be the radius of the base and H be the height of the right circular cylinder that can be inscribed in the right circular cone.
In the figure, ∠GAO = α, OG = r, OA = h, OE = R, CE = H.
We have, \(\frac{r}{h}\) = tan α
∴ r = h tan α ……(1)
Since ∆AOG and ∆CEG are similar.
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q16.1
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q16.2
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q16.3
Hence, the height of the right circular cylinder is one-third of that of the cone.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

Question 17.
A wire of length l is cut into two parts. One part is bent into a circle and the other into a square. Show that the sum of the areas of the circle and the square is the least if the radius of the circle is half of the side of the square.
Solution:
Let r be the radius of the circle and x be the length of the side of the square. Then
(circumference of the circle) + (perimeter of the square) = l
∴ 2πr + 4x = l
∴ r = \(\frac{l-4 x}{2 \pi}\)
A = (area of the circle) + (area of the square)
= πr2 + x2
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q17
This shows that the sum of the areas of circle and square is least when the radius of the circle = (\(\frac{1}{2}\)) side of the square.

Question 18.
A rectangular sheet of paper of fixed perimeter with the sides having their lengths in the ratio 8 : 15 converted into an open rectangular box by folding after removing the squares of the equal area from all comers. If the total area of the removed squares is 100, the resulting box has maximum volume. Find the lengths of the rectangular sheet of paper.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q18
The sides of the rectangular sheet of paper are in the ratio 8 : 15.
Let the sides of the rectangular sheet of paper be 8k and 15k respectively.
Let x be the side of the square which is removed from the comers of the sheet of paper.
The total area of removed squares is 4x2, which is given to be 100.
4x2 = 100
⇒ x2 = 25
⇒x = 5 ……[x > 0]
Now, the length, breadth, and height of the rectangular box are 15k – 2x, 8k – 2x, and x respectively.
Let V be the volume of the box.
Then V = (15k – 2x) (8k – 2x) . x
⇒ V = (120k2 – 16kx – 30kx + 4x2) . x
⇒ V = 4x3 – 46kx2 + 120k2x
\(\frac{d V}{d x}=\frac{d}{d x}\) (4x3 – 46kx2 + 120k2x)
= 4 × 3x2 – 46k × 2x + 120k2 × 1
= 12x2 – 92kx + 120k2
Since, volume is maximum when the square of side x = 5 is removed from the corners,
\(\left(\frac{d V}{d x}\right)_{\text {at } x=5}=0\)
⇒ 12(5)2 – 92k(5) + 120k2 = 0
⇒ 60 – 92k + 24k2 = 0
⇒ 6k2 – 23k + 15 = 0
⇒ 6k2 – 18k – 5k + 15 = 0
⇒ 6k(k – 3) – 5 (k – 3) = 0
⇒ (k – 3)(6k – 5) = 0
⇒ k = 3 or k = \(\frac{5}{6}\)
If k = \(\frac{5}{6}\), then
8k – 2x = 8k – 10 < 0
∴ k ≠ \(\frac{5}{6}\)
∴ k = 3
∴ 8k = 8 × 3 = 24 and 15k = 15 × 3 = 45
Hence, the lengths of the rectangular sheet are 24 and 45.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

Question 19.
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is \(\frac{4 r}{3}\).
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q19
Let x be the radius of the base and h be the height of the cone which is inscribed in a sphere of radius r.
In the figure, AD = h and CD = x = BD
Since, ΔABD and ΔBDE are similar,
\(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{BD}}{\mathrm{DE}}\)
BD2 = AD . DE = AD (AE – AD)
x2 = h(2r – h) …… (1)
Let V be the volume of the cone.
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q19.1
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q19.2
∴ V is maximum when h = \(\frac{4 r}{3}\)
Hence, the altitude (i.e. height) of the right circular cone of maximum volume = \(\frac{4 r}{3}\).

Question 20.
Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is \(\frac{2 R}{\sqrt{3}}\). Also, find the maximum Volume.
Solution:
Let R be the radius and h be the height of the cylinder which is inscribed in a sphere of radius r cm.
Then from the figure,
\(R^{2}+\left(\frac{h}{2}\right)^{2}=r^{2}\)
∴ R2 = r2 – \(\frac{h^{2}}{4}\) ………(1)
Let V be the volume of the cylinder.
Then V = πR2h
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q20
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q20.1
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2 II Q20.2
Hence, the volume of the largest cylinder inscribed in a sphere of radius ‘r’ cm = \(\frac{4 R^{3}}{3 \sqrt{3}}\) cu cm.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

Question 21.
Find the maximum and minimum values of the function f(x) = cos2x + sin x.
Solution:
f(x) = cos2x + sin x
∴ f'(x) = \(\frac{d}{d x}\) (cos2x + sin x)
= 2 cos x . \(\frac{d}{d x}\) (cos x) + cos x
= 2 cos x(-sin x) + cos x
= -sin 2x + cos x
and f”(x) = \(\frac{d}{d x}\) (-sin 2x + cos x)
= -cos 2x . \(\frac{d}{d x}\) (2x) – sin x
= -cos 2x × 2 – sin x
= -2 cos 2x – sin x
For extreme values of f(x), f'(x) = 0
-sin 2x + cos x = 0
-2 sin x cos x + cos x = 0
cos x (-2 sin x + 1) = 0
cos x = 0 or -2 sin x + 1 = 0
cos x = cos \(\frac{\pi}{2}\) or sin x = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
∴ x = \(\frac{\pi}{2}\) or x = \(\frac{\pi}{6}\)

(i) f”(\(\frac{\pi}{2}\)) = -2 cos π – sin \(\frac{\pi}{2}\)
= -2(-1) – 1
= 1 > 0
∴ by the second derivative test, f is minimum at x = \(\frac{\pi}{2}\) and minimum value of f at x = \(\frac{\pi}{2}\)
= f(\(\frac{\pi}{2}\))
= \(\cos ^{2} \frac{\pi}{2}+\sin \frac{\pi}{2}\)
= 0 + 1
= 1

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Miscellaneous Exercise 2

(ii) f”(\(\frac{\pi}{6}\)) = \(-2 \cos \frac{\pi}{3}-\sin \frac{\pi}{6}\)
= \(-2\left(\frac{1}{2}\right)-\frac{1}{2}\)
= \(-\frac{3}{2}\) < 0
∴ by the second derivative test, f is maximum at x = \(\frac{\pi}{6}\) and maximum value of f at x = \(\frac{\pi}{6}\)
= f(\(\frac{\pi}{6}\))
= \(\cos ^{2} \frac{\pi}{6}+\sin \frac{\pi}{6}\)
= \(\left(\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{2}\)
= \(\frac{5}{4}\)
Hence, the maximum and minimum values of the function f(x) are \(\frac{5}{4}\) and 1 respectively.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Applications of Derivatives Ex 2.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

Question 1.
Test whether the following functions are increasing or decreasing.
(i) f(x) = x3 – 6x2 + 12x – 16, x ∈ R.
Solution:
f(x) = x3 – 6x2 + 12x – 16
∴ f'(x) = \(\frac{d}{d x}\) (x3 – 6x2 + 12x – 16)
= 3x2 – 6 × 2x + 12 × 1 – 0
= 3x2 – 12x + 12
= 3(x2 – 4x + 4)
= 3(x – 2)2 ≥ 0 for all x ∈ R
∴ f(x) ≥ 0 for all x ∈ R
∴ f is increasing for all x ∈ R.

(ii) f(x) = 2 – 3x + 3x2 – x3, x ∈ R.
Solution:
f(x) = 2 – 3x + 3x2 – x3
∴ f'(x) = \(\frac{d}{d x}\) (2 – 3x + 3x2 – x3)
= 0 – 3 × 1 + 3 × 2x – 3x2
= -3 + 6x – 3x2
= -3(x2 – 2x + 1)
= -3(x – 1)2 ≤ 0 for all x ∈ R
∴ f'(x) ≤ 0 for all x ∈ R
∴ f is decreasing for all x ∈ R.

(iii) f(x) = x – \(\frac{1}{x}\), x ∈ R, x ≠ 0.
Solution:
f(x) = x – \(\frac{1}{x}\)
f'(x) = \(\frac{d}{d x}\left(x-\frac{1}{x}\right)=1-\left(\frac{-1}{x^{2}}\right)\)
= \(1+\frac{1}{x^{2}}\) > 0 for all x ∈ R, x ≠ 0
∴ f'(x) > 0 for all x ∈ R, where x ≠ 0
∴ f is increasing for all x ∈ R, where x ≠ 0.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

Question 2.
Find the values of x for which the following functions are strictly increasing:
(i) f(x) = 2x3 – 3x2 – 12x + 6
Solution:
f(x) = 2x3 – 3x2 – 12x + 6
∴ f'(x) = \(\frac{d}{d x}\) (2x3 – 3x2 – 12x + 6)
= 2 × 3x2 – 3 × 2x – 12 × 1 + 0
= 6x2 – 6x – 12
= 6(x2 – x – 2)
f is strictly increasing if f'(x) > 0
i.e. if 6(x2 – x – 2) > 0
i.e. if x2 – x – 2 > 0
i.e. if x2 – x > 2
i.e. if x2 – x + \(\frac{1}{4}\) > 2 + \(\frac{1}{4}\)
i.e. if \(\left(x-\frac{1}{2}\right)^{2}>\frac{9}{4}\)
i.e. if x – \(\frac{1}{2}\) > \(\frac{3}{2}\) or x – \(\frac{1}{2}\) < \(\frac{-3}{2}\) i.e. if x > 2 or x < -1
∴ f is strictly increasing if x < -1 or x > 2.

(ii) f(x) = 3 + 3x – 3x2 + x3
Solution:
f(x) = 3 + 3x – 3x2 + x3
∴ f'(x) = \(\frac{d}{d x}\) (3 + 3x – 3x2 + x3)
= 0 + 3 × 1 – 3 × 2x + 3x2
= 3 – 6x + 3x2
= 3(x2 – 2x + 1)
f is strictly increasing if f'(x) > 0
i.e. if 3(x2 – 2x + 1) > 0
i.e. if x2 – 2x + 1 > 0
i.e. if (x – 1)2 > 0
This is possible if x ∈ R and x ≠ 1
i.e. x ∈ R – {1}
∴ f is strictly increasing if x ∈ R – {1}.

(iii) f(x) = x3 – 6x2 – 36x + 7
Solution:
f(x) = x3 – 6x2 – 36x + 7
∴ f'(x) = \(\frac{d}{d x}\) (x3 – 6x2 – 36x + 7)
= 3x2 – 6 × 2x – 36 × 1 + 0
= 3x2 – 12x – 36
= 3(x2 – 4x – 12)
f is strictly increasing if f'(x) > 0
i.e. if 3(x2 – 4x – 12) > 0
i.e. if x2 – 4x – 12 > 0
i.e. if x2 – 4x > 12
i.e. if x2 – 4x + 4 > 12 + 4
i.e. if (x – 2)2 > 16
i.e. if x – 2 > 4 or x – 2 < -4 i.e. if x > 6 or x < -2
∴ f is strictly increasing if x < -2 or x > 6.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

Question 3.
Find the values of x for which the following functions are strictly decreasiong:
(i) f(x) = 2x3 – 3x2 – 12x + 6
Solution:
f(x) = 2x3 – 3x2 – 12x + 6
∴ f'(x) = \(\frac{d}{d x}\) (2x3 – 3x2 – 12x + 6)
= 2 × 3x2 – 3 × 2x – 12 × 1 + 0
= 6x2 – 6x – 12
= 6(x2 – x – 2)
f is strictly decreasing if f'(x) < 0
i.e. if 6(x2 – x – 2) < 0
i.e. if x2 – x – 2 < 0
i.e. if x2 – x < 2
i.e. if x2 – x + \(\frac{1}{4}\) < 2 + \(\frac{1}{4}\)
i.e. if \(\left(x-\frac{1}{2}\right)^{2}<\frac{9}{4}\)
i.e. if \(-\frac{3}{2}<x-\frac{1}{2}<\frac{3}{2}\)
i.e. if \(-\frac{3}{2}+\frac{1}{2}<x-\frac{1}{2}+\frac{1}{2}<\frac{3}{2}+\frac{1}{2}\)
i.e. if -1 < x < 2
∴ f is strictly decreasing if -1 < x < 2.

(ii) f(x) = x + \(\frac{25}{x}\)
Solution:
f(x) = x + \(\frac{25}{x}\), x ≠ 0
∴ f'(x) = \(\frac{d}{d x}\left(x+\frac{25}{x}\right)\)
= 1 + 25(-1) x-2
= 1 – \(\frac{25}{x^{2}}\)
f is is strictly decreasing if f'(x) < 0
i.e. if 1 – \(\frac{25}{x^{2}}\) < 0
i.e. if 1 < \(\frac{25}{x^{2}}\)
i.e. if x2 < 25
i.e. if -5 < x < 5, x ≠ 0
i.e. if x ∈ (-5, 5) – {0}
∴ f is strictly decreasing if x ∈ (-5, 5) – {0}.

(iii) f(x) = x3 – 9x2 + 24x + 12
Solution:
f(x) = x3 – 9x2 + 24x + 12
∴ f'(x) = \(\frac{d}{d x}\) (x3 – 9x2 + 24x + 12)
= 3x2 – 9 × 2x + 24 × 1 + 0
= 3x2 – 18x + 24
= 3(x2 – 6x + 8)
f is strictly decreasing if f'(x) < 0
i.e. if 3(x2 – 6x + 8) < 0
i.e. if x2 – 6x + 8 < 0
i.e. if x2 – 6x < -8
i.e. if x2 – 6x + 9 < -8 + 9
i.e. if (x – 3)2 < 1
i.e. if -1 < x – 3 < 1
i.e. if -1 + 3 < x – 3 + 3 < 1 + 3
i.e. if 2 < x < 4
i.e., if x ∈ (2, 4)
∴ f is strictly decreasing if x ∈ (2, 4)

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

Question 4.
Find the values of x for which the function f(x) = x3 – 12x2 – 144x + 13
(a) increasing
(b) decreasing.
Solution:
f(x) = x3 – 12x2 – 144x + 13
∴ f'(x) = \(\frac{d}{d x}\) (x3 – 12x2 – 144x + 13)
= 3x2 – 12 × 2x – 144 × 1 + 0
= 3x2 – 24x – 144
= 3(x2 – 8x – 48)

(a) f is increasing if f'(x) ≥ 0
i.e. if 3(x2 – 8x – 48) ≥ 0
i.e. if x2 – 8x – 48 ≥ 0
i.e. if x2 – 8x ≥ 48
i.e. if x2 – 8x + 16 ≥ 48 + 16
i.e. if (x – 4)2 ≥ 64
i.e. if x – 4 ≥ 8 or x – 4 ≤ -8
i.e. if x > 12 or x ≤ -4
∴ f is increasing if x ≤ -4 or x ≥ 12,
i.e. x ∈ (-∞, -4] ∪ [12, ∞).

(b) f is decreasing if f'(x) ≤ 0
i.e. if 3(x2 – 8x – 48) ≤ 0
i.e. if x2 – 8x – 48 ≤ 0
i.e. if x2 – 8x ≤ 48
i.e. if x2 – 8x + 16 ≤ 48 + 16
i.e. if (x – 4)2 ≤ 64
i.e. if -8 ≤ x – 4 ≤ 8
i.e. if -4 ≤ x ≤ 12
∴ f is decreasing if -4 ≤ x ≤ 12, i.e. x ∈ [-4, 12].

Question 5.
Find the values of x for which f(x) = 2x3 – 15x2 – 144x – 7 is
(a) strictly increasing
(b) strictly decreasing.
Solution:
f(x) = 2x3 – 15x2 – 144x – 7
f'(x) = \(\frac{d}{d x}\) (2x3 – 15x2 – 144x – 7)
= 2 × 3x2 – 15 × 2x – 144 × 1 – 0
= 6x2 – 30x – 144
= 6(x2 – 5x – 24)
(a) f is strictly increasing if f'(x) > 0
i.e. if 6(x2 – 5x – 24) > 0
i.e. if x2 – 5x – 24 > 0
i.e. if x2 – 5x > 24
i.e. if x2 – 5x + \(\frac{25}{4}\) > 24 + \(\frac{25}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}>\frac{121}{4}\)
i.e. if \(x-\frac{5}{2}>\frac{11}{2} \text { or } x-\frac{5}{2}<-\frac{11}{2}\) i.e. if x > 8 or x < -3
∴ f is strictly increasing, if x < -3 or x > 8.

(b) f is strictly decreasing if f'(x) < 0
i.e. if 6(x2 – 5x – 24) < 0
i.e. if x2 – 5x – 24 < 0
i.e. if x2 – 5x < 24
i.e. if x2 – 5x + \(\frac{25}{4}\) < 24 + \(\frac{25}{4}\)
i.e. if \(\left(x-\frac{5}{2}\right)^{2}<\frac{121}{4}\)
i.e. if \(-\frac{11}{2}<x-\frac{5}{2}<\frac{11}{2}\)
i.e. if \(-\frac{11}{2}+\frac{5}{2}<x-\frac{5}{2}+\frac{5}{2}<\frac{11}{2}+\frac{5}{2}\)
i.e. if -3 < x < 8
∴ f is strictly decreasing, if -3 < x < 8.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

Question 6.
Find the values of x for which f(x) = \(\frac{\boldsymbol{x}}{x^{2}+1}\) is
(a) strictly increasing
(b) strictly decreasing.
Solution:
f(x) = \(\frac{\boldsymbol{x}}{x^{2}+1}\)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q6
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q6.1

(a) f is strictly increasing if f'(x) > 0
i.e. if \(\frac{1-x^{2}}{\left(x^{2}+1\right)^{2}}\) > 0
i.e. if 1 – x2 > 0 ……..[∵ (x2 + 1)2 > 0]
i.e. if 1 > x2
i.e. if x2 < 1
i.e. if -1 < x < 1
∴ f is strictly increasing if -1 < x < 1

(b) f is strictly decreasing if f'(x) < 0
i.e. if \(\frac{1-x^{2}}{\left(x^{2}+1\right)^{2}}\) < 0
i.e. if 1 – x2 < 0 ……..[∵ (x2 + 1)2 > 0]
i.e. if 1 < x2 i.e. if x2 > 1
i.e. if x > 1 or x < -1
∴ f is strictly decreasing if x < -1 or x > 1
i.e. x ∈ (-∞, -1) ∪ (1, ∞).

Question 7.
Show that f(x) = 3x + \(\frac{1}{3 x}\) is increasing in (\(\frac{1}{3}\), 1) and decreasing in (\(\frac{1}{9}\), \(\frac{1}{3}\))
Solution:
f(x) = 3x + \(\frac{1}{3 x}\)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q7
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q7.1

Question 8.
Show that f(x) = x – cos x is increasing for all x.
Solution:
f(x) = x – cos x
∴ f'(x) = \(\frac{d}{d x}\) (x – cos x)
= 1 – (-sin x)
= 1 + sin x
Now, -1 ≤ sin x ≤ 1 for all x ∈ R
∴ -1 + 1 ≤ 1 + sin x ≤ 1 for all x ∈ R
∴ 0 ≤ f'(x) ≤ 1 for all x ∈ R
∴ f'(x) ≥ 0 for all x ∈ R
∴ f is increasing for all x.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

Question 9.
Find the maximum and minimum of the following functions:
(i) y = 5x3 + 2x2 – 3x
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q9 (i)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q9 (i).1
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q9 (i).2
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q9 (i).3
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q9 (i).4
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q9 (i).5
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q9 (i).6

(ii) f(x) = 2x3 – 21x2 + 36x – 20
Solution:
f(x) = 2x3 – 21x2 + 36x – 20
∴ f'(x) = \(\frac{d}{d x}\) (2x3 – 21x2 + 36x – 20)
= 2 × 3x2 – 21 × 2x + 36 × 1 – 0
= 6x2 – 42x + 36
and f”(x) = \(\frac{d}{d x}\) (6x2 – 42x + 36)
= 6 × 2x – 42 × 1 + 0
= 12x – 42
f'(x) = 0 gives 6x2 – 42x + 36 = 0
∴ x2 – 7x + 6 = 0
∴ (x – 1)(x – 6) = 0
the roots of f'(x) = 0 are x1 = 1 and x2 = 6.

Method 1 (Second Derivative Test):
(a) f”(1) = 12(1) – 42 = -30 < 0
∴ by the second derivative test, f has maximum at x = 1
and maximum value of f at x = 1
f(1) = 2(1)3 – 21(1)2 + 36(1) – 20
= 2 – 21 + 36 – 20
= -3

(b) f”(6) = 12(6) – 42 = 30 > 0
∴ by the second derivative test, f has minimum at x = 6
and minimum value of f at x = 6
f(6) = 2(6)3 – 21(6)2 + 36(6) – 20
= 432 – 756 + 216 – 20
= -128.
Hence, the function f has maximum value -3 at x = 1 and minimum value -128 at x = 6.

Method 2 (First Derivative Test):
(a) f'(x) = 6(x – 1)(x – 6)
Consider x = 1
Let h be a small positive number. Then
f'(1 – h) = 6(1 – h – 1)(1 – h – 6)
= 6(-h)(-5 – h)
= 6h(5 + h)> 0
and f'(1 + h) = 6(1 + h – 1)(1 + h – 6)
= 6h(h – 5) < 0, as h is small positive number.
∴ by the first derivative test, f has maximum at x = 1 and maximum value of f at x = 1
f(1) = 2(1)3 – 21(1)2 + 36(1) – 20
= 2 – 21 + 36 – 20
= -3

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

(b) f'(x) = 6(x – 1)(x – 6)
Consider x = 6
Let h be a small positive number. Then
f'(6 – h) = 6(6 – h – 1)(6 – h – 6)
= 6(5 – h)(-h)
= -6h(5 – h) < 0, as h is small positive number
and f'(6 + h) = 6(6 + h – 1)(6 + h – 6) = 6(5 + h)(h) > 0
∴ by the first derivative test, f has minimum at x = 6
and minimum value of f at x = 6
f(6) = 2(6)3 – 21(6)2 + 36(6) – 20
= 432 – 756 + 216 – 20
= -128
Hence, the function f has maximum value -3 at x = 1
and minimum value -128 at x = 6.

(iii) f(x) = x3 – 9x2 + 24x
Solution:
f(x) = x3 – 9x2 + 24x
∴ f'(x) = \(\frac{d}{d x}\) (x3 – 9x2 + 24x)
= 3x2 – 9 × 2x + 24 × 1
= 3x2 – 18x + 24
and f”(x) = \(\frac{d}{d x}\) (3x2 – 18x + 24)
= 3 × 2x – 18 × 1 + 0
= 6x – 18
f'(x) = 0 gives 3x2 – 18x + 24 = 0
∴ x2 – 6x + 8 = 0
∴ (x – 2)(x – 4) = 0
∴ the roots of f'(x) = 0 are x1 = 2 and x2 = 4.

(a) f”(2) = 6(2) – 18 = -6 < 0
∴ by the second derivative test, f has maximum at x = 2
and maximum value of f at x = 2
f(2) = (2)3 – 9(2)2 + 24(2)
= 8 – 36 + 48
= 20

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

(b) f”(4) = 6(4) – 18 = 6 > 0
∴ by the second derivative test, f has minimum at x = 4
and minimum value of f at x = 4
f(4) = (4)3 – 9(4)2 + 24(4)
= 64 – 144 + 96
= 16
Hence, the function f has maximum value 20 at x = 2 and minimum value 16 at x = 4.

(iv) f(x) = x2 + \(\frac{16}{x^{2}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q9 (iv)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q9 (iv).1

(v) f(x) = x log x
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q9 (v)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q9 (v).1

(vi) f(x) = \(\frac{\log x}{x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q9 (vi)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q9 (vi).1

Question 10.
Divide the number 30 into two parts such that their product is maximum.
Solution:
Let the first part of 30 be x.
Then the second part is 30 – x.
∴ their product = x(30 – x) = 30x – x2 = f(x) ……(Say)
∴ f'(x) = \(\frac{d}{d x}\) (30x – x2)
= 30 × 1 – 2x
= 30 – 2x
and f”(x) = \(\frac{d}{d x}\) (30 – 2x)
= 0 – 2 × 1
= -2
The root of the equation f(x) = 0,
i.e. 30 – 2x = 0 is x = 15 and f”(15) = -2 < 0
∴ by the second derivative test, f is maximum at x = 15.
Hence, the required parts of 30 are 15 and 15.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

Question 11.
Divide the number 20 into two parts such that the sum of their squares is minimum.
Solution:
Let the first part of 20 be x.
Then the second part is 20 – x.
∴ sum of their squares = x2 + (20 – x)2 = f(x) …… (Say)
∴ f'(x) = \(\frac{d}{d x}\) [x2 + (20 – x)2]
= 2x + 2(20 – x) . \(\frac{d}{d x}\) (20 – x)
= 2x + 2(20 – x) × (0 – 1)
= 2x – 40 + 2x
= 4x – 40
and f”(x) = \(\frac{d}{d x}\) (4x – 40)
= 4 × 1 – 0
= 4
The root of the equation f'(x) = 0,
i.e. 4x – 40 = 0 is x = 10 and f”(10) = 4 > 0
∴ by the second derivative test, f is minimum at x = 10.
Hence, the required parts of 20 are 10 and 10.

Question 12.
A wire of length 36 meters is bent in the form of a rectangle. Find its dimensions if the area of the rectangle is maximum.
Solution:
Let x metres and y metres be the length and breadth of the rectangle.
Then its perimeter is 2(x + y) = 36
x + y = 18
y = 18 – x
Area of the rectangle = xy = x (18 – x)
Let f(x) = x(18 – x) = 18x – x2
∴ f'(x) = \(\frac{d}{d x}\) (18x – x2) = 18 – 2x
and f”(x) = \(\frac{d}{d x}\) (18 – 2x) = 0 – 2 × 1 = -2
Now, f'(x) = 0, if 18 – 2x = 0
i.e. if x = 9
and f”(9) = -2 < 0
∴ by the second derivative test, f has maximum value at x = 9.
When x = 9, y = 18 – 9 = 9
∴ x = 9 cm, y = 9 cm
∴ the rectangle is a square of side 9 metres.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

Question 13.
A ball is thrown in the air. Its height at any time t is given by h = 3 + 14t – 5t2. Find the maximum height it can reach.
Solution:
The height h at any t is given by h = 3 + 14t – 5t2
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q13
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q13.1
Hence, the maximum height the ball can reach = 12.8 units.

Question 14.
Find the largest size of a rectangle that can be inscribed in a semicircle of radius 1 unit, so that two vertices lie on the diameter.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q14
Let ABCD be the rectangle inscribed in a semicircle of radius 1 unit such that the vertices A and B lie on the diameter.
Let AB = DC = x and BC = AD = y.
Let O be the centre of the semicircle.
Join OC and OD. Then OC = OD = radius = 1.
Also, AD = BC and m∠A = m∠B = 90°.
∴ OA = OB
∴ OB = \(\frac{1}{2}\) AB = \(\frac{x}{2}\)
In right angled triangle OBC,
OB2 + BC2 = OC2
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q14.1
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q14.2
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q14.3
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q14.4
Hence, the area of the rectangle is maximum (i.e. rectangle has the largest size) when its length is √2 units and breadth is \(\frac{1}{\sqrt{2}}\) unit.

Question 15.
An open cylindrical tank whose base is a circle is to be constructed of metal sheet so as to contain a volume of πa3 cu cm of water. Find the dimensions so that the quantity of the metal sheet required is minimum.
Solution:
Let x be the radius of the base, h be the height, V be the volume and S be the total surface area of the cylindrical tank.
Then V = πa3 … (Given)
∴ πx2h = πa3
∴ h = \(\frac{a^{3}}{x^{2}}\) ……..(1)
Now, S = 2πxh + πx2
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q15
∴ by the second derivative test, S is minimum when x = a
When x = a, from (1)
h = \(\frac{a^{3}}{a^{2}}\) = a
Hence, the quantity of metal sheet is minimum when radius height = a cm.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

Question 16.
The perimeter of a triangle is 10 cm. If one of the sides is 4 cm. What are the other two sides of the triangle for its maximum area?
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q16
Let ABC be the triangle such that the side BC = a = 4 cm.
Also, the perimeter of the triangle is 10 cm.
i.e. a + b + c = 10
∴ 2s = 10
∴ s = 5
Also, 4 + b + c = 10
∴ b + c = 6
∴ b = 6 – c
Let ∆ be the area of the triangle.
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q16.1
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q16.2
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q16.3
∴ by the second derivative test, ∆ is maximum when c = 3.
When c = 3, b = 6 – c = 6 – 3 = 3
Hence, the area of the triangle is maximum when the other two sides are 3 cm and 3 cm.

Question 17.
A box with a square base is to have an open top. The surface area of the box is 192 sq cm. What should be its dimensions in order that the volume is largest?
Solution:
Let x cm be the side of square base and h cm be its height.
Then x2 + 4xh = 192
∴ h = \(\frac{192-x^{2}}{4 x}\) …… (1)
Let V be the volume of the box.
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q17
∴ by the second derivative test, V is maximum at x = 8.
If x = 8, h = \(\frac{192-64}{4(8)}=\frac{128}{32}\) = 4
Hence, the volume of the box is largest, when the side of square base is 8 cm and its height is 4 cm.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

Question 18.
The profit function P (x) of a firm, selling x items per day is given by P(x) = (150 – x)x – 1625. Find the number of items the firm should manufacture to get maximum profit. Find the maximum profit.
Solution:
Profit function P (x) is given by
P(x) = (150 – x)x – 1625 = 150x – x2 -1625
∴ P'(x) = \(\frac{d}{d x}\) (150x – x2 – 1625)
= 150 × 1 – 2x – 0
= 150 – 2x
and P”(x) = \(\frac{d}{d x}\) (150 – 2x)
= 0 – 2 × 1
= -2
Now, P'(x) = 0 gives, 150 – 2x = 0
∴ x = 75
and P”(75) = -2 < 0
∴ by the second derivative test, P(x) is maximum when x = 75
Maximum profit = P(75)
= (150 – 75)75 – 1625
= 75 × 75 – 1625
= 4000
Hence, the profit will be maximum, if the manufacturer manufactures 75 items and the maximum profit is 4000.

Question 19.
Find two numbers whose sum is 15 and when the square of one multiplied by the cube of the other is maximum.
Solution:
Let the two numbers be x and y.
Then x + y = 15
∴ y = 15 – x
Let P is the product of square of y and cube of x.
Then P = x3y2
= x3(15 – x)2
= x3(225 – 30x + x2)
= x5 – 30x4 + 225x3
∴ \(\frac{d P}{d x}\) = \(\frac{d}{d x}\) (x5 – 30x4 + 225x3)
= 5x4 – 30 × 4x3 + 225 × 3x2
= 5x4 – 120x3 + 675x2
and \(\frac{d^{2} P}{d x^{2}}\) = \(\frac{d}{d x}\) (5x4 – 120x3 + 675x2)
= 5 × 4x3 – 120 × 3x2 + 675 × 2x
= 20x3 – 360x2 + 1350x
= 10x(2x2 – 36x + 135)
Now, \(\frac{d P}{d x}\) = 0 gives 5x4 – 120x3 + 675x2 = 0
∴ 5x2(x2 – 24x +135) = 0
∴ 5x2(x2 – 15x – 9x + 135) = 0
∴ 5x2[x(x – 15) – 9(x – 35)] = 0
∴ 5x2(x – 15)(x – 9) = 0
∴ the roots of \(\frac{d P}{d x}\) = 0 are x1 = 0, x2 = 15 and x3 = 9
If x = 0, then y = 15 – 0 = 15
If x = 15, then y = 15 – 15 = 0
In both cases, product x3y2 is zero, which is not maximum.
∴ x ≠ 0 and x ≠ 15
∴ x = 6
Now, \(\left(\frac{d^{2} P}{d x^{2}}\right)_{\text {at } x=6}\) = 10(6)[2(6)2 – 36 × 6 + 135]
= 60[72 – 216 + 135]
= 60(-9)
= -540 < 0
∴ P is maximum when x = 6
If x = 6, then y = 15 – 6 = 9
Hence, the required numbers are 6 and 9.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

Question 20.
Show that among rectangles of given area, the square has least perimeter.
Solution:
Let x be the length and y be the breadth of the rectangle whose area is A sq units (which is given as constant).
Then xy = A
∴ y = \(\frac{A}{x}\) ………(1)
Let P be the perimeter of the rectangle.
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q20
x = y
∴ rectangle is a square.
Hence, among rectangles of given area, the square has least perimeter.

Question 21.
Show that the height of a closed right circular cylinder of given volume and least surface area is equal to its diameter.
Solution:
Let x be the radius of base, h be the height and S be the surface area of the closed right circular cylinder whose volume is V which is given to be constant.
Then πr2h = V
∴ h = \(\frac{V}{\pi r^{2}}=\frac{A}{x^{2}}\) …….(1)
where A = \(\frac{V}{\pi}\), which is constant.
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q21
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q21.1
Hence, the surface area is least when height of the closed right circular cylinder is equal to its diameter.

Question 22.
Find the volume of the largest cylinder that can be inscribed in a sphere of radius ‘r’ cm.
Solution:
Let R be the radius and h be the height of the cylinder which is inscribed in a sphere of radius r cm.
Then from the figure,
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q22
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q22.1
Hence, the volume of the largest cylinder inscribed in a sphere of radius ‘r’ cm = \(\frac{4 \pi r^{3}}{3 \sqrt{3}}\) cu cm.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4

Question 23.
Show that y = log(1 + x) – \(\frac{2 x}{2+x}\), x > -1 is an increasing function on its domain.
Solution:
y = log(1 + x) – \(\frac{2 x}{2+x}\), x > -1
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q23
Hence, the given function is increasing function on its domain.

Question 24.
Prove that y = \(\frac{4 \sin \theta}{2+\cos \theta}\) – θ is an increasing function if θ ∈ [0, \(\frac{\pi}{2}\)]
Solution:
y = \(\frac{4 \sin \theta}{2+\cos \theta}\) – θ
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q24
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.4 Q24.1

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Applications of Derivatives Ex 2.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3

Question 1.
Check the validity of the Rolle’s theorem for the following functions.
(i) f(x) = x2 – 4x + 3, x ∈ [1, 3]
Solution:
The function f given as f(x) = x2 – 4x + 3 is polynomial function.
Hence, it is continuous on [1, 3] and differentiable on (1, 3).
Now, f(1) = 12 – 4(1) + 3 = 1 – 4 + 3 = 0
and f(3) = 32 – 4(3) + 3 = 9 – 12 + 3 = 0
∴ f(1) = f(3)
Thus, the function f satisfies all the conditions of Rolle’s theorem.

(ii) f(x) = e-x sin x, x ∈ [0, π].
Solution:
The functions e-x and sin x are continuous and differentiable on their domains.
∴ f(x) = e-x sin x is continuous on [0, π] and differentiable on (0, π).
Now, f(0) = e0 sin 0 = 1 × 0 = 0
and f(π) = e sin π = e × 0 = 0
∴ f(0) = f(π)
Thus, the function f satisfies all the conditions of the Rolle’s theorem.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3

(iii) f(x) = 2x2 – 5x + 3, x ∈ [1, 3].
Solution:
The function f given as f(x) = 2x2 – 5x + 3 is a polynomial function.
Hence, it is continuous on [1, 3] and differentiable on (1, 3).
Now, f(1) = 2(1)2 – 5(1) + 3 = 2 – 5 + 3 = 0
and f(3) = 2(3)2 – 5(3) + 3 = 18 – 15 + 3 = 6
∴ f(1) ≠ f(3)
Hence, the conditions of Rolle’s theorem are not satisfied.

(iv) f(x) = sin x – cos x + 3, x ∈ [0, 2π].
Solution:
The functions sin x, cos x and 3 are continuous and differentiable on their domains.
∴ f(x) = sin x – cos x + 3 is continuous on [0, 2π] and differentiable on (0, 2π).
Now, f(0) = sin 0 – cos 0 + 3 = 0 – 1 + 3 = 2
and f(2π) = sin 2π – cos 2π + 3 = 0 – 1 + 3 = 2
∴ f(0) = f(2π)
Thus, the function f satisfies all the conditions of the Rolle’s theorem.

(v) f(x) = x2, if 0 ≤ x ≤ 2
= 6 – x, if 2 < x ≤ 6.
Solution:
f(x) = x2, if 0 ≤ x ≤ 2
= 6 – x, if 2 < x ≤ 6
∴ f(x) = \(\frac{d}{d x}\left(x^{2}\right)\) = 2x, if 0 ≤ x ≤ 2
= \(\frac{d}{d x}(6-x)\) = -1, if 2 < x ≤ 6
∴ Lf'(2) = 2(2) = 4 and Rf'(2) = -1
∴ Lf'(2) ≠ Rf'(2)
∴ f is not differentiable at x = 2 and 2 ∈ (0, 6).
∴ f is not differentiable at all the points on (0, 6).
Hence, the conditions of Rolle’s theorem are not satisfied.

(vi) f(x) = \(x^{\frac{2}{3}}\), x ∈ [-1, 1].
Solution:
f(x) = \(x^{\frac{2}{3}}\)
∴ \(f^{\prime}(x)=\frac{d}{d x}\left(x^{\frac{2}{3}}\right)=\frac{2}{3} x^{-\frac{1}{3}}\) = \(\frac{2}{3 \sqrt[3]{x}}\)
This does not exist at x = 0 and 0 ∈ (-1, 1)
∴ f is not differentiable on the interval (-1, 1).
Hence, the conditions of Rolle’s theorem are not satisfied.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3

Question 2.
Given an interval [a, b] that satisfies hypothesis of Rolle’s theorem for the function f(x) = x4 + x2 – 2. It is known that a = -1. Find the value of b.
Solution:
f(x) = x4 + x2 – 2
Since the hypothesis of Rolle’s theorem are satisfied by f in the interval [a, b], we have
f(a) = f(b), where a = -1
Now, f(a) = f(-1) = (-1)4 + (-1)2 – 2 = 1 + 1 – 2 = 0
and f(b) = b4 + b2 – 2
∴ f(a) = f(b) gives
0 = b4 + b2 – 2 i.e. b4 + b2 – 2 = 0.
Since, b = 1 satisfies this equation, b = 1 is one of the roots of this equation.
Hence, b = 1.

Question 3.
Verify Rolle’s theorem for the following functions.
(i) f(x) = sin x + cos x + 7, x ∈ [0, 2π]
Solution:
The functions sin x, cos x and 7 are continuous and differentiable on their domains.
∴ f(x) = sin x + cos x + 7 is continuous on [0, 2π] and differentiable on (0, 2π)
Now, f(0) = sin 0 + cos 0 + 7 = 0 + 1 + 7 = 8
and f(2π) = sin 2π + cos 2π + 7 = 0 + 1 + 7 = 8
∴ f(0) = f(2π)
Thus, the function f satisfies all the conditions of Rolle’s theorem.
∴ there exists c ∈ (0, 2π) such that f'(c) = 0.
Now, f(x) = sin x + cos x + 7
∴ f'(x) = \(\frac{d}{d x}\) (sin x + cos x + 7)
= cos x – sin x + 0
= cos x – sin x
∴ f'(c) = cos c – sin c
∴ f'(c) = 0 gives, cos c – sin c = 0
∴ cos c = sin c
∴ c = \(\frac{\pi}{4}, \frac{5 \pi}{4}, \frac{9 \pi}{4}, \ldots\)
But \(\frac{\pi}{4}, \frac{5 \pi}{4}\) ∈ (0, 2π)
∴ c = \(\frac{\pi}{4} \text { or } \frac{5 \pi}{4}\)
Hence, the Rolle’s theorem is verified.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3

(ii) f(x) = sin(\(\frac{x}{2}\)), x ∈ [0, 2π]
Solution:
The function f(x) = sin(\(\frac{x}{2}\)) is continuous on [0, 2π] and differentiable on (0, 2π).
Now, f(0) = sin 0 = 0
and f(2π) = sin π = 0
∴ f(0) = f(2π)
Thus, the function f satisfies all the conditions of Rolle’s theorem.
∴ there exists c ∈ (0, 2π) such that f'(c) = 0.
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3 Q3 (ii)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3 Q3 (ii).1
Hence, Rolle’s theorem is verified.

(iii) f(x) = x2 – 5x + 9, x ∈ [1, 4].
Solution:
The function f given as f(x) = x2 – 5x + 9 is a polynomial function.
Hence it is continuous on [1, 4] and differentiable on (1, 4).
Now, f(1) = 12 – 5(1) + 9 = 1 – 5 + 9 = 5
and f(4) = 42 – 5(4) + 9 = 16 – 20+ 9 = 5
∴ f(1) = f(4)
Thus, the function f satisfies all the conditions of the Rolle’s theorem.
∴ there exists c ∈ (1, 4) such that f'(c) = 0.
Now, f(x) = x2 – 5x + 9
∴ f'(x) = \(\frac{d}{d x}\) (x2 – 5x + 9)
= 2x – 5 × 1 + 0
= 2x – 5
∴ f'(c) = 2c – 5
∴ f'(c) = 0 gives, 2c – 5 = 0
∴ c = 5/2 ∈ (1, 4)
Hence, the Rolle’s theorem is verified.

Question 4.
If Rolle’s theorem holds for the function f(x) = x3 + px2 + qx + 5, x ∈ [1, 3] with c = 2 + \(\frac{1}{\sqrt{3}}\), find the values of p and q.
Solution:
The Rolle’s theorem holds for the function f(x) = x3 + px2 + qx + 5, x ∈ [1, 3]
∴ f(1) = f(3)
∴ 13 + p(1)2 + q(1) + 5 = 33 + p (3)2 + q(3) + 5
∴ 1 + p + q + 5 = 27 + 9p + 3q + 5
∴ 8p + 2q = -26
∴ 4p + q = -13 ….. (1)
Also, there exists at least one point c ∈ (1, 3) such that f'(c) = 0.
Now, f'(x) = \(\frac{d}{d x}\) (x3 + px2 + qx + 5)
= 3x2 + p × 2x + q × 1 + 0
= 3x2 + 2px + q
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3 Q4
But f'(c) = 0
∴ \(4 p+\frac{2 p}{\sqrt{3}}+q+13+\frac{12}{\sqrt{3}}=0\)
∴ (4√3 + 2)p + √3q + (13√3 + 12) = 0
∴ (4√3 + 2)p + √3q = -13√3 – 12 ……. (2)
Multiplying equation (1) by √3, we get
4√3p + √3q= -13√3
Subtracting this equation from (2), we get
2p = -12 ⇒ p= -6
∴ from (1), 4(-6) + q = -13 ⇒ q = 11
Hence, p = -6 and q = 11.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3

Question 5.
If Rolle’s theorem holds for the function f(x) = (x – 2) log x, x ∈ [1, 2], show that the equation x log x = 2 – x is satisfied by at least one value of x in (1, 2).
Solution:
The Rolle’s theorem holds for the function f(x) = (x – 2) log x, x ∈ [1, 2].
∴ there exists at least one real number c ∈ (1, 2) such that f'(c) = 0.
Now, f(x) = (x – 2) log x
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3 Q5
∴ f'(c) = 0 gives 1 – \(\frac{2}{c}\) + log c = 0
∴ c – 2 + c log c = 0
∴ c log c = 2 – c, where c ∈ (1, 2)
∴ c satisfies the equation x log x = 2 – x, c ∈ (1, 2).
Hence, the equation x log x = 2 – x is satisfied by at least one value of x in (1, 2).

Question 6.
The function f(x) = \(x(x+3) e^{-\frac{x}{2}}\) satisfies all the conditions of Rolle’s theorem on [-3, 0]. Find the value of c such that f'(c) = 0.
Solution:
The function f(x) satisfies all the conditions of Rolle’s theorem, therefore there exist c ∈ (-3, 0) such that f'(c) = 0.
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3 Q6
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3 Q6.1

Question 7.
Verify Lagrange’s mean value theorem for the following functions:
(i) f(x) = log x on [1, e].
Solution:
The function f given as f(x) = log x is a logarithmic function that is continuous for all positive real numbers.
Hence, it is continuous on [1, e] and differentiable on (1, e).
Thus, the function f satisfies the conditions of Lagrange’s mean value theorem.
∴ there exists c ∈ (1, e) such that
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3 Q7 (i)
Hence, Lagrange’s mean value theorem is verified.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3

(ii) f(x) = (x – 1)(x – 2)(x – 3) on [0, 4].
Solution:
The function f given as
f(x) = (x – 1)(x – 2)(x – 3)
= (x – 1)(x2 – 5x + 6)
= x3 – 5x2 + 6x – x2 + 5x – 6
= x3 – 6x2 + 11x – 6 is a polynomial function.
Hence, it is continuous on [0, 4] and differentiable on (0, 4).
Thus, the function f satisfies the conditions of Lagrange’s, mean value theorem.
∴ there exists c ∈ (0, 4) such that
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3 Q7 (ii)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3 Q7 (ii).1
Hence, Lagrange’s mean value theorem is verified.

(iii) f(x) = x2 – 3x – 1, x ∈ \(\left[\frac{-11}{7}, \frac{13}{7}\right]\)
Solution:
The function f given as f(x) = x2 – 3x – 1 is a polynomial function.
Hence, it is continuous on \(\left[\frac{-11}{7}, \frac{13}{7}\right]\) and differentiable on \(\left(\frac{-11}{7}, \frac{13}{7}\right)\).
Thus, the function f satisfies the conditions of LMVT.
∴ there exists c ∈ \(\left(\frac{-11}{7}, \frac{13}{7}\right)\) such that
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3 Q7 (iii)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3 Q7 (iii).1
Hence, Lagrange’s mean value theorem is verified.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3

(iv) f(x) = 2x – x2, x ∈ [0, 1].
Solution:
The function f given as f(x) = 2x – x2 is a polynomial function.
Hence, it is continuous on [0, 1] and differentiable on (0, 1).
Thus, the function f satisfies the conditions of Lagrange’s mean value theorem.
∴ there exists c ∈ (0, 1) such that
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3 Q7 (iv)
Hence, Lagrange’s mean value theorem is verified.

(v) f(x) = \(\frac{x-1}{x-3}\) on [4, 5].
Solution:
The function f given as
f(x) = \(\frac{x-1}{x-3}\) is a rational function which is continuous except at x = 3.
But 3 ∉ [4, 5]
Hence, it is continuous on [4, 5] and differentiable on (4, 5).
Thus, the function f satisfies the conditions of Lagrange’s mean value theorem.
∴ there exists c ∈ (4, 5) such that
f'(c) = \(\frac{f(5)-f(4)}{5-4}\) ……..(1)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.3 Q7 (v)
Hence, Lagrange’s mean value theorem is verified.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Applications of Derivatives Ex 2.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2

Question 1.
Find the approximate value of given functions, at required points.
(i) √8.95
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q1 (i)
√8.95 = 2.9917

(ii) \(\sqrt[3]{28}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q1 (ii)

(iii) \(\sqrt[5]{31.98}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q1 (iii)

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2

(iv) (3.97)4
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q1 (iv)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q1 (iv).1

(v) (4.01)3
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q1 (v)

Question 2.
Find the approximate values of:
(i) sin 61°, given that 1° = 0.0174c, √3 = 1.732.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q2 (i)

(ii) sin(29° 30′), given that 1° = 0.0175c, √3 = 1.732.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q2 (ii)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q2 (ii).1

(iii) cos(60° 30′), given that 1° = 0.0175c, √3 = 1.732.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q2 (iii)

(iv) tan (45° 40′), given that 1° = 0.0175c.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q2 (iv)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q2 (iv).1

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2

Question 3.
Find the approximate values of
(i) tan-1(0.999).
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q3 (i)

(ii) cot-1(0.999).
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q3 (ii)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q3 (ii).1

(iii) tan-1(1.001).
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q3 (iii)

Question 4.
Find the approximate values of:
(i) e0.995, given that e = 2.7183.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q4 (i)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q4 (i).1

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2

(ii) e2.1, given that e2 = 7.389.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q4 (ii)

(iii) 32.01, given that log 3 = 1.0986.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q4 (iii)

Question 5.
Find the approximate values of:
(i) loge(101), given that loge10 = 2.3026.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q5 (i)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q5 (i).1

(ii) loge(9.01), given that log 3 = 1.0986.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q5 (ii)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q5 (ii).1

(iii) log10(1016), given that log10e = 0.4343.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q5 (iii)

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2

Question 6.
Find the approximate values of:
(i) f(x) = x3 – 3x + 5 at x = 1.99
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q6 (i)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q6 (i).1

(ii) f(x) = x3 + 5x2 – 7x +10 at x = 1.12
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.2 Q6 (ii)

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Applications of Derivatives Ex 2.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1

Question 1.
Find the equations of tangents and normals to the curve at the point on it.
(i) y = x2 + 2ex + 2 at (0, 4)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q1 (i)

(ii) x3 + y3 – 9xy = 0 at (2, 4)
Solution:
x3 + y3 – 9xy = 0
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q1 (ii)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q1 (ii).1
Hence, the equations of tangent and normal are 4x – 5y + 12 = 0 and 5x + 4y – 26 = 0 respectively.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1

(iii) x2 – √3xy + 2y2 = 5 at (√3, 2)
Solution:
x2 – √3xy + 2y2 = 5
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q1 (iii)
the slope of normal at (√3, 2) does not exist.
normal is parallel to Y-axis.
equation of the normal is of the form x = k
Since, it passes through the point (√3, 2), k = √3
equation of the normal is x = √3.
Hence, the equations of tangent and normal are y = 2 and x = √3 respectively.

(iv) 2xy + π sin y = 2π at (1, \(\frac{\pi}{2}\))
Solution:
2xy + π sin y = 2π
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q1 (iv)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q1 (iv).1
Hence, the equations of tangent and normal are πx + 2y – 2π = 0 and 4x – 2πy + π2 – 4 = 0 respectively.

(v) x sin 2y = y cos 2x at (\(\frac{\pi}{4}\), \(\frac{\pi}{2}\))
Solution:
x sin 2y = y cos 2x
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q1 (v)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q1 (v).1
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q1 (v).2
Hence, the equations of the tangent and normal are 2x – y = 0 and 4x + 8y – 5π = 0 respectively.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1

(vi) x = sin θ and y = cos 2θ at θ = \(\frac{\pi}{6}\)
Solution:
When θ = \(\frac{\pi}{6}\), x = sin\(\frac{\pi}{6}\) and y = cos\(\frac{\pi}{3}\)
∴ x = \(\frac{1}{2}\) and y = \(\frac{1}{2}\)
Hence, the point at which we want to find the equations of tangent and normal is (\(\frac{1}{2}\), \(\frac{1}{2}\))
Now, x = sin θ, y = cos 2θ
Differentiating x and y w.r.t. θ, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q1 (vi)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q1 (vi).1
2y – 1 = x – \(\frac{1}{2}\)
4y – 2 = 2x – 1
2x – 4y + 1 = 0
Hence, equations of the tangent and normal are 4x + 2y – 3 = 0 and 2x – 4y + 1 = 0 respectively.

(vii) x = √t, y = t – \(\frac{1}{\sqrt{t}}\), at t = 4.
Solution:
When t = 4, x = √4 and y = 4 – \(\frac{1}{\sqrt{4}}\)
∴ x = 2 and y = 4 – \(\frac{1}{2}\) = \(\frac{7}{2}\)
Hence, the point at which we want to find the equations of tangent and normal is (2, \(\frac{7}{2}\)).
Now, x = √t, y = t – \(\frac{1}{\sqrt{t}}\)
Differentiating x and y w.r.t. t, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q1 (vii)
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q1 (vii).1
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q1 (vii).2
Hence, the equations of tangent and normal are 17x – 4y – 20 = 0 and 8x + 34y – 135 = 0 respectively.

Question 2.
Find the point of the curve y = \(\sqrt{x-3}\) where the tangent is perpendicular to the line 6x + 3y – 5 = 0.
Solution:
Let the required point on the curve y = \(\sqrt{x-3}\) be P(x1, y1).
Differentiating y = \(\sqrt{x-3}\) w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q2
Hence, the required points are (4, 1) and (4, -1).

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1

Question 3.
Find the points on the curve y = x3 – 2x2 – x where the tangents are parallel to 3x – y + 1 = 0.
Solution:
Let the required point on the curve y = x3 – 2x2 – x be P(x1, y1).
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q3
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q3.1

Question 4.
Find the equations of the tangents to the curve x2 + y2 – 2x – 4y + 1 = 0 which are parallel to the X-axis.
Solution:
Let P (x1, y1) be the point on the curve x2 + y2 – 2x – 4y + 1 = 0 where the tangent is parallel to X-axis.
Differentiating x2 + y2 – 2x – 4y + 1 = 0 w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q4
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q4.1
the coordinates of the points are (1, 0) or (1, 4)
Since the tangents are parallel to X-axis, their equations are of the form y = k
If it passes through the point (1, 0), k = 0, and if it passes through the point (1, 4), k = 4
Hence, the equations of the tangents are y = 0 and y = 4.

Question 5.
Find the equations of the normals to the curve 3x2 – y2 = 8, which are parallel to the line x + 3y = 4.
Solution:
Let P(x1, y1) be the foot of the required normal to the curve 3x2 – y2 = 8.
Differentiating 3x2 – y2 = 8 w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q5
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q5.1
Hence, the equations of the normals are x + 3y – 8 = 0 and x + 3y + 8 = 0.

Question 6.
If the line y = 4x – 5 touches the curve y2 = ax3 + b at the point (2, 3), find a and b.
Solution:
y2 = ax3 + b
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q6
= slope of the tangent at (2, 3)
Since, the line y = 4x – 5 touches the curve at the point (2, 3), slope of the tangent at (2, 3) is 4.
2a = 4 ⇒ a = 2
Since (2, 3) lies on the curve y2 = ax3 + b
(3)2 = a(2)3 + b
9 = 8a + b
9 = 8(2) + b …… [∵ a = 2]
b = -7
Hence, a = 2 and b = -7.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1

Question 7.
A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which y-coordinate is changing 8 times as fast as the x-coordinate.
Solution:
Let P(x1, y1) be the point on the curve 6y = x3 + 2 whose y-coordinate is changing 8 times as fast as the x-coordinate.
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q7

Question 8.
A spherical soap bubble is expanding so that its radius is increasing at the rate of 0.02 cm/sec. At what rate is the surface area increasing, when its radius is 5 cm?
Solution:
Let r be the radius and S be the surface area of the soap bubble at any time t.
Then S = 4πr2
Differentiating w.r.t. t, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q8
Hence, the surface area of the soap bubble is increasing at the rate of 0.87c cm2 / sec.

Question 9.
The surface area of a spherical balloon is increasing at the rate of 2 cm2/sec. At what rate is the volume of the balloon is increasing, when the radius of the balloon is 6 cm?
Solution:
Let r be the radius, S be the surface area and V be the volume of the spherical balloon at any time t.
Then S = 4πr2 and V = \(\frac{4}{3} \pi r^{3}\)
Differentiating w.r.t. t, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q9
Hence, the volume of the spherical balloon is increasing at the rate of 6 cm3 / sec.

Question 10.
If each side of an equilateral triangle increases at the rate of √2 cm/sec, find the rate of increase of its area when its side of length is 3 cm.
Solution:
If x cm is the side of the equilateral triangle and A is its area, then \(A=\frac{\sqrt{3}}{4} x^{2}\)
Differentiating w.r.t. f, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q10
Hence, rate of increase of the area of equilateral triangle = \(\frac{3 \sqrt{6}}{2}\) cm2 / sec.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1

Question 11.
The volume of a sphere increases at the rate of 20 cm3/sec. Find the rate of change of its surface area, when its radius is 5 cm.
Solution:
Let r be the radius, S be the surface area and V be the volume of the sphere at any time t.
Then S = 4πr2 and V = \(\frac{4}{3} \pi r^{3}\)
Differentiating w.r.t. t, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q11
Hence, the surface area of the sphere is changing at the rate of 8 cm2/sec.

Question 12.
The edge of a cube is decreasing at the rate of 0.6 cm/sec. Find the rate at which its volume is decreasing, when the edge of the cube is 2 cm.
Solution:
Let x be the edge of the cube and V be its volume at any time t.
Then V = x3
Differentiating both sides w.r.t. t, we get
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q12
Hence, the volume of the cube is decreasing at the rate of 7.2 cm3/sec.

Question 13.
A man of height 2 meters walks at a uniform speed of 6 km/hr away from a lamp post of 6 meters high. Find the rate at which the length of the shadow is increasing.
Solution:
Let OA be the lamp post, MN the man, MB = x, his shadow, and OM = y, the distance of the man from the lamp post at time t.
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q13
Then \(\frac{d y}{d t}\) = 6 km/hr is the rate at which the man is moving at away from the lamp post.
\(\frac{d x}{d t}\) is the rate at which his shadow is increasing.
From the figure,
\(\frac{x}{2}=\frac{x+y}{6}\)
6x = 2x + 2y
4x = 2y
x = \(\frac{1}{2}\) y
\(\frac{d x}{d t}=\frac{1}{2} \frac{d y}{d t}=\frac{1}{2} \times 6=3 \mathrm{~km} / \mathrm{hr}\)
Hence, the length of the shadow is increasing at the rate of 3 km/hr.

Question 14.
A man of height 1.5 meters walks towards a lamp post of height 4.5 meters, at the rate of (\(\frac{3}{4}\)) meter/sec.
Find the rate at which
(i) his shadow is shortening
(ii) the tip of the shadow is moving.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q14
Let OA be the lamp post, MN the man, MB = x his shadow and OM = y the distance of the man from lamp post at time t.
Then \(\frac{d y}{d t}=\frac{3}{4}\) is the rate at which the man is moving towards the lamp post.
\(\frac{d x}{d t}\) is the rate at which his shadow is shortening.
B is the tip of the shadow and it is at a distance of x + y from the post.
\(\frac{d}{d t}(x+y)=\frac{d x}{d t}+\frac{d y}{d t}\) is the rate at which the tip of the shadow is moving.
From the figure,
\(\frac{x}{1.5}=\frac{x+y}{4.5}\)
45x = 15x + 15y
30x = 15y
x = \(\frac{1}{2}\)y
\(\frac{d x}{d t}=\frac{1}{2} \cdot \frac{d y}{d t}=\frac{1}{2}\left(\frac{3}{4}\right)=\left(\frac{3}{8}\right) \text { metre/sec }\)
and \(\frac{d x}{d t}+\frac{d y}{d t}=\frac{3}{8}+\frac{3}{4}=\left(\frac{9}{8}\right) \text { metres } / \mathrm{sec}\)
Hence (i) the shadow is shortening at the rate of (\(\frac{3}{8}\)) metre/sec, and
(ii) the tip of shadow is moving at the rate of (\(\frac{9}{8}\)) metres/sec.

Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1

Question 15.
A ladder 10 metres long is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at the rate of 1.2 metres per second, find how fast the top of the ladder is sliding down the wall, when the bottom is 6 metres away from the wall.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q15
Let AB be the ladder, where AB = 10 metres.
Let at time t seconds, the end A of the ladder be x metres from the wall and the end B be y metres from the ground.
Since, OAB is a right angled triangle, by Pythagoras’ theorem
x2 + y2 = 102 i.e. y2 = 100 – x2
Differentiating w.r.t. t, we get
2y \(\frac{d y}{d t}\) = 0 – 2x \(\frac{d x}{d t}\)
∴ \(\frac{d y}{d t}=-\frac{x}{y} \cdot \frac{d x}{d t}\) ……..(1)
Now, \(\frac{d x}{d t}\) = 1.2 metres/sec is the rate at which the bottom at of the ladder is pulled horizontally and \(\frac{d y}{d t}\) is the rate at which the top of ladder B is sliding.
When x = 6, y2 = 100 – 36 = 64
y = 8
(1) gives \(\frac{d y}{d t}=-\frac{6}{8}(1.2)=-\frac{6}{8} \times \frac{12}{10}\)
\(=-\frac{9}{10}=-0.9\)
Hence, the top of the ladder is sliding down the wall, at the rate of 0.9 metre/sec.

Question 16.
If water is poured into an inverted hollow cone whose semi-vertical angle is 30° so that its depth (measured along the axis) increases at the rate of 1 cm/sec. Find the rate at which the volume of water increases when the depth is 2 cm.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q16
Let r be the radius, h be the height, θ be the semi-vertical angle and V be the volume of the water at any time t.
Maharashtra Board 12th Maths Solutions Chapter 2 Applications of Derivatives Ex 2.1 Q16.1
Hence, the volume of water is increasing at the rate of \(\left(\frac{4 \pi}{3}\right)\) cm3/sec.

Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4

I : Choose correct alternatives.
Question 1.
If the equation 4x2 + hxy + y2 = 0 represents two coincident lines, then h = _________.
(A) ± 2
(B) ± 3
(C) ± 4
(D) ± 5
Solution:
(C) ± 4

Question 2.
If the lines represented by kx2 – 3xy + 6y2 = 0 are perpendicular to each other then _________.
(A) k = 6
(B) k = -6
(C) k = 3
(D) k = -3
Solution:
(B) k = -6

Question 3.
Auxiliary equation of 2x2 + 3xy – 9y2 = 0 is _________.
(A) 2m2 + 3m – 9 = 0
(B) 9m2 – 3m – 2 = 0
(C) 2m2 – 3m + 9 = 0
(D) -9m2 – 3m + 2 = 0
Solution:
(B) 9m2 – 3m – 2 = 0

Question 4.
The difference between the slopes of the lines represented by 3x2 – 4xy + y2 = 0 is _________.
(A) 2
(B) 1
(C) 3
(D) 4
Solution:
(A) 2

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 5.
If the two lines ax2 +2hxy+ by2 = 0 make angles α and β with X-axis, then tan (α + β) = _____.
(A) \(\frac{h}{a+b}\)
(B) \(\frac{h}{a-b}\)
(C) \(\frac{2 h}{a+b}\)
(D) \(\frac{2 h}{a-b}\)
Solution:
(D) \(\frac{2 h}{a-b}\)
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 1

Question 6.
If the slope of one of the two lines \(\frac{x^{2}}{a}+\frac{2 x y}{h}+\frac{y^{2}}{b}\) = 0 is twice that of the other, then ab:h2 = ___.
(A) 1 : 2
(B) 2 : 1
(C) 8 : 9
(D) 9 : 8
Solution:
(D) 9 : 8
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 2

Question 7.
The joint equation of the lines through the origin and perpendicular to the pair of lines 3x2 + 4xy – 5y2 = 0 is _________.
(A) 5x2 + 4xy – 3y2 = 0
(B) 3x2 + 4xy – 5y2 = 0
(C) 3x2 – 4xy + 5y2 = 0
(D) 5x2 + 4xy + 3y2 = 0
Solution:
(A) 5x2 + 4xy – 3y2 = 0

Question 8.
If acute angle between lines ax2 + 2hxy + by2 = 0 is, \(\frac{\pi}{4}\) then 4h2 = _________.
(A) a2 + 4ab + b2
(B) a2 + 6ab + b2
(C) (a + 2b)(a + 3b)
(D) (a – 2b)(2a + b)
Solution:
(B) a2 + 6ab + b2

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 9.
If the equation 3x2 – 8xy + qy2 + 2x + 14y + p = 1 represents a pair of perpendicular lines then
the values of p and q are respectively _________.
(A) -3 and -7
(B) -7 and -3
(C) 3 and 7
(D) -7 and 3
Solution:
(B) -7 and -3

Question 10.
The area of triangle formed by the lines x2 + 4xy + y2 = 0 and x – y – 4 = 0 is _________.
(A) \(\frac{4}{\sqrt{3}}\) Sq. units
(B) \(\frac{8}{\sqrt{3}}\) Sq. units
(C) \(\frac{16}{\sqrt{3}}\) Sq. units
(D)\(\frac{15}{\sqrt{3}}\) Sq. units
Solution:
(B) \(\frac{8}{\sqrt{3}}\) Sq. units
[Hint : Area = \(\frac{p^{2}}{\sqrt{3}}\), where p is the length of perpendicular from the origin to x – y – 4 = 0]

Question 11.
The combined equation of the co-ordinate axes is _________.
(A) x + y = 0
(B) x y = k
(C) xy = 0
(D) x – y = k
Solution:
(C) xy = 0

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 12.
If h2 = ab, then slope of lines ax2 + 2hxy + by2 = 0 are in the ratio _________.
(A) 1 : 2
(B) 2 : 1
(C) 2 : 3
(D) 1 : 1
Solution:
(D) 1 : 1
[Hint: If h2 = ab, then lines are coincident. Therefore slopes of the lines are equal.]

Question 13.
If slope of one of the lines ax2 + 2hxy + by2 = 0 is 5 times the slope of the other, then 5h2 = _________.
(A) ab
(B) 2 ab
(C) 7 ab
(D) 9 ab
Solution:
(D) 9 ab

Question 14.
If distance between lines (x – 2y)2 + k(x – 2y) = 0 is 3 units, then k =
(A) ± 3
(B) ± 5\(\sqrt {5}\)
(C) 0
(D) ± 3\(\sqrt {5}\)
Solution:
(D) ± 3\(\sqrt {5}\)
[Hint: (x – 2y)2 + k(x – 2y) = 0
∴ (x – 2y)(x – 2y + k) = 0
∴ equations of the lines are x – 2y = 0 and x – 2y + k = 0 which are parallel to each other.
∴ \(\left|\frac{k-0}{\sqrt{1+4}}\right|\) = 3
∴ k = ± 3\(\sqrt {5}\)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

II. Solve the following.
Question 1.
Find the joint equation of lines:
(i) x – y = 0 and x + y = 0
Solution:
The joint equation of the lines x – y = 0 and
x + y = 0 is
(x – y)(x + y) = 0
∴ x2 – y2 = 0.

(ii) x + y – 3 = 0 and 2x + y – 1 = 0
Solution:
The joint equation of the lines x + y – 3 = 0 and 2x + y – 1 = 0 is
(x + y – 3)(2x + y – 1) = 0
∴ 2x2 + xy – x + 2xy + y2 – y – 6x – 3y + 3 = 0
∴ 2x2 + 3xy + y2 – 7x – 4y + 3 = 0.

(iii) Passing through the origin and having slopes 2 and 3.
Solution:
We know that the equation of the line passing through the origin and having slope m is y = mx. Equations of the lines passing through the origin and having slopes 2 and 3 are y = 2x and y = 3x respectively.
i.e. their equations are
2x – y = 0 and 3x – y = 0 respectively.
∴ their joint equation is (2x – y)(3x – y) = 0
∴ 6x2 – 2xy – 3xy + y2 = 0
∴ 6x2 – 5xy + y2 = 0.

(iv) Passing through the origin and having inclinations 60° and 120°.
Solution:
Slope of the line having inclination θ is tan θ .
Inclinations of the given lines are 60° and 120°
∴ their slopes are m1 = tan60° = \(\sqrt {3}\) and
m2 = tan 120° = tan (180° – 60°)
= -tan 60° = –\(\sqrt {3}\)
Since the lines pass through the origin, their equa-tions are
y = \(\sqrt {3}\)x and y= –\(\sqrt {3}\)x
i.e., \(\sqrt {3}\)x – y = 0 and \(\sqrt {3}\)x + y = 0
∴ the joint equation of these lines is
(\(\sqrt {3}\)x – y)(\(\sqrt {3}\)x + y) = 0
∴ 3x2 – y2 = 0.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(v) Passing through (1, 2) amd parallel to the co-ordinate axes.
Solution:
Equations of the coordinate axes are x = 0 and y = 0
∴ the equations of the lines passing through (1, 2) and parallel to the coordinate axes are x = 1 and y =1
i.e. x – 1 = 0 and y – 2 0
∴ their combined equation is
(x – 1)(y – 2) = 0
∴ x(y – 2) – 1(y – 2) = 0
∴ xy – 2x – y + 2 = 0

(vi) Passing through (3, 2) and parallel to the line x = 2 and y = 3.
Solution:
Equations of the lines passing through (3, 2) and parallel to the lines x = 2 and y = 3 are x = 3 and y = 2.
i.e. x – 3 = 0 and y – 2 = 0
∴ their joint equation is
(x – 3)(y – 2) = 0
∴ xy – 2x – 3y + 6 = 0.

(vii) Passing through (-1, 2) and perpendicular to the lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0.
Solution:
Let L1 and L2 be the lines passing through the origin and perpendicular to the lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0 respectively.
Slopes of the lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0 are \(-\frac{1}{2}\) and \(-\frac{3}{-4}=\frac{3}{4}\) respectively.
∴ slopes of the lines L1and L2 are 2 and \(\frac{-4}{3}\) respectively.
Since the lines L1 and L2 pass through the point (-1, 2), their equations are
∴ (y – y1) = m(x – x1)
∴ (y – 2) = 2(x + 1)
⇒ y – 1 = 2x + 2
⇒ 2x – y + 4 = 0 and
∴ (y – 2) = \(\left(\frac{-4}{3}\right)\)(x + 1)
⇒ 3y – 6 = (-4)(x + 1)
⇒ 3y – 6 = -4x + 4
⇒ 4x + 3y – 6 + 4 = 0
⇒ 4x + 3y – 2 = 0
their combined equation is
∴ (2x – y + 4)(4x + 3y – 2) = 0
∴ 8x2 + 6xy – 4x – 4xy – 3y2 + 2y + 16x + 12y – 8 = 0
∴ 8x2 + 2xy + 12x – 3y2 + 14y – 8 = 0

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(viii) Passing through the origin and having slopes 1 + \(\sqrt {3}\) and 1 – \(\sqrt {3}\)
Solution:
Let l1 and l2 be the two lines. Slopes of l1 is 1 + \(\sqrt {3}\) and that of l2 is 1 – \(\sqrt {3}\)
Therefore the equation of a line (l1) passing through the origin and having slope is
y = (1 + \(\sqrt {3}\))x
∴ (1 + \(\sqrt {3}\))x – y = 0 ..(1)
Similarly, the equation of the line (l2) passing through the origin and having slope is
y = (1 – \(\sqrt {3}\))x
∴ (1 – \(\sqrt {3}\))x – y = 0 …(2)
From (1) and (2) the required combined equation is
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 4
∴ (1 – 3)x2 – 2xy + y2 = 0
∴ -2x2 – 2xy + y2 = 0
∴ 2x2 + 2xy – y2 = 0
This is the required combined equation.

(ix) Which are at a distance of 9 units from the Y – axis.
Solution:
Equations of the lines, which are parallel to the Y-axis and at a distance of 9 units from it, are x = 9 and x = -9
i.e. x – 9 = 0 and x + 9 = 0
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 3
∴ their combined equation is
(x – 9)(x + 9) = 0
∴ x2 – 81 = 0.

(x) Passing through the point (3, 2), one of which is parallel to the line x – 2y = 2 and other is perpendicular to the line y = 3.
Solution:
Let L1 be the line passes through (3, 2) and parallel to the line x – 2y = 2 whose slope is \(\frac{-1}{-2}=\frac{1}{2}\)
∴ slope of the line L1 is \(\frac{1}{2}\).
∴ equation of the line L1 is
y – 2 = \(\frac{1}{2}\)(x – 3)
∴ 2y – 4 = x – 3 ∴ x – 2y + 1 = 0
Let L2 be the line passes through (3, 2) and perpendicular to the line y = 3.
∴ equation of the line L2 is of the form x = a.
Since L2 passes through (3, 2), 3 = a
∴ equation of the line L2 is x = 3, i.e. x – 3 = 0
Hence, the equations of the required lines are
x – 2y + 1 = 0 and x – 3 = 0
∴ their joint equation is
(x – 2y + 1)(x – 3) = 0
∴ x2 – 2xy + x – 3x + 6y – 3 = 0
∴ x2 – 2xy – 2x + 6y – 3 = 0.

(xi) Passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18.
Solution:
Let L1 and L2 be the lines passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18 respectively.
Slopes of the lines x + 2y = 19 and 3x + y = 18 are \(-\frac{1}{2}\) and \(-\frac{3}{1}\) = -3 respectively.
Since the lines L1 and L2 pass through the origin, their equations are
y = 2x and y = \(\frac{1}{3}\)x
i.e. 2x – y = 0 and x – 3y = 0
∴ their combined equation is
(2x – y)(x – 3y) = 0
∴ 2x2 – 6xy – xy + 3y2 = 0
∴ 2x2 – 7xy + 3y2 = 0.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 2.
Show that each of the following equation represents a pair of lines.
(i) x2 + 2xy – y2 = 0
Solution:
Comparing the equation x2 + 2xy – y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 2, i.e. h = 1 and b = -1
∴ h2 – ab = (1)2 – 1(-1) = 1 + 1=2 > 0
Since the equation x2 + 2xy – y2 = 0 is a homogeneous equation of second degree and h2 – ab > 0, the given equation represents a pair of lines which are real and distinct.

(ii) 4x2 + 4xy + y2 = 0
Solution:
Comparing the equation 4x2 + 4xy + y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 4, 2h = 4, i.e. h = 2 and b = 1
∴ h2 – ab = (2)2 – 4(1) = 4 – 4 = 0
Since the equation 4x2 + 4xy + y2 = 0 is a homogeneous equation of second degree and h2 – ab = 0, the given equation represents a pair of lines which are real and coincident.

(iii) x2 – y2 = 0
Solution:
Comparing the equation x2 – y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 0, i.e. h = 0 and b = -1
∴ h2 – ab = (0)2 – 1(-1) = 0 + 1 = 1 > 0
Since the equation x2 – y2 = 0 is a homogeneous equation of second degree and h2 – ab > 0, the given equation represents a pair of lines which are real and distinct.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) x2 + 7xy – 2y2 = 0
Solution:
Comparing the equation x2 + 7xy – 2y2 = 0
a = 1, 2h = 7 i.e., h = \(\frac{7}{2}\) and b = -2
∴ h2 – ab = \(\left(\frac{7}{2}\right)^{2}\) – 1(-2)
= \(\frac{49}{4}\) + 2
= \(\frac{57}{4}\) i.e. 14.25 = 14 > 0
Since the equation x2 + 7xy – 2y2 = 0 is a homogeneous equation of second degree and h2 – ab > 0, the given equation represents a pair of lines which are real and distinct.

(v) x2 – 2\(\sqrt {3}\) xy – y2 = 0
Solution:
Comparing the equation x2 – 2\(\sqrt {3}\) xy – y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h= -2\(\sqrt {3}\), i.e. h = –\(\sqrt {3}\) and b = 1
∴ h2 – ab = (-\(\sqrt {3}\))2 – 1(1) = 3 – 1 = 2 > 0
Since the equation x2 – 2\(\sqrt {3}\)xy – y2 = 0 is a homo¬geneous equation of second degree and h2 – ab > 0, the given equation represents a pair of lines which are real and distinct.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 3.
Find the separate equations of lines represented by the following equations:
(i) 6x2 – 5xy – 6y2 = 0
Solution:
6x2 – 5xy – 6y2 = 0
∴ 6x2 – 9xy + 4xy – 6y2 = 0
∴ 3x(2x – 3y) + 2y(2x – 3y) = 0
∴ (2x – 3y)(3x + 2y) = 0
∴ the separate equations of the lines are
2x – 3y = 0 and 3x + 2y = 0.

(ii) x2 – 4y2 = 0
Solution:
x2 – 4y2 = 0
∴ x2 – (2y)2 = 0
∴(x – 2y)(x + 2y) = 0
∴ the separate equations of the lines are
x – 2y = 0 and x + 2y = 0.

(iii) 3x2 – y2 = 0
Solution:
3x2 – y2 = 0
∴ (\(\sqrt {3}\) x)2 – y2 = 0
∴ (\(\sqrt {3}\)x – y)(\(\sqrt {3}\)x + y) = 0
∴ the separate equations of the lines are
\(\sqrt {3}\)x – y = 0 and \(\sqrt {3}\)x + y = 0.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) 2x2 + 2xy – y2 = 0
Solution:
2x2 + 2xy – y2 = 0
∴ The auxiliary equation is -m2 + 2m + 2 = 0
∴ m2 – 2m – 2 = 0
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 5
m1 = 1 + \(\sqrt {3}\) and m2 = 1 – \(\sqrt {3}\) are the slopes of the lines.
∴ their separate equations are
y = m1x and y = m2x
i.e. y = (1 + \(\sqrt {3}\))x and y = (1 – \(\sqrt {3}\))x
i.e. (\(\sqrt {3}\) + 1)x – y = 0 and (\(\sqrt {3}\) – 1)x + y = 0.

Question 4.
Find the joint equation of the pair of lines through the origin and perpendicular to the lines
given by :
(i) x2 + 4xy – 5y2 = 0
Solution:
Comparing the equation x2 + 4xy – 5y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 4, b= -5
Let m1 and m2 be the slopes of the lines represented by x2 + 4xy – 5y2 = 0.
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 6
Now, required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m1y = -x and m2y = -x
i.e. x + m1y = 0 and x + m2y = 0
∴ their combined equation is
(x + m1x + m2y) = 0
∴ x2 + (m1 + m2)xy + m1m2y2 = 0
∴ x2 + \(\frac{4}{5}\)xy – \(\frac{1}{5}\)y2 = 0 …[By (1)]
∴ 5x2 + 4xy – y2 = 0

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) 2x2 – 3xy – 9y2 = 0
Solution:
Comparing the equation 2x2 – 3xy – 9y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 2, 2h = -3, b = -9
Let m1 and m2 be the slopes of the lines represented by 2x2 – 3xy – 9y2 = 0
∴ m1 + m2 =\(\frac{-2 h}{b}=-\frac{3}{9}\) and m1m2 = \(\frac{a}{b}=-\frac{2}{9}\) …(1)
Now, required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m1y = -x and m2y = -x
i.e. x + m1y = 0 and x + m2y = 0
∴ their combined equation is
(x + m1y)(x + m2y) = 0
∴ x2 + (m1 + m2)xy + m1m2y2 = 0
∴ x2 + \(\left(-\frac{3}{9}\right)\)xy + \(\left(-\frac{2}{9}\right)\)y2 = 0 …[By (1)]
∴ 9x2 – 3xy – 2y2 = 0

(iii) x2 + xy – y2 = 0
Solution:
Comparing the equation x2+ xy – y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 1, b = -1
Let m1 and m2 be the slopes of the lines represented by x2 + xy – y2 = 0
∴ m1 + m2 = \(\frac{-2 h}{b}=\frac{-1}{-1}\) and m1m2 = \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{-1}\) = -1 ..(1)
Now, required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(\frac{-1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m1y = -x and m2y = -x
i.e. x + m1y = 0 and x + m2y = 0
∴ their combined equation is
(x + m1y)(x + m2y) = 0
∴ x2 + (m1 + m2) + m1m2y2 = 0
∴ x2 + 1xy + (-1)y2 = 0 …[By (1)]
∴ x2 + xy – y2 = 0

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 5.
Find k if
(i) The sum of the slopes of the lines given by 3x2 + kxy – y2 = 0 is zero.
Solution:
Comparing the equation 3x2 + kxy – y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 3, 2h = k, b = -1
Let m1 and m2 be the slopes of the lines represented by 3x2 + kxy – y2 = 0.
∴ m1 + m2 = \(\frac{-2 h}{b}=\frac{-k}{-1}\) = k
Now, m1 + m2 = 0 … (Given)
∴ k = 0.

(ii) The sum of slopes of the lines given by 2x2 + kxy – 3y2 = 0 is equal to their product.
Question is modified.
The sum of slopes of the lines given by x2 + kxy – 3y2 = 0 is equal to their product.
Solution:
Comparing the equation x2 + kxy – 3y2 = 0, with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = k, b = -3
Let m1 and m2 be the slopes of the lines represented by x2 + kxy – 3y2 = 0.
∴ m1 + m2 = \(-\frac{2 h}{b}=\frac{-k}{-3}=\frac{k}{3}\)
and m1m2 = \(\frac{a}{b}=\frac{1}{-3}=\frac{-1}{3}\)
Now, m1 + m2 = m1m2 … (Given)
∴ \(\frac{k}{3}=\frac{-1}{3}\)
∴ k = -1.

(iii) The slope of one of the lines given by 3x2 – 4xy + ky2 = 0 is 1.
Solution:
The auxiliary equation of the lines given by 3x2 – 4xy + ky2 = 0 is km2 – 4m + 3 = 0.
Given, slope of one of the lines is 1.
∴ m = 1 is the root of the auxiliary equation km2 – 4m + 3 = 0.
∴ k(1)2 – 4(1) + 3 = 0
∴ k – 4 + 3 = 0
∴ k = 1.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) One of the lines given by 3x2 – kxy + 5y2 = 0 is perpendicular to the 5x + 3y = 0.
Solution:
The auxiliary equation of the lines represented by 3x2 – kxy + 5y2 = 0 is 5m2 – km + 3 = 0.
Now, one line is perpendicular to the line 5x + 3y = 0, whose slope is \(-\frac{5}{3}\).
∴ slope of that line = m = \(\frac{3}{5}\)
∴ m = \(\frac{3}{5}\) is the root of the auxiliary equation 5
5m2 – km + 3 = 0.
∴ 5\(\left(\frac{3}{5}\right)^{2}\) – k\(\left(\frac{3}{5}\right)\) + 3 = 0
∴ \(\frac{9}{5}-\frac{3 k}{5}\) + 3 = 0
∴ 9 – 3k + 15 = 0
∴ 3k = 24
∴ k = 8.

(v) The slope of one of the lines given by 3x2 + 4xy + ky2 = 0 is three times the other.
Solution:
3x2 + 4xy + ky2 = 0
∴ divide by x2
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 7
∴ y = mx
∴ \(\frac{\mathrm{y}}{\mathrm{x}}\) = m
put \(\frac{\mathrm{y}}{\mathrm{x}}\) = m in equation (1)
Comparing the equation km2 + 4m + 3 = 0 with ax2 + 2hxy+ by2 = 0, we get,
a = k, 2h = 4, b = 3
m1 = 3m2 ..(given condition)
m1 + m2 = \(\frac{-2 h}{k}=-\frac{4}{k}\)
m1m2 = \(\frac{a}{b}=\frac{3}{k}\)
m1 + m2 = \(-\frac{4}{\mathrm{k}}\)
4m2 = \(-\frac{4}{\mathrm{k}}\) …(m1 = 3m2)
m2 = \(-\frac{1}{\mathrm{k}}\)
m1m2 = \(\frac{3}{k}\)
\(3 \mathrm{~m}_{2}^{2}=\frac{3}{\mathrm{k}}\) …(m1 = 3m2)
\(3\left(-\frac{1}{\mathrm{k}}\right)^{2}=\frac{3}{\mathrm{k}}\) …(m2 = \(-\frac{1}{k}\))
\(\frac{1}{k^{2}}=\frac{1}{k}\)
k2 = k
k = 1 or k = 0

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(vi) The slopes of lines given by kx2 + 5xy + y2 = 0 differ by 1.
Solution:
Comparing the equation kx2 + 5xy +y2 = 0 with ax2 + 2hxy + by2
a = k, 2h = 5 i.e. h = \(\frac{5}{2}\)
m1 + m2 = \(\frac{-2 h}{b}=-\frac{5}{1}\) = -5
and m1m2 = \(\frac{a}{b}=\frac{k}{1}\) = k
the slope of the line differ by (m1 – m2) = 1 …(1)
∴ (m1 – m2)2 = (m1 + m2)2 – 4m1m2
(m1 – m2)2 = (-5)2 – 4(k)
(m1 – m2)2 = 25 – 4k
1 = 25 – 4k ..[By (1)]
4k = 24
k = 6

(vii) One of the lines given by 6x2 + kxy + y2 = 0 is 2x + y = 0.
Solution:
The auxiliary equation of the lines represented by 6x2 + kxy + y2 = 0 is
m2 + km + 6 = 0.
Since one of the line is 2x + y = 0 whose slope is m = -2.
∴ m = -2 is the root of the auxiliary equation m2 + km + 6 = 0.
∴ (-2)2 + k(-2) + 6 = 0
∴ 4 – 2k + 6 = 0
∴ 2k = 10 ∴ k = 5

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 6.
Find the joint equation of the pair of lines which bisect angle between the lines given by x2 + 3xy + 2y2 = 0
Solution:
x2 + 3xy + 2y2 = 0
∴ x2 + 2xy + xy + 2y2 = 0
∴ x(x + 2y) + y(x + 2y) = 0
∴ (x + 2y)(x + y) = 0
∴ separate equations of the lines represented by x2 + 3xy + 2y2 = 0 are x + 2y = 0 and x + y = 0.
Let P (x, y) be any point on one of the angle bisector. Since the points on the angle bisectors are equidistant from both the lines,
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 8
the distance of P (x, y) from the line x + 2y = 0
= the distance of P(x, y) from the line x + y = 0
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 9
∴ 2(x + 2y)2 = 5(x + y)2
∴ 2(x2 + 4xy + 4y2) = 5(x2 + 2xy + y2)
∴ 2x2 + 8xy + 8y2 = 5x2 + 10xy + 5y2
∴ 3x2 + 2xy – 3y2 = 0.
This is the required joint equation of the lines which bisect the angles between the lines represented by x2 + 3xy + 2y2 = 0.

Question 7.
Find the joint equation of the pair of lies through the origin and making equilateral triangle with the line x = 3.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 10
Let OA and OB be the lines through the origin making an angle of 60° with the line x = 3.
∴ OA and OB make an angle of 30° and 150° with the positive direction of X-axis
∴ slope of OA = tan 30° = 1/\(\sqrt {3}\)
∴ equation of the line OA is y = \(\frac{1}{\sqrt{3}}\)x
∴ \(\sqrt {3}\)y = x ∴ x – \(\sqrt {3}\)y = 0
Slope of OB = tan 150° = tan (180° – 30°)
= – tan 30°= -1/\(\sqrt {3}\)
∴ equation of the line OB is y = \(\frac{-1}{\sqrt{3}}\)x
∴ \(\sqrt {3}\)y = -x ∴ x + \(\sqrt {3}\)y = 0
∴ required combined equation of the lines is
(x – \(\sqrt {3}\)y) (x + \(\sqrt {3}\)y) = 0
i.e. x2 – 3y2 = 0.

Question 8.
Show that the lines x2 – 4xy + y2 = 0 and x + y = 10 contain the sides of an equilateral triangle. Find the area of the triangle.
Solution:
We find the joint equation of the pair of lines OA and OB through origin, each making an angle of 60° with x + y = 10 whose slope is -1.
Let OA (or OB) has slope m.
∴ its equation is y = mx … (1)
Also, tan 60° = \(\left|\frac{m-(-1)}{1+m(-1)}\right|\)
∴ \(\sqrt {3}\) = \(\left|\frac{m+1}{1-m}\right|\)
Squaring both sides, we get,
3 = \(\frac{(m+1)^{2}}{(1-m)^{2}}\)
∴ 3(1 – 2m + m2) = m2 + 2m + 1
∴ 3 – 6m + 3m2 = m2 + 2m + 1
∴ 2m2 – 8m + 2 = 0
∴ m2 – 4m + 1 = 0
∴ \(\left(\frac{y}{x}\right)^{2}\) – 4\(\left(\frac{y}{x}\right)\) + 1 = 0 …[By (1)]
∴ y2 – 4xy + x2 = 0
∴ x2 – 4xy + y\left(\frac{y}{x}\right) = 0 is the joint equation of the two lines through the origin each making an angle of 60° with x + y = 10
∴ x2 – 4xy + y2 = 0 and x + y = 10 form a triangle OAB which is equilateral.
Let seg OM ⊥r line AB whose question is x + y = 10
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 11

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 9.
If the slope of one of the lines represented by ax2 + 2hxy + by2 = 0 is three times the other then prove that 3h2 = 4ab.
Solution:
Let m1 and m2 be the slopes of the lines represented by ax2 + 2hxy + by2 = 0.
∴ m1 + m2 = \(-\frac{2 h}{b}\) and m1m2 = \(\frac{a}{b}\)
We are given that m2 = 3m1
∴ m1 + 3m1 = \(-\frac{2 h}{b}\) 4m1 = \(-\frac{2 h}{b}\)
∴ m1 = \(-\frac{h}{2 b}\) …(1)
Also, m1(3m1) = \(\frac{a}{b}\) ∴ 3m12 = \(\frac{a}{b}\)
∴ 3\(\left(-\frac{h}{2 b}\right)^{2}\) = \(\frac{a}{b}\) ….[By (1)]
∴ \(\frac{3 h^{2}}{4 b^{2}}=\frac{a}{b}\)
∴ 3h2 = 4ab, as b ≠0.

Question 10.
Find the combined equation of the bisectors of the angles between the lines represented by 5x2 + 6xy – y2 = 0.
Solution:
Comparing the equation 5x2 + 6xy – y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = 5, 2h = 6, b = -1
Let m1 and m2 be the slopes of the lines represented by 5x2 + 6xy – y2 = 0.
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 12
The separate equations of the lines are
y = m1x and y = m2x, where m1 ≠ m2
i.e. m1x – y = 0 and m1x – y = 0.
Let P (x, y) be any point on one of the bisector of the angles between the lines.
∴ the distance of P from the line m1x – y = 0 is equal to the distance of P from the line m2x – y = 0.
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 13
∴ (m22 + 1)(m1x – y)2 = (m12 + 1)(m2x – y)2
∴ (m22 + 1)(m12x2 – 2m1xy + y2) = (m12 + 1)(m22x2 – 2m2xy + y2)
∴ m12m22x2 – 2m1m12y2xy + m22y2 + m12x2 – 2m12xy + y2
= m12m22x2 – 2m12m2xy + m12y2 + m22x2 – 2m2xy + y2
∴ (m12 – m22)x2 + 2m1m2(m1 – m2)xy – 2(m1 – m2)xy – (m12 – m22)y2 = 0
Dividing throughout by m1 – m2 (≠0), we get,
(m1 + m2)x2 + 2m1m2xy – 2xy – (m1 + m2)y2 = 0
∴ 6x2 – 10xy – 2xy – 6y2 = 0 …[By (1)]
∴ 6x2 – 12xy – 6y2 = 0
∴ x2 – 2xy – y2 = 0
This is the joint equation of the bisectors of the angles between the lines represented by 5x2 + 6xy – y2 = 0.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 11.
Find a, if the sum of the slopes of the lines represented by ax2 + 8xy + 5y2 = 0 is twice their product.
Solution :
Comparing the equation ax2 + 8xy + 5y2 = 0 with ax2 + 2hxy + by2 = 0,
we get, a = a, 2h = 8, b = 5
Let m1 and m2 be the slopes of the lines represented by ax2 + 8xy + 5y2 = 0.
∴ m1 + m2 = \(\frac{-2 h}{b}=-\frac{8}{5}\)
and m1m2 = \(\frac{a}{b}=\frac{a}{5}\)
Now, (m1 + m2) = 2(m1m2)
\(-\frac{8}{5}\) = \(2\left(\frac{a}{5}\right)\)
a = -4

Question 12.
If the line 4x – 5y = 0 coincides with one of the lines given by ax2 + 2hxy + by2 = 0, then show that 25a + 40h +16b = 0.
Solution :
The auxiliary equation of the lines represented by ax2 + 2hxy + by2 = 0 is bm2 + 2hm + a = 0
Given that 4x – 5y = 0 is one of the lines represented by ax2 + 2hxy + by2 = 0.
The slope of the line 4x – 5y = 0 is \(\frac{-4}{-5}=\frac{4}{5}\)
∴ m = \(\frac{4}{5}\) is a root of the auxiliary equation bm2 + 2hm + a = 0.
∴ b\(\left(\frac{4}{5}\right)^{2}\) + 2h\(\left(\frac{4}{5}\right)\) + a = 0
∴ \(\frac{16 b}{25}+\frac{8 h}{5}\) + a = 0
∴ 16b + 40h + 25a = 0 i.e.
∴ 25a + 40h + 16b = 0

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 13.
Show that the following equations represent a pair of lines. Find the acute angle between them :
(i) 9x2 – 6xy + y2 + 18x – 6y + 8 = 0
Solution:
Comparing this equation with
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get,
a = 9, h = -3, b = 1, g = 9, f = -3 and c = 8.
∴ D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
9 & -3 & 9 \\
-3 & 1 & -3 \\
9 & -3 & 8
\end{array}\right|\)
= 9(8 – 9) + 3(-24 + 27) + 9(9 – 9)
= 9(-1) + 3(3) + 9(0)
= -9 + 9 + 0 = 0
and h2 – ab = (-3)2 – 9(1) = 9 – 9 = 0
∴ the given equation represents a pair of lines.
Let θ be the acute angle between the lines.
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 14
∴ tan θ = tan0°
∴ θ = 0°.

(ii) 2x2 + xy – y2 + x + 4y – 3 = 0
Solution:
Comparing this equation with
ax2 + 2hxy + by2 + 2gx + 2fy+ c = 0, we get,
a = 2, h = \(\frac{1}{2}\), b = -1, g = \(\frac{1}{2}\), f = 2 and c = -3
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 15
= -2 + 1 + 1
= -2 + 2= 0
∴ the given equation represents a pair of lines.
Let θ be the acute angle between the lines.
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 16
∴ tan θ = tan 3
∴ θ = tan-1(3)

(iii) (x – 3)2 + (x – 3)(y – 4) – 2(y – 4)2 = 0.
Solution :
Put x – 3 = X and y – 4 = Y in the given equation, we get,
X2 + XY – 2Y2 = 0
Comparing this equation with ax2 + 2hxy + by2 = 0, we get,
a = 1, h = \(\frac{1}{2}\), b = -2
This is the homogeneous equation of second degreeand h2 – ab = \(\left(\frac{1}{2}\right)^{2}\) – 1(-2)
= \(\frac{1}{4}\) + 2 = \(\frac{9}{4}\) > 0
Hence, it represents a pair of lines passing through the new origin (3, 4).
Let θ be the acute angle between the lines.
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 17
∴ tanθ = 3 ∴ θ = tan-1(3)

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 14.
Find the combined equation of pair of lines through the origin each of which makes angle of 60° with the Y-axis.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 18
Let OA and OB be the lines through the origin making an angle of 60° with the Y-axis.
Then OA and OB make an angle of 30° and 150° with the positive direction of X-axis.
∴ slope of OA = tan 30° = \(\frac{1}{\sqrt{3}}\)
∴ equation of the line OA is
y = \(\frac{1}{\sqrt{3}}\) = x, i.e. x – \(\sqrt {3}\)y = 0
Slope of OB = tan 150° = tan (180° – 30°)
= tan 30° = \(-\frac{1}{\sqrt{3}}\)
∴ equation of the line OB is
y = \(-\frac{1}{\sqrt{3}}\)x, i.e. x + \(\sqrt {3}\) y = 0
∴ required combined equation is
(x – \(\sqrt {3}\)y)(x + \(\sqrt {3}\)y) = 0
i.e. x2 – 3y2 = 0.

Question 15.
If lines representedby ax2 + 2hxy + by2 = 0 make angles of equal measures with the co-ordinate
axes then show that a = ± b.
OR
Show that, one of the lines represented by ax2 + 2hxy + by2 = 0 will make an angle of the same measure with the X-axis as the other makes with the Y-axis, if a = ± b.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 19
Let OA and OB be the two lines through the origin represented by ax2 + 2hxy + by2 = 0.
Since these lines make angles of equal measure with the coordinate axes, they make angles ∝ and \(\frac{\pi}{2}\) – ∝ with the positive direction of X-axis or ∝ and \(\frac{\pi}{2}\) + ∝ with thepositive direction of X-axis.
∴ slope of the line OA = m1 = tan ∝
and slope of the line OB = m2
= tan(\(\frac{\pi}{2}\) – ∝) or tan(\(\frac{\pi}{2}\) + ∝)
i.e. m2 = cot ∝ or m2 = -cot ∝
∴ m1m2 – tan ∝ x cot ∝ = 1
OR m1m2 = tan ∝ (-cot ∝) = -1
i.e. m1m2 = ± 1
But m1m2 = \(\frac{a}{b}\)
∴ \(\frac{a}{b}\)= ±1 ∴ a = ±b
This is the required condition.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 16.
Show that the combined equation of a pair of lines through the origin and each making an angle of ∝ with the line x + y = 0 is x2 + 2(sec 2∝) xy + y2 = 0.
Solution:
Let OA and OB be the required lines.
Let OA (or OB) has slope m.
∴ its equation is y = mx … (1)
It makes an angle ∝ with x + y = 0 whose slope is -1. m +1
∴ tan ∝ = \(\left|\frac{m+1}{1+m(-1)}\right|\)
Squaring both sides, we get,
tan2∝ = \(\frac{(m+1)^{2}}{(1-m)^{2}}\)
∴ tan2∝(1 – 2m + m2) = m2 + 2m + 1
∴ tan2∝ – 2m tan2∝ + m2tan2∝ = m2 + 2m + 1
∴ (tan2∝ – 1)m2 – 2(1 + tan2∝)m + (tan2∝ – 1) = 0
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 20
∴ y2 + 2xysec2∝ + x2 = 0
∴ x2 + 2(sec2∝)xy + y2 = 0 is the required equation.

Question 17.
Show that the line 3x + 4y+ 5 = 0 and the lines (3x + 4y)2 – 3(4x – 3y)2 =0 form an equilateral triangle.
Solution:
The slope of the line 3x + 4y + 5 = 0 is \(\frac{-3}{4}\)
Let m be the slope of one of the line making an angle of 60° with the line 3x + 4y + 5 = 0. The angle between the lines having slope m and m1 is 60°.
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 21
On squaring both sides, we get,
3 = \(\frac{(4 m+3)^{2}}{(4-3 m)^{2}}\)
∴ 3 (4 – 3m)2 = (4m + 3)2
∴ 3(16 – 24m + 9m2) = 16m2 + 24m + 9
∴ 48 – 72m + 27m2 = 16m2 + 24m + 9
∴ 11m2 – 96m + 39 = 0
This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = \(\frac{y}{x}\).
∴ the combined equation of the two lines is
11\(\left(\frac{y}{x}\right)^{2}\) – 96\(\left(\frac{y}{x}\right)\) + 39 = 0
∴ \(\frac{11 y^{2}}{x^{2}}-\frac{96 y}{x}\) + 39 = 0
∴ 11y2 – 96xy + 39x2 = 0
∴ 39x2 – 96xy + 11y2 = 0.
∴ 39x2 – 96xy + 11y2 = 0 is the joint equation of the two lines through the origin each making an angle of 60° with the line 3x + 4y + 5 = 0.
The equation 39x2 – 96xy + 11y2 = 0 can be written as :
-39x2 + 96xy – 11y2 = 0
i.e., (9x2 – 48x2) + (24xy + 72xy) + (16y2 – 27y2) = 0
i.e. (9x2 + 24xy + 16y2) – (48x2 – 72xy + 27y2) = 0
i.e. (9x2 + 24xy + 16y2) – 3(16x2 – 24xy + 9y2) = 0
i.e. (3x + 4y)2 – 3(4x – 3y)2 = 0
Hence, the line 3x + 4y + 5 = 0 and the lines
(3x + 4y)2 – 3(4x – 3y)2 form the sides of an equilateral triangle.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 18.
Show that lines x2 – 4xy + y2 = 0 and x + y = \(\sqrt {6}\) form an equilateral triangle. Find its area and perimeter.
Solution:
x2 – 4xy + y2 = 0 and x + y = \(\sqrt {6}\) form a triangle OAB which is equilateral.
Let OM be the perpendicular from the origin O to AB whose equation is x + y = \(\sqrt {6}\)
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 22
In right angled triangle OAM,
sin 60° = \(\frac{\mathrm{OM}}{\mathrm{OA}}\) ∴ \(\frac{\sqrt{3}}{2}\) = \(\frac{\sqrt{3}}{\mathrm{OA}}\)
∴ OA = 2
∴ length of the each side of the equilateral triangle OAB = 2 units.
∴ perimeter of ∆ OAB = 3 × length of each side
= 3 × 2 = 6 units.

Question 19.
If the slope of one of the lines given by ax2 + 2hxy + by2 = 0 is square of the other then show that a2b + ab2 + 8h3 = 6abh.
Solution:
Let m be the slope of one of the lines given by ax2 + 2hxy + by2 = 0.
Then the other line has slope m2
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 23
Multiplying by b3, we get,
-8h3 = ab2 + a2b – 6abh
∴ a2b + ab2 + 8h3 = 6abh
This is the required condition.

Question 20.
Prove that the product of lengths of perpendiculars drawn from P (x1, y1) to the lines repersented by ax2 + 2hxy + by2 = 0 is \(\left|\frac{a x_{1}^{2}+2 h x_{1} y_{1}+b y_{1}^{2}}{\sqrt{(a-b)^{2}+4 h^{2}}}\right|\)
Solution:
Let m1 and m2 be the slopes of the lines represented by ax2 + 2hxy + by2 = 0.
∴ m1 + m2 = \(-\frac{2 h}{b}\) and m1m2 = \(\frac{a}{b}\) …(1)
The separate equations of the lines represented by
ax2 + 2hxy + by2 = 0 are
y = m1x and y = m2x
i.e. m1x – y = 0 and m2x – y = 0
Length of perpendicular from P(x1, 1) on
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 24
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 25

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 21.
Show that the difference between the slopes of lines given by (tan2θ + cos2θ )x2 – 2xytanθ + (sin2θ )y2 = 0 is two.
Solution:
Comparing the equation (tan2θ + cos2θ)x2 – 2xy tan θ + (sin2θ) y2 = 0 with ax2 + 2hxy + by2 = 0, we get,
a = tan2θ + cos2θ, 2h = -2 tan θ and b = sin2θ
Let m1 and m2 be the slopes of the lines represented by the given equation.
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 26
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 27

Question 22.
Find the condition that the equation ay2 + bxy + ex + dy = 0 may represent a pair of lines.
Solution:
Comparing the equation
ay2 + bxy + ex + dy = 0 with
Ax2 + 2Hxy + By2 + 2Gx + 2Fy + C = 0, we get,
A = 0, H = \(\frac{b}{2}\), B = a,G = \(\frac{e}{2}\), F = \(\frac{d}{2}\), C = 0
The given equation represents a pair of lines,
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 28
i.e. if bed – ae2 = 0
i.e. if e(bd – ae) = 0
i.e. e = 0 or bd – ae = 0
i.e. e = 0 or bd = ae
This is the required condition.

Question 23.
If the lines given by ax2 + 2hxy + by2 = 0 form an equilateral triangle with the line lx + my = 1 then show that (3a + b)(a + 3b) = 4h2.
Solution:
Since the lines ax2 + 2hxy + by2 = 0 form an equilateral triangle with the line lx + my = 1, the angle between the lines ax2 + 2hxy + by2 = 0 is 60°.
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 29
∴ 3(a + b)2 = 4(h2 – ab)
∴ 3(a2 + 2ab + b2) = 4h2 – 4ab
∴ 3a2 + 6ab + 3b2 + 4ab = 4h2
∴ 3a2 + 10ab + 3b2 = 4h2
∴ 3a2 + 9ab + ab + 3b2 = 4h2
∴ 3a(a + 3b) + b(a + 3b) = 4h2
∴ (3a + b)(a + 3b) = 4h2
This is the required condition.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 24.
If line x + 2 = 0 coincides with one of the lines represented by the equation x2 + 2xy + 4y + k = 0 then show that k = -4.
Solution:
One of the lines represented by
x2 + 2xy + 4y + k = 0 … (1)
is x + 2 = 0.
Let the other line represented by (1) be ax + by + c = 0.
∴ their combined equation is (x + 2)(ax + by + c) = 0
∴ ax2 + bxy + cx + 2ax + 2by + 2c = 0
∴ ax2 + bxy + (2a + c)x + 2by + 2c — 0 … (2)
As the equations (1) and (2) are the combined equations of the same two lines, they are identical.
∴ by comparing their corresponding coefficients, we get,
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 30
∴ 1 = \(\frac{-4}{k}\)
∴ k = -4.

Question 25.
Prove that the combined equation of the pair of lines passing through the origin and perpendicular to the lines represented by ax2 + 2hxy + by2 = 0 is bx2 – 2hxy + ay2 = 0
Solution:
Let m1 and m2 be the slopes of the lines represented by ax2 + 2hxy + by2 = 0.
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 31
Now, required lines are perpendicular to these lines.
∴ their slopes are and \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(-\frac{1}{m_{1}}\)x and y = \(-\frac{1}{m_{2}}\)x
i.e. m1y= -x and m2y = -x
i.e. x + m1y = 0 and x + m2y = 0
∴ their combined equation is
(x + m1y)(x + m2y) = 0
∴ x2 + (m1 + m2)xy + m1m2y2 = 0
∴ x2\(\frac{-2 h}{b}\)x + \(\frac{a}{b}\)y2 = 0
∴ bx2 – 2hxy + ay2 = 0.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 26.
If equation ax2 – y2 + 2y + c = 1 represents a pair of perpendicular lines then find a and c.
Solution:
The given equation represents a pair of lines perpendicular to each other.
∴ coefficient of x2 + coefficient of y2 = 0
∴ a – 1 = 0 ∴ a = 1
With this value of a, the given equation is
x2 – y2 + 2y + c – 1 = 0
Comparing this equation with
Ax2 + 2Hxy + By2 + 2Gx + 2Fy + C = 0, we get,
A = 1, H = 0, B = -1, G = 0, F = 1, C = c – 1
Since the given equation represents a pair of lines,
D = \(\left|\begin{array}{ccc}
A & H & G \\
H & B & F \\
G & F & C
\end{array}\right|\) = 0
∴ \(\left|\begin{array}{rrr}
1 & 0 & 0 \\
0 & -1 & 1 \\
0 & 1 & c-1
\end{array}\right|\) = 0
∴ 1(-c + 1 – 1) – 0 + 0 = 0
∴ -c = 0
∴ c = 0.
Hence, a = 1, c = 0.

Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.3

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Ex 4.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.3

Question 1.
Find the joint equation of the pair of lines:
(i) Through the point (2, -1) and parallel to lines represented by 2x2 + 3xy – 9y2 = 0
Solution:
The combined equation of the given lines is
2x2 + 3 xy – 9y2 = 0
i.e. 2x2 + 6xy – 3xy – 9y2 = 0
i.e. 2x(x + 3y) – 3y(x + 3y) = 0
i.e. (x + 3y)(2x – 3y) = 0
∴ their separate equations are
x + 3y = 0 and 2x – 3y = 0
∴ their slopes are m1 = \(\frac{-1}{3}\) and m2 = \(\frac{-2}{-3}=\frac{2}{3}\).
The slopes of the lines parallel to these lines are m1 and m2, i.e. \(-\frac{1}{3}\) and \(\frac{2}{3}\).
∴ the equations of the lines with these slopes and through the point (2, -1) are
y + 1 = \(-\frac{1}{3}\) (x – 2) and y + 1 = \(\frac{2}{3}\)(x – 2)
i.e. 3y + 3= -x + 2 and 3y + 3 = 2x – 4
i.e. x + 3y + 1 = 0 and 2x – 3y – 7 = 0
∴ the joint equation of these lines is
(x + 3y + 1)(2x – 3y – 7) = 0
∴ 2x2 – 3xy – 7x + 6xy – 9y2 – 21y + 2x – 3y – 7 = 0
∴ 2x2 + 3xy – 9y2 – 5x – 24y – 7 = 0.

(ii) Through the point (2, -3) and parallel to lines represented by x2 + xy – y2 = 0
Solution:
Comparing the equation
x2 + xy – y2 = 0 … (1)
with ax2 + 2hxy + by2 = 0, we get,
a = 1, 2h = 1, b = -1
Let m1 and m2 be the slopes of the lines represented by (1).
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.3 1
The slopes of the lines parallel to these lines are m1 and m2.
∴ the equations of the lines with these slopes and through the point (2, -3) are
y + 3 = m1(x – 2) and y + 3 = m2(x – 2)
i.e. m1(x – 2) – (y + 3) = 0 and m2(x – 2) – (y + 3) = 0
∴ the joint equation of these lines is
[m1(x – 2) – (y + 3)][m2(x – 2) – (y + 3)] = 0
∴ m1m2(x – 2)2 – m1(x – 2)(y + 3) – m2(x – 2)(y + 3) + (y + 3)2 = o
∴ m1m2(x – 2)2 – (m1 + m2)(x – 2)(y + 3) + (y + 3)3 = 0
∴ -(x – 2)2 – (x – 2)(y + 3) + (y + 3)2 = 0 …… [By (2)]
∴ (x – 2)2 + (x – 2)(y + 3) – (y + 3)2 = 0
∴ (x2 – 4x + 4) + (xy + 3x – 2y – 6) – (y2 + 6y + 9) = 0
∴ x2 – 4x + 4 + xy + 3x – 2y – 6 – y2 – 6y – 9 = 0
∴ x2 + xy – y2 – x – 8y – 11 = 0.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 2.
Show that equation x2 + 2xy+ 2y2 + 2x + 2y + 1 = 0 does not represent a pair of lines.
Solution:
Comparing the equation
x2 + 2xy + 2y2 + 2x + 2y + 1 = 0 with
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get,
a = 1, h = 1, b = 2, g = 1, f = 1, c = 1.
The given equation represents a pair of lines, if
D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|\) = 0 and h2 – ab ≥ 0
Now, D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 1
\end{array}\right|\)
= 1 (2 – 1) – 1(1 – 1) + 1 (1 – 2)
= 1 – 0 – 1 = 0
and h2 – ab = (1)2 – 1(2) = -1 < 0
∴ given equation does not represent a pair of lines.

Question 3.
Show that equation 2x2 – xy – 3y2 – 6x + 19y – 20 = 0 represents a pair of lines.
Solution:
Comparing the equation
2x2 – xy – 3y2 – 6x + 19y – 20 = 0
with ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get,
a = 2, h = \(-\frac{1}{2}\), b = -3, g = -3, f = \(\frac{19}{2}\), c = -20.
∴ D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
2 & -\frac{1}{2} & -3 \\
-\frac{1}{2} & -3 & \frac{19}{2} \\
-3 & \frac{19}{2} & -20
\end{array}\right|\)
Taking \(\frac{1}{2}\) common from each row, we get,
D = \(\frac{1}{8}\left|\begin{array}{rrr}
4 & -1 & -6 \\
-1 & -6 & 19 \\
-6 & 19 & -40
\end{array}\right|\)
= \(\frac{1}{8}\)[4(240 – 361) + 1(40 + 114) – 6(-19 – 36)]
= \(\frac{1}{8}\)[4(-121) + 154 – 6(-55)]
= \(\frac{11}{8}\)[4(-11) + 14 – 6(-5)]
= \(\frac{1}{8}\)(-44 + 14 + 30) = 0
Also h2 – ab = \(\left(-\frac{1}{2}\right)^{2}\) – 2(-3) = \(\frac{1}{4}\) + 6 = \(\frac{25}{4}\) > 0
∴ the given equation represents a pair of lines.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 4.
Show the equation 2x2 + xy – y2 + x + 4y – 3 = 0 represents a pair of lines. Also find the acute angle between them.
Solution:
Comparing the equation
2x2 + xy — y2 + x + 4y — 3 = 0 with
ax2 + 2hxy + by2 + 2gx + 2fy + c — 0, we get,
a = 2, h = \(\frac{1}{2}\), b = -1, g = \(\frac{1}{2}\), f = 2, c = – 3.
∴ D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{lrr}
2 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & -1 & 2 \\
\frac{1}{2} & 2 & -3
\end{array}\right|\)
Taking \(\frac{1}{2}\) common from each row, we get,
D = \(\frac{1}{8}\left|\begin{array}{rrr}
4 & 1 & 1 \\
1 & -2 & 4 \\
1 & 4 & -6
\end{array}\right|\)
= \(\frac{1}{8}\)[4(12 —16) — 1( —6 — 4) + 1(4 + 2)]
= \(\frac{1}{8}\)[4( – 4) – 1(-10) + 1(6)]
= \(\frac{1}{8}\)(—16 + 10 + 6) = 0
Also, h2 – ab = \(\left(\frac{1}{2}\right)^{2}\) – 2(-1) = \(\frac{1}{4}\) + 2 = \(\frac{9}{4}\) > 0
∴ the given equation represents a pair of lines. Let θ be the acute angle between the lines
∴ tan θ = \(\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\)
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.3 2

Question 5.
Find the separate equation of the lines represented by the following equations :
(i) (x – 2)2 – 3(x – 2)(y + 1) + 2(y + 1)2 = 0
Solution:
(x – 2)2 – 3(x – 2)(y + 1) + 2(y + 1)2 = 0
∴ (x – 2)2 – 2(x – 2)(y + 1) – (x – 2)(y + 1) + 2(y + 1)2 = 0
∴ (x – 2) [(x – 2) – 2(y + 1)] – (y + 1)[(x – 2) – 2(y + 1)] = 0
∴ (x – 2)(x – 2 – 2y – 2) – (y + 1)(x – 2 – 2y – 2) = 0
∴ (x – 2)(x – 2y – 4) – (y + 1)(x – 2y – 4) = 0
∴ (x – 2y – 4)(x – 2 – y – 1) = 0
∴ (x – 2y – 4)(x – y – 3) = 0
∴ the separate equations of the lines are
x – 2y – 4 = 0 and x – y – 3 = 0.
Alternative Method :
(x – 2)2 – 3(x – 2)(y + 1) + 2(y + 1)2 = 0 … (1)
Put x – 2 = X and y + 1 = Y
∴ (1) becomes,
X2 – 3XY + 2Y2 = 0
∴ X2 – 2XY – XY + 2Y2 = 0
∴ X(X – 2Y) – Y(X – 2Y) = 0
∴ (X – 2Y)(X – Y) = 0
∴ the separate equations of the lines are
∴ X – 2Y = 0 and X – Y = 0
∴ (x – 2) – 2(y + 1) = 0 and (x – 2) – (y +1) = 0
∴ x – 2y – 4 = 0 and x – y – 3 = 0.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) 10(x + 1)2 + (x + 1)( y – 2) – 3(y – 2)2 = 0
Solution:
10(x + 1)2 + (x + 1)( y – 2) – 3(y – 2)2 = 0 …(1)
Put x + 1 = X and y – 2 = Y
∴ (1) becomes
10x2 + xy – 3y2 = 0
10x2 + 6xy – 5xy – 3y2 = 0
2x(5x + 3y) – y(5x + 3y) = 0
(2x – y)(5x + 3y) = 0
5x + 3y = 0 and 2x – y = 0
5x + 3y = 0
5(x + 1) + 3(y – 2) = 0
5x + 5 + 3y – 6 = 0
∴ 5x + 3y – 1 = 0
2x – y = 0
2(x + 1) – (y – 2) = 0
2x + 2 – y + 2 = 0
∴ 2x – y + 4 = 0

Question 6.
Find the value of k if the following equations represent a pair of lines :
(i) 3x2 + 10xy + 3y2 + 16y + k = 0
Solution:
Comparing the given equation with
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0,
we get, a = 3, h = 5, b = 3, g = 0, f= 8, c = k.
Now, given equation represents a pair of lines.
∴ abc + 2fgh – af2 – bg2 – ch2 = 0
∴ (3)(3)(k) + 2(8)(0)(5) – 3(8)2 – 3(0)2 – k(5)2 = 0
∴ 9k + 0 – 192 – 0 – 25k = 0
∴ -16k – 192 = 0
∴ – 16k = 192
∴ k= -12.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) kxy + 10x + 6y + 4 = 0
Solution:
Comparing the given equation with
ax2 + 2 hxy + by2 + 2gx + 2fy + c = 0,
we get, a = 0, h = \(\frac{k}{2}\), b = 0, g = 5, f = 3, c = 4
Now, given equation represents a pair of lines.
∴ abc + 2fgh – af2 – bg2 – ch2 = 0
∴ (0)(0)(4) + 2(3)(5)\(\left(\frac{k}{2}\right)\) – 0(3)2 – 0(5)2 – 4\(\left(\frac{k}{2}\right)^{2}\) = 0
∴ 0 + 15k – 0 – 0 – k2 = 0
∴ 15k – k2 = 0
∴ -k(k – 15) = 0
∴ k = 0 or k = 15.
If k = 0, then the given equation becomes
10x + 6y + 4 = 0 which does not represent a pair of lines.
∴ k ≠ o
Hence, k = 15.

(iii) x2 + 3xy + 2y2 + x – y + k = 0
Solution:
Comparing the given equation with
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0,
we get, a = 1, h = \(\frac{3}{2}\), b = 2, g = \(\frac{1}{2}\), f= \(-\frac{1}{2}\), c = k.
Now, given equation represents a pair of lines.
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.3 3
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.3 4
∴ 2(8k – 1) – 3(6k + 1) + 1(-3 – 4) = 0
∴ 16k – 2 – 18k – 3 – 7 = 0
∴ -2k – 12 = 0
∴ -2k = 12 ∴ k = -6.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 7.
Find p and q if the equation px2 – 8xy + 3y2 + 14x + 2y + q = 0 represents a pair of perpendicular lines.
Solution:
The given equation represents a pair of lines perpendicular to each other
∴ (coefficient of x2) + (coefficient of y2) = 0
∴ p + 3 = 0 p = -3
With this value of p, the given equation is
– 3x2 – 8xy + 3y2 + 14x + 2y + q = 0.
Comparing this equation with
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we have,
a = -3, h = -4, b = 3, g = 7, f = 1 and c = q.
D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
-3 & -4 & 7 \\
-4 & 3 & 1 \\
7 & 1 & q
\end{array}\right|\)
= -3(3q – 1) + 4(-4q – 7) + 7(-4 – 21)
= -9q + 3 – 16q – 28 – 175
= -25q – 200= -25(q + 8)
Since the given equation represents a pair of lines, D = 0
∴ -25(q + 8) = 0 ∴ q= -8.
Hence, p = -3 and q = -8.

Question 8.
Find p and q if the equation 2x2 + 8xy + py2 + qx + 2y – 15 = 0 represents a pair of parallel lines.
Solution:
The given equation is
2x2 + 8xy + py2 + qx + 2y – 15 = 0
Comparing it with ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get,
a = 2, h = 4, b = p, g = \(\frac{q}{2}\), f = 1, c = – 15
Since the lines are parallel, h2 = ab
∴ (4)2 = 2p ∴ P = 8
Since the given equation represents a pair of lines
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.3 5
i.e. – 242 + 240 + 2q + 2q – 2q2 = 0
i.e. -2q2 + 4q – 2 = 0
i.e. q2 – 2q + 1 = 0
i.e. (q – 1)2 = 0 ∴ q – 1 = 0 ∴ q = 1.
Hence, p = 8 and q = 1.

Question 9.
Equations of pairs of opposite sides of a parallelogram are x2 – 7x+ 6 = 0 and y2 – 14y + 40 = 0. Find the joint equation of its diagonals.
Solution:
Let ABCD be the parallelogram such that the combined equation of sides AB and CD is x2 – 7x + 6 = 0 and the combined equation of sides BC and AD is y2 – 14y + 40 = 0.
The separate equations of the lines represented by x2 – 7x + 6 = 0, i.e. (x – 1)(x – 6) = 0 are x – 1 = 0 and x – 6 = 0.
Let equation of the side AB be x – 1 = 0 and equation of side CD be x – 6 = 0.
The separate equations of the lines represented by y2 – 14y + 40 = 0, i.e. (y – 4)(y – 10) = 0 are y – 4 = 0 and y – 10 = 0.
Let equation of the side BC be y – 4 = 0 and equation of side AD be y – 10 = 0.
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.3 6
Coordinates of the vertices of the parallelogram are A(1, 10), B(1, 4), C(6, 4) and D(6, 10).
∴ equation of the diagonal AC is
\(\frac{y-10}{x-1}\) = \(\frac{10-4}{1-6}\) = \(\frac{6}{-5}\)
∴ -5y + 50 = 6x – 6
∴ 6x + 5y – 56 = 0
and equation of the diagonal BD is
\(\frac{y-4}{x-1}\) = \(\frac{4-10}{1-6}\) = \(\frac{-6}{-5}\) = \(\frac{6}{5}\)
∴ 5y – 20 = 6x – 6
∴ 6x – 5y + 14 = 0
Hence, the equations of the diagonals are 6x + 5y – 56 = 0 and 6x – 5y + 14 = 0.
∴ the joint equation of the diagonals is (6x + 5y – 56)(6x – 5y + 14) = 0
∴ 36x2 – 30xy + 84x + 30xy – 25y2 + 70y – 336x + 280y – 784 = 0
∴ 36x2 – 25y2 – 252x + 350y – 784 = 0.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 10.
∆OAB is formed by lines x2 – 4xy + y2 = 0 and the line 2x + 3y – 1 = 0. Find the equation of the median of the triangle drawn from O.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.3 7
Let D be the midpoint of seg AB where A is (x1, y1) and B is (x2, y2).
Then D has coordinates \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\).
The joint (combined) equation of the lines OA and OB is x2 – 4xy + y2 = 0 and the equation of the line AB is 2x + 3y – 1 = 0.
∴ points A and B satisfy the equations 2x + 3y – 1 = 0
and x2 – 4xy + y2 = 0 simultaneously.
We eliminate x from the above equations, i.e.,
put x = \(\frac{1-3 y}{2}\) in the equation x2 – 4xy + y2 = 0, we get,
∴ \(\left(\frac{1-3 y}{2}\right)^{2}\) – 4\(\left(\frac{1-3 y}{2}\right)\)y + y2 = 0
∴ (1 – 3y)2 – 8(1 – 3y)y + 4y2 = 0
∴1 – 6y + 9y2 – 8y + 24y2 + 4y2 = 0
∴ 37y2 – 14y + 1 = 0
The roots y1 and y2 of the above quadratic equation are the y-coordinates of the points A and B.
∴ y1 + y2 = \(\frac{-b}{a}=\frac{14}{37}\)
∴ y-coordinate of D = \(\frac{y_{1}+y_{2}}{2}=\frac{7}{37}\).
Since D lies on the line AB, we can find the x-coordinate of D as
2x + 3\(\left(\frac{7}{37}\right)\) – 1 = 0
∴ 2x = 1 – \(\frac{21}{37}=\frac{16}{37}\)
∴ x = \(\frac{8}{37}\)
∴ D is (8/37, 7/37)
∴ equation of the median OD is \(\frac{x}{8 / 37}=\frac{y}{7 / 37}\),
i.e., 7x – 8y = 0.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 11.
Find the co-ordinates of the points of intersection of the lines represented by x2 – y2 – 2x + 1 = 0.
Solution:
Consider, x2 – y2 – 2x + 1 = 0
∴ (x2 – 2x + 1) – y2 = 0
∴ (x – 1)2 – y2 = 0
∴ (x – 1 + y)(x – 1 – y) = 0
∴ (x + y – 1)(x – y – 1) = 0
∴ separate equations of the lines are
x + y – 1 = 0 and x – y +1 = 0.
To find the point of intersection of the lines, we have to solve
x + y – 1 = 0 … (1)
and x – y + 1 = 0 … (2)
Adding (1) and (2), we get,
2x = 0 ∴ x = 0
Substituting x = 0 in (1), we get,
0 + y – 1 = 0 ∴ y = 1
∴ coordinates of the point of intersection of the lines are (0, 1).

Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Miscellaneous Exercise 2B Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B

I) Choose the correct answer from the given alternatives in each of the following questions :
Question 1.
If A = \(\left(\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right)\), adj = \(\left(\begin{array}{ll}
4 & a \\
-3 & b
\end{array}\right)\) then the values of a and b are,
(a) a = – 2, b = 1
(b) a = 2, b = 4
(c) a = 2, b = –1
(d) a = 1, b = –2
Solution:
(a) a = – 2, b = 1

Question 2.
The inverse of \(\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right)\) is
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 1
Solution:
\(\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right)\)

Question 3.
If A = \(\left(\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right)\) and A(adj A) = k 1, then the value of k is
(a) 1
(b) -1
(c) 0
(d) -3
Solution:
(d) -3 [Hint : A(adj A) = |A| ∙ I]

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 4.
If A = \(\left(\begin{array}{ll}
2 & -4 \\
3 & 1
\end{array}\right)\), then the adjoint of matrix A is
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 2
Solution:\(\left(\begin{array}{ll}
1 & 4 \\
-3 & 2
\end{array}\right)\)

Question 5.
If A = \(\left(\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right)\) and A(adj A) = kI, then the value of k is
(a) 2
(b) -2
(c) 10
(d) -10
Solution:
(b) -2

Question 6.
If A = \(\left(\begin{array}{rr}
\lambda & 1 \\
-1 & -\lambda
\end{array}\right)\), then A-1 does not exist if λ = ………..
(a) 0
(b) ± 1
(c) 2
(d) 3
Solution:
(b) ± 1

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 7.
If A = \(\left[\begin{array}{ll}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) then A-1 = ….
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 3
Solution:
\(\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\)

Question 8.
If F (∝) = \(\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\) where ∝ ∈ R then [F(∝)]-1 is =
(a) F(-∝)
(b) F(∝-1)
(c) F(2∝)
(d) None of these
Solution:
(a) F(-∝)

Question 9.
The inverse of A = \(\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right]\)
(a) I
(b) A
(c) A’
(d) -I
Solution:
(b) A

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 10.
The inverse of a symmetric matrix is
(a) Symmetric
(b) Non-symmetric
(c) Null matrix
(d) Diagonal matrix
Solution:
(a) Symmetric

Question 11.
For a 2 × 2 matrix A, if A(adjA) = \(\left(\begin{array}{ll}
10 & 0 \\
0 & 10
\end{array}\right)\) then determinant A equals
(a) 20
(b) 10
(c) 30
(d) 40
Solution:
(b) 10

Question 12.
If A2 = \(-\frac{1}{2}\left[\begin{array}{cc}
1 & -4 \\
-1 & 2
\end{array}\right]\) then A =
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 4
Solution:
\(-\frac{1}{2}\left[\begin{array}{cc}
2 & 4 \\
1 & 1
\end{array}\right]\)

II) Solve the following equations by the methods of inversion.
(i) 2x – y = -2 , 3x + 4y = 5
Solution:
The given equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 5
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 6
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 7
By equality of matrices,
x = \(-\frac{5}{11}\), y = \(\frac{12}{11}\) is the required solution.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) x + y + z = 1, 2x + 3y + 2z = 2 and ax + ay + 2az = 4, a ≠ 0.
Solution:
The given equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 8
= 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a)
= 4a – 2a – a = a ≠ 0 ∴ A-1 exists.
Consider AA-1 = I
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 9
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 10
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 11
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 12

(iii) 5x – y +4z = 5, 2x + 3y + 5z = 2 and 5x – 2y + 6z = -1
Solution:
The given equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 13
= 5(18 + 10) + 1 (12 – 25) + 4( -4 – 15)
= 140 – 13 – 76 = 51 #0
∴ A-1 exists.
Now, we have to find the cofactor matrix
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 14
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 15
Now, premultiply AX = B by A-1, we get,
A-1(AX) = A-1B
∴ (A-1A)X = A-1B
∴ IX = A-1B
∴ X = \(\frac{1}{51}\left[\begin{array}{rrr}
28 & -2 & -17 \\
13 & 10 & -17 \\
-19 & 5 & 17
\end{array}\right]\left[\begin{array}{r}
5 \\
2 \\
-1
\end{array}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 16
By equality of matrices,
x = 3, y = 2, z = -2 is the required solution.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) 2x + 3y = -5, 3x + y = 3
Solution:

(v) x + y + z = -1, y + z = 2 and x + y – z = 3
Solution:
The given equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 17
= 1(-1 – 1) – 1 (0 – 1) + 1(0 – 1)
= -2 + 1 – 1 = -2 ≠ 0 ∴ A-1 exists.
Consider AA-1 = I
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 18
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 19
Now, premultiply AX = B by A-1, we get,
A-1(AX) = A-1B
∴ (A-1A)X = A-1B
∴ IX = A-1B
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 20
∴ by equality of the matrices, x= -3, y = 4, z = -2 is the required solution.

Question 2.
Express the following equation in matrix from and solve them by the method of reduction.
(i) x – y + z = 1, 2x – y = 1, 3x + 3y – 4z = 2
Solution:
The given equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 21
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 22
By equality of matrices,
x – y + z = 1 ……(1)
y – 2z = -1 …..(2)
5z = 5 ….(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
y – 2 = -1 ∴ y = 1
Substituting y = 1, z = 1 in (1), we get,
x – 1 + 1 = 1
∴ x = 1
Hence, x = 1, y = 1, z = 1 is the required solution.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) x + y = 1, y + z = \(\frac{5}{3}\), z + x = \(\frac{4}{3}\).
Solution:
The given equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 23
By equality of matrices,
x + y = 1 ……(1)
y + z = \(\frac{5}{3}\) …(2)
2z = 2 ……..(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
y + 1 = \(\frac{5}{3}\) ∴ y = \(\frac{2}{3}\)
Substituting y = \(\frac{2}{3}\) in (1), we get,
x + \(\frac{2}{3}\) = 1 ∴ x = \(\frac{1}{3}\)
Hence, x = \(\frac{1}{3}\), y = \(\frac{2}{3}\), z = 1 is the required solution.

(iii) 2x – y + z = 1, x + 2y + 3z = 8 and 3x + y – 4z = 1
Solution:
The given equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 24
∴ \(\left[\begin{array}{r}
x+2 y+3 z \\
0-5 y-5 z \\
0+0-8 z
\end{array}\right]\) = \(\left[\begin{array}{r}
8 \\
-15 \\
-8
\end{array}\right]\)
By equality of matrices,
x + 2y + 3z = 8 …..(1)
-5y – 5z = -15 ….(2)
-8z = -8 …..(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
-5y – 5 = -15
-5y = -10
∴ y = 2
Substituting y = 2, z = 1 in (1), we get,
x + 4 + 3 = 8 ∴ x = 1
Hence, x = 1, y = 2, z = 1 is the required solution.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) x + y + z = 6, 3x – y + 3z =10 and 5x + 5y – 4z = 3.
Solution:

(v) x + 2y + z = 8, 2x + 3y – z =11 and 3x – y – 2z = 5
Solution:
The given equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 25
By equality of matrices,
x + 2y + z = 8 … (1)
-y – 3z = -5 … (2)
16z = 16 … (3)
From (3), z = 1
Substituting z = 1 in (2), we get,
-y – 3 = -5, ∴ y = 2
Substituting y = 2, z = 1 in (1), we get,
x + 4 + 1 = 8 ∴ x = 3
Hence, x = 3, y = 2, z = 1 is the required solution.

(vi) x + 3y + 2z = 6, 3x – 2y + 5z =5 and 2x – 3y + 6z = 7.
Solution:
The given equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 26
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 27
By equality of matrices,
x + 3y + 2z = 6 …(1)
y + \(\frac{3}{2}\) z = 4 …(2)
\(\frac{31}{2}\)z = 31 …..(3)
From (3), z = 2
Substituting z = 2 in (2), we get,
y + \(\frac{3}{2}\)z = 4
y + \(\frac{3}{2}\)(2) = 4
y + 3 = 4
y = 1
Substituting y = 1, z = 2 in (2), we get,
x + 3y + 2z = 6
x + 3(1) + 2(2) = 6
x + 3 + 4 = 6
x = -1
Hence, x = -1, y = 1, z = 2 is the required solution.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 3.
The sum of three numbers is 6. If we multiply third number by 3 and add it to the second number we get 11. By adding first and the third numbers we get a number which is double the second number. Use this information and find a system of linear equations. Find the three numbers using matrices.
Solution:
Let the three numbers be x, y and z. According to the given conditions,
x + y + z = 6.
3z + y = 11, i.e., y + 3z = 11 and x + z = 2y,
i.e., x – 2y + z = 0
Hence, the system of the linear equations is
x + y + z = 6
y + 3z = 11
x – 2y + z = 0
These equations can be written in the matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 28
By equality of matrices,
x + y + z = 6 …(1)
y + 3z = 11 …(2)
-3y = -6 …(3)
From (3), y = 2
Substituting y = 2 in (2), we get,
2 + 3z = 11
∴ 3z = 9 ∴ z = 3
Put y = 2, z — 3 in (1), we get,
x + 2 + 3 = 6 ∴ x = 1
∴ x = 1, y = 2, z = 3
Hence, the required numbers are 1, 2 and 3.

Question 4.
The cost of 4 pencils, 3 pens and 2 books is ₹ 150. The cost of 1 pencil, 2 pens and 3 books is ₹ 125. The cos of 6 pencils, 2 pens and 3 books is ₹ 175. Fild the cost of each item by using Matrices.
Solution:
Let the cost of 1 pencil, 1 pen and 1 book be ₹x, ₹ y, ₹ z respectively.
According to the given conditions,
4x + 3y + 2z = 150
x + 2y + 3z = 125
6x + 2y + 3z = 175
The equations can be written in matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 29
By equality of matrices,
x + 2y + 3z = 125 …(1)
-5y – 10z = -350 …(2)
5z = 125 …(3)
From (3), z = 25
Substituting z = 25 in (2), we get
-5y – 10(25) = -350
∴ -5y = -350 + 250 = -100
∴ y = 20
Substituting y = 20, z = 25 in (1), we get
x + 2(20) + 3(25) = 125
∴ x = 125 – 40 – 75 = 10
∴ x = 10, y = 20, z = 25
Hence, the cost of 1 pencil is ₹ 10, 1 pen is ₹ 20 and 1 book is ₹ 25.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 5.
The sum of three numbers is 6. Thrice the third number when added to the first number, gives 7. On adding three times first number to the sum of second and third number, we get 12. Find the three numbers by using Matrices.
Solution:
Let the numbers be x, y and z.
According to the given conditions,
x + y + z = 6
3z + x = 7, i.e., x + 3z = 7
and 3x + y + z = 12
Hence, the system of linear equations is
x + y + z = 6
x + 3z = 7
3x + y + z = 12
These equations can be written in matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 30
By equality of matrices,
x + y + z = 6 …(1)
-y + 2z = 1 …(2)
-3y = -5 …(3)
From (3), y = \(\frac{5}{3}\)
Substituting y = \(\frac{5}{3}\) in (2), we get,
–\(\frac{5}{3}\) + 2z = 1
∴ 2z = 1 + \(\frac{5}{3}\) = \(\frac{8}{3}\)
∴ z = \(\frac{4}{3}\)
Substituting y =\(\frac{5}{3}\), z = \(\frac{5}{3}\) in (1), we get,
x + \(\frac{5}{3}+\frac{4}{3}\) = 6
∴ x = 3
∴ x = 3, y = \(\frac{5}{3}\), z = \(\frac{4}{3}\)
Hence, the required numbers are 3, \(\frac{5}{3}\) and \(\frac{4}{3}\).

Question 6.
The sum of three numbers is 2. If twice the second number is added to the sum of first and third number, we get 1 adding five times the first number to the sum of second and third we get 6. Find the three numbers by using matrices.
Solution:
Let the three numbers be x, y and z.
According to the question,
x + y + 2
x + 2y + z = 1
5x + y + z = 6
The given system of equations can be written in matrix form as follows:
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 33
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 34
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 35

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 7.
An amount of ₹ 5000 is invested in three types of investments, at interest rates 6%, 7%, 8% per annum respectively. The total annual income from these investimest is ₹ 350. If the total annual income from first two investment is ₹ 70 more than the income from the third, find the amount of each investment using matrix method.
Solution:
Let the amounts in three investments by ₹ x, ₹ y and ₹ z respectively.
Then x + y + z = 5000
Since the rate of interest in these investments are 6%, 7% and 8% respectively, the annual income of the three investments are \(\frac{6 x}{100}\), \(\frac{7 y}{100}\) and \(\frac{8 z}{100}\) respectively.
According to the given conditions,
\(\frac{6 x}{100}+\frac{7 y}{100}+\frac{8 z}{100}\) = 350
i.e. 6x + 7y + 8z = 35000
Also, \(\frac{6 x}{100}+\frac{7 y}{100}\) = \(\frac{8 z}{100}\) + 70
i.e. 6x + 7y – 8z = 7000
Hence, the system of linear equation is
x + y + z = 5000
6x + 7y + 8z = 35000
6x + 7y – 8z = 7000
These equations can be written in matrix form as :
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 31
Maharashtra Board 12th Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2B 32
By equality of matrices,
x + y + z = 5000 …(1)
y + 2z = 5000 …(2)
-16z = -28000 ….(3)
From (3), z = 1750
Substituting z = 1750 in (2), we get,
y + 2(1750) = 5000
∴ y = 5000 – 3500 = 1500
Substituting y = 1500, z = 1750 in (1), we get,
x + 1500 + 1750 = 5000
∴ x = 5000 – 3250 = 1750
∴ x = 1750, y = 1500, z = 1750
Hence, the amounts of the three investments are ₹ 1750, ₹ 1500 and ₹ 1750 respectively.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 8.
The sum of the costs of one ook each of Mathematics, Physics and Chemistry is ₹ 210. Total cost of a mathematics book, 2 physics books, and a chemistry book is ₹ 240 Also the total cost of a Mathematics book, 3 physics book and chemistry books is Rs. 300/-. Find the cost of each book, using Matrices.
Solution:

Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Ex 7.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1

Question 1.
Solve graphically :
(i) x ≥ 0
Solution:
Consider the line whose equation is x = 0. This represents the Y-axis.
To find the solution set, we have to check any point other than origin.
Let us check the point (1, 1)
When x = 1, x ≥ 0
∴ (1, 1) lies in the required region
Therefore, the solution set is the Y-axis and the right
side of the Y-axis which is shaded in the graph.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 1

(ii) x ≤ 0
Solution:
Consider the line whose equation is x = 0.
This represents the Y-axis.
To find the solution set, we have to check any point other than origin.
Let us check the point (1, 1).
When x = 1, x ≰ 0
∴ (1, 1) does not lie in the required region.
Therefore, the solution set is the Y-axis and the left side of the Y-axis which is shaded in the graph.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 2

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) y ≥ 0
Solution:
Consider the line whose equation is y = 0. This represents the X-axis. To find the solution set, we have to check any point other than origin. Let us check the point (1, 1).
When y = 1, y ≥ 0
∴ (1, 1) lies in the required region.
Therefore, the solution set is the X-axis and above the X-axis which is shaded in the graph.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 3

(iv) y ≤ 0
Solution:
(iv) Consider the line whose equation is y = 0. This represents the X-axis.
To find the solution set, we have to check any point other than origin.
Let us check the point (1, 1).
When y = 1, y ≰ 0.
∴ (1, 1) does not lie in the required region.
Therefore, the solution set is the X-axis and below the X-axis which is shaded in the graph.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 4

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 2.
Solve graphically :
(i) x ≥ 0 and y ≥ 0
Solution:
Consider the lines whose equations are x = 0, y = 0.
These represents the equations of Y-axis and X-axis respectively, which divide the plane into four parts.
(i) Since x ≥ 0, y ≥ 0, the solution set is in the first quadrant which is shaded in the graph.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 5

(ii) x ≤ 0 and y ≥ 0
Solution:
Since x ≤ 0, y ≥ 0, the solution set is in the second quadrant which is shaded in the graph.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 6

(iii) x ≤ 0 and y ≤ 0
Solution:
Since x ≤ 0, y ≤ 0, the solution set is in the third quadrant which is shaded in the graph.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 7

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) x ≥ 0 and y ≤ 0
Solution:
Since x ≥ 0, y ≤ 0, the solution set is in the fourth ! quadrant which is shaded in the graph.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 8

Question 3.
Solve graphically :
(i) 2x – 3 ≥ 0
Solution:
Consider the line whose equation is 2x – 3 = 0,
i.e. x = \(\frac{3}{2}\)
This represents a line parallel to Y-axis passing through the point (\(\frac{3}{2}\), 0)
Draw the line x =\(\frac{3}{2}\).
To find the solution set, we have to check the position of the origin (0, 0).
When x = 0, 2x – 3 = 2 × 0 – 3 = -3 ≱ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line x = \(\frac{3}{2}\) and the non-origin side of the line which is shaded in the graph.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 9

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) 2y – 5 ≥ 0
Solution:
Consider the line whose equation is 2y – 5 = 0, i.e. y = \(\frac{5}{2}\)
This represents a line parallel to X-axis passing through the point (0, \(\frac{5}{2}\)).
Draw the line y = \(\frac{5}{2}\).
To find the solution set, we have to check the position of the origin (0, 0).
When y = 0, 2y – 5 = 2 × 0 – 5 = -5 ≱ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line y = \(\frac{5}{2}\) and the non-origin side of the line which is shaded in the graph.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 10

(iii) 3x + 4 ≤ 0
Solution:
(iii) Consider the line whose equation is 3x + 4 = 0,
i.e. x = \(-\frac{4}{3}\)
This represents a line parallel to Y-axis passing through the point (\(-\frac{4}{3}\), 0).
Draw the line x = \(-\frac{4}{3}\).
To find the solution set, we have to check the position of the origin (0, 0).
When x = 0, 3x + 4 = 3 × 0 + 4= 4 ≰ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line x = \(-\frac{4}{3}\) and the non-origin side of the line which is shaded in the graph.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 11

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iv) 5y + 3 ≤ 0
Solution:
(iv) Consider the line whose equation is 5y + 3 = 0,
i.e. y = \(\frac{-3}{5}\)
This represents a line parallel to X-axis passing through the point (0, \(\frac{-3}{5}\))
Draw the line y = \(\frac{-3}{5}\).
To find the solution set, we have to check the position of the origin (0, 0).
When y = 0, 5y + 3 = 5 × 0 + 3 = 3 ≰ 0
∴ the coordinates of the origin does not satisfy the given inequality.
∴ the solution set consists of the line y = \(\frac{-3}{5}\) and the non-origin side of the line which is shaded in the graph.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 12

Question 4.
Solve graphically :
(i) x + 2y ≤ 6
Solution:
Consider the line whose equation is x + 2y = 6.
To find the points of intersection of this line with the coordinate axes.
Put y = 0, we get x = 6.
∴ A = (6, 0) is a point on the line.
Put x = 0, we get 2y = 6, i.e. y = 3
∴ B = (0, 3) is another point on the line.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 13
Draw the line AB joining these points. This line divide the line into two parts.
1. Origin side 2. Non-origin side
To find the solution set, we have to check the position of the origin (0, 0) with respect to the line.
When x = 0, y = 0, then x + 2y = 0 which is less than 6.
∴ x + 2y ≤ 6 in this case.
Hence, origin lies in the required region. Therefore, the given inequality is the origin side which is
shaded in the graph.
This is the solution set of x + 2y ≤ 6.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(ii) 2x – 5y ≥ 10
Solution:
Consider the line whose equation is 2x – 5y = 10.
To find the points of intersection of this line with the coordinate axes.
Put y = 0, we get 2x = 10, i.e. x = 5.
∴ A = (5, 0) is a point on the line.
Put x = 0, we get -5y = 10, i.e. y = -2
∴ B = (0, -2) is another point on the line.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 14
Draw the line AB joining these points. This line J divide the plane in two parts.
1. Origin side 2. Non-origin side
To find the solution set, we have to check the position of the origin (0, 0) with respect to the line. When x = 0, y = 0, then 2x – 5y = 0 which is neither greater nor equal to 10.
∴ 2x – 5y ≱ 10 in this case.
Hence (0, 0) will not lie in the required region.
Therefore, the given inequality is the non-origin side, which is shaded in the graph.
This is the solution set of 2x – 5y ≥ 10.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) 3x + 2y ≥ 0
Solution:
Consider the line whose equation is 3x + 2y = 0.
The constant term is zero, therefore this line is passing through the origin.
∴ one point on the line is O ≡ (0, 0).
To find the another point, we can give any value of x and get the corresponding value of y.
Put x = 2, we get 6 + 2y = 0 i.e. y = – 3
∴ A = (2, -3) is another point on the line.
Draw the line OA.
To find the solution set, we cannot check (0, 0) as it is already on the line.
We can check any other point which is not on the line.
Let us check the point (1, 1)
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 15
When x = 1, y = 1, then 3x + 2y = 3 + 2 = 5 which is greater than zero.
∴ 3x + 2y > 0 in this case.
Hence (1, 1) lies in the required region. Therefore, the required region is the upper side which is shaded in the graph.
This is the solution set of 3x + 2y ≥ 0.

(iv) 5x – 3y ≤ 0
Solution:
Consider the line whose equation is 5x – 3y = 0. The constant term is zero, therefore this line is passing through the origin.
∴ one point on the line is the origin O = (0, 0).
To find the other point, we can give any value of x and get the corresponding value of y.
Put x = 3, we get 15 – 3y = 0, i.e. y = 5
∴ A ≡ (3, 5) is another point on the line.
Draw the line OA.
To find the solution set, we cannot check 0(0, 0), as it is already on the line. We can check any other point which is not on the line.
Let us check the point (1, -1).
When x = 1, y = -1 then 5x – 3y = 5 + 3 = 8
which is neither less nor equal to zero.
∴ 5x – 3y ≰ 0 in this case.
Hence (1, -1) will not lie in the required region. Therefore, the required region is the upper side which is shaded in the graph.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 16
This is the solution set of 5x – 3y ≤ 0.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 5.
Solve graphically :
(i) 2x + y ≥ 2 and x – y ≤ 1
Solution:
First we draw the lines AB and AC whose equations are 2x + y = 2 and x – y = 1 respectively.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 17
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 18
The solution set of the given system of inequalities is shaded in the graph.

(ii) x – y ≤ 2 and x + 2y ≤ 8
Solution:
First we draw the lines AB and CD whose equations are x – y = 2 and x + 2y = 8 respectively.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 19
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 20
The solution set of the given system of inequalities is shaded in the graph.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(iii) x + y ≥ 6 and x + 2y ≤ 10
Solution:
First we draw the lines AB and CD whose equations are x + y = 6 and x + 2y = 10 respectively.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 21
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 22
The solution set of the given system of inequalities is shaded in the graph.

(iv) 2x + 3y ≤ 6 and x + 4y ≥ 4
Solution:
First we draw the lines AB and CD whose equations are 2x + 3y = 6 and x + 4y = 4 respectively.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 23
The solution set of the given system of inequalities is shaded in the graph.

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

(v) 2x + y ≥ 5 and x – y ≤ 1
Solution:
First we draw the lines AB and CD whose equations are 2x + y = 5 and x – y = 1 respectively.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 24
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.1 25
The solution set of the given system of inequations is shaded in the graph.