Category: Class 12

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Ex 5.5 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Ex 5.5

Question 1.
Find \(\bar{a}\)∙(\(\bar{b}\) × \(\bar{c}\)), if \(\bar{a}\) = \(3 \hat{i}-\hat{j}+4 \hat{k}\), \(\bar{b}\) = \(2 \hat{i}+3 \hat{j}-\hat{k}\) and \(\bar{c}\) = \(-5 \hat{i}+2 \hat{j}+3 \hat{k}\)
Solution:
\(\bar{a}\)∙(\(\bar{b}\) × \(\bar{c}\)) = \(\left|\begin{array}{rrr}
3 & -1 & 4 \\
2 & 3 & -1 \\
-5 & 2 & 3
\end{array}\right|\)
= 3(9 + 2) + 1 (6 – 5) + 4(4 + 15)
= 33 + 1 + 76
= 110.

Question 2.
If the vectors \(3 \hat{i}+5 \hat{k}, 4 \hat{i}+2 \hat{j}-3 \hat{k}\) and \(3 \hat{i}+\hat{j}+4 \hat{k}\) are to co-terminus edges of the parallelo piped, then find the volume of the parallelopiped.
Solution:

= 3(8 + 3) – 0(16 + 9) + 5(4 – 6)
= 33 – 0 – 10 = 23
∴ volume of the parallelopiped = \([\bar{a} \bar{b} \bar{c}]\)
= 23 cubic units.

Question 3.
If the vectors \(-3 \hat{i}+4 \hat{j}-2 \hat{k}, \hat{i}+2 \hat{k}\) and\(\hat{i}-p \hat{j}\) are coplanar, then find the value of p.
Solution:

∴ -3(0 + 2p) – 4(0 – 2) – 2(-p – 0) = 0
∴ -6p + 8 + 2p = 0
∴ -4p = -8
P = 2.

Question 4.
Prove that :
(i) [latex]\bar{a} \bar{b}+\bar{c} \bar{a}+\bar{b}+\bar{c}[/latex] = 0
Solution:

(ii) (\(\bar{a}\) – \(2 \bar{b}\) – \(\bar{c}\))∙[(\(\bar{a}\) – \(\bar{b}\)) × \(\bar{a}\) – \(\bar{b}\) – \(\bar{c}\)] = 3[\(\bar{a}\) – \(\bar{b}\) – \(\bar{c}\)]
Question is modified.
(\(\bar{a}\) – \(2 \bar{b}\) – \(\bar{c}\)) [(\(\bar{a}\) – \(\bar{b}\)) × \(\bar{a}\) – \(\bar{b}\) – \(\bar{c}\)] = 3[\(\bar{a}\) \(\bar{b}\) \(\bar{c}\)]
Solution:

Question 5.
If \(\bar{c}\) =3\(\bar{a}\) – 2\(\bar{b}\) prove that [\(\bar{a}\) \(\bar{b}\) \(\bar{c}\)] = 0
Solution:
We use the results :\(\bar{b}\) × \(\bar{b}\) = 0 and if in a scalar triple product, two vectors are equal, then the scalar triple product is zero.

Question 6.
If u = \(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{k}\), \(\bar{v}\) = \(3 \hat{\mathbf{i}}+\hat{k}\) and \(\bar{w}\) = \(\hat{\mathrm{j}}-\hat{\mathrm{k}}\) are given vectors, then find
(i) [\(\bar{u}\) + \(\bar{w}\)]∙[(\(\bar{w}\) × \(\bar{r}\)) × (\(\bar{r}\) × \(\bar{w}\))]
Question is modified.
If \(\bar{u}\) = \(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{k}\), \(\bar{r}\) = \(3 \hat{\mathbf{i}}+\hat{k}\) and \(\bar{w}\) = \(\hat{\mathrm{j}}-\hat{\mathrm{k}}\) are given vectors, then find [\(\bar{u}\) + \(\bar{w}\)]∙[(\(\bar{u}\) × \(\bar{r}\)) × (\(\bar{r}\) × \(\bar{w}\))]
Solution:

= 1(6 – 18) + 1 (-6 + 6) + 0
= -12 + 0 + 0 = -12.

Question 7.
Find the volume of a tetrahedron whose vertices are A( -1, 2, 3) B(3, -2, 1), C (2, 1, 3) and D(-1, -2, 4).
Solution:
The position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d }\) of the points A, B, C and D w.r.t. the origin are \(\bar{a}\) = \(-\hat{i}+2 \hat{j}+3 \hat{k}\), \(\bar{b}\) = \(3 \hat{i}-2 \hat{j}+\hat{k}\), \(\bar{c}\) = \(2 \hat{i}+\hat{j}+3 \hat{k}\) and

Question 8.
If \(\bar{a}\) = \(\hat{i}+2 \hat{j}+3\), \(\bar{b}\) = \(3 \hat{i}+2 \hat{j}\) and \(\bar{c}\) = ,\(2 \hat{i}+\hat{j}+3\) then verify that \(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\)) = (\(\bar{a}\) ⋅ \(\bar{c}\))\(\bar{b}\) – (\(\bar{a}\) ⋅ \(\bar{b}\))\(\bar{c}\)
Solution:


From (1) and (2), we get
\(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\)) = (\(\bar{a}\) ⋅ \(\bar{c}\))\(\bar{b}\) – (\(\bar{a}\) ⋅ \(\bar{b}\))\(\bar{c}\)

Question 9.
If, \(\bar{a}\) = \(\hat{i}-2 \hat{j}\), \(\bar{b}\) = \(\hat{i}+2 \hat{j}\) and \(\bar{c}\) =\(2 \hat{i}+\hat{j}-2\) then find
(i) \(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\))
Solution:

(ii) (\(\bar{a}\) × \(\bar{b}\)) × \(\bar{c}\) Are the results same? Justify.
Solution:

\(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\)) ≠ (\(\bar{a}\) × \(\bar{b}\)) × \(\bar{c}\)

Question 10.
Show that \(\bar{a}\) × (\(\bar{b}\) × \(\bar{c}\)) + \(\bar{b}\) × (\(\bar{c}\) × \(\bar{a}\)) + \(\bar{c}\) × (\(\bar{a}\) × \(\bar{b}\)) = 0
Solution:

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Ex 5.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Ex 5.4

Question 1.
If \(\bar{a}\) = \(2 \hat{i}+3 \hat{j}-\hat{k}\), \(\bar{b}\) = \(\hat{i}-4 \hat{j}+2 \hat{k}\) find (\(\bar{a}\) + \(\bar{b}\)) × (\(\bar{a}\) – \(\bar{b}\))
Solution:

Question 2.
Find a unit vector perpendicular to the vectors \(\hat{j}+2 \hat{k}\) and \(\hat{i}+\hat{j}\).
Solution:
Let \(\bar{a}\) = \(\hat{j}+2 \hat{k}\), \(\bar{b}\) = \(\hat{i}+\hat{j}\)

Question 3.
If \(\bar{a} \cdot \bar{b}\) = \(\sqrt {3}\) and \(\bar{a} \times \bar{b}\) = \(2 \hat{i}+\hat{j}+2 \hat{k}\), find the angle between \(\bar{a}\) and \(\bar{b}\).
Solution:
Let θ be the angle between \(\bar{a}\) and \(\bar{b}\)

∴ θ = 60°.

Question 4.
If \(\bar{a}\) = \(2 \hat{i}+\hat{j}-3 \hat{k}\) and \(\bar{b}\) = \(\hat{i}-2 \hat{j}+\hat{k}\), find a vector of magnitude 5 perpendicular to both \(\bar{a}\) and \(\bar{b}\).
Solution:
Given : \(\bar{a}\) = \(2 \hat{i}+\hat{j}-3 \hat{k}\) and \(\bar{b}\) = \(\hat{i}-2 \hat{j}+\hat{k}\)

∴ unit vectors perpendicular to both the vectors \(\bar{a}\) and \(\bar{b}\).

∴ required vectors of magnitude 5 units
= ±\(\frac{5}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\)

Question 5.
Find
(i) \(\bar{u}\)∙\(\bar{v}\) if \(|\bar{u}|\) = 2, \(|\vec{v}|\) = 5, \(|\bar{u} \times \bar{v}|\) = 8
Solution:
Let θ be the angle between \(\bar{u}\) and \(\bar{v}\).
Then \(|\bar{u} \times \bar{v}|\) = 8 gives
\(|\bar{u}||\bar{v}|\) sin θ = 8
∴ 2 × 5 × sin θ = 8

(ii) \(|\bar{u} \times \bar{v}|\) if \(|\bar{u}|\) = 10, \(|\vec{v}|\) = 2, \(\bar{u} \cdot \bar{v}\) = 12
Solution:
Let θ be the angle between \(\bar{u}\) and \(\bar{v}\).
Then \(\bar{u} \cdot \bar{v}\) = 12 gives
\(|\bar{u} \| \bar{v}|\)cos θ = 12
∴ 10 × 2 × cos θ = 12

Question 6.
Prove that 2(\(\bar{a}\) – \(\bar{b}\)) × 2(\(\bar{a}\) + \(\bar{b}\)) = 8(\(\bar{a}\) × \(\bar{b}\))
Solution:

Question 7.
If \(\bar{a}\) = \(\hat{i}-2 \hat{j}+3 \hat{k}\), \(\bar{b}\) = \(4 \hat{i}-3 \hat{j}+\hat{k}\), and \(\bar{c}\) = \(\hat{i}-\hat{j}+2 \hat{k}\), verify that \(\bar{a}\) × (\(\bar{b}\) + \(\bar{c}\)) = \(\bar{a}\) × \(\bar{b}\) + \(\bar{a}\) × \(\bar{c}\)
Solution:
Given : \(\bar{a}\) = \(\hat{i}-2 \hat{j}+3 \hat{k}\), \(\bar{b}\) = \(4 \hat{i}-3 \hat{j}+\hat{k}\), \(\bar{c}\) = \(\hat{i}-\hat{j}+2 \hat{k}\)

Question 8.
Find the area of the parallelogram whose adjacent sides are the vectors \(\bar{a}\) = \(2 \hat{i}-2 \hat{j}+\hat{k}\) and \(\bar{b}\) = \(\hat{i}-3 \hat{j}-3 \hat{k}\).
Solution:
Given : \(\bar{a}\) = \(2 \hat{i}-2 \hat{j}+\hat{k}\), \(\bar{b}\) = \(\hat{i}-3 \hat{j}-3 \hat{k}\)

Area of the parallelogram whose adjacent sides are \(\bar{a}\) and \(\bar{b}\) is \(|\bar{a} \times \bar{b}|\) =\(\sqrt {146}\) sq units.

Question 9.
Show that vector area of a quadrilateral ABCD is \(\frac{1}{2}\) (\(\overline{A C}\) × \(\overline{B D}\)), where AC and BD are its diagonals.
Solution:
Let ABCD be a parallelogram.
Then \(\overline{\mathrm{AC}}\) = \(\overline{\mathrm{AB}}\) + \(\overline{\mathrm{BC}}\)

Question 10.
Find the area of parallelogram whose diagonals are determined by the vectors \(\bar{a}\) = \(3 i-\hat{j}-2 \hat{k}\), and \(\bar{b}\) = \(-\hat{i}+3 \hat{j}-3 \hat{k}\)
Solution:
Given: \(\bar{a}\) = \(3 i-\hat{j}-2 \hat{k}\), \(\bar{b}\) = \(-\hat{i}+3 \hat{j}-3 \hat{k}\)

Question 11.
If \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d}\) are four distinct vectors such that \(\bar{a} \times \bar{b}=\bar{c} \times \bar{d}\) and \(\bar{a} \times \bar{c}=\bar{b} \times \bar{d}\), prove that \(\bar{a}\) – \(\bar{d}\) is parallel to \(\bar{b}\) – \(\bar{c}\).
Solution:
\(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d}\) are four distinct vectors

Question 12.
If \(\bar{a}\) = \(\hat{i}+\hat{j}+\hat{k}\) and, \(\bar{c}\) = \(\hat{j}-\hat{k}\), find a vector \(\bar{b}\) satisfying \(\bar{a}\) × \(\bar{b}\) = \(\bar{c}\) and \(\bar{a} \cdot \bar{b}\) = 3
Solution:
Given \(\bar{a}\) = \(\hat{i}+\hat{j}+\hat{k}\), \(\bar{c}\) = \(\hat{j}-\hat{k}\)

By equality of vectors,
z – y = 0 ….(2)
x – z = 1 ……(3)
y – x = -1 ……(4)
From (2), y = z.
From (3), x = 1 + z
Substituting these values of x and y in (1), we get
1 + z + z + z = 3 ∴ z = \(\frac{2}{3}\)
∴ y = z = \(\frac{2}{3}\)
∴ x = 1 + z =1 + \(\frac{2}{3}=\frac{5}{3}\)

Question 13.
Find \(\bar{a}\), if \(\bar{a} \times \hat{i}+2 \bar{a}-5 \hat{j}=\overline{0}\).
Solution:

By equality of vectors
2x = 0 i.e. x = 0
2y + z – 5 = 0 … (1)
2z – y = 0 … (2)
From (2), y = 2z
Substituting y = 2z in (1), we get
4z + z = 5 ∴ z = 1
∴ y = 2z = 2(1) = 2
∴ x = 0, y = 2, z = 1
∴ \(\bar{a}=2 \hat{j}+\hat{k}\)

Question 14.
If \(|\bar{a} \cdot \bar{b}|\) = \(|\bar{a} \times \bar{b}|\) and \(\bar{a} \cdot \bar{b}\) < 0, then find the angle between \(\bar{a}\) and \(\bar{b}\)
Solution:
Let θ be the angle between \(\bar{a}\) and \(\bar{b}\).
Then \(|\bar{a} \cdot \bar{b}|\) = \(|\bar{a} \times \bar{b}|\) gives

Hence, the angle between \(\bar{a}\) and \(\bar{b}\) is \(\frac{3 \pi}{4}\).

Question 15.
Prove by vector method that sin (α + β) = sinα∙cosβ+cosα∙sinβ.
Solution:

Let ∠XOP and ∠XOQ be in standard position and m∠XOP = -α, m∠XOQ = β.
Take a point A on ray OP and a point B on ray OQ such that
OA = OB = 1.
Since cos (-α) = cos α
and sin (-α) = -sin α,
A is (cos (-α), sin (-α)),
i.e. (cos α, – sin α)
B is (cos β, sin β)

The angle between \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) is α + β.
Also \(\overline{\mathrm{OA}}\), \(\overline{\mathrm{OB}}\) lie in the XY-plane.
∴ the unit vector perpendicular to \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) is \(\bar{k}\).
∴ \(\overline{\mathrm{OA}}\) × \(\overline{\mathrm{OB}}\) = [OA∙OB sin (α + β)]\(\bar{k}\)
= sin(α + β)∙\(\bar{k}\) …(2)
∴ from (1) and (2),
sin (α + β) = sin α cos β + cos α sin β.

Question 16.
Find the direction ratios of a vector perpendicular to the two lines whose direction ratios are
(i) -2, 1, -1 and -3, -4, 1
Solution:
Let a, b, c be the direction ratios of the vector which is perpendicular to the two lines whose direction ratios are -2, 1, -1 and -3, -4, 1
∴ -2a + b – c = 0 and -3a – 4b + c = 0

∴ the required direction ratios are -3, 5, 11
Alternative Method:
Let \(\bar{a}\) and \(\bar{b}\) be the vectors along the lines whose direction ratios are -2, 1, -1 and -3, -4, 1 respectively.
Then \(\bar{a}\) = \(-2 \hat{i}+\hat{j}-\hat{k}\) and \(\bar{b}\) = \(-3 \hat{i}-4 \hat{j}+\hat{k}\)
The vector perpendicular to both \(\bar{a}\) and \(\bar{b}\) is given by

Hence, the required direction ratios are -3, 5, 11.

(ii) 1, 3, 2 and -1, 1, 2
Solution:

Question 17.
Prove that two vectors whose direction cosines are given by relations al + bm + cn = 0 and fmn + gnl + hlm = 0 are perpendicular if \(\frac{f}{a}+\frac{g}{b}+\frac{h}{c}\) = 0
Solution:
Given, al + bm + cn = 0 …(1)
and fmn + gnl + hlm = 0 …..(2)
From (1), n = \(-\left(\frac{a l+b m}{c}\right)\) …..(3)
Substituting this value of n in equation (2), we get
(fm + gl)∙[latex]-\left(\frac{a l+b m}{c}\right)[/latex] + hlm = 0
∴ -(aflm + bfm² + agl² + bglm) + chlm = 0
∴ agl² + (af + bg – ch)lm + bfm² = 0 … (4)
Note that both l and m cannot be zero, because if l = m = 0, then from (3), we get
n = 0, which is not possible as l² + m² + n² = 1.
Let us take m # 0.
Dividing equation (4) by m², we get
ag\(\left(\frac{l}{m}\right)^{2}\) + (af + bg – ch)\(\left(\frac{l}{m}\right)\) + bf = 0 … (5)
This is quadratic equation in \(\left(\frac{l}{m}\right)\).
If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of the two lines given by the equation (1) and (2), then \(\frac{l_{1}}{m_{1}}\) and \(\frac{l_{2}}{m_{2}}\) are the roots of the equation (5).
From the quadratic equation (5), we get

i.e. if \(\frac{f}{a}+\frac{g}{b}+\frac{h}{c}\) = 0.

Question 18.
If A(1, 2, 3) and B(4, 5, 6) are two points, then find the foot of the perpendicular from the point B to the line joining the origin and point A.
Solution:

Let M be the foot of the perpendicular drawn from B to the line joining O and A.
Let M = (x, y, z)
OM has direction ratios x – 0, y – 0, z – 0 = x, y, z
OA has direction ratios 1 – 0, 2 – 0, 3 – 0 = 1, 2, 3
But O, M, A are collinear.
∴ \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) = k …(Let)
∴ x = k, y = 2k, z = 3k
∴ M = (k, 2k, 3k)
∵ BM has direction ratios
k – 4, 2k – 5, 3k – 6
BM is perpendicular to OA
∴ (l)(k – 4) + 2(2k – 5) + 3(3k – 6)
∴ = k – 4 + 4k – 10 + 9k – 18 = 0
∴ 14k = 32
∴ k = \(\frac{16}{7}\)
∴ M = (k, 2k, 3k) = (\(\frac{16}{7}, \frac{32}{7}, \frac{48}{7}\))

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

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

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

Question 1.
\(\frac{(\log x)^{n}}{x}\)
Solution:

Question 2.
\(\frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^{2}}}\)
Solution:
Let I = \(\int \frac{\left(\sin ^{-1} x\right)^{\frac{3}{2}}}{\sqrt{1-x^{2}}} d x\)

Question 3.
\(\frac{1+x}{x \cdot \sin (x+\log x)}\)
Solution:

Question 4.
\(\frac{x \cdot \sec ^{2}\left(x^{2}\right)}{\sqrt{\tan ^{3}\left(x^{2}\right)}}\)
Solution:

Question 5.
\(\frac{e^{3 x}}{e^{3 x}+1}\)
Solution:

Question 6.
\(\frac{\left(x^{2}+2\right)}{\left(x^{2}+1\right)} \cdot a^{x+\tan ^{-1} x}\)
Solution:

Question 7.
\(\frac{e^{x} \cdot \log \left(\sin e^{x}\right)}{\tan \left(e^{x}\right)}\)
Solution:

Question 8.
\(\frac{e^{2 x}+1}{e^{2 x}-1}\)
Solution:

Question 9.
sin⁴x . cos³x
Solution:

Question 10.
\(\frac{1}{4 x+5 x^{-11}}\)
Solution:

Question 11.
x⁹ . sec²(x¹⁰)
Solution:

Question 12.
\(e^{3 \log x} \cdot\left(x^{4}+1\right)^{-1}\)
Solution:

Question 13.
\(\frac{\sqrt{\tan x}}{\sin x \cdot \cos x}\)
Solution:
Let I = \(\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} d x\)
Dividing numerator and denominator by cos²x, we get

Question 14.
\(\frac{(x-1)^{2}}{\left(x^{2}+1\right)^{2}}\)
Solution:

Question 15.
\(\frac{2 \sin x \cos x}{3 \cos ^{2} x+4 \sin ^{2} x}\)
Solution:

Question 16.
\(\frac{1}{\sqrt{x}+\sqrt{x^{3}}}\)
Solution:

Question 17.
\(\frac{10 x^{9}+10^{x} \cdot \log 10}{10^{x}+x^{10}}\)
Solution:

Question 18.
\(\frac{x^{n-1}}{\sqrt{1+4 x^{n}}}\)
Solution:

Question 19.
(2x + 1) \(\sqrt{x+2}\)
Solution:

Question 20.
\(x^{5} \sqrt{a^{2}+x^{2}}\)
Solution:

Question 21.
\((5-3 x)(2-3 x)^{-\frac{1}{2}}\)
Solution:

Question 22.
\(\frac{7+4 x+5 x^{2}}{(2 x+3)^{\frac{3}{2}}}\)
Solution:

Question 23.
\(\frac{x^{2}}{\sqrt{9-x^{6}}}\)
Solution:

Question 24.
\(\frac{1}{x\left(x^{3}-1\right)}\)
Solution:

Question 25.
\(\frac{1}{x \cdot \log x \cdot \log (\log x)}\)
Solution:

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

Question 1.
\(\frac{\cos 3 x-\cos 4 x}{\sin 3 x+\sin 4 x}\)
Solution:

Question 2.
\(\frac{\cos x}{\sin (x-a)}\)
Solution:

Question 3.
\(\frac{\sin (x-a)}{\cos (x+b)}\)
Solution:

Question 4.
\(\frac{1}{\sin x \cdot \cos x+2 \cos ^{2} x}\)
Solution:
Let I = \(\int \frac{1}{\sin x \cdot \cos x+2 \cos ^{2} x} d x\)
Dividing numerator and denominator of cos²x, we get

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

Question 6.
\(\frac{1}{2+3 \tan x}\)
Solution:
Let I = \(\int \frac{1}{2+3 \tan x} d x\)

Numerator = A (Denominator) + B [\(\frac{d}{d x}\) (Denominator)]
∴ cos x = A(2 cos x + 3 sin x) + B [\(\frac{d}{d x}\) (2 cos x + 3 sin x)]
= A (2 cos x + 3 sin x) + B (-2 sin x + 3 cos x)
∴ cos x = (2A + 3B) cos x + (3A – 2B) sin x
Equating the coefficients of cosx and sinx on both the sides, we get
2A + 3B = 1 …… (1)
and 3A – 2B = 0 ……. (2)
Multiplying equation (1) by 2 and equation (2) by 3, we get
4A + 6B = 2
9A – 6B = 0
On adding, we get

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

Question 8.
\(\frac{20+12 e^{x}}{3 e^{x}+4}\)
Solution:

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

Question 10.
cos⁸ x . cot x
Solution:

Question 11.
tan⁵x
Solution:
Let I = ∫ tan⁵x dx

Question 12.
cos⁷x
Solution:

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

Question 14.
sin⁵x cos⁸x
Solution:

Question 15.
\(3^{\cos ^{2} x \cdot} \sin 2 x\)
Solution:

Question 16.
\(\frac{\sin 6 x}{\sin 10 x \sin 4 x}\)
Solution:

Question 17.
\(\frac{\sin x \cos ^{3} x}{1+\cos ^{2} x}\)
Solution:

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

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

I. Choose the correct option from the given alternatives:

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

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

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

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

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

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

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

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

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

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

Hint: x² + 2xy – 3y² = 0

= slope of the tangent at (1, 1)
∴ equation of the tangent at (1, 1) is -1
∴ equation of the normal is
y – 1= -1 (x – 1) = -x + 1
∴ x + y = 2
∴ y = 2 – x
Substituting y = 2 – x in x² + 2xy – 3y² = 0, we get
x² + 2x(2 – x) – 3 (2 – x)² = 0
⇒ x² + 4x – 2x² – 3(4 – 4x + x²) = 0
⇒ x² – 4x + 3 = 0
⇒ (x – 1)(x – 3) = 0
⇒ x = 1, x = 3
When x = 1, y = 2 – 1 = 1
When x = 3, y = 2 – 3 = -1
∴ the normal at (1, 1) meets the curve at (3, -1) which is in the fourth quadrant.

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

Question 9.
If the tangent at (1, 1) on y² = x(2 – x)² meets the curve again at P, then P is
(a) (4, 4)
(b) (-1, 2)
(c) (3, 6)
(d) \(\left(\frac{9}{4}, \frac{3}{8}\right)\)
Answer:
(d) \(\left(\frac{9}{4}, \frac{3}{8}\right)\)
Hint: y² = x(2 – x)²
= x(4 – 4x + x²)
= x³ – 4x² + 4x

= slope of the tangent at (1, 1)
∴ equation of the tangent at (1, 1) is
y – 1 = –\(\frac{1}{2}\) (x – 1)
∴ 2y – 2 = -x + 1
∴ x + 2y = 3
Only the coordinates \(\left(\frac{9}{4}, \frac{3}{8}\right)\) satisfy both the equations y² = x(2 – x)² and x + 2y = 3
∴ P is \(\left(\frac{9}{4}, \frac{3}{8}\right)\)

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

II. Solve the following:

Question 1.
If the curves ax² + by² = 1 and a’x² + b’y² = 1, intersect orthogonally, then prove that \(\frac{1}{a}-\frac{1}{b}=\frac{1}{a^{\prime}}-\frac{1}{b^{\prime}}\)
Solution:
Let P(x₁, y₁) be the point of intersection of the curves.

Question 2.
Determine the area of the triangle formed by the tangent to the graph of the function y = 3 – x² drawn at the point (1, 2) and the coordinate axes.
Solution:
y = 3 – x²

= slope of the tangent at (1, 2)
∴ equation of the tangent at (1, 2) is
y – 2= -2(x – 1)
⇒ y – 2= -2x + 2
⇒ 2x + y = 4
Let this tangent cuts the coordinate axes at A(a, 0) and B(0, b).
∴ 2a + 0 = 4 and 2(0) + b = 4
∴ a = 2 and b = 4
∴ area of required triangle = \(\frac{1}{2}\) × l(OA) × l(OB)
= \(\frac{1}{2}\) ab
= \(\frac{1}{2}\) (2)(4)
= 4 sq units.

Question 3.
Find the equation of the tangent and normal drawn to the curve y⁴ – 4x⁴ – 6xy = 0 at the point M (1, 2).
Solution:
y⁴ – 4x⁴ – 6xy = 0
Differentiating both sides w.r.t. x, we get

= slope of the tangent at (1, 2)
∴ the equation of normal at M (1, 2) is
y – 2 = \(\frac{14}{13}\) (x – 1)
∴ 13y – 26 = 14x – 14
∴ 14x – 13y + 12 = 0
The slope of normal at (1, 2)
\(=\frac{-1}{\left(\frac{d y}{d x}\right)_{\mathrm{at}(1,2)}}=\frac{-1}{\left(\frac{14}{13}\right)}=-\frac{13}{14}\)
∴ the equation of normal at M (1, 2) is
y – 2 = \(\frac{-13}{14}\) (x – 1)
14y – 28 = -13x + 13
13x + 14y – 41 = 0
Hence, the equations of tangent and normal are 14x – 13y + 12 = 0 and 13x + 14y – 41 = 0 respectively.

Question 4.
A water tank in the form of an inverted cone is being emptied at the rate of 2 cubic feet per second. The height of the cone is 8 feet and the radius is 4 feet. Find the rate of change of the water level when the depth is 6 feet.
Solution:

Let r be the radius of the base, h be the height and V be the volume of the water level at any time t.
Since, the height of the cone is 8 feet and the radius is 4 feet,
\(\frac{r}{h}=\frac{4}{8}=\frac{1}{2}\)
r = \(\frac{h}{2}\) ……..(1)

Hence, the rate of change of water level is \(\left(\frac{2}{9 \pi}\right)\) ft/sec.

Question 5.
Find all points on the ellipse 9x² + 16y² = 400, at which the y-coordinate is decreasing and the x-coordinate is increasing at the same rate.
Solution:
Let P(x₁, y₁) be the point on the ellipse 9x² + 16y² = 400 whose y-coordinate decreasing x-coordinate is increasing at the same rate.

Question 6.
Verify Rolle’s theorem for the function f(x) = \(\frac{2}{e^{x}+e^{-x}}\) on [-1, 1].
Solution:
The functions e^x, e^-x, and 2 are continuous and differentiable in their respective domains.
∴ f(x) = \(\frac{2}{e^{x}+e^{-x}}\) is continuous on [-1, 1] and differentiable on (-1, 1), because e^x + e^-x ≠ 0 for all x ∈ [-1, 1].
Now, f(-1) = \(\frac{2}{e^{-1}+e}=\frac{2}{e+e^{-1}}\) and f(1) = \(\frac{2}{e+e^{-1}}\)
∴ f(-1) = f(1)
Thus, the function f satisfies all the conditions of the Rolle’s theorem.
∴ there exist c ∈ (-1, 1) such that f'(c) = 0
Now, f(x) = \(\frac{2}{e^{x}+e^{-x}}\)

Hence, Rolle’s theorem is verified.

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

Question 8.
Find the approximate value of the function f(x) = \(\sqrt{x^{2}+3 x}\) at x = 1.02.
Solution:
f(x) = \(\sqrt{x^{2}+3 x}\)

Question 9.
Find the approximate value of cos^-1(0.51), given π = 3.1416, \(\frac{2}{\sqrt{3}}\) = 1.1547.
Solution:
Let f(x) = cos^-1 x

The formula for approximation is f(a + h)= f(a) + h . f'(a)
∴ cos^-1 (0.51) = f(0.51)
= f(0.5 + 0.01)
= f(0.5) + (0.01) f'(0.5)
= \(\frac{\pi}{3}\) + 0.01 × (-1.1547)
= \(\frac{3.1416}{3}\) – 0.011547
= 1.0472 – 0.011547
= 1.035653
∴ cos^-1 (0.51) = 1.035653.

Question 10.
Find the intervals on which the function y = x^x, (x > 0) is increasing and decreasing.
Solution:
y = x^x
∴ log y = log x^x = x log x
Differentiating both sides w.r.t. x, we get

y is increasing if \(\frac{d y}{d x}\) ≥ 0
i.e. if x^x (1 + log x) ≥ 0
i.e. if 1 + log x ≥ 0 ……[∵ x > 0]
i.e. if log x ≥ -1
i.e. if log x ≥ -log e …….[∵ log e = 1]
i.e. if logx ≥ log \(\frac{1}{e}\)
i.e. if x ≥ \(\frac{1}{e}\)
∴ y is increasing in \(\left[\frac{1}{e^{\prime}}, \infty\right)\)
y is decreasing if \(\frac{d y}{d x}\) ≤ 0
i.e. if x^x (1 + log x) ≤ 0
i.e. if 1 + log x ≤ 0 ……[∵ x > 0]
i.e. if log x ≤ -1
i.e. if log x ≤ -log e
i.e. if log x ≤ log \(\frac{1}{e}\)
i.e. if x ≤ \(\frac{1}{e}\) where x > 0
∴ y is decreasing is \(\left(0, \frac{1}{e}\right]\)
Hence, the given function is increasing in \(\left[\frac{1}{e^{\prime}}, \infty\right)\) and decreasing in \(\left(0, \frac{1}{e}\right]\)

Question 11.
Find the intervals on which the function f(x) = \(\frac{x}{\log x}\) is increasing and decreasing.
Solution:
f(x) = \(\frac{x}{\log x}\)

f is increasing if f'(x) ≥ 0
i.e. if \(\frac{\log x-1}{(\log x)^{2}}\) ≥ 0
i.e. if log x – 1 ≥ 0 ……..[∵ (log x)² > 0]
i.e. if log x ≥ 1
i.e. if log x ≥ log e ………[∵ log e = 1]
i.e. if x ≥ e
∴ f is increasing on [e, ∞)
f is decreasing if f'(x) ≤ 0
i.e. if \(\frac{\log x-1}{(\log x)^{2}}\) ≤ 0
i.e. if log x – 1 ≤ 0 ……..[∵ (log x)² > 0]
i.e. if log x ≤ 1
i.e. if log x ≤ log e
i.e. if x ≤ e
Also, x > 0 and x ≠ 1 because f(x) = \(\frac{x}{\log x}\) is not defined at x = 1.
∴ f is decreasing in (0, e] – {1}
Hence, f is increasing in [e, ∞) and decreasing in (0, e] – {1}.

Question 12.
An open box with a square base is to be made out of the given quantity of sheet of area a². Show that the maximum volume of the box is \(\frac{a^{3}}{6 \sqrt{3}}\).
Solution:
Let x be the side of square base and h be the height of the box.
Then x² + 4xh = a²
∴ h = \(\frac{a^{2}-x^{2}}{4 x}\) …….(1)
Let V be the volume of the box.
Then V = x²h


Hence, the maximum volume of the box is \(\frac{a^{3}}{6 \sqrt{3}}\) cu units.

Question 13.
Show that of all rectangles inscribed in a given circle, the square has the maximum area.
Solution:

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

Question 14.
Show that a closed right circular cylinder of a given surface area has maximum volume if its height equals the diameter of its base.
Solution:
Let r be the radius of the base, h be the height and V be the volume of the closed right circular cylinder, whose surface area is a² sq units (which is given).
2πrh + 2πr² = a²
⇒ 2πr(h + r) = a²
⇒ h = \(\frac{a^{2}}{2 \pi r}\) – r ……(1)


Hence, the volume of the cylinder is maximum if its height is equal to the diameter of the base.

Question 15.
A window is in the form of a rectangle surmounted by a semicircle. If the perimeter is 30 m, find the dimensions so that the greatest possible amount of light may be admitted.
Solution:

Let x be the length, y be the breadth of the rectangle and r be the radius of the semicircle.
Then perimeter of the window = x + 2y + πr, where x = 2r
This is given to be 30 m
⇒ 2r + 2y + πr = 30
⇒ 2y = 30 – (π + 2)r
⇒ y = 15 – \(\frac{(\pi+2) r}{2}\) ……..(1)
The greatest possible amount of light may be admitted if the area of the window is maximized.
Let A be the area of the window.


Hence, the required dimensions of the window are as follows:
Length of rectangle = \(\left(\frac{60}{\pi+4}\right)\) metres
breadth of rectangle = \(\left(\frac{30}{\pi+4}\right)\) metres
radius of the semicircle = \(\left(\frac{30}{\pi+4}\right)\) metres

Question 16.
Show that the height of a right circular cylinder of greatest volume that can be inscribed in a right circular cone is one-third of that of the cone.
Solution:

Given the right circular cone of fixed height h and semi-vertical angle a.
Let R be the radius of the base and H be the height of the right circular cylinder that can be inscribed in the right circular cone.
In the figure, ∠GAO = α, OG = r, OA = h, OE = R, CE = H.
We have, \(\frac{r}{h}\) = tan α
∴ r = h tan α ……(1)
Since ∆AOG and ∆CEG are similar.



Hence, the height of the right circular cylinder is one-third of that of the cone.

Question 17.
A wire of length l is cut into two parts. One part is bent into a circle and the other into a square. Show that the sum of the areas of the circle and the square is the least if the radius of the circle is half of the side of the square.
Solution:
Let r be the radius of the circle and x be the length of the side of the square. Then
(circumference of the circle) + (perimeter of the square) = l
∴ 2πr + 4x = l
∴ r = \(\frac{l-4 x}{2 \pi}\)
A = (area of the circle) + (area of the square)
= πr² + x²

This shows that the sum of the areas of circle and square is least when the radius of the circle = (\(\frac{1}{2}\)) side of the square.

Question 18.
A rectangular sheet of paper of fixed perimeter with the sides having their lengths in the ratio 8 : 15 converted into an open rectangular box by folding after removing the squares of the equal area from all comers. If the total area of the removed squares is 100, the resulting box has maximum volume. Find the lengths of the rectangular sheet of paper.
Solution:

The sides of the rectangular sheet of paper are in the ratio 8 : 15.
Let the sides of the rectangular sheet of paper be 8k and 15k respectively.
Let x be the side of the square which is removed from the comers of the sheet of paper.
The total area of removed squares is 4x², which is given to be 100.
4x² = 100
⇒ x² = 25
⇒x = 5 ……[x > 0]
Now, the length, breadth, and height of the rectangular box are 15k – 2x, 8k – 2x, and x respectively.
Let V be the volume of the box.
Then V = (15k – 2x) (8k – 2x) . x
⇒ V = (120k² – 16kx – 30kx + 4x²) . x
⇒ V = 4x³ – 46kx² + 120k²x
\(\frac{d V}{d x}=\frac{d}{d x}\) (4x³ – 46kx² + 120k²x)
= 4 × 3x² – 46k × 2x + 120k² × 1
= 12x² – 92kx + 120k²
Since, volume is maximum when the square of side x = 5 is removed from the corners,
\(\left(\frac{d V}{d x}\right)_{\text {at } x=5}=0\)
⇒ 12(5)² – 92k(5) + 120k² = 0
⇒ 60 – 92k + 24k² = 0
⇒ 6k² – 23k + 15 = 0
⇒ 6k² – 18k – 5k + 15 = 0
⇒ 6k(k – 3) – 5 (k – 3) = 0
⇒ (k – 3)(6k – 5) = 0
⇒ k = 3 or k = \(\frac{5}{6}\)
If k = \(\frac{5}{6}\), then
8k – 2x = 8k – 10 < 0
∴ k ≠ \(\frac{5}{6}\)
∴ k = 3
∴ 8k = 8 × 3 = 24 and 15k = 15 × 3 = 45
Hence, the lengths of the rectangular sheet are 24 and 45.

Question 19.
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is \(\frac{4 r}{3}\).
Solution:

Let x be the radius of the base and h be the height of the cone which is inscribed in a sphere of radius r.
In the figure, AD = h and CD = x = BD
Since, ΔABD and ΔBDE are similar,
\(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{BD}}{\mathrm{DE}}\)
BD² = AD . DE = AD (AE – AD)
x² = h(2r – h) …… (1)
Let V be the volume of the cone.


∴ V is maximum when h = \(\frac{4 r}{3}\)
Hence, the altitude (i.e. height) of the right circular cone of maximum volume = \(\frac{4 r}{3}\).

Question 20.
Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is \(\frac{2 R}{\sqrt{3}}\). Also, find the maximum Volume.
Solution:
Let R be the radius and h be the height of the cylinder which is inscribed in a sphere of radius r cm.
Then from the figure,
\(R^{2}+\left(\frac{h}{2}\right)^{2}=r^{2}\)
∴ R² = r² – \(\frac{h^{2}}{4}\) ………(1)
Let V be the volume of the cylinder.
Then V = πR²h



Hence, the volume of the largest cylinder inscribed in a sphere of radius ‘r’ cm = \(\frac{4 R^{3}}{3 \sqrt{3}}\) cu cm.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 6.
Find the values of x for which f(x) = \(\frac{\boldsymbol{x}}{x^{2}+1}\) is
(a) strictly increasing
(b) strictly decreasing.
Solution:
f(x) = \(\frac{\boldsymbol{x}}{x^{2}+1}\)

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

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

Question 7.
Show that f(x) = 3x + \(\frac{1}{3 x}\) is increasing in (\(\frac{1}{3}\), 1) and decreasing in (\(\frac{1}{9}\), \(\frac{1}{3}\))
Solution:
f(x) = 3x + \(\frac{1}{3 x}\)

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

Question 9.
Find the maximum and minimum of the following functions:
(i) y = 5x³ + 2x² – 3x
Solution:

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

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

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

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

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

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

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

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

(iv) f(x) = x² + \(\frac{16}{x^{2}}\)
Solution:

(v) f(x) = x log x
Solution:

(vi) f(x) = \(\frac{\log x}{x}\)
Solution:

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

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

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

Question 13.
A ball is thrown in the air. Its height at any time t is given by h = 3 + 14t – 5t². Find the maximum height it can reach.
Solution:
The height h at any t is given by h = 3 + 14t – 5t²


Hence, the maximum height the ball can reach = 12.8 units.

Question 14.
Find the largest size of a rectangle that can be inscribed in a semicircle of radius 1 unit, so that two vertices lie on the diameter.
Solution:

Let ABCD be the rectangle inscribed in a semicircle of radius 1 unit such that the vertices A and B lie on the diameter.
Let AB = DC = x and BC = AD = y.
Let O be the centre of the semicircle.
Join OC and OD. Then OC = OD = radius = 1.
Also, AD = BC and m∠A = m∠B = 90°.
∴ OA = OB
∴ OB = \(\frac{1}{2}\) AB = \(\frac{x}{2}\)
In right angled triangle OBC,
OB² + BC² = OC²




Hence, the area of the rectangle is maximum (i.e. rectangle has the largest size) when its length is √2 units and breadth is \(\frac{1}{\sqrt{2}}\) unit.

Question 15.
An open cylindrical tank whose base is a circle is to be constructed of metal sheet so as to contain a volume of πa³ cu cm of water. Find the dimensions so that the quantity of the metal sheet required is minimum.
Solution:
Let x be the radius of the base, h be the height, V be the volume and S be the total surface area of the cylindrical tank.
Then V = πa³ … (Given)
∴ πx²h = πa³
∴ h = \(\frac{a^{3}}{x^{2}}\) ……..(1)
Now, S = 2πxh + πx²

∴ by the second derivative test, S is minimum when x = a
When x = a, from (1)
h = \(\frac{a^{3}}{a^{2}}\) = a
Hence, the quantity of metal sheet is minimum when radius height = a cm.

Question 16.
The perimeter of a triangle is 10 cm. If one of the sides is 4 cm. What are the other two sides of the triangle for its maximum area?
Solution:

Let ABC be the triangle such that the side BC = a = 4 cm.
Also, the perimeter of the triangle is 10 cm.
i.e. a + b + c = 10
∴ 2s = 10
∴ s = 5
Also, 4 + b + c = 10
∴ b + c = 6
∴ b = 6 – c
Let ∆ be the area of the triangle.



∴ by the second derivative test, ∆ is maximum when c = 3.
When c = 3, b = 6 – c = 6 – 3 = 3
Hence, the area of the triangle is maximum when the other two sides are 3 cm and 3 cm.

Question 17.
A box with a square base is to have an open top. The surface area of the box is 192 sq cm. What should be its dimensions in order that the volume is largest?
Solution:
Let x cm be the side of square base and h cm be its height.
Then x² + 4xh = 192
∴ h = \(\frac{192-x^{2}}{4 x}\) …… (1)
Let V be the volume of the box.

∴ by the second derivative test, V is maximum at x = 8.
If x = 8, h = \(\frac{192-64}{4(8)}=\frac{128}{32}\) = 4
Hence, the volume of the box is largest, when the side of square base is 8 cm and its height is 4 cm.

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

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

Question 20.
Show that among rectangles of given area, the square has least perimeter.
Solution:
Let x be the length and y be the breadth of the rectangle whose area is A sq units (which is given as constant).
Then xy = A
∴ y = \(\frac{A}{x}\) ………(1)
Let P be the perimeter of the rectangle.

x = y
∴ rectangle is a square.
Hence, among rectangles of given area, the square has least perimeter.

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


Hence, the surface area is least when height of the closed right circular cylinder is equal to its diameter.

Question 22.
Find the volume of the largest cylinder that can be inscribed in a sphere of radius ‘r’ cm.
Solution:
Let R be the radius and h be the height of the cylinder which is inscribed in a sphere of radius r cm.
Then from the figure,


Hence, the volume of the largest cylinder inscribed in a sphere of radius ‘r’ cm = \(\frac{4 \pi r^{3}}{3 \sqrt{3}}\) cu cm.

Question 23.
Show that y = log(1 + x) – \(\frac{2 x}{2+x}\), x > -1 is an increasing function on its domain.
Solution:
y = log(1 + x) – \(\frac{2 x}{2+x}\), x > -1

Hence, the given function is increasing function on its domain.

Question 24.
Prove that y = \(\frac{4 \sin \theta}{2+\cos \theta}\) – θ is an increasing function if θ ∈ [0, \(\frac{\pi}{2}\)]
Solution:
y = \(\frac{4 \sin \theta}{2+\cos \theta}\) – θ

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

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

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

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

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

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

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

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

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

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

(ii) f(x) = sin(\(\frac{x}{2}\)), x ∈ [0, 2π]
Solution:
The function f(x) = sin(\(\frac{x}{2}\)) is continuous on [0, 2π] and differentiable on (0, 2π).
Now, f(0) = sin 0 = 0
and f(2π) = sin π = 0
∴ f(0) = f(2π)
Thus, the function f satisfies all the conditions of Rolle’s theorem.
∴ there exists c ∈ (0, 2π) such that f'(c) = 0.


Hence, Rolle’s theorem is verified.

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

Question 4.
If Rolle’s theorem holds for the function f(x) = x³ + px² + qx + 5, x ∈ [1, 3] with c = 2 + \(\frac{1}{\sqrt{3}}\), find the values of p and q.
Solution:
The Rolle’s theorem holds for the function f(x) = x³ + px² + qx + 5, x ∈ [1, 3]
∴ f(1) = f(3)
∴ 1³ + p(1)² + q(1) + 5 = 3³ + p (3)² + q(3) + 5
∴ 1 + p + q + 5 = 27 + 9p + 3q + 5
∴ 8p + 2q = -26
∴ 4p + q = -13 ….. (1)
Also, there exists at least one point c ∈ (1, 3) such that f'(c) = 0.
Now, f'(x) = \(\frac{d}{d x}\) (x³ + px² + qx + 5)
= 3x² + p × 2x + q × 1 + 0
= 3x² + 2px + q

But f'(c) = 0
∴ \(4 p+\frac{2 p}{\sqrt{3}}+q+13+\frac{12}{\sqrt{3}}=0\)
∴ (4√3 + 2)p + √3q + (13√3 + 12) = 0
∴ (4√3 + 2)p + √3q = -13√3 – 12 ……. (2)
Multiplying equation (1) by √3, we get
4√3p + √3q= -13√3
Subtracting this equation from (2), we get
2p = -12 ⇒ p= -6
∴ from (1), 4(-6) + q = -13 ⇒ q = 11
Hence, p = -6 and q = 11.

Question 5.
If Rolle’s theorem holds for the function f(x) = (x – 2) log x, x ∈ [1, 2], show that the equation x log x = 2 – x is satisfied by at least one value of x in (1, 2).
Solution:
The Rolle’s theorem holds for the function f(x) = (x – 2) log x, x ∈ [1, 2].
∴ there exists at least one real number c ∈ (1, 2) such that f'(c) = 0.
Now, f(x) = (x – 2) log x

∴ f'(c) = 0 gives 1 – \(\frac{2}{c}\) + log c = 0
∴ c – 2 + c log c = 0
∴ c log c = 2 – c, where c ∈ (1, 2)
∴ c satisfies the equation x log x = 2 – x, c ∈ (1, 2).
Hence, the equation x log x = 2 – x is satisfied by at least one value of x in (1, 2).

Question 6.
The function f(x) = \(x(x+3) e^{-\frac{x}{2}}\) satisfies all the conditions of Rolle’s theorem on [-3, 0]. Find the value of c such that f'(c) = 0.
Solution:
The function f(x) satisfies all the conditions of Rolle’s theorem, therefore there exist c ∈ (-3, 0) such that f'(c) = 0.

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

Hence, Lagrange’s mean value theorem is verified.

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


Hence, Lagrange’s mean value theorem is verified.

(iii) f(x) = x² – 3x – 1, x ∈ \(\left[\frac{-11}{7}, \frac{13}{7}\right]\)
Solution:
The function f given as f(x) = x² – 3x – 1 is a polynomial function.
Hence, it is continuous on \(\left[\frac{-11}{7}, \frac{13}{7}\right]\) and differentiable on \(\left(\frac{-11}{7}, \frac{13}{7}\right)\).
Thus, the function f satisfies the conditions of LMVT.
∴ there exists c ∈ \(\left(\frac{-11}{7}, \frac{13}{7}\right)\) such that


Hence, Lagrange’s mean value theorem is verified.

(iv) f(x) = 2x – x², x ∈ [0, 1].
Solution:
The function f given as f(x) = 2x – x² is a polynomial function.
Hence, it is continuous on [0, 1] and differentiable on (0, 1).
Thus, the function f satisfies the conditions of Lagrange’s mean value theorem.
∴ there exists c ∈ (0, 1) such that

Hence, Lagrange’s mean value theorem is verified.

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

Hence, Lagrange’s mean value theorem is verified.

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

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

Question 1.
Find the approximate value of given functions, at required points.
(i) √8.95
Solution:

√8.95 = 2.9917

(ii) \(\sqrt[3]{28}\)
Solution:

(iii) \(\sqrt[5]{31.98}\)
Solution:

(iv) (3.97)⁴
Solution:

(v) (4.01)³
Solution:

Question 2.
Find the approximate values of:
(i) sin 61°, given that 1° = 0.0174^c, √3 = 1.732.
Solution:

(ii) sin(29° 30′), given that 1° = 0.0175^c, √3 = 1.732.
Solution:

(iii) cos(60° 30′), given that 1° = 0.0175^c, √3 = 1.732.
Solution:

(iv) tan (45° 40′), given that 1° = 0.0175^c.
Solution:

Question 3.
Find the approximate values of
(i) tan^-1(0.999).
Solution:

(ii) cot^-1(0.999).
Solution:

(iii) tan^-1(1.001).
Solution:

Question 4.
Find the approximate values of:
(i) e^0.995, given that e = 2.7183.
Solution:

(ii) e^2.1, given that e² = 7.389.
Solution:

(iii) 3^2.01, given that log 3 = 1.0986.
Solution:

Question 5.
Find the approximate values of:
(i) log_e(101), given that log_e10 = 2.3026.
Solution:

(ii) log_e(9.01), given that log 3 = 1.0986.
Solution:

(iii) log₁₀(1016), given that log₁₀e = 0.4343.
Solution:

Question 6.
Find the approximate values of:
(i) f(x) = x³ – 3x + 5 at x = 1.99
Solution:

(ii) f(x) = x³ + 5x² – 7x +10 at x = 1.12
Solution:

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

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

Question 1.
Find the equations of tangents and normals to the curve at the point on it.
(i) y = x² + 2e^x + 2 at (0, 4)
Solution:

(ii) x³ + y³ – 9xy = 0 at (2, 4)
Solution:
x³ + y³ – 9xy = 0
Differentiating both sides w.r.t. x, we get


Hence, the equations of tangent and normal are 4x – 5y + 12 = 0 and 5x + 4y – 26 = 0 respectively.

(iii) x² – √3xy + 2y² = 5 at (√3, 2)
Solution:
x² – √3xy + 2y² = 5
Differentiating both sides w.r.t. x, we get

the slope of normal at (√3, 2) does not exist.
normal is parallel to Y-axis.
equation of the normal is of the form x = k
Since, it passes through the point (√3, 2), k = √3
equation of the normal is x = √3.
Hence, the equations of tangent and normal are y = 2 and x = √3 respectively.

(iv) 2xy + π sin y = 2π at (1, \(\frac{\pi}{2}\))
Solution:
2xy + π sin y = 2π
Differentiating both sides w.r.t. x, we get


Hence, the equations of tangent and normal are πx + 2y – 2π = 0 and 4x – 2πy + π² – 4 = 0 respectively.

(v) x sin 2y = y cos 2x at (\(\frac{\pi}{4}\), \(\frac{\pi}{2}\))
Solution:
x sin 2y = y cos 2x
Differentiating both sides w.r.t. x, we get



Hence, the equations of the tangent and normal are 2x – y = 0 and 4x + 8y – 5π = 0 respectively.

(vi) x = sin θ and y = cos 2θ at θ = \(\frac{\pi}{6}\)
Solution:
When θ = \(\frac{\pi}{6}\), x = sin\(\frac{\pi}{6}\) and y = cos\(\frac{\pi}{3}\)
∴ x = \(\frac{1}{2}\) and y = \(\frac{1}{2}\)
Hence, the point at which we want to find the equations of tangent and normal is (\(\frac{1}{2}\), \(\frac{1}{2}\))
Now, x = sin θ, y = cos 2θ
Differentiating x and y w.r.t. θ, we get


2y – 1 = x – \(\frac{1}{2}\)
4y – 2 = 2x – 1
2x – 4y + 1 = 0
Hence, equations of the tangent and normal are 4x + 2y – 3 = 0 and 2x – 4y + 1 = 0 respectively.

(vii) x = √t, y = t – \(\frac{1}{\sqrt{t}}\), at t = 4.
Solution:
When t = 4, x = √4 and y = 4 – \(\frac{1}{\sqrt{4}}\)
∴ x = 2 and y = 4 – \(\frac{1}{2}\) = \(\frac{7}{2}\)
Hence, the point at which we want to find the equations of tangent and normal is (2, \(\frac{7}{2}\)).
Now, x = √t, y = t – \(\frac{1}{\sqrt{t}}\)
Differentiating x and y w.r.t. t, we get



Hence, the equations of tangent and normal are 17x – 4y – 20 = 0 and 8x + 34y – 135 = 0 respectively.

Question 2.
Find the point of the curve y = \(\sqrt{x-3}\) where the tangent is perpendicular to the line 6x + 3y – 5 = 0.
Solution:
Let the required point on the curve y = \(\sqrt{x-3}\) be P(x₁, y₁).
Differentiating y = \(\sqrt{x-3}\) w.r.t. x, we get

Hence, the required points are (4, 1) and (4, -1).

Question 3.
Find the points on the curve y = x³ – 2x² – x where the tangents are parallel to 3x – y + 1 = 0.
Solution:
Let the required point on the curve y = x³ – 2x² – x be P(x₁, y₁).

Question 4.
Find the equations of the tangents to the curve x² + y² – 2x – 4y + 1 = 0 which are parallel to the X-axis.
Solution:
Let P (x₁, y₁) be the point on the curve x² + y² – 2x – 4y + 1 = 0 where the tangent is parallel to X-axis.
Differentiating x² + y² – 2x – 4y + 1 = 0 w.r.t. x, we get


the coordinates of the points are (1, 0) or (1, 4)
Since the tangents are parallel to X-axis, their equations are of the form y = k
If it passes through the point (1, 0), k = 0, and if it passes through the point (1, 4), k = 4
Hence, the equations of the tangents are y = 0 and y = 4.

Question 5.
Find the equations of the normals to the curve 3x² – y² = 8, which are parallel to the line x + 3y = 4.
Solution:
Let P(x₁, y₁) be the foot of the required normal to the curve 3x² – y² = 8.
Differentiating 3x² – y² = 8 w.r.t. x, we get


Hence, the equations of the normals are x + 3y – 8 = 0 and x + 3y + 8 = 0.

Question 6.
If the line y = 4x – 5 touches the curve y² = ax³ + b at the point (2, 3), find a and b.
Solution:
y² = ax³ + b
Differentiating both sides w.r.t. x, we get

= slope of the tangent at (2, 3)
Since, the line y = 4x – 5 touches the curve at the point (2, 3), slope of the tangent at (2, 3) is 4.
2a = 4 ⇒ a = 2
Since (2, 3) lies on the curve y² = ax³ + b
(3)² = a(2)³ + b
9 = 8a + b
9 = 8(2) + b …… [∵ a = 2]
b = -7
Hence, a = 2 and b = -7.

Question 7.
A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which y-coordinate is changing 8 times as fast as the x-coordinate.
Solution:
Let P(x₁, y₁) be the point on the curve 6y = x³ + 2 whose y-coordinate is changing 8 times as fast as the x-coordinate.

Question 8.
A spherical soap bubble is expanding so that its radius is increasing at the rate of 0.02 cm/sec. At what rate is the surface area increasing, when its radius is 5 cm?
Solution:
Let r be the radius and S be the surface area of the soap bubble at any time t.
Then S = 4πr²
Differentiating w.r.t. t, we get

Hence, the surface area of the soap bubble is increasing at the rate of 0.87c cm² / sec.

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

Hence, the volume of the spherical balloon is increasing at the rate of 6 cm³ / sec.

Question 10.
If each side of an equilateral triangle increases at the rate of √2 cm/sec, find the rate of increase of its area when its side of length is 3 cm.
Solution:
If x cm is the side of the equilateral triangle and A is its area, then \(A=\frac{\sqrt{3}}{4} x^{2}\)
Differentiating w.r.t. f, we get

Hence, rate of increase of the area of equilateral triangle = \(\frac{3 \sqrt{6}}{2}\) cm² / sec.

Question 11.
The volume of a sphere increases at the rate of 20 cm³/sec. Find the rate of change of its surface area, when its radius is 5 cm.
Solution:
Let r be the radius, S be the surface area and V be the volume of the sphere at any time t.
Then S = 4πr² and V = \(\frac{4}{3} \pi r^{3}\)
Differentiating w.r.t. t, we get

Hence, the surface area of the sphere is changing at the rate of 8 cm²/sec.

Question 12.
The edge of a cube is decreasing at the rate of 0.6 cm/sec. Find the rate at which its volume is decreasing, when the edge of the cube is 2 cm.
Solution:
Let x be the edge of the cube and V be its volume at any time t.
Then V = x³
Differentiating both sides w.r.t. t, we get

Hence, the volume of the cube is decreasing at the rate of 7.2 cm³/sec.

Question 13.
A man of height 2 meters walks at a uniform speed of 6 km/hr away from a lamp post of 6 meters high. Find the rate at which the length of the shadow is increasing.
Solution:
Let OA be the lamp post, MN the man, MB = x, his shadow, and OM = y, the distance of the man from the lamp post at time t.

Then \(\frac{d y}{d t}\) = 6 km/hr is the rate at which the man is moving at away from the lamp post.
\(\frac{d x}{d t}\) is the rate at which his shadow is increasing.
From the figure,
\(\frac{x}{2}=\frac{x+y}{6}\)
6x = 2x + 2y
4x = 2y
x = \(\frac{1}{2}\) y
\(\frac{d x}{d t}=\frac{1}{2} \frac{d y}{d t}=\frac{1}{2} \times 6=3 \mathrm{~km} / \mathrm{hr}\)
Hence, the length of the shadow is increasing at the rate of 3 km/hr.

Question 14.
A man of height 1.5 meters walks towards a lamp post of height 4.5 meters, at the rate of (\(\frac{3}{4}\)) meter/sec.
Find the rate at which
(i) his shadow is shortening
(ii) the tip of the shadow is moving.
Solution:

Let OA be the lamp post, MN the man, MB = x his shadow and OM = y the distance of the man from lamp post at time t.
Then \(\frac{d y}{d t}=\frac{3}{4}\) is the rate at which the man is moving towards the lamp post.
\(\frac{d x}{d t}\) is the rate at which his shadow is shortening.
B is the tip of the shadow and it is at a distance of x + y from the post.
\(\frac{d}{d t}(x+y)=\frac{d x}{d t}+\frac{d y}{d t}\) is the rate at which the tip of the shadow is moving.
From the figure,
\(\frac{x}{1.5}=\frac{x+y}{4.5}\)
45x = 15x + 15y
30x = 15y
x = \(\frac{1}{2}\)y
\(\frac{d x}{d t}=\frac{1}{2} \cdot \frac{d y}{d t}=\frac{1}{2}\left(\frac{3}{4}\right)=\left(\frac{3}{8}\right) \text { metre/sec }\)
and \(\frac{d x}{d t}+\frac{d y}{d t}=\frac{3}{8}+\frac{3}{4}=\left(\frac{9}{8}\right) \text { metres } / \mathrm{sec}\)
Hence (i) the shadow is shortening at the rate of (\(\frac{3}{8}\)) metre/sec, and
(ii) the tip of shadow is moving at the rate of (\(\frac{9}{8}\)) metres/sec.

Question 15.
A ladder 10 metres long is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at the rate of 1.2 metres per second, find how fast the top of the ladder is sliding down the wall, when the bottom is 6 metres away from the wall.
Solution:

Let AB be the ladder, where AB = 10 metres.
Let at time t seconds, the end A of the ladder be x metres from the wall and the end B be y metres from the ground.
Since, OAB is a right angled triangle, by Pythagoras’ theorem
x² + y² = 10² i.e. y² = 100 – x²
Differentiating w.r.t. t, we get
2y \(\frac{d y}{d t}\) = 0 – 2x \(\frac{d x}{d t}\)
∴ \(\frac{d y}{d t}=-\frac{x}{y} \cdot \frac{d x}{d t}\) ……..(1)
Now, \(\frac{d x}{d t}\) = 1.2 metres/sec is the rate at which the bottom at of the ladder is pulled horizontally and \(\frac{d y}{d t}\) is the rate at which the top of ladder B is sliding.
When x = 6, y² = 100 – 36 = 64
y = 8
(1) gives \(\frac{d y}{d t}=-\frac{6}{8}(1.2)=-\frac{6}{8} \times \frac{12}{10}\)
\(=-\frac{9}{10}=-0.9\)
Hence, the top of the ladder is sliding down the wall, at the rate of 0.9 metre/sec.

Question 16.
If water is poured into an inverted hollow cone whose semi-vertical angle is 30° so that its depth (measured along the axis) increases at the rate of 1 cm/sec. Find the rate at which the volume of water increases when the depth is 2 cm.
Solution:

Let r be the radius, h be the height, θ be the semi-vertical angle and V be the volume of the water at any time t.

Hence, the volume of water is increasing at the rate of \(\left(\frac{4 \pi}{3}\right)\) cm3/sec.

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

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

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

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

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

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

Question 5.
If the two lines ax² +2hxy+ by² = 0 make angles α and β with X-axis, then tan (α + β) = _____.
(A) \(\frac{h}{a+b}\)
(B) \(\frac{h}{a-b}\)
(C) \(\frac{2 h}{a+b}\)
(D) \(\frac{2 h}{a-b}\)
Solution:
(D) \(\frac{2 h}{a-b}\)

Question 6.
If the slope of one of the two lines \(\frac{x^{2}}{a}+\frac{2 x y}{h}+\frac{y^{2}}{b}\) = 0 is twice that of the other, then ab:h² = ___.
(A) 1 : 2
(B) 2 : 1
(C) 8 : 9
(D) 9 : 8
Solution:
(D) 9 : 8

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(viii) Passing through the origin and having slopes 1 + \(\sqrt {3}\) and 1 – \(\sqrt {3}\)
Solution:
Let l₁ and l₂ be the two lines. Slopes of l₁ is 1 + \(\sqrt {3}\) and that of l₂ is 1 – \(\sqrt {3}\)
Therefore the equation of a line (l₁) passing through the origin and having slope is
y = (1 + \(\sqrt {3}\))x
∴ (1 + \(\sqrt {3}\))x – y = 0 ..(1)
Similarly, the equation of the line (l₂) passing through the origin and having slope is
y = (1 – \(\sqrt {3}\))x
∴ (1 – \(\sqrt {3}\))x – y = 0 …(2)
From (1) and (2) the required combined equation is

∴ (1 – 3)x² – 2xy + y² = 0
∴ -2x² – 2xy + y² = 0
∴ 2x² + 2xy – y² = 0
This is the required combined equation.

(ix) Which are at a distance of 9 units from the Y – axis.
Solution:
Equations of the lines, which are parallel to the Y-axis and at a distance of 9 units from it, are x = 9 and x = -9
i.e. x – 9 = 0 and x + 9 = 0

∴ their combined equation is
(x – 9)(x + 9) = 0
∴ x² – 81 = 0.

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

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

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

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

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

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

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

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

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

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

(iv) 2x² + 2xy – y² = 0
Solution:
2x² + 2xy – y² = 0
∴ The auxiliary equation is -m² + 2m + 2 = 0
∴ m² – 2m – 2 = 0

m₁ = 1 + \(\sqrt {3}\) and m₂ = 1 – \(\sqrt {3}\) are the slopes of the lines.
∴ their separate equations are
y = m₁x and y = m₂x
i.e. y = (1 + \(\sqrt {3}\))x and y = (1 – \(\sqrt {3}\))x
i.e. (\(\sqrt {3}\) + 1)x – y = 0 and (\(\sqrt {3}\) – 1)x + y = 0.

Question 4.
Find the joint equation of the pair of lines through the origin and perpendicular to the lines
given by :
(i) x² + 4xy – 5y² = 0
Solution:
Comparing the equation x² + 4xy – 5y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 4, b= -5
Let m₁ and m₂ be the slopes of the lines represented by x² + 4xy – 5y² = 0.

Now, required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m₁y = -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁x + m₂y) = 0
∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x² + \(\frac{4}{5}\)xy – \(\frac{1}{5}\)y² = 0 …[By (1)]
∴ 5x² + 4xy – y² = 0

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

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

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

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

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

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

(v) The slope of one of the lines given by 3x² + 4xy + ky² = 0 is three times the other.
Solution:
3x² + 4xy + ky² = 0
∴ divide by x²

∴ y = mx
∴ \(\frac{\mathrm{y}}{\mathrm{x}}\) = m
put \(\frac{\mathrm{y}}{\mathrm{x}}\) = m in equation (1)
Comparing the equation km² + 4m + 3 = 0 with ax² + 2hxy+ by² = 0, we get,
a = k, 2h = 4, b = 3
m₁ = 3m₂ ..(given condition)
m₁ + m₂ = \(\frac{-2 h}{k}=-\frac{4}{k}\)
m₁m₂ = \(\frac{a}{b}=\frac{3}{k}\)
m₁ + m₂ = \(-\frac{4}{\mathrm{k}}\)
4m₂ = \(-\frac{4}{\mathrm{k}}\) …(m₁ = 3m₂)
m₂ = \(-\frac{1}{\mathrm{k}}\)
m₁m₂ = \(\frac{3}{k}\)
\(3 \mathrm{~m}_{2}^{2}=\frac{3}{\mathrm{k}}\) …(m₁ = 3m₂)
\(3\left(-\frac{1}{\mathrm{k}}\right)^{2}=\frac{3}{\mathrm{k}}\) …(m₂ = \(-\frac{1}{k}\))
\(\frac{1}{k^{2}}=\frac{1}{k}\)
k² = k
k = 1 or k = 0

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

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

Question 6.
Find the joint equation of the pair of lines which bisect angle between the lines given by x² + 3xy + 2y² = 0
Solution:
x² + 3xy + 2y² = 0
∴ x² + 2xy + xy + 2y² = 0
∴ x(x + 2y) + y(x + 2y) = 0
∴ (x + 2y)(x + y) = 0
∴ separate equations of the lines represented by x² + 3xy + 2y² = 0 are x + 2y = 0 and x + y = 0.
Let P (x, y) be any point on one of the angle bisector. Since the points on the angle bisectors are equidistant from both the lines,

the distance of P (x, y) from the line x + 2y = 0
= the distance of P(x, y) from the line x + y = 0

∴ 2(x + 2y)² = 5(x + y)²
∴ 2(x² + 4xy + 4y²) = 5(x² + 2xy + y²)
∴ 2x² + 8xy + 8y² = 5x² + 10xy + 5y²
∴ 3x² + 2xy – 3y² = 0.
This is the required joint equation of the lines which bisect the angles between the lines represented by x² + 3xy + 2y² = 0.

Question 7.
Find the joint equation of the pair of lies through the origin and making equilateral triangle with the line x = 3.
Solution:

Let OA and OB be the lines through the origin making an angle of 60° with the line x = 3.
∴ OA and OB make an angle of 30° and 150° with the positive direction of X-axis
∴ slope of OA = tan 30° = 1/\(\sqrt {3}\)
∴ equation of the line OA is y = \(\frac{1}{\sqrt{3}}\)x
∴ \(\sqrt {3}\)y = x ∴ x – \(\sqrt {3}\)y = 0
Slope of OB = tan 150° = tan (180° – 30°)
= – tan 30°= -1/\(\sqrt {3}\)
∴ equation of the line OB is y = \(\frac{-1}{\sqrt{3}}\)x
∴ \(\sqrt {3}\)y = -x ∴ x + \(\sqrt {3}\)y = 0
∴ required combined equation of the lines is
(x – \(\sqrt {3}\)y) (x + \(\sqrt {3}\)y) = 0
i.e. x² – 3y² = 0.

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

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

Question 10.
Find the combined equation of the bisectors of the angles between the lines represented by 5x² + 6xy – y² = 0.
Solution:
Comparing the equation 5x² + 6xy – y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 5, 2h = 6, b = -1
Let m₁ and m₂ be the slopes of the lines represented by 5x² + 6xy – y² = 0.

The separate equations of the lines are
y = m₁x and y = m₂x, where m1 ≠ m2
i.e. m₁x – y = 0 and m₁x – y = 0.
Let P (x, y) be any point on one of the bisector of the angles between the lines.
∴ the distance of P from the line m₁x – y = 0 is equal to the distance of P from the line m₂x – y = 0.

∴ (m₂² + 1)(m₁x – y)² = (m₁² + 1)(m₂x – y)²
∴ (m₂² + 1)(m₁²x² – 2m₁xy + y²) = (m₁² + 1)(m₂²x² – 2m₂xy + y²)
∴ m₁²m₂²x² – 2m₁m₁²y²xy + m₂²y² + m₁²x² – 2m₁²xy + y²
= m₁²m₂²x² – 2m₁²m₂xy + m₁²y² + m₂²x² – 2m₂xy + y²
∴ (m₁² – m₂²)x² + 2m₁m₂(m₁ – m₂)xy – 2(m₁ – m₂)xy – (m₁² – m₂²)y² = 0
Dividing throughout by m₁ – m₂ (≠0), we get,
(m₁ + m₂)x² + 2m₁m₂xy – 2xy – (m₁ + m₂)y² = 0
∴ 6x² – 10xy – 2xy – 6y² = 0 …[By (1)]
∴ 6x² – 12xy – 6y² = 0
∴ x² – 2xy – y² = 0
This is the joint equation of the bisectors of the angles between the lines represented by 5x² + 6xy – y² = 0.

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

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

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

∴ tan θ = tan0°
∴ θ = 0°.

(ii) 2x² + xy – y² + x + 4y – 3 = 0
Solution:
Comparing this equation with
ax² + 2hxy + by² + 2gx + 2fy+ c = 0, we get,
a = 2, h = \(\frac{1}{2}\), b = -1, g = \(\frac{1}{2}\), f = 2 and c = -3

= -2 + 1 + 1
= -2 + 2= 0
∴ the given equation represents a pair of lines.
Let θ be the acute angle between the lines.

∴ tan θ = tan 3
∴ θ = tan^-1(3)

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

∴ tanθ = 3 ∴ θ = tan^-1(3)

Question 14.
Find the combined equation of pair of lines through the origin each of which makes angle of 60° with the Y-axis.
Solution:

Let OA and OB be the lines through the origin making an angle of 60° with the Y-axis.
Then OA and OB make an angle of 30° and 150° with the positive direction of X-axis.
∴ slope of OA = tan 30° = \(\frac{1}{\sqrt{3}}\)
∴ equation of the line OA is
y = \(\frac{1}{\sqrt{3}}\) = x, i.e. x – \(\sqrt {3}\)y = 0
Slope of OB = tan 150° = tan (180° – 30°)
= tan 30° = \(-\frac{1}{\sqrt{3}}\)
∴ equation of the line OB is
y = \(-\frac{1}{\sqrt{3}}\)x, i.e. x + \(\sqrt {3}\) y = 0
∴ required combined equation is
(x – \(\sqrt {3}\)y)(x + \(\sqrt {3}\)y) = 0
i.e. x² – 3y² = 0.

Question 15.
If lines representedby ax² + 2hxy + by² = 0 make angles of equal measures with the co-ordinate
axes then show that a = ± b.
OR
Show that, one of the lines represented by ax² + 2hxy + by² = 0 will make an angle of the same measure with the X-axis as the other makes with the Y-axis, if a = ± b.
Solution:

Let OA and OB be the two lines through the origin represented by ax² + 2hxy + by² = 0.
Since these lines make angles of equal measure with the coordinate axes, they make angles ∝ and \(\frac{\pi}{2}\) – ∝ with the positive direction of X-axis or ∝ and \(\frac{\pi}{2}\) + ∝ with thepositive direction of X-axis.
∴ slope of the line OA = m₁ = tan ∝
and slope of the line OB = m₂
= tan(\(\frac{\pi}{2}\) – ∝) or tan(\(\frac{\pi}{2}\) + ∝)
i.e. m₂ = cot ∝ or m₂ = -cot ∝
∴ m₁m₂ – tan ∝ x cot ∝ = 1
OR m₁m₂ = tan ∝ (-cot ∝) = -1
i.e. m₁m₂ = ± 1
But m₁m₂ = \(\frac{a}{b}\)
∴ \(\frac{a}{b}\)= ±1 ∴ a = ±b
This is the required condition.

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

∴ y² + 2xysec2∝ + x² = 0
∴ x² + 2(sec2∝)xy + y² = 0 is the required equation.

Question 17.
Show that the line 3x + 4y+ 5 = 0 and the lines (3x + 4y)² – 3(4x – 3y)² =0 form an equilateral triangle.
Solution:
The slope of the line 3x + 4y + 5 = 0 is \(\frac{-3}{4}\)
Let m be the slope of one of the line making an angle of 60° with the line 3x + 4y + 5 = 0. The angle between the lines having slope m and m₁ is 60°.

On squaring both sides, we get,
3 = \(\frac{(4 m+3)^{2}}{(4-3 m)^{2}}\)
∴ 3 (4 – 3m)² = (4m + 3)²
∴ 3(16 – 24m + 9m²) = 16m² + 24m + 9
∴ 48 – 72m + 27m² = 16m² + 24m + 9
∴ 11m² – 96m + 39 = 0
This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = \(\frac{y}{x}\).
∴ the combined equation of the two lines is
11\(\left(\frac{y}{x}\right)^{2}\) – 96\(\left(\frac{y}{x}\right)\) + 39 = 0
∴ \(\frac{11 y^{2}}{x^{2}}-\frac{96 y}{x}\) + 39 = 0
∴ 11y² – 96xy + 39x² = 0
∴ 39x² – 96xy + 11y² = 0.
∴ 39x² – 96xy + 11y² = 0 is the joint equation of the two lines through the origin each making an angle of 60° with the line 3x + 4y + 5 = 0.
The equation 39x² – 96xy + 11y² = 0 can be written as :
-39x² + 96xy – 11y² = 0
i.e., (9x² – 48x²) + (24xy + 72xy) + (16y² – 27y²) = 0
i.e. (9x² + 24xy + 16y²) – (48x² – 72xy + 27y²) = 0
i.e. (9x² + 24xy + 16y²) – 3(16x² – 24xy + 9y²) = 0
i.e. (3x + 4y)² – 3(4x – 3y)² = 0
Hence, the line 3x + 4y + 5 = 0 and the lines
(3x + 4y)² – 3(4x – 3y)² form the sides of an equilateral triangle.

Question 18.
Show that lines x² – 4xy + y² = 0 and x + y = \(\sqrt {6}\) form an equilateral triangle. Find its area and perimeter.
Solution:
x² – 4xy + y² = 0 and x + y = \(\sqrt {6}\) form a triangle OAB which is equilateral.
Let OM be the perpendicular from the origin O to AB whose equation is x + y = \(\sqrt {6}\)

In right angled triangle OAM,
sin 60° = \(\frac{\mathrm{OM}}{\mathrm{OA}}\) ∴ \(\frac{\sqrt{3}}{2}\) = \(\frac{\sqrt{3}}{\mathrm{OA}}\)
∴ OA = 2
∴ length of the each side of the equilateral triangle OAB = 2 units.
∴ perimeter of ∆ OAB = 3 × length of each side
= 3 × 2 = 6 units.

Question 19.
If the slope of one of the lines given by ax² + 2hxy + by² = 0 is square of the other then show that a²b + ab² + 8h³ = 6abh.
Solution:
Let m be the slope of one of the lines given by ax² + 2hxy + by² = 0.
Then the other line has slope m²

Multiplying by b³, we get,
-8h³ = ab² + a²b – 6abh
∴ a²b + ab² + 8h³ = 6abh
This is the required condition.

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

Question 21.
Show that the difference between the slopes of lines given by (tan²θ + cos²θ )x² – 2xytanθ + (sin²θ )y² = 0 is two.
Solution:
Comparing the equation (tan²θ + cos²θ)x² – 2xy tan θ + (sin²θ) y² = 0 with ax² + 2hxy + by² = 0, we get,
a = tan²θ + cos²θ, 2h = -2 tan θ and b = sin2θ
Let m₁ and m₂ be the slopes of the lines represented by the given equation.

Question 22.
Find the condition that the equation ay² + bxy + ex + dy = 0 may represent a pair of lines.
Solution:
Comparing the equation
ay² + bxy + ex + dy = 0 with
Ax² + 2Hxy + By² + 2Gx + 2Fy + C = 0, we get,
A = 0, H = \(\frac{b}{2}\), B = a,G = \(\frac{e}{2}\), F = \(\frac{d}{2}\), C = 0
The given equation represents a pair of lines,

i.e. if bed – ae² = 0
i.e. if e(bd – ae) = 0
i.e. e = 0 or bd – ae = 0
i.e. e = 0 or bd = ae
This is the required condition.

Question 23.
If the lines given by ax² + 2hxy + by² = 0 form an equilateral triangle with the line lx + my = 1 then show that (3a + b)(a + 3b) = 4h².
Solution:
Since the lines ax² + 2hxy + by² = 0 form an equilateral triangle with the line lx + my = 1, the angle between the lines ax² + 2hxy + by² = 0 is 60°.

∴ 3(a + b)² = 4(h² – ab)
∴ 3(a² + 2ab + b²) = 4h² – 4ab
∴ 3a² + 6ab + 3b² + 4ab = 4h²
∴ 3a² + 10ab + 3b² = 4h²
∴ 3a² + 9ab + ab + 3b² = 4h²
∴ 3a(a + 3b) + b(a + 3b) = 4h²
∴ (3a + b)(a + 3b) = 4h²
This is the required condition.

Question 24.
If line x + 2 = 0 coincides with one of the lines represented by the equation x² + 2xy + 4y + k = 0 then show that k = -4.
Solution:
One of the lines represented by
x² + 2xy + 4y + k = 0 … (1)
is x + 2 = 0.
Let the other line represented by (1) be ax + by + c = 0.
∴ their combined equation is (x + 2)(ax + by + c) = 0
∴ ax² + bxy + cx + 2ax + 2by + 2c = 0
∴ ax² + bxy + (2a + c)x + 2by + 2c — 0 … (2)
As the equations (1) and (2) are the combined equations of the same two lines, they are identical.
∴ by comparing their corresponding coefficients, we get,

∴ 1 = \(\frac{-4}{k}\)
∴ k = -4.

Question 25.
Prove that the combined equation of the pair of lines passing through the origin and perpendicular to the lines represented by ax² + 2hxy + by² = 0 is bx² – 2hxy + ay² = 0
Solution:
Let m₁ and m₂ be the slopes of the lines represented by ax² + 2hxy + by² = 0.

Now, required lines are perpendicular to these lines.
∴ their slopes are and \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(-\frac{1}{m_{1}}\)x and y = \(-\frac{1}{m_{2}}\)x
i.e. m₁y= -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁y)(x + m₂y) = 0
∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x²\(\frac{-2 h}{b}\)x + \(\frac{a}{b}\)y² = 0
∴ bx² – 2hxy + ay² = 0.

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

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

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

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

(ii) Through the point (2, -3) and parallel to lines represented by x² + xy – y² = 0
Solution:
Comparing the equation
x² + xy – y² = 0 … (1)
with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 1, b = -1
Let m₁ and m₂ be the slopes of the lines represented by (1).

The slopes of the lines parallel to these lines are m₁ and m₂.
∴ the equations of the lines with these slopes and through the point (2, -3) are
y + 3 = m₁(x – 2) and y + 3 = m₂(x – 2)
i.e. m₁(x – 2) – (y + 3) = 0 and m₂(x – 2) – (y + 3) = 0
∴ the joint equation of these lines is
[m₁(x – 2) – (y + 3)][m₂(x – 2) – (y + 3)] = 0
∴ m₁m₂(x – 2)² – m₁(x – 2)(y + 3) – m₂(x – 2)(y + 3) + (y + 3)² = o
∴ m₁m₂(x – 2)² – (m₁ + m₂)(x – 2)(y + 3) + (y + 3)³ = 0
∴ -(x – 2)² – (x – 2)(y + 3) + (y + 3)² = 0 …… [By (2)]
∴ (x – 2)² + (x – 2)(y + 3) – (y + 3)² = 0
∴ (x² – 4x + 4) + (xy + 3x – 2y – 6) – (y² + 6y + 9) = 0
∴ x² – 4x + 4 + xy + 3x – 2y – 6 – y² – 6y – 9 = 0
∴ x² + xy – y² – x – 8y – 11 = 0.

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

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

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

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

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

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

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

(iii) x² + 3xy + 2y² + x – y + k = 0
Solution:
Comparing the given equation with
ax² + 2hxy + by² + 2gx + 2fy + c = 0,
we get, a = 1, h = \(\frac{3}{2}\), b = 2, g = \(\frac{1}{2}\), f= \(-\frac{1}{2}\), c = k.
Now, given equation represents a pair of lines.


∴ 2(8k – 1) – 3(6k + 1) + 1(-3 – 4) = 0
∴ 16k – 2 – 18k – 3 – 7 = 0
∴ -2k – 12 = 0
∴ -2k = 12 ∴ k = -6.

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

Question 8.
Find p and q if the equation 2x² + 8xy + py² + qx + 2y – 15 = 0 represents a pair of parallel lines.
Solution:
The given equation is
2x² + 8xy + py² + qx + 2y – 15 = 0
Comparing it with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get,
a = 2, h = 4, b = p, g = \(\frac{q}{2}\), f = 1, c = – 15
Since the lines are parallel, h² = ab
∴ (4)² = 2p ∴ P = 8
Since the given equation represents a pair of lines

i.e. – 242 + 240 + 2q + 2q – 2q² = 0
i.e. -2q² + 4q – 2 = 0
i.e. q² – 2q + 1 = 0
i.e. (q – 1)² = 0 ∴ q – 1 = 0 ∴ q = 1.
Hence, p = 8 and q = 1.

Question 9.
Equations of pairs of opposite sides of a parallelogram are x² – 7x+ 6 = 0 and y² – 14y + 40 = 0. Find the joint equation of its diagonals.
Solution:
Let ABCD be the parallelogram such that the combined equation of sides AB and CD is x² – 7x + 6 = 0 and the combined equation of sides BC and AD is y² – 14y + 40 = 0.
The separate equations of the lines represented by x² – 7x + 6 = 0, i.e. (x – 1)(x – 6) = 0 are x – 1 = 0 and x – 6 = 0.
Let equation of the side AB be x – 1 = 0 and equation of side CD be x – 6 = 0.
The separate equations of the lines represented by y² – 14y + 40 = 0, i.e. (y – 4)(y – 10) = 0 are y – 4 = 0 and y – 10 = 0.
Let equation of the side BC be y – 4 = 0 and equation of side AD be y – 10 = 0.

Coordinates of the vertices of the parallelogram are A(1, 10), B(1, 4), C(6, 4) and D(6, 10).
∴ equation of the diagonal AC is
\(\frac{y-10}{x-1}\) = \(\frac{10-4}{1-6}\) = \(\frac{6}{-5}\)
∴ -5y + 50 = 6x – 6
∴ 6x + 5y – 56 = 0
and equation of the diagonal BD is
\(\frac{y-4}{x-1}\) = \(\frac{4-10}{1-6}\) = \(\frac{-6}{-5}\) = \(\frac{6}{5}\)
∴ 5y – 20 = 6x – 6
∴ 6x – 5y + 14 = 0
Hence, the equations of the diagonals are 6x + 5y – 56 = 0 and 6x – 5y + 14 = 0.
∴ the joint equation of the diagonals is (6x + 5y – 56)(6x – 5y + 14) = 0
∴ 36x² – 30xy + 84x + 30xy – 25y² + 70y – 336x + 280y – 784 = 0
∴ 36x² – 25y² – 252x + 350y – 784 = 0.

Question 10.
∆OAB is formed by lines x² – 4xy + y² = 0 and the line 2x + 3y – 1 = 0. Find the equation of the median of the triangle drawn from O.
Solution:

Let D be the midpoint of seg AB where A is (x₁, y₁) and B is (x₂, y₂).
Then D has coordinates \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\).
The joint (combined) equation of the lines OA and OB is x² – 4xy + y² = 0 and the equation of the line AB is 2x + 3y – 1 = 0.
∴ points A and B satisfy the equations 2x + 3y – 1 = 0
and x² – 4xy + y² = 0 simultaneously.
We eliminate x from the above equations, i.e.,
put x = \(\frac{1-3 y}{2}\) in the equation x² – 4xy + y² = 0, we get,
∴ \(\left(\frac{1-3 y}{2}\right)^{2}\) – 4\(\left(\frac{1-3 y}{2}\right)\)y + y² = 0
∴ (1 – 3y)² – 8(1 – 3y)y + 4y² = 0
∴1 – 6y + 9y² – 8y + 24y² + 4y² = 0
∴ 37y² – 14y + 1 = 0
The roots y₁ and y₂ of the above quadratic equation are the y-coordinates of the points A and B.
∴ y₁ + y₂ = \(\frac{-b}{a}=\frac{14}{37}\)
∴ y-coordinate of D = \(\frac{y_{1}+y_{2}}{2}=\frac{7}{37}\).
Since D lies on the line AB, we can find the x-coordinate of D as
2x + 3\(\left(\frac{7}{37}\right)\) – 1 = 0
∴ 2x = 1 – \(\frac{21}{37}=\frac{16}{37}\)
∴ x = \(\frac{8}{37}\)
∴ D is (8/37, 7/37)
∴ equation of the median OD is \(\frac{x}{8 / 37}=\frac{y}{7 / 37}\),
i.e., 7x – 8y = 0.

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