Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

Question 1.
In each of the following examples verify that the given expression is a solution of the corresponding differential equation.
(i) xy = log y + c; \(\frac{d y}{d x}=\frac{y^{2}}{1-x y}\)
Solution:
xy = log y + c
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q1 (i)
Hence, xy = log y + c is a solution of the D.E.
\(\frac{d y}{d x}=\frac{y^{2}}{1-x y^{\prime}}, x y \neq 1\)

(ii) y = (sin-1x)2 + c; (1 – x2) \(\frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=2\)
Solution:
y = (sin-1 x)2 + c …….(1)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q1 (ii)
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q1 (ii).1

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(iii) y = e-x + Ax + B; \(e^{x} \frac{d^{2} y}{d x^{2}}=1\)
Solution:
y = e-x + Ax + B
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q1 (iii)
∴ \(e^{x} \frac{d^{2} y}{d x^{2}}=1\)
Hence, y = e-x + Ax + B is a solution of the D.E.
\(e^{x} \frac{d^{2} y}{d x^{2}}=1\)

(iv) y = xm; \(x^{2} \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m y=0\)
Solution:
y = xm
Differentiating twice w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q1 (iv)
This shows that y = xm is a solution of the D.E.
\(x^{2} \frac{d^{2} y}{d x^{2}}-m x \frac{d y}{d x}+m y=0\)

(v) y = a + \(\frac{b}{x}\); \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)
Solution:
y = a + \(\frac{b}{x}\)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q1 (v)
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q1 (v).1
Hence, y = a + \(\frac{b}{x}\) is a solution of the D.E.
\(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}=0\)

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(vi) y = eax; x \(\frac{d y}{d x}\) = y log y
Solution:
y = eax
log y = log eax = ax log e
log y = ax …….(1) ……..[∵ log e = 1]
Differentiating w.r.t. x, we get
\(\frac{1}{y} \cdot \frac{d y}{d x}\) = a × 1
∴ \(\frac{d y}{d x}\) = ay
∴ x \(\frac{d y}{d x}\) = (ax)y
∴ x \(\frac{d y}{d x}\) = y log y ………[By (1)]
Hence, y = eax is a solution of the D.E.
x \(\frac{d y}{d x}\) = y log y.

Question 2.
Solve the following differential equations.
(i) \(\frac{d y}{d x}=\frac{1+y^{2}}{1+x^{2}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q2 (i)

(ii) log(\(\frac{d y}{d x}\)) = 2x + 3y
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q2 (ii)

(iii) y – x \(\frac{d y}{d x}\) = 0
Solution:
y – x \(\frac{d y}{d x}\) = 0
∴ x \(\frac{d y}{d x}\) = y
∴ \(\frac{1}{x} d x=\frac{1}{y} d y\)
Integrating both sides, we get
\(\int \frac{1}{x} d x=\int \frac{1}{y} d y\)
∴ log |x| = log |y| + log c
∴ log |x| = log |cy|
∴ x = cy
This is the general solution.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(iv) sec2x . tan y dx + sec2y . tan x dy = 0
Solution:
sec2x . tan y dx + sec2y . tan x dy = 0
∴ \(\frac{\sec ^{2} x}{\tan x} d x+\frac{\sec ^{2} y}{\tan y} d y=0\)
Integrating both sides, we get
\(\int \frac{\sec ^{2} x}{\tan x} d x+\int \frac{\sec ^{2} y}{\tan y} d y=c_{1}\)
Each of these integrals is of the type
\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log |f(x)| + c
∴ the general solution is
∴ log|tan x| + log|tan y | = log c, where c1 = log c
∴ log |tan x . tan y| = log c
∴ tan x . tan y = c
This is the general solution.

(v) cos x . cos y dy – sin x . sin y dx = 0
Solution:
cos x . cos y dy – sin x . sin y dx = 0
\(\frac{\cos y}{\sin y} d y-\frac{\sin x}{\cos x} d x=0\)
Integrating both sides, we get
∫cot y dy – ∫tan x dx = c1
∴ log|sin y| – [-log|cos x|] = log c, where c1 = log c
∴ log |sin y| + log|cos x| = log c
∴ log|sin y . cos x| = log c
∴ sin y . cos x = c
This is the general solution.

(vi) \(\frac{d y}{d x}\) = -k, where k is a constant.
Solution:
\(\frac{d y}{d x}\) = -k
∴ dy = -k dx
Integrating both sides, we get
∫dy = -k∫dx
∴ y = -kx + c
This is the general solution.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(vii) \(\frac{\cos ^{2} y}{x} d y+\frac{\cos ^{2} x}{y} d x=0\)
Solution:
\(\frac{\cos ^{2} y}{x} d y+\frac{\cos ^{2} x}{y} d x=0\)
∴ y cos2y dy + x cos2x dx = 0
∴ \(x\left(\frac{1+\cos 2 x}{2}\right) d x+y\left(1+\frac{\cos 2 y}{2}\right) d y=0\)
∴ x(1 + cos 2x) dx + y(1 + cos 2y) dy = 0
∴ x dx + x cos 2x dx + y dy+ y cos 2y dy = 0
Integrating both sides, we get
∫x dx + ∫y dy + ∫x cos 2x dx + ∫y cos 2y dy = c1 ……..(1)
Using integration by parts
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q2 (vii)
Multiplying throughout by 4, this becomes
2x2 + 2y2 + 2x sin 2x + cos 2x + 2y sin 2y + cos 2y = 4c1
∴ 2(x2 + y2) + 2(x sin 2x + y sin 2y) + cos 2y + cos 2x + c = 0, where c = -4c1
This is the general solution.

(viii) \(y^{3}-\frac{d y}{d x}=x^{2} \frac{d y}{d x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q2 (viii)

(ix) 2ex+2y dx – 3 dy = 0
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q2 (ix)

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(x) \(\frac{d y}{d x}\) = ex+y + x2 ey
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q2 (x)
∴ 3ex + 3e-y + x3 = -3c1
∴ 3ex + 3e-y + x3 = c, where c = -3c1
This is the general solution.

Question 3.
For each of the following differential equations, find the particular solution satisfying the given condition:
(i) 3ex tan y dx + (1 + ex) sec2y dy = 0, when x = 0, y = π
Solution:
3ex tan y dx + (1 + ex) sec2y dy = 0
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q3 (i)

(ii) (x – y2x) dx – (y + x2y) dy = 0, when x = 2, y = 0
Solution:
(x – y2x) dx – (y + x2y) dy = 0
∴ x(1 – y2) dx – y(1 + x2) dy = 0
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q3 (ii)
When x = 2, y = 0, we have
(1 + 4)(1 – 0) = c
∴ c = 5
∴ the particular solution is (1 + x2)(1 – y2) = 5.

(iii) y(1 + log x) \(\frac{d x}{d y}\) – x log x = 0, y = e2, when x = e
Solution:
y(1 + log x) \(\frac{d x}{d y}\) – x log x = 0
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q3 (iii)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q3 (iii).1

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(iv) (ey + 1) cos x + ey sin x \(\frac{d y}{d x}\) = 0, when x = \(\frac{\pi}{6}\), y = 0
Solution:
(ey + 1) cos x + ey sin x \(\frac{d y}{d x}\) = 0
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q3 (iv)
\(\int \frac{f^{\prime}(x)}{f(x)} d x\) = log|f(x)| + c
∴ from (1), the general solution is
log|sin x| + log|ey + 1| = log c, where c1 = log c
∴ log|sin x . (ey + 1)| = log c
∴ sin x . (ey + 1) = c
When x = \(\frac{\pi}{4}\), y = 0, we get
\(\left(\sin \frac{\pi}{4}\right)\left(e^{0}+1\right)=c\)
∴ c = \(\frac{1}{\sqrt{2}}\)(1 + 1) = √2
∴ the particular solution is sin x . (ey + 1) = √2

(v) (x + 1) \(\frac{d y}{d x}\) – 1 = 2e-y, y = 0, when x = 1
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q3 (v)
This is the general solution.
Now, y = 0, when x = 1
∴ 2 + e0 = c(1 + 1)
∴ 3 = 2c
∴ c = \(\frac{3}{2}\)
∴ the particular solution is 2 + ey = \(\frac{3}{2}\) (x + 1)
∴ 2(2 + ey) = 3(x + 1).

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(vi) cos(\(\frac{d y}{d x}\)) = a, a ∈ R, y (0) = 2
Solution:
cos(\(\frac{d y}{d x}\)) = a
∴ \(\frac{d y}{d x}\) = cos-1 a
∴ dy = (cos-1 a) dx
Integrating both sides, we get
∫dy = (cos-1 a) ∫dx
∴ y = (cos-1 a) x + c
∴ y = x cos-1 a + c
This is the general solution.
Now, y(0) = 2, i.e. y = 2,
when x = 0, 2 = 0 + c
∴ c = 2
∴ the particular solution is
∴ y = x cos-1 a + 2
∴ y – 2 = x cos-1 a
∴ \(\frac{y-2}{x}\) = cos-1a
∴ cos(\(\frac{y-2}{x}\)) = a

Question 4.
Reduce each of the following differential equations to the variable separable form and hence solve:
(i) \(\frac{d y}{d x}\) = cos(x + y)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q4 (i)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q4 (i).1

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(ii) (x – y)2 \(\frac{d y}{d x}\) = a2
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q4 (ii)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q4 (ii).1

(iii) x + y \(\frac{d y}{d x}\) = sec(x2 + y2)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q4 (iii)
Integrating both sides, we get
∫cos u du = 2 ∫dx
∴ sin u = 2x + c
∴ sin(x2 + y2) = 2x + c
This is the general solution.

(iv) cos2(x – 2y) = 1 – 2 \(\frac{d y}{d x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q4 (iv)
Integrating both sides, we get
∫dx = ∫sec2u du
∴ x = tan u + c
∴ x = tan(x – 2y) + c
This is the general solution.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3

(v) (2x – 2y + 3) dx – (x – y + 1) dy = 0, when x = 0, y = 1
Solution:
(2x – 2y + 3) dx – (x – y + 1) dy = 0
∴ (x – y + 1) dy = (2x – 2y + 3) dx
∴ \(\frac{d y}{d x}=\frac{2(x-y)+3}{(x-y)+1}\) ………(1)
Put x – y = u, Then \(1-\frac{d y}{d x}=\frac{d u}{d x}\)
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.3 Q4 (v)
∴ u – log|u + 2| = -x + c
∴ x – y – log|x – y + 2| = -x + c
∴ (2x – y) – log|x – y + 2| = c
This is the general solution.
Now, y = 1, when x = 0.
∴ (0 – 1) – log|0 – 1 + 2| = c
∴ -1 – o = c
∴ c = -1
∴ the particular solution is
(2x – y) – log|x – y + 2| = -1
∴ (2x – y) – log|x – y + 2| + 1 = 0

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

Question 1.
Obtain the differential equation by eliminating the arbitrary constants from the following equations:
(i) x3 + y3 = 4ax
Solution:
x3 + y3 = 4ax ……..(1)
Differentiating both sides w.r.t. x, we get
3x2 + 3y2 \(\frac{d y}{d x}\) = 4a × 1
∴ 3x2 + 3y2 \(\frac{d y}{d x}\) = 4a
Substituting the value of 4a in (1), we get
x3 + y3 = (3x2 + 3y2 \(\frac{d y}{d x}\)) x
∴ x3 + y3 = 3x3 + 3xy2 \(\frac{d y}{d x}\)
∴ 2x3 + 3xy2 \(\frac{d y}{d x}\) – y3 = 0
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

(ii) Ax2 + By2 = 1
Solution:
Ax2 + By2 = 1
Differentiating both sides w.r.t. x, we get
A × 2x + B × 2y \(\frac{d y}{d x}\) = 0
∴ Ax + By \(\frac{d y}{d x}\) = 0 ……..(1)
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (ii)
Substituting the value of A in (1), we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (ii).1
This is the required D.E.

Alternative Method:
Ax2 + By2 = 1 ……..(1)
Differentiating both sides w.r.t. x, we get
A × 2x + B × 2y \(\frac{d y}{d x}\) = 0
∴ Ax + By \(\frac{d y}{d x}\) = 0 ……….(2)
Differentiating again w.r.t. x, we get,
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (ii).2
The equations (1), (2) and (3) are consistent in A and B.
∴ determinant of their consistency is zero.
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (ii).3
This is the required D.E.

(iii) y = A cos(log x) + B sin(log x)
Solution:
y = A cos(log x) + B sin (log x) ……. (1)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (iii)

(iv) y2 = (x + c)3
Solution:
y2 = (x + c)3
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (iv)
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

(v) y = Ae5x + Be-5x
Solution:
y = Ae5x + Be-5x ……….(1)
Differentiating twice w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (v)
This is the required D.E.

(vi) (y – a)2 = 4(x – b)
Solution:
(y – a)2 = 4(x – b)
Differentiating both sides w.r.t. x, we get
2(y – a) . \(\frac{d}{d x}\)(y – a) = 4 \(\frac{d}{d x}\)(x – b)
∴ 2(y – a) . (\(\frac{d y}{d x}\) – 0) = 4(1 – 0)
∴ 2(y – a) \(\frac{d y}{d x}\) = 4
∴ (y – a) \(\frac{d y}{d x}\) = 2 ……..(1)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (vi)
This is the required D.E.

(vii) y = a + \(\frac{a}{x}\)
Solution:
y = a + \(\frac{a}{x}\)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (vii)
Substituting the value of a in (1), we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (vii).1
This is the required D.E.

(viii) y = c1e2x + c2e5x
Solution:
y = c1e2x + c2e5x ………(1)
Differentiating twice w.r.t. x, we get
\(\frac{d y}{d x}\) = c1e2x × 2 + c2e5x × 5
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (viii)
The equations (1), (2) and (3) are consistent in c1e2x and c2e5x
∴ determinant of their consistency is zero.
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (viii).1
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

Alternative Method:
y = c1e2x + c2e5x
Dividing both sides by e5x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (viii).2
This is the required D.E.

(ix) c1x3 + c2y2 = 5.
Solution:
c1x3 + c2y2 = 5 ……….(1)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (ix)
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (ix).1
The equations (1), (2) and (3) in c1, c2 are consistent.
∴ determinant of their consistency is zero.
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (ix).2
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

(x) y = e-2x(A cos x + B sin x)
Solution:
y = e-2x(A cos x + B sin x)
∴ e2x . y = A cos x + B sin x ………(1)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (x)
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (x).1
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q1 (x).2
This is the required D.E.

Question 2.
Form the differential equation of family of lines having intercepts a and b on the coordinate axes respectively.
Solution:
The equation of the line having intercepts a and b on the coordinate axes respectively, is
\(\frac{x}{a}+\frac{y}{b}=1\) ……….(1)
where a and b are arbitrary constants.
[For different values of a and b, we get, different lines. Hence (1) is the equation of family of lines.]
Differentiating (1) w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q2
Differentiating again w.r.t. x, we get \(\frac{d^{2} y}{d x^{2}}=0\)
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

Question 3.
Find the differential equation all parabolas having length of latus rectum 4a and axis is parallel to the X-axis.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q3
Let A(h, k) be the vertex of the parabola whose length of latus rectum is 4a.
Then the equation of the parabola is (y – k)2 = 4a (x – h), where h and k are arbitrary constants.
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q3.1
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q3.2
This is the required D.E.

Question 4.
Find the differential equation of the ellipse whose major axis is twice its minor axis.
Solution:
Let 2a and 2b be lengths of major axis and minor axis of the ellipse.
Then 2a = 2(2b)
∴ a = 2b
∴ equation of the ellipse is
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
i.e., \(\frac{x^{2}}{(2 b)^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ \(\frac{x^{2}}{4 b^{2}}+\frac{y^{2}}{b^{2}}=1\)
∴ x2 + 4y2 = 4b2
Differentiating w.r.t. x, we get
2x + 4 × 2y \(\frac{d y}{d x}\) = 0
∴ x + 4y \(\frac{d y}{d x}\) = 0
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

Question 5.
Form the differential equation of family of lines parallel to the line 2x + 3y + 4 = 0.
Solution:
The equation of the line parallel to the line 2x + 3y + 4 = 0 is 2x + 3y + c = 0, where c is an arbitrary constant.
Differentiating w.r.t. x, we get
2 × 1 + 3 \(\frac{d y}{d x}\) + 0 = 0
∴ 3 \(\frac{d y}{d x}\) + 2 = 0
This is the required D.E.

Question 6.
Find the differential equation of all circles having radius 9 and centre at point (h, k).
Solution:
Equation of the circle having radius 9 and centre at point (h, k) is
(x – h)2 + (y – k)2 = 81 …… (1)
where h and k are arbitrary constant.
Differentiating (1) w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q6
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q6.1
From (2), x – h = -(y – k) \(\frac{d y}{d x}\)
Substituting the value of (x – h) in (1), we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q6.2
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q6.3
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2

Question 7.
Form the differential equation of all parabolas whose axis is the X-axis.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q7
The equation of the parbola whose axis is the X-axis is
y2 = 4a(x – h) …… (1)
where a and h are arbitrary constants.
Differentiating (1) w.r.t. x, we get
2y \(\frac{d y}{d x}\) = 4a(1 – 0)
∴ y \(\frac{d y}{d x}\) = 2a
Differentiating again w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.2 Q7.1
This is the required D.E.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 6 Differential Equations Ex 6.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1

1. Determine the order and degree of each of the following differential equations:

Question (i).
\(\frac{d y}{d x^{2}}+X\left(\frac{d y}{d x}\right)+y=2 \sin x\)
Solution:
The given D.E. is \(\frac{d y}{d x^{2}}+X\left(\frac{d y}{d x}\right)+y=2 \sin x\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. is of order 2 and degree 1.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1

Question (ii).
\(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\sqrt[3]{1+\left(\frac{d y}{d x}\right)^{2}}=\frac{d^{2} y}{d x^{2}}\)
On cubing both sides, we get
\(1+\left(\frac{d y}{d x}\right)^{2}=\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 3.
∴ the given D.E. is of order 2 and degree 3.

Question (iii).
\(\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}\)
Solution:
The given D.E. is \(\frac{d y}{d x}=\frac{2 \sin x+3}{\frac{d y}{d x}}\)
∴ \(\left(\frac{d y}{d x}\right)^{2}\) = 2 sin x + 3
This D.E. has highest order derivative \(\frac{d y}{d x}\) with power 2.
∴ the given D.E. is of order 1 and degree 2.

Question (iv).
\(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^{3} y}{d x^{3}}}\)
Solution:
The given D.E. is \(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x=\sqrt{1+\frac{d^{3} y}{d x^{3}}}\)
On squaring both sides, we get
\(\left(\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}+x\right)^{2}=1+\frac{d^{3} y}{d x^{3}}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 1.
∴ the given D.E. has order 3 and degree 1.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1

Question (v).
\(\frac{d^{2} y}{d t^{2}}+\left(\frac{d y}{d t}\right)^{2}+7 x+5=0\)
Solution:
The given D.E. is \(\frac{d^{2} y}{d t^{2}}+\left(\frac{d y}{d t}\right)^{2}+7 x+5=0\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. has order 2 and degree 1.

Question (vi).
(y”‘)2 + 3y” + 3xy’ + 5y = 0
Solution:
The given D.E. is (y”‘)2 + 3y” + 3xy’ + 5y = 0
This can be written as:
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{2}+3 \frac{d^{2} y}{d x^{2}}+3 x \frac{d y}{d x}+5 y=0\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 2.
∴ The given D.E. has order 3 and degree 2.

Question (vii).
\(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0\)
Solution:
The given D.E. is \(\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\cos \left(\frac{d y}{d x}\right)=0\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\)
∴ order = 2
Since this D.E. cannot be expressed as a polynomial in differential coefficients, the degree is not defined.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1

Question (viii).
\(\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}=8 \frac{d^{2} y}{d x^{2}}\)
Solution:
The given D.E. is \(\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}=8 \frac{d^{2} y}{d x^{2}}\)
On squaring both sides, we get
\(\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3}=8^{2} \cdot\left(\frac{d^{2} y}{d x^{2}}\right)^{2}\)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 2.
∴ the given D.E. has order 2 and degree 2.

Question (ix).
\(\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{2}} \cdot\left(\frac{d y}{d x}\right)^{\frac{1}{3}}=20\)
Solution:
The given D.E. is \(\left(\frac{d^{3} y}{d x^{3}}\right)^{\frac{1}{2}} \cdot\left(\frac{d y}{d x}\right)^{\frac{1}{3}}=20\)
∴ \(\left(\frac{d^{3} y}{d x^{3}}\right)^{3} \cdot\left(\frac{d y}{d x}\right)^{2}=20^{6}\)
This D.E. has highest order derivative \(\frac{d^{3} y}{d x^{3}}\) with power 3.
∴ the given D.E. has order 3 and degree 3.

Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1

Question (x).
\(x+\frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}\)
Solution:
The given D.E. is \(x+\frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d^{2} y}{d x^{2}}\right)^{2}}\)
On squaring both sides, we get
Maharashtra Board 12th Maths Solutions Chapter 6 Differential Equations Ex 6.1 (x)
This D.E. has highest order derivative \(\frac{d^{2} y}{d x^{2}}\) with power 1.
∴ the given D.E. has order 2 and degree 1.

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

I. Choose the correct option from the given alternatives:

Question 1.
The area bounded by the region 1 ≤ x ≤ 5 and 2 ≤ y ≤ 5 is given by
(a) 12 sq units
(b) 8 sq units
(c) 25 sq units
(d) 32 sq units
Answer:
(a) 12 sq units

Question 2.
The area of the region enclosed by the curve y = \(\frac{1}{x}\), and the lines x = e, x = e2 is given by
(a) 1 sq unit
(b) \(\frac{1}{2}\) sq units
(c) \(\frac{3}{2}\) sq units
(d) \(\frac{5}{2}\) sq units
Answer:
(a) 1 sq unit

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

Question 3.
The area bounded by the curve y = x3, the X-axis and the lines x = -2 and x = 1 is
(a) -9 sq units
(b) \(-\frac{15}{4}\) sq units
(c) \(\frac{15}{4}\) sq units
(d) \(\frac{17}{4}\) sq units
Answer:
(c) \(\frac{15}{4}\) sq units

Question 4.
The area enclosed between the parabola y2 = 4x and line y = 2x is
(a) \(\frac{2}{3}\) sq units
(b) \(\frac{1}{3}\) sq units
(c) \(\frac{1}{4}\) sq units
(d) \(\frac{3}{4}\) sq units
Answer:
(b) \(\frac{1}{3}\) sq units

Question 5.
The area of the region bounded between the line x = 4 and the parabola y2 = 16x is
(a) \(\frac{128}{3}\) sq units
(b) \(\frac{108}{3}\) sq units
(c) \(\frac{118}{3}\) sq units
(d) \(\frac{218}{3}\) sq units
Answer:
(a) \(\frac{128}{3}\) sq units

Question 6.
The area of the region bounded by y = cos x, Y-axis and the lines x = 0, x = 2π is
(a) 1 sq unit
(b) 2 sq units
(c) 3 sq units
(d) 4 sq units
Answer:
(d) 4 sq units

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

Question 7.
The area bounded by the parabola y2 = 8x, the X-axis and the latus rectum is
(a) \(\frac{31}{3}\) sq units
(b) \(\frac{32}{3}\) sq units
(c) \(\frac{32 \sqrt{2}}{3}\) sq units
(d) \(\frac{16}{3}\) sq units
Answer:
(b) \(\frac{32}{3}\) sq units

Question 8.
The area under the curve y = 2√x, enclosed between the lines x = 0 and x = 1 is
(a) 4 sq units
(b) \(\frac{3}{4}\) sq units
(c) \(\frac{2}{3}\) sq units
(d) \(\frac{4}{3}\) sq units
Answer:
(d) \(\frac{4}{3}\) sq units

Question 9.
The area of the circle x2 + y2 = 25 in first quadrant is
(a) \(\frac{25 \pi}{3}\) sq units
(b) 5π sq units
(c) 5 sq units
(d) 3 sq units
Answer:
(a) \(\frac{25 \pi}{3}\) sq units

Question 10.
The area of the region bounded by the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is
(a) ab sq units
(b) πab sq units
(c) \(\frac{\pi}{a b}\) sq units ab
(d) πa2 sq units
Answer:
(b) πab sq units

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

Question 11.
The area bounded by the parabola y2 = x and the line 2y = x is
(a) \(\frac{4}{3}\) sq units
(b) 1 sq unit
(c) \(\frac{2}{3}\) sq unit
(d) \(\frac{1}{3}\) sq unit
Answer:
(a) \(\frac{4}{3}\) sq units

Question 12.
The area enclosed between the curve y = cos 3x, 0 ≤ x ≤ \(\frac{\pi}{6}\) and the X-axis is
(a) \(\frac{1}{2}\) sq unit
(b) 1 sq unit
(c) \(\frac{2}{3}\) sq unit
(d) \(\frac{1}{3}\) sq unit
Answer:
(d) \(\frac{1}{3}\) sq unit

Question 13.
The area bounded by y = √x and line x = 2y + 3, X-axis in first quadrant is
(a) 2√3 sq units
(b) 9 sq units
(c) \(\frac{34}{3}\) sq units
(d) 18 sq units
Answer:
(b) 9 sq units

Question 14.
The area bounded by the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) and the line \(\frac{x}{a}+\frac{y}{b}=1\) is
(a) (πab – 2ab) sq units
(b) \(\frac{\pi a b}{4}-\frac{a b}{2}\) sq units
(c) (πab – ab) sq units
(d) πab sq units
Answer:
(b) \(\frac{\pi a b}{4}-\frac{a b}{2}\) sq units

Question 15.
The area bounded by the parabola y = x2 and the line y = x is
(a) \(\frac{1}{2}\) sq unit
(b) \(\frac{1}{3}\) sq unit
(c) \(\frac{1}{6}\) sq unit
(d) \(\frac{1}{12}\) sq unit
Answer:
(c) \(\frac{1}{6}\) sq unit

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

Question 16.
The area enclosed between the two parabolas y2 = 4x and y = x is
(a) \(\frac{8}{3}\) sq units
(b) \(\frac{32}{3}\) sq units
(c) \(\frac{16}{3}\) sq units
(d) \(\frac{4}{3}\) sq units
Answer:
(c) \(\frac{16}{3}\) sq units

Question 17.
The area bounded by the curve y = tan x, X-axis and the line x = \(\frac{\pi}{4}\) is
(a) \(\frac{1}{3}\) log 2 sq units
(b) log 2 sq units
(c) 2 log 2 sq units
(d) 3 log 2 sq units
Answer:
(a) \(\frac{1}{3}\) log 2 sq units

Question 18.
The area of the region bounded by x2 = 16y, y = 1, y = 4 and x = 0 in the first quadrant, is
(a) \(\frac{7}{3}\) sq units
(b) \(\frac{8}{3}\) sq units
(c) \(\frac{64}{3}\) sq units
(d) \(\frac{56}{3}\) sq units
Answer:
(d) \(\frac{56}{3}\) sq units

Question 19.
The area of the region included between the parabolas y2 = 4ax and x2 = 4ay, (a > 0) is given by
(a) \(\frac{16 a^{2}}{3}\) sq units
(b) \(\frac{8 a^{2}}{3}\) sq units
(c) \(\frac{4 a^{2}}{3}\) sq units
(d) \(\frac{32 a^{2}}{3}\) sq units
Answer:
(a) \(\frac{16 a^{2}}{3}\) sq units

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

Question 20.
The area of the region included between the line x + y = 1 and the circle x2 + y2 = 1 is
(a) \(\frac{\pi}{2}-1\) sq units
(b) π – 2 sq units
(c) \(\frac{\pi}{4}-\frac{1}{2}\) sq units
(d) π – \(\frac{1}{2}\) sq units
Answer:
(c) \(\frac{\pi}{4}-\frac{1}{2}\) sq units

(II) Solve the following:

Question 1.
Find the area of the region bounded by the following curve, the X-axis and the given lines:
(i) 0 ≤ x ≤ 5, 0 ≤ y ≤ 2
(ii) y = sin x, x = 0, x = π
(iii) y = sin x, x = 0, x = \(\frac{\pi}{3}\)
Solution:
(i) Required area = \(\int_{0}^{5} y d x\), where y = 2
= \(\int_{0}^{5} 2 d x\)
= \([2 x]_{0}^{5}\)
= 2 × 5 – 0
= 10 sq units.

(ii) The curve y = sin x intersects the X-axis at x = 0 and x = π between x = 0 and x = π.
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q1(ii)
Two bounded regions A1 and A2 are obtained. Both the regions have equal areas.
∴ required area = A1 + A2 = 2A1
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q1(ii).1

(iii) Required area = \(\int_{0}^{\pi / 3} y d x\), where y = sin x
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q1(iii)

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

Question 2.
Find the area of the circle x2 + y2 = 9, using integration.
Solution:
By the symmetry of the circle, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 3.
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q2
From the equation of the circle, y2 = 9 – x2.
In the first quadrant, y > 0
∴ y = \(\sqrt{9-x^{2}}\)
∴ area of the circle = 4 (area of the region OABO)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q2.1

Question 3.
Find the area of the ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\) using integration.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q3
By the symmetry of the ellipse, its area is equal to 4 times the area of the region OABO.
Clearly, for this region, the limits of integration are 0 and 5.
From the equation of the ellipse
\(\frac{y^{2}}{16}=1-\frac{x^{2}}{25}=\frac{25-x^{2}}{25}\)
∴ y2 = \(\frac{16}{25}\) (25 – x2)
In the first quadrant y > 0
∴ y = \(\frac{4}{5} \sqrt{25-x^{2}}\)
∴ area of the ellipse = 4(area of the region OABO)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q3.1

Question 4.
Find the area of the region lying between the parabolas:
(i) y2 = 4x and x2 = 4y
(ii) 4y2 = 9x and 3x2 = 16y
(iii) y2 = x and x2 = y.
Solution:
(i)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q4(i)
For finding the points of intersection of the two parabolas, we equate the values of y2 from their equations.
From the equation x2 = 4y, y = \(\frac{x^{2}}{4}\)
y = \(\frac{x^{4}}{16}\)
\(\frac{x^{4}}{16}\) = 4x
∴ x4 – 64x = 0
∴ x(x3 – 64) = 0
∴ x = 0 or x3 = 64 i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = \(\frac{4^{2}}{4}\) = 4
∴ the points of intersection are 0(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y2 = 4x, i.e. y = 2√x between x = 0 and x = 4
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q4(i).1

(ii)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q4(ii)
For finding the points of intersection of the two parabolas, we equate the values of 4y2 from their equations.
From the equation 3x2 = 16y, y = \(\frac{3 x^{2}}{16}\)
∴ y = \(\frac{3 x^{4}}{256}\)
∴ \(\frac{3 x^{4}}{256}\) = 9x
∴ 3x4 – 2304x = 0
∴ x(x3 – 2304) = 0
∴ x = 0 or x3 = 2304 i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = \(\frac{4^{2}}{4}\)
∴ the points of intersection are O(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y2 = 4x,
i.e. y = 2√x between x = 0 and x = 4
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q4(ii).1
Area of the region ODABO = area under the rabola x2 = 4y,
i.e. y = \(\frac{x^{2}}{4}\) between x = 0 and x = 4
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q4(ii).2

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

(iii)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q4(iii)
For finding the points of intersection of the two parabolas, we equate the values of y2 from their equations.
From the equation x2 = y, y = \(\frac{x^{2}}{y}\)
∴ y = \(\frac{x^{2}}{y}\)
∴ \(\frac{x^{2}}{y}\) = x
∴ x2 – y = 0
∴ x(x3 – y) = 0
∴ x = 0 or x3 = y
i.e. x = 0 or x = 4
When x = 0, y = 0
When x = 4, y = \(\frac{4^{2}}{4}\) = 4
∴ the points of intersection are O(0, 0) and A(4, 4).
Required area = area of the region OBACO = [area of the region ODACO] – [area of the region ODABO]
Now, area of the region ODACO = area under the parabola y2 = 4x,
i.e. y = 2√x between x = 0 and x = 4
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q4(iii).1
Area ofthe region ODABO = area under the rabola x2 = 4y,
i.e. y = \(\frac{x^{2}}{3}\) between x = 0 and x = 4
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q4(iii).2

Question 5.
Find the area of the region in the first quadrant bounded by the circle x2 + y2 = 4 and the X-axis and the line x = y√3.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q5
For finding the points of intersection of the circle and the line, we solve
x2 + y2 = 4 ………(1)
and x = y√3 ……..(2)
From (2), x2 = 3y2
From (1), x2 = 4 – y2
3y2 = 4 – y2
4y2 = 4
y2 = 1
y = 1 in the first quadrant.
When y = 1, r = 1 × √3 = √3
∴ the circle and the line intersect at A(√3, 1) in the first quadrant
Required area = area of the region OCAEDO = area of the region OCADO + area of the region DAED
Now, area of the region OCADO = area under the line x = y√3, i.e. y = \(\frac{x}{\sqrt{3}}\) between x = 0
and x = √3
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q5.1

Question 6.
Find the area of the region bounded by the parabola y2 = x and the line y = x in the first quadrant.
Solution:
To obtain the points of intersection of the line and the parabola, we equate the values of x from both equations.
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q6
∴ y2 = y
∴ y2 – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = 0
When y = 1, x = 1
∴ the points of intersection are O(0, 0) and A(1, 1).
Required area = area of the region OCABO = area of the region OCADO – area of the region OBADO
Now, area of the region OCADO = area under the parabola y2 = x i.e. y = +√x (in the first quadrant) between x = 0 and x = 1
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q6.1
Area of the region OBADO = area under the line y = x between x = 0 and x = 1
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q6.2

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

Question 7.
Find the area enclosed between the circle x2 + y2 = 1 and the line x + y = 1, lying in the first quadrant.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q7
Required area = area of the region ACBPA = (area of the region OACBO) – (area of the region OADBO)
Now, area of the region OACBO = area under the circle x2 + y2 = 1 between x = 0 and x = 1
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q7.1
Area of the region OADBO = area under the line x + y = 1 between x = 0 and x = 1
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q7.2
∴ required area = \(\left(\frac{\pi}{4}-\frac{1}{2}\right)\) sq units.

Question 8.
Find the area of the region bounded by the curve (y – 1)2 = 4(x + 1) and the line y = (x – 1).
Solution:
The equation of the curve is (y – 1)2 = 4(x + 1)
This is a parabola with vertex at A (-1, 1).
To find the points of intersection of the line y = x – 1 and the parabola.
Put y = x – 1 in the equation of the parabola, we get
(x – 1 – 1)2 = 4(x + 1)
∴ x2 – 4x + 4 = 4x + 4
∴ x2 – 8x = 0
∴ x(x – 8) = 0
∴ x = 0, x = 8
When x = 0, y = 0 – 1 = -1
When x = 8, y = 8 – 1 = 7
∴ the points of intersection are B (0, -1) and C (8, 7).
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q8
To find the points where the parabola (y – 1)2 = 4(x + 1) cuts the Y-axis.
Put x = 0 in the equation of the parabola, we get
(y – 1)2 = 4(0 + 1) = 4
∴ y – 1 = ±2
∴ y – 1 = 2 or y – 1 = -2
∴ y = 3 or y = -1
∴ the parabola cuts the Y-axis at the points B(0, -1) and F(0, 3).
To find the point where the line y = x – 1 cuts the X-axis.
Put y = 0 in the equation of the line, we get
x – 1 = 0
∴ x = 1
∴ the line cuts the X-axis at the point G (1, 0).
Required area = area of the region BFAB + area of the region OGDCEFO + area of the region OBGO
Now, area of the region BFAB = area under the parabola (y – 1)2 = 4(x + 1), Y-axis from y = -1 to y = 3
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q8.1
Since, the area cannot be negative,
Area of the region BFAB = \(\left|-\frac{8}{3}\right|=\frac{8}{3}\) sq units.
Area of the region OGDCEFO = area of the region OPCEFO – area of the region GPCDG
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q8.2
Since, area cannot be negative,
area of the region = \(\left|-\frac{1}{2}\right|=\frac{1}{2}\) sq units.
∴ required area = \(\frac{8}{3}+\frac{109}{6}+\frac{1}{2}\)
= \(\frac{16+109+3}{6}\)
= \(\frac{128}{6}\)
= \(\frac{64}{3}\) sq units.

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

Question 9.
Find the area of the region bounded by the straight line 2y = 5x + 7, X-axis and x = 2, x = 5.
Solution:
The equation of the line is
2y = 5x + 7, i.e., y = \(\frac{5}{2} x+\frac{7}{2}\)
Required area = area of the region ABCDA = area under the line y = \(\frac{5}{2} x+\frac{7}{2}\) between x = 2 and x = 5
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q9

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5

Question 10.
Find the area of the region bounded by the curve y = 4x2, Y-axis and the lines y = 1, y = 4.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q10
By symmetry of the parabola, the required area is 2 times the area of the region ABCD.
From the equation of the parabola, x2 = \(\frac{y}{4}\)
In the first quadrant, x > 0
∴ x = \(\frac{1}{2} \sqrt{y}\)
∴ required area = \(\int_{1}^{4} x d y\)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Miscellaneous Exercise 5 II Q10.1

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Application of Definite Integration Ex 5.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1

1. Find the area of the region bounded by the following curves, X-axis, and the given lines:

(i) y = 2x, x = 0, x = 5.
Solution:
Required area = \(\int_{0}^{5} y d x\), where y = 2x
= \(\int_{0}^{5} 2x d x\)
= \(\left[\frac{2 x^{2}}{2}\right]_{0}^{5}\)
= 25 – 0
= 25 sq units.

(ii) x = 2y, y = 0, y = 4.
Solution:
Required area = \(\int_{0}^{4} x d y\), where x = 2y
= \(\int_{0}^{4} 2 y d y\)
= \(\left[\frac{2 y^{2}}{2}\right]_{0}^{4}\)
= 16 – 0
= 16 sq units.

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1

(iii) x = 0, x = 5, y = 0, y = 4.
Solution:
Required area = \(\int_{0}^{5} y d x\), where y = 4
= \(\int_{0}^{5} 4 d x\)
= \([4 x]_{0}^{5}\)
= 20 – 0
= 20 sq units.

(iv) y = sin x, x = 0, x = \(\frac{\pi}{2}\)
Solution:
Required area = \(\int_{0}^{\pi / 2} y d x\), where y = sin x
= \(\int_{0}^{\pi / 2} \sin x d x\)
= \([-\cos x]_{0}^{\pi / 2}\)
= -cos \(\frac{\pi}{2}\) + cos 0
= 0 + 1
= 1 sq unit.

(v) xy = 2, x = 1, x = 4.
Solution:
For xy = 2, y = \(\frac{2}{x}\)
Required area = \(\int_{1}^{4} y d x\), where y = \(\frac{2}{x}\)
= \(\int_{1}^{4} \frac{2}{x} d x\)
= \([2 \log |x|]_{1}^{4}\)
= 2 log 4 – 2 log 1
= 2 log 4 – 0
= 2 log 4 sq units.

(vi) y2 = x, x = 0, x = 4.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q1 (vi)
The required area consists of two bounded regions A1 and A2 which are equal in areas.
For y2 = x, y = √x
Required area = A1 + A2 = 2A1
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q1 (vi).1

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1

(vii) y2 = 16x, x = 0, x = 4.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q1 (vii)
The required area consists of two bounded regions A1 and A2 which are equal in areas.
For y2 = x, y = √x
Required area = A1 + A2 = 2A1
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q1 (vii).1

2. Find the area of the region bounded by the parabola:

(i) y2 = 16x and its latus rectum.
Solution:
Comparing y2 = 16x with y2 = 4ax, we get
4a = 16
∴ a = 4
∴ focus is S(a, 0) = (4, 0)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q2 (i)
For y2 = 16x, y = 4√x
Required area = area of the region OBSAO
= 2 [area of the region OSAO]
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q2 (i).1

(ii) y = 4 – x2 and the X-axis.
Solution:
The equation of the parabola is y = 4 – x2
∴ x2 = 4 – y
i.e. (x – 0)2 = -(y – 4)
It has vertex at P(0, 4)
For points of intersection of the parabola with X-axis,
we put y = 0 in its equation.
∴ 0 = 4 – x2
∴ x2 = 4
∴ x = ± 2
∴ the parabola intersect the X-axis at A(-2, 0) and B(2, 0)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q2 (ii)
Required area = area of the region APBOA
= 2[area of the region OPBO]
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q2 (ii).1

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1

3. Find the area of the region included between:

(i) y2 = 2x and y = 2x.
Solution:
The vertex of the parabola y2 = 2x is at the origin O = (0, 0).
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (i)
To find the points of intersection of the line and the parabola, equaling the values of 2x from both the equations we get,
y2 = y
∴ y2 – y = 0
∴ y = 0 or y = 1
When y = 0, x = \(\frac{0}{2}\) = 0
When y = 1, x = \(\frac{1}{2}\)
∴ the points of intersection are 0(0, 0) and B(\(\frac{1}{2}\), 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO = area under the parabola y2 = 2x between x = 0 and x = \(\frac{1}{2}\)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (i).1
Area of the region OCBDO = area under the line y = 2x between x = 0 and x = \(\frac{1}{2}\)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (i).2

(ii) y2 = 4x and y = x.
Solution:
The vertex of the parabola y2 = 4x is at the origin O = (0, 0).
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (ii).jpg
To find the points of intersection of the line and the parabola, equaling the values of 4x from both the equations we get,
∴ y2 = y
∴ y2 – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = \(\frac{0}{2}\) = 0
When y = 1, x = \(\frac{1}{2}\)
∴ the points of intersection are O(0, 0) and B(\(\frac{1}{2}\), 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO = area under the parabola y2 = 4x between x = 0 and x = \(\frac{1}{2}\)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (ii).1
Area of the region OCBDO = area under the line y = 2x between x = 0 and x = \(\frac{1}{2}\)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (ii).2

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1

(iii) y = x2 and the line y = 4x.
Solution:
The vertex of the parabola y = x2 is at the origin 0(0, 0)
To find the points of the intersection of a line and the parabola.
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (iii)
Equating the values of y from the two equations, we get
x2 = 4x
∴ x2 – 4x = 0
∴ x(x – 4) = 0
∴ x = 0, x = 4
When x = 0, y = 4(0) = 0
When x = 4, y = 4(4) = 16
∴ the points of intersection are 0(0, 0) and B(4, 16)
Required area = area of the region OABCO = (area of the region ODBCO) – (area of the region ODBAO)
Now, area of the region ODBCO = area under the line y = 4x between x = 0 and x = 4
= \(\int_{0}^{4} y d x\), where y = 4x
= \(\int_{0}^{4} 4 x d x\)
= 4\(\int_{0}^{4} x d x\)
= 4\([latex]\int_{0}^{4} x d x\)[/latex]
= 2(16 – 0)
= 32
Area of the region ODBAO = area under the parabola y = x2 between x = 0 and x = 4
= \(\int_{0}^{4} y d x\), where y = x2
= \(\int_{0}^{4} x^{2} d x\)
= \(\left[\frac{x^{3}}{3}\right]_{0}^{4}\)
= \(\frac{1}{3}\) (64 – 0)
= \(\frac{64}{3}\)
∴ required area = 32 – \(\frac{64}{3}\) = \(\frac{32}{3}\) sq units.

(iv) y2 = 4ax and y = x.
Solution:
The vertex of the parabola y2 = 4ax is at the origin O = (0, 0).
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (iv).jpg
To find the points of intersection of the line and the parabola, equaling the values of 4ax from both the equations we get,
∴ y2 = y
∴ y2 – y = 0
∴ y(y – 1) = 0
∴ y = 0 or y = 1
When y = 0, x = \(\frac{0}{2}\) = 0
When y = 1, x = \(\frac{1}{2}\)
∴ the points of intersection are O(0, 0) and B(\(\frac{1}{2}\), 1)
Required area = area of the region OABCO = area of the region OABDO – area of the region OCBDO
Now, area of the region OABDO
= area under the parabola y2 = 4ax between x = 0 and x = \(\frac{1}{2}\)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (iv).1
Area of the region OCBDO
= area under the line y
= 4ax between x = 0 and x = \(\frac{1}{4 a x}\)
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (iv).2

Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1

(v) y = x2 + 3 and y = x + 3.
Solution:
The given parabola is y = x2 + 3, i.e. (x – 0)2 = y – 3
∴ its vertex is P(0, 3).
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (v)
To find the points of intersection of the line and the parabola.
Equating the values of y from both the equations, we get
x2 + 3 = x + 3
∴ x2 – x = 0
∴ x(x – 1) = 0
∴ x = 0 or x = 1
When x = 0, y = 0 + 3 = 3
When x = 1, y = 1 + 3 = 4
∴ the points of intersection are P(0, 3) and B(1, 4)
Required area = area of the region PABCP = area of the region OPABDO – area of the region OPCBDO
Now, area of the region OPABDO
= area under the line y = x + 3 between x = 0 and x = 1
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (v).1
Area of the region OPCBDO = area under the parabola y = x2 + 3 between x = 0 and x = 1
Maharashtra Board 12th Maths Solutions Chapter 5 Application of Definite Integration Ex 5.1 Q3 (v).2

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Miscellaneous Exercise 4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

I. Choose the correct option from the given alternatives:

Question 1.
\(\int_{2}^{3} \frac{d x}{x\left(x^{3}-1\right)}=\)
(a) \(\frac{1}{3} \log \left(\frac{208}{189}\right)\)
(b) \(\frac{1}{3} \log \left(\frac{189}{208}\right)\)
(c) \(\log \left(\frac{208}{189}\right)\)
(d) \(\log \left(\frac{189}{208}\right)\)
Answer:
(a) \(\frac{1}{3} \log \left(\frac{208}{189}\right)\)

Question 2.
\(\int_{0}^{\pi / 2} \frac{\sin ^{2} x \cdot d x}{(1+\cos x)^{2}}=\)
(a) \(\frac{4-\pi}{2}\)
(b) \(\frac{\pi-4}{2}\)
(c) 4 – \(\frac{\pi}{2}\)
(d) \(\frac{4+\pi}{2}\)
Answer:
(a) \(\frac{4-\pi}{2}\)

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

Question 3.
\(\int_{0}^{\log 5} \frac{e^{x} \sqrt{e^{x}-1}}{e^{x}+3} \cdot d x=\)
(a) 3 + 2π
(b) 4 – π
(c) 2 + π
(d) 4 + π
Answer:
(b) 4 – π

Question 4.
\(\int_{0}^{\pi / 2} \sin ^{6} x \cos ^{2} x \cdot d x=\)
(a) \(\frac{7 \pi}{256}\)
(b) \(\frac{3 \pi}{256}\)
(c) \(\frac{5 \pi}{256}\)
(d) \(\frac{-5 \pi}{256}\)
Answer:
(c) \(\frac{5 \pi}{256}\)

Question 5.
If \(\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{X}}=\frac{k}{3}\), then k is equal to
(a) √2(2√2 – 2)
(b) \(\frac{\sqrt{2}}{3}\)(2 – 2√2)
(c) \(\frac{2 \sqrt{2}-2}{3}\)
(d) 4√2
Answer:
(d) 4√2

Question 6.
\(\int_{1}^{2} \frac{1}{x^{2}} e^{\frac{1}{x}} \cdot d x=\)
(a) √e + 1
(b) √e − 1
(c) √e(√e − 1)
(d) \(\frac{\sqrt{e}-1}{e}\)
Answer:
(c) √e(√e − 1)

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

Question 7.
If \(\int_{2}^{e}\left[\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right] \cdot d x=a+\frac{b}{\log 2}\), then
(a) a = e, b = -2
(b) a = e, b = 2
(c) a = -e, b = 2
(d) a = -e, b = -2
Answer:
(a) a = e, b = -2

Question 8.
Let \(\mathrm{I}_{1}=\int_{e}^{e^{2}} \frac{d x}{\log x}\) and \(\mathrm{I}_{2}=\int_{1}^{2} \frac{e^{x}}{\boldsymbol{X}} \cdot d x\), then
(a) I1 = \(\frac{1}{3}\) I2
(b) I1 + I2 = 0
(c) I1 = 2I2
(d) I1 = I2
Answer:
(d) I1 = I2

Question 9.
\(\int_{0}^{9} \frac{\sqrt{X}}{\sqrt{X}+\sqrt{9-X}} \cdot d x=\)
(a) 9
(b) \(\frac{9}{2}\)
(c) 0
(d) 1
Answer:
(b) \(\frac{9}{2}\)

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

Question 10.
The value of \(\int_{-\pi / 4}^{\pi / 4} \log \left(\frac{2+\sin \theta}{2-\sin \theta}\right) \cdot d \theta\) is
(a) 0
(b) 1
(c) 2
(d) π
Answer:
(a) 0

II. Evaluate the following:

Question 1.
\(\int_{0}^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x\)
Solution:
Let I = \(\int_{0}^{\pi / 2} \frac{\cos x}{3 \cos x+\sin x} d x\)
Put Numerator = A(Denominator) + B[\(\frac{d}{d x}\)(Denominator)]
∴ cos x = A(3 cos x + sin x) + B[\(\frac{d}{d x}\)(3 cos x + sin x)]
= A(3 cos x + sin x) + B(-3 sin x + cos x)
∴ cos x + 0 . sin x = (3A + B) cos x + (A – 3B) sin x
Comparing the coefficients of sinx and cos x on both the sides, we get
3A + B = 1 ………. (1)
A – 3B = 0 ………. (2)
Multiplying equation (1) by 3, we get
9A + 3B = 3 ………(3)
Adding (2) and (3), we get
10A = 3
∴ A = \(\frac{3}{10}\)
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q1.1

Question 2.
\(\int_{\pi / 4}^{\pi / 2} \frac{\cos \theta}{\left[\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right]^{3}} d \theta\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q2
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q2.1

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

Question 3.
\(\int_{0}^{1} \frac{1}{1+\sqrt{x}} d x\)
Solution:
Let I = \(\int_{0}^{1} \frac{1}{1+\sqrt{x}} d x\)
Put √x = t
∴ x = t2 and dx = 2t . dt
When x = 0, t = 0
When x = 1, t = 1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q3

Question 4.
\(\int_{0}^{\pi / 4} \frac{\tan ^{3} x}{1+\cos 2 x} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q4

Question 5.
\(\int_{0}^{1} t^{5} \sqrt{1-t^{2}} d t\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q5

Question 6.
\(\int_{0}^{1}\left(\cos ^{-1} x\right)^{2} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q6
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q6.1

Question 7.
\(\int_{-1}^{1} \frac{1+x^{3}}{9-x^{2}} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q7
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q7.1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q7.2

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

Question 8.
\(\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{4} x d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q8
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q8.1

Question 9.
\(\int_{0}^{\pi} \frac{x}{1+\sin ^{2} x} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q9
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q9.1

Question 10.
\(\int_{1}^{\infty} \frac{1}{\sqrt{x}(1+x)} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q10
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 II Q10.1

III. Evaluate the following:

Question 1.
\(\int_{0}^{1}\left(\frac{1}{1+x^{2}}\right) \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q1

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

Question 2.
\(\int_{0}^{\pi / 2} \frac{1}{6-\cos x} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q2
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q2.1

Question 3.
\(\int_{0}^{a} \frac{1}{a^{2}+a x-x^{2}} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q3
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q3.1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q3.2

Question 4.
\(\int_{\pi / 5}^{3 \pi / 10} \frac{\sin x}{\sin x+\cos x} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q4
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q4.1

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

Question 5.
\(\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)
Solution:
Let I = \(\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x\)
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q5

Question 6.
\(\int_{0}^{\pi / 4} \frac{\cos 2 x}{1+\cos 2 x+\sin 2 x} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q6
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q6.1

Question 7.
\(\int_{0}^{\pi / 2}[2 \log (\sin x)-\log (\sin 2 x)] d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q7

Question 8.
\(\int_{0}^{\pi}\left(\sin ^{-1} x+\cos ^{-1} x\right)^{3} \sin ^{3} x d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q8
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q8.1

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

Question 9.
\(\int_{0}^{4}\left[\sqrt{x^{2}+2 x+3}\right]^{-1} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q9

Question 10.
\(\int_{-2}^{3}|x-2| d x\)
Solution:
|x – 2|= 2 – x, if x < 2
= x – 2, if x ≥ 2
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 III Q10

IV. Evaluate the following:

Question 1.
If \(\int_{a}^{a} \sqrt{x} d x=2 a \int_{0}^{\pi / 2} \sin ^{3} x d x\), find the value of \(\int_{a}^{a+1} x d x\).
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 IV Q1

Question 2.
If \(\int_{0}^{k} \frac{1}{2+8 x^{2}} \cdot d x=\frac{\pi}{16}\), find k.
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 IV Q2
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 IV Q2.1

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4

Question 3.
If f(x) = a + bx + cx2, show that \(\int_{0}^{1} f(x) d x=\frac{1}{6}\left[f(0)+4 f\left(\frac{1}{2}\right)+f(1)\right]\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 IV Q3
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Miscellaneous Exercise 4 IV Q3.1

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Ex 4.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

I. Evaluate:

Question 1.
\(\int_{1}^{9} \frac{x+1}{\sqrt{x}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q1

Question 2.
\(\int_{2}^{3} \frac{1}{x^{2}+5 x+6} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q2

Question 3.
\(\int_{0}^{\pi / 4} \cot ^{2} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q3
The integral does not exist since cot 0 is not defined.

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

Question 4.
\(\int_{-\pi / 4}^{\pi / 4} \frac{1}{1-\sin x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q4
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q4.1

Question 5.
\(\int_{3}^{5} \frac{1}{\sqrt{2 x+3}-\sqrt{2 x-3}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q5

Question 6.
\(\int_{0}^{1} \frac{x^{2}-2}{x^{2}+1} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q6
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q6.1

Question 7.
\(\int_{0}^{\pi / 4} \sin 4 x \sin 3 x \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q7

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

Question 8.
\(\int_{0}^{\pi / 4} \sqrt{1+\sin 2 x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q8

Question 9.
\(\int_{0}^{\pi / 4} \sin ^{4} x \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q9
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q9.1

Question 10.
\(\int_{-4}^{2} \frac{1}{x^{2}+4 x+13} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q10

Question 11.
\(\int_{0}^{4} \frac{1}{\sqrt{4 x-x^{2}}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q11

Question 12.
\(\int_{0}^{1} \frac{1}{\sqrt{3+2 x-x^{2}}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q12

Question 13.
\(\int_{0}^{\pi / 2} x \cdot \sin x \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q13
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q13.1

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

Question 14.
\(\int_{0}^{1} x \cdot \tan ^{-1} x \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q14

Question 15.
\(\int_{0}^{\infty} x \cdot e^{-x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q15
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 I Q15.1

II. Evaluate:

Question 1.
\(\int_{0}^{\frac{1}{\sqrt{2}}} \frac{\sin ^{-1} x}{\left(1-x^{2}\right)^{\frac{3}{2}}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q1.1

Question 2.
\(\int_{0}^{\pi / 4} \frac{\sec ^{2} x}{3 \tan ^{2} x+4 \tan x+1} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q2

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

Question 3.
\(\int_{0}^{4 \pi} \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q3
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q3.1

Question 4.
\(\int_{0}^{2 \pi} \sqrt{\cos x} \cdot \sin ^{3} x \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q4

Question 5.
\(\int_{0}^{\pi / 2} \frac{1}{5+4 \cos x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q5

Question 6.
\(\int_{0}^{\pi / 4} \frac{\cos x}{4-\sin ^{2} x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q6

Question 7.
\(\int_{0}^{\pi / 2} \frac{\cos X}{(1+\sin x)(2+\sin x)} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q7
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q7.1

Question 8.
\(\int_{-1}^{1} \frac{1}{a^{2} e^{x}+b^{2} e^{-x}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q8

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

Question 9.
\(\int_{0}^{\pi} \frac{1}{3+2 \sin x+\cos x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q9
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q9.1

Question 10.
\(\int_{0}^{\pi / 4} \sec ^{4} x \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q10

Question 11.
\(\int_{0}^{1} \sqrt{\frac{1-x}{1+x}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q11
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q11.1

Question 12.
\(\int_{0}^{\pi} \sin ^{3} x(1+2 \cos x)(1+\cos x)^{2} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q12
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q12.1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q12.2

Question 13.
\(\int_{0}^{\pi / 2} \sin 2 x \cdot \tan ^{-1}(\sin x) \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q13
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q13.1

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

Question 14.
\(\int_{\frac{1}{\sqrt{2}}}^{1} \frac{\left(e^{\cos ^{-1} x}\right)\left(\sin ^{-1} x\right)}{\sqrt{1-x^{2}}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q14
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q14.1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q14.2

Question 15.
\(\int_{2}^{3} \frac{\cos (\log x)}{x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 II Q15

III. Evaluate:

Question 1.
\(\int_{0}^{a} \frac{1}{x+\sqrt{a^{2}-x^{2}}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q1.1

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

Question 2.
\(\int_{0}^{\pi / 2} \log \tan x \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q2
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q2.1

Question 3.
\(\int_{0}^{1} \log \left(\frac{1}{x}-1\right) \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q3

Question 4.
\(\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cdot \cos x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q4
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q4.1

Question 5.
\(\int_{0}^{3} x^{2}(3-x)^{\frac{5}{2}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q5

Question 6.
\(\int_{-3}^{3} \frac{x^{3}}{9-x^{2}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q6

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

Question 7.
\(\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2+\sin x}{2-\sin x}\right) \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q7
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q7.1

Question 8.
\(\int_{-\pi / 4}^{\pi / 4} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q8
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q8.1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q8.2

Question 9.
\(\int_{-\pi / 4}^{\pi / 4} x^{3} \cdot \sin ^{4} x \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q9

Question 10.
\(\int_{0}^{1} \frac{\log (x+1)}{x^{2}+1} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q10
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q10.1

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

Question 11.
\(\int_{-1}^{1} \frac{x^{3}+2}{\sqrt{x^{2}+4}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q11
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q11.1

Question 12.
\(\int_{-a}^{a} \frac{x+x^{3}}{16-x^{2}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q12

Question 13.
\(\int_{0}^{1} t^{2} \sqrt{1-t} \cdot d t\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q13

Question 14.
\(\int_{0}^{\pi} x \cdot \sin x \cdot \cos ^{2} x \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q14
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q14.1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q14.2

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2

Question 15.
\(\int_{0}^{1} \frac{\log x}{\sqrt{1-x^{2}}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q15
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q15.1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q15.2
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q15.3
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.2 III Q15.4

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Definite Integration Ex 4.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1

I. Evaluate the following integrals as a limit of a sum.

Question 1.
\(\int_{1}^{3}(3 x-4) \cdot d x\)
Solution:
Let f(x) = 3x – 4, for 1 ≤ x ≤ 3
Divide the closed interval [1, 3] into n subintervals each of length h at the points
1, 1 + h, 1 + 2h, 1 + rh, ….., 1 + nh = 3
∴ nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here, a = 1
∴ f(a + rh) = f(1 + rh)
= 3(1 + rh) – 4
= 3rh – 1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1 Q1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1 Q1.1

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1

Question 2.
\(\int_{0}^{4} x^{2} d x\)
Solution:
Let f(x) = x2, for 0 ≤ x ≤ 4
Divide the closed interval [0, 4] into n subintervals each of length h at the points
0, 0 + h, 0 + 2h, ….., 0 + rh, ….., 0 + nh = 4
i.e. 0, h, 2h, ….., rh, ….., nh = 4
∴ h = \(\frac{4}{n}\) as n → ∞, h → 0
Here, a = 0
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1 Q2

Question 3.
\(\int_{0}^{2} e^{x} d x\)
Solution:
Let f(x) = ex, for 0 ≤ x ≤ 2
Divide the closed interval [0, 2] into n equal subntervals each of length h at the points
0, 0 + h, 0 + 2h, ….., 0 + rh, ….., 0 + nh = 2
i.e. 0, h, 2h, ….., rh, ….., nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here, a = 0
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1 Q3
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1 Q3.1

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1

Question 4.
\(\int_{0}^{2}\left(3 x^{2}-1\right) d x\)
Solution:
Let f(x) = 3x2 – 1, for 0 ≤ x ≤ 2
Divide the closed interval [0, 2] into n subintervals each of length h at the points.
0, 0 + h, 0 + 2h, ….., 0 + rh, ……, 0 + nh = 2
i.e. 0, h, 2h, ….., rh, ….., nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here, a = 0
∴ f(a + rh) = f(0 + rh)
= f(rh)
= 3(rh)2 – 1
= 3r2h2 – 1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1 Q4

Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1

Question 5.
\(\int_{1}^{3} x^{3} d x\)
Solution:
Let f(x) = x3, for 1 ≤ x ≤ 3.
Divide the closed interval [1, 3] into n equal su bintervals each of length h at the points
1, 1 + h, 1 + 2h, ……, 1 + rh, ……, 1 + nh = 3
∴ nh = 2
∴ h = \(\frac{2}{n}\) and as n → ∞, h → 0
Here a = 1
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1 Q5
Maharashtra Board 12th Maths Solutions Chapter 4 Definite Integration Ex 4.1 Q5.1

Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Differentiation Miscellaneous Exercise 1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1

(I) Choose the correct option from the given alternatives:

Question 1.
Let f(1) = 3, f'(1) = \(-\frac{1}{3}\), g(1) = -4 and g'(1) = \(-\frac{8}{3}\). The derivative of \(\sqrt{[f(x)]^{2}+[g(x)]^{2}}\) w.r.t. x at x = 1 is
(a) \(-\frac{29}{15}\)
(b) \(\frac{7}{3}\)
(c) \(\frac{31}{15}\)
(d) \(\frac{29}{15}\)
Answer:
(d) \(\frac{29}{15}\)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q1

Question 2.
If y = sec(tan-1 x), then \(\frac{d y}{d x}\) at x = 1, is equal to
(a) \(\frac{1}{2}\)
(b) 1
(c) \(\frac{1}{\sqrt{2}}\)
(d) 2
Answer:
(c) \(\frac{1}{\sqrt{2}}\)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q2
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q2.1

Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1

Question 3.
If f(x) = \(\sin ^{-1}\left(\frac{4^{x+\frac{1}{2}}}{1+2^{4 x}}\right)\), which of the following is not the derivative of f(x)?
(a) \(\frac{2 \cdot 4^{x} \log 4}{1+4^{2 x}}\)
(b) \(\frac{4^{x+1} \log 2}{1+4^{2 x}}\)
(c) \(\frac{4^{x+1} \log 4}{1+4^{4 x}}\)
(d) \(\frac{2^{2(x+1)} \log 2}{1+2^{4 x}}\)
Answer:
(c) \(\frac{4^{x+1} \log 4}{1+4^{4 x}}\)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q3

Question 4.
If xy = yx, then\(\frac{d y}{d x}\) = _______
(a) \(\frac{x(x \log y-y)}{y(y \log x-x)}\)
(b) \(\frac{y(y \log x-x)}{x(x \log y-y)}\)
(c) \(\frac{y^{2}(1-\log x)}{x^{2}(1-\log y)}\)
(d) \(\frac{y(1-\log x)}{x(1-\log y)}\)
Answer:
(b) \(\frac{y(y \log x-x)}{x(x \log y-y)}\)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q4

Question 5.
If y = sin (2 sin-1 x), then \(\frac{d y}{d x}\) = _______
(a) \(\frac{2-4 x^{2}}{\sqrt{1-x^{2}}}\)
(b) \(\frac{2+4 x^{2}}{\sqrt{1-x^{2}}}\)
(c) \(\frac{4 x^{2}-1}{\sqrt{1-x^{2}}}\)
(d) \(\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}\)
Answer:
(a) \(\frac{2-4 x^{2}}{\sqrt{1-x^{2}}}\)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q5

Question 6.
If y = \(\tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^{2}}}\right)+\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]\), then \(\frac{d y}{d x}\) = _______
(a) \(\frac{x}{\sqrt{1-x^{2}}}\)
(b) \(\frac{1-2 x}{\sqrt{1-x^{2}}}\)
(c) \(\frac{1-2 x}{2 \sqrt{1-x^{2}}}\)
(d) \(\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}\)
Answer:
(c) \(\frac{1-2 x}{2 \sqrt{1-x^{2}}}\)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q6
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q6.1

Question 7.
If y is a function of x and log(x + y) = 2xy, then the value of y'(0) = _______
(a) 2
(b) 0
(c) -1
(d) 1
Answer:
(d) 1
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q7
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q7.1

Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1

Question 8.
If g is the inverse of function f and f'(x) = \(\frac{1}{1+x^{7}}\), then the value of g'(x) is equal to:
(a) 1 + x7
(b) \(\frac{1}{1+[g(x)]^{7}}\)
(c) 1 + [g(x)]7
(d) 7x6
Answer:
(c) 1 + [g(x)]7
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q8

Question 9.
If \(x \sqrt{y+1}+y \sqrt{x+1}=0\) and x ≠ y, then \(\frac{d y}{d x}\) = _______
(a) \(\frac{1}{(1+x)^{2}}\)
(b) \(-\frac{1}{(1+x)^{2}}\)
(c) (1 + x)2
(d) \(-\frac{x}{x+1}\)
Answer:
(b) \(-\frac{1}{(1+x)^{2}}\)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q9

Question 10.
If y = \(\tan ^{-1}\left(\sqrt{\frac{a-x}{a+x}}\right)\), where -a < x < a, then \(\frac{d y}{d x}\) = _______
(a) \(\frac{x}{\sqrt{a^{2}-x^{2}}}\)
(b) \(\frac{a}{\sqrt{a^{2}-x^{2}}}\)
(c) \(-\frac{1}{2 \sqrt{a^{2}-x^{2}}}\)
(d) \(\frac{1}{2 \sqrt{a^{2}-x^{2}}}\)
Answer:
(c) \(-\frac{1}{2 \sqrt{a^{2}-x^{2}}}\)
[Hint: Put x = a cos θ]

Question 11.
If x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), then \(\left[\frac{d^{2} y}{d x^{2}}\right]_{\theta=\frac{\pi}{4}}\) = _______
(a) \(\frac{8 \sqrt{2}}{a \pi}\)
(b) \(-\frac{8 \sqrt{2}}{a \pi}\)
(c) \(\frac{a \pi}{8 \sqrt{2}}\)
(d) \(\frac{4 \sqrt{2}}{a \pi}\)
Answer:
(a) \(\frac{8 \sqrt{2}}{a \pi}\)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q11
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q11.1

Question 12.
If y = a cos (log x) and \(A \frac{d^{2} y}{d x^{2}}+B \frac{d y}{d x}+C=0\), then the values of A, B, C are _______
(a) x2, -x, -y
(b) x2, x, y
(c) x2, x, -y
(d) x2, -x, y
Answer:
(b) x2, x, y
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 I Q12

(II) Solve the following:

Question 1.
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q1
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q1.1
Let u(x) = f[g(x)], v(x) = g[f(x)] and w(x) = g[g(x)]. Find each derivative at x = 1, if it exists i.e. find u'(1), v'(1) and w'(1). if it doesn’t exist then explain why?
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q1.2
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q1.3
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q1.4
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q1.5
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q1.6

Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1

Question 2.
The values of f(x), g(x), f'(x) and g'(x) are given in the following table:
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q2
Match the following:
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q2.1
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q2.2
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q2.3

Question 3.
Suppose that the functions f and g and their derivatives with respect to x have the following values at x = 0 and x = 1.
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q3
(i) The derivative of f[g(x)] w.r.t. x at x = 0 is _______
(ii) The derivative of g[f(x)] w.r.t. x at x = 0 is _______
(iii) The value of \(\left[\frac{d}{d x}\left[x^{10}+f(x)\right]^{-2}\right]_{x=1}\) is _______
(iv) The derivative of f[(x+g(x))] w.r.t. x at x = 0 is _______
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q3.1
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q3.2
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q3.3
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q3.4

Question 4.
Differentiate the following w.r.t. x:
(i) \(\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]\)
Solution:
Let y = \(\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]\)
Put x = cos θ, Then θ = cos-1 x and
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q4 (i)

(ii) \(\sin ^{2}\left[\cot ^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right]\)
Solution:
Let y = \(\sin ^{2}\left[\cot ^{-1}\left(\sqrt{\frac{1+x}{1-x}}\right)\right]\)
Put x = cos θ. Then θ = cos-1 x and
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q4 (ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q4 (ii).1

(iii) \(\tan ^{-1}\left[\frac{\sqrt{x}(3-x)}{1-3 x}\right]\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q4 (iii)

(iv) \(\cos ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)\)
Solution:
Let y = \(\cos ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{2}\right)\)
Put x = cos θ. Then θ = cos-1 x and
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q4 (iv)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q4 (iv).1

Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1

(v) \(\tan ^{-1}\left(\frac{x}{1+6 x^{2}}\right)+\cot ^{-1}\left(\frac{1-10 x^{2}}{7 x}\right)\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q4 (v)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q4 (v).1

(vi) \(\tan ^{-1}\left[\sqrt{\frac{\sqrt{1+x^{2}+x}}{\sqrt{1+x^{2}}-x}}\right]\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q4 (vi)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q4 (vi).1
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q4 (vi).2

Question 5.
(i) If \(\sqrt{y+x}+\sqrt{y-x}=c\), show that \(\frac{d y}{d x}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}\)
Solution:
\(\sqrt{y+x}+\sqrt{y-x}=c\)
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q5 (i)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q5 (i).1

(ii) If \(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1\), then show that \(\frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}\)
Solution:
\(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1\)
\(y \sqrt{1-x^{2}}+x \sqrt{1-y^{2}}=1\)
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q5 (ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q5 (ii).1

(iii) If x sin(a + y) + sin a cos(a + y) = 0, then show \(\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}\)
Solution:
x sin(a + y) + sin a . cos (a + y) = 0 ….. (1)
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q5 (iii)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q5 (iii).1
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q5 (iii).2

(iv) If sin y = x sin(a + y), then show that \(\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q5 (iv)

(v) If x = \(e^{\frac{x}{y}}\), then show that \(\frac{d y}{d x}=\frac{x-y}{x \log x}\)
Solution:
x = \(e^{\frac{x}{y}}\)
\(\frac{x}{y}\) = log x …..(1)
y = \(\frac{x}{\log x}\)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q5 (v)

Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1

(vi) If y = f(x) is a differentiable function of x, then show that \(\frac{d^{2} x}{d y^{2}}=-\left(\frac{d y}{d x}\right)^{-3} \cdot \frac{d^{2} y}{d x^{2}}\)
Solution:
If y = f(x) is a differentiable function of x such that inverse function x = f-1(y) exists,
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q5 (vi)

Question 6.
(i) Differentiate \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) w.r.t. \(\tan ^{-1}\left(\frac{2 x \sqrt{1-x^{2}}}{1-2 x^{2}}\right)\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q6 (i)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q6 (i).1
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q6 (i).2
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q6 (i).3
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q6 (i).4

(ii) Differentiate \(\log \left[\frac{\sqrt{1+x^{2}}+x}{\sqrt{1+x^{2}}-x}\right]\) w.r.t. cos(log x)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q6 (ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q6 (ii).1
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q6 (ii).2

(iii) Differentiate \(\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}-1}{x}\right)\) w.r.t. \(\cos ^{-1}\left(\sqrt{\frac{1+\sqrt{1+x^{2}}}{2 \sqrt{1+x^{2}}}}\right)\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q6 (iii)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q6 (iii).1
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q6 (iii).2

Question 7.
(i) If y2 = a2 cos2x + b2 sin2x, show that \(y+\frac{d^{2} y}{d x^{2}}=\frac{a^{2} b^{2}}{y^{3}}\)
Solution:
y2 = a2 cos2x + b2 sin2x …… (1)
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q7 (i)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q7 (i).1
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q7 (i).2

(ii) If log y = log(sin x) – x2, show that \(\frac{d^{2} y}{d x^{2}}+4 x \frac{d y}{d x}+\left(4 x^{2}+3\right) y=0\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q7 (ii)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q7 (ii).1

Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1

(iii) If x = a cos θ, y = b sin θ, show that \(a^{2}\left[y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}\right]+b^{2}=0\)
Solution:
x = a cos θ, y = b sin θ
Differentiating x and y w.r.t. θ, we get
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q7 (iii)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q7 (iii).1

(iv) If y = A cos(log x) + B sin(log x), show that x2y2 + xy1 + y = o.
Solution:
y = A cos (log x) + B sin (log x) …… (1)
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q7 (iv)
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q7 (iv).1

(v) If y = A emx + B enx, show that y2 – (m + n) y1 + (mn) y = 0.
Solution:
y = A emx + B enx
Differentiating w.r.t. x, we get
Maharashtra Board 12th Maths Solutions Chapter 1 Differentiation Miscellaneous Exercise 1 II Q7 (v)

Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Ex 7.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4

Question 1.
Maximize : z = 11x + 8y subject to x ≤ 4, y ≤ 6,
x + y ≤ 6, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and ED whose equations are x = 4, y = 6 and x + y = 6 respectively.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 1
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 2
The feasible region is shaded portion OAPDO in the graph.
The vertices of the feasible region are O (0, 0), A (4, 0), P and D (0, 6)
P is point of intersection of lines x + y = 6 and x = 4.
Substituting x = 4 in x + y = 6, we get
4 + y = 6 ∴ y = 2 ∴ P is (4, 2).
∴ the corner points of feasible region are O (0, 0), A (4, 0), P(4, 2) and D(0 ,6).
The values of the objective function z = 11x + 8y at these vertices are
z (O) = 11(0) + 8(0) = 0 + 0 = 0
z(a) = 11(4) + 8(0) = 44 + 0 = 44
z (P) = 11(4) + 8(2) = 44 + 16 = 60
z (D) = 11(0) + 8(2) = 0 + 16 = 16
∴ z has maximum value 60, when x = 4 and y = 2.

Question 2.
Maximize : z = 4x + 6y subject to 3x + 2y ≤ 12,
x + y ≥ 4, x, y ≥ 0.
Solution:
First we draw the lines AB and AC whose equations are 3x + 2y = 12 and x + y = 4 respectively.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 3
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 4
The feasible region is the ∆ABC which is shaded in the graph.
The vertices of the feasible region (i.e. corner points) are A (4, 0), B (0, 6) and C (0, 4).
The values of the objective function z = 4x + 6y at these vertices are
z(a) = 4(4) + 6(0) = 16 + 0 = 16
z(B) = 4(0)+ 6(6) = 0 + 36 = 36
z(C) = 4(0) + 6(4) = 0 + 24 = 24
∴ has maximum value 36, when x = 0, y = 6.

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

Question 3.
Maximize : z = 7x + 11y subject to 3x + 5y ≤ 26
5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB and CD whose equations are 3x + 5y = 26 and 5x + 3y = 30 respectively.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 5
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 6
The feasible region is OCPBO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (6, 0), p and B(0, \(\frac{26}{5}\))
The vertex P is the point of intersection of the lines
3x + 5y = 26 … (1)
and 5x + 3y = 30 … (2)
Multiplying equation (1) by 3 and equation (2) by 5, we get
9x + 15y = 78
and 25x + 15y = 150
On subtracting, we get
16x = 72 ∴ x = \(\frac{72}{16}=\frac{9}{2}\) = 4.5
Substituting x = 4.5 in equation (2), we get
5(4.5) + 3y = 30
22.5 + 3y = 30
∴ 3y = 7.5 ∴ y = 2.5
∴ P is (4.5, 2.5)
The values of the objective function z = 7x + 11y at these corner points are
z (O) = 7(0) + 11(0) = 0 + 0 = 0
z (C) = 7(6) + 11(0) = 42 + 0 = 42
z (P) = 7(4.5) + 11 (2.5) = 31.5 + 27.5 = 59.0 = 59
z(B) = 7(0) + 11\(\left(\frac{26}{5}\right)=\frac{286}{5}\) = 57.2
∴ z has maximum value 59, when x = 4.5 and y = 2.5.

Question 4.
Maximize : z = 10x + 25y subject to 0 ≤ x ≤ 3,
0 ≤ y ≤ 3, x + y ≤ 5 also find maximum value of z.
Solution:
First we draw the lines AB, CD and EF whose equations are x = 3, y = 3 and x + y = 5 respectively.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 7
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 8
The feasible region is OAPQDO which is shaded in the i graph.
The vertices of the feasible region are O (0, 0), A (3, 0), P, Q and D(0, 3).
t P is the point of intersection of the lines x + y = 5 and x = 3.
Substituting x = 3 in x + y = 5, we get
3 + y = 5 ∴ y = 2
∴ P is (3, 2)
Q is the point of intersection of the lines x + y = 5 and y = 3
Substituting y = 3 in x + y = 5, we get
x + 3 = 5 ∴ x = 2
∴ Q is (2, 3)
The values of the objective function z = 10x + 25y at these vertices are
z(O) = 10(0) + 25(0) = 0 + 0 = 0
z(a) = 10(3) + 25(0) = 30 + 0 = 30
z(P) = 10(3) + 25(2) = 30 + 50 = 80
z(Q) = 10(2) + 25(3) = 20 + 75 = 95
z(D) = 10(0)+ 25(3) = 0 + 75 = 75
∴ z has maximum value 95, when x = 2 and y = 3.

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

Question 5.
Maximize : z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21,
x + y ≤ 9, x ≥ 0, y ≥ 0 also find maximum value of z.
Solution:
First we draw the lines AB, CD and EF whose equations are x + 4y = 24, 3x + y = 21 and x + y = 9 respectively.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 9
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 10
The feasible region is OCPQBO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (7, 0), P, Q and B (0, 6).
P is the point of intersection of the lines
3x + y = 21 … (1)
and x + y = 9 … (2)
On subtracting, we get 2x = 12 ∴ x = 6
Substituting x = 6 in equation (2), we get
6 + y = 9 ∴ y = 3
∴ P = (6, 3)
Q is the point of intersection of the lines
x + 4y = 24 … (3)
and x + y = 9 … (2)
On subtracting, we get
3y = 15 ∴ y = 5
Substituting y = 5 in equation (2), we get
x + 5= 9 ∴ x = 4
∴ Q = (4, 5)
∴ the corner points of the feasible region are 0(0,0), C(7, 0), P (6, 3), Q (4, 5) and B (0, 6).
The values of the objective function 2 = 3x + 5y at these corner points are
z(O) = 3(0)+ 5(0) = 0 + 0 = 0
z(C) = 3(7) + 5(0) = 21 + 0 = 21
z(P) = 3(6) + 5(3) = 18 + 15 = 33
z(Q) = 3(4) + 5(5) = 12 + 25 = 37
z(B) = 3(0)+ 5(6) = 0 + 30 = 30
∴ z has maximum value 37, when x = 4 and y = 5.

Question 6.
Minimize : z = 7x + y subject to 5x + y ≥ 5, x + y ≥ 3,
x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB and CD whose equations are 5x + y = 5 and x + y = 3 respectively.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 11
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 12
The feasible region is XCPBY which is shaded in the graph.
The vertices of the feasible region are C (3, 0), P and B (0, 5).
P is the point of the intersection of the lines
5x + y = 5
and x + y = 3
On subtracting, we get
4x = 2 ∴ x = \(\frac{1}{2}\)
Substituting x = \(\frac{1}{2}\) in x + y = 3, we get
\(\frac{1}{2}\) + y = 3
∴ y = \(\frac{5}{2}\) ∴ P = \(\left(\frac{1}{2}, \frac{5}{2}\right)\)
The values of the objective function z = 7x + y at these vertices are
z(C) = 7(3) + 0 = 21
z(B) = 7(0) + 5 = 5
∴ z has minimum value 5, when x = 0 and y = 5.

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

Question 7.
Minimize : z = 8x + 10y subject to 2x + y ≥ 7, 2x + 3y ≥ 15,
y ≥ 2, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and EF whose equations are 2x + y = 7, 2x + 3y = 15 and y = 2 respectively.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 13
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 14
The feasible region is EPQBY which is shaded in the graph. The vertices of the feasible region are P, Q and B(0,7). P is the point of intersection of the lines 2x + 3y = 15 and y = 2.
Substituting y – 2 in 2x + 3y = 15, we get 2x + 3(2) = 15
∴ 2x = 9 ∴ x = 4.5 ∴ P = (4.5, 2)
Q is the point of intersection of the lines
2x + 3y = 15 … (1)
and 2x + y = 7 … (2)
On subtracting, we get
2y = 8 ∴ y = 4
∴ from (2), 2x + 4 = 7
∴ 2x = 3 ∴ x = 1.5
∴ Q = (1.5, 4)
The values of the objective function z = 8x + 10y at these vertices are
z(P) = 8(4.5) + 10(2) = 36 + 20 = 56
z(Q) = 8(1.5) + 10(4) = 12 + 40 = 52
z(B) = 8(0) +10(7) = 70
∴ z has minimum value 52, when x = 1.5 and y = 4

Question 8.
Minimize : z = 6x + 21y subject to x + 2y ≥ 3, x + 4y ≥ 4,
3x + y ≥ 3, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and EF whose equations are x + 2y = 3, x + 4y = 4 and 3x + y = 3 respectively.
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 15
Maharashtra Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4 16
The feasible region is XCPQFY which is shaded in the graph.
The vertices of the feasible region are C (4, 0), P, Q and F(0, 3).
P is the point of intersection of the lines x + 4y = 4 and x + 2y = 3
On subtracting, we get
2y = 1 ∴ y = \(\frac{1}{2}\)
Substituting y = \(\frac{1}{2}\) in x + 2y = 3, we get
x + 2\(\left(\frac{1}{2}\right)\) = 3
∴ x = 2
∴ P = (2, \(\frac{1}{2}\))
Q is the point of intersection of the lines
x + 2y = 3 … (1)
and 3x + y = 3 ….(2)
Multiplying equation (1) by 3, we get 3x + 6y = 9
Subtracting equation (2) from this equation, we get
5y = 6
∴ y = \(\frac{6}{5}\)
∴ from (1), x + 2\(\left(\frac{6}{5}\right)\) = 3
∴ x = 3 – \(\frac{12}{5}=\frac{3}{5}\)
Q ≡ \(\left(\frac{3}{5}, \frac{6}{5}\right)\)
The values of the objective function z = 6x + 21y at these vertices are
z(C) = 6(4) + 21(0) = 24
z(P) = 6(2) + 21\(\left(\frac{1}{2}\right)\)
= 12 + 10.5 = 22.5
z(Q)= 6\(\left(\frac{3}{5}\right)\) + 21\(\left(\frac{6}{5}\right)\)
= \(\frac{18}{5}+\frac{126}{5}=\frac{144}{5}\) = 28.8
2 (F) = 6(0) + 21(3) = 63
∴ z has minimum value 22.5, when x = 2 and y = \(\frac{1}{2}\).