Indefinite Integration Class 12 Maths 2 Exercise 3.3 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.3 Questions and Answers.

12th Maths Part 2 Indefinite Integration Exercise 3.3 Questions And Answers Maharashtra Board

I. Evaluate the following:

Question 1.
∫x2 log x dx
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q1

Question 2.
∫x2 sin 3x dx
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q2
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q2.1

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3

Question 3.
∫x tan-1 x dx
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q3

Question 4.
∫x2 tan-1 x dx
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q4

Question 5.
∫x3 tan-1 x dx
Solution:
Let I = ∫x3 tan-1 x dx
= ∫(tan-1 x) . x3 dx
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q5

Question 6.
∫(log x)2 dx
Solution:
Let I = ∫(log x)2 dx
Put log x = t
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q6
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q6.1

Question 7.
∫sec3 x dx
Solution:
Let I = ∫sec3 x dx
= ∫sec x sec2 x dx
= sec x ∫sec2 x dx – ∫[\(\frac{d}{d x}\)(sec x) ∫sec2 x dx] dx
= sec x tan x – ∫(sec x tan x)(tan x) dx
= sec x tan x – ∫sec x tan2 x dx
= sec x tan x – ∫sec x (sec2 x – 1) dx
= sec x tan x – ∫sec3 x dx + ∫sec x dx
∴ I = sec x tan x – I + log|sec x + tan x|
∴ 2I = sec x tan x + log|sec x + tan x|
∴ I = \(\frac{1}{2}\) [sec x tan x + log|sec x + tan x|] + c.

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3

Question 8.
∫x . sin2 x dx
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q8

Question 9.
∫x3 log x dx
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q9

Question 10.
∫e2x cos 3x dx
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q10
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q10.1

Question 11.
∫x sin-1 x dx
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q11
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q11.1

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3

Question 12.
∫x2 cos-1 x dx
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q12

Question 13.
\(\int \frac{\log (\log x)}{x} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q13
= t(log t – 1) + c
= (log x) . [log(log x) – 1] + c.

Question 14.
\(\int \frac{t \cdot \sin ^{-1} t}{\sqrt{1-t^{2}}} d t\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q14

Question 15.
∫cos√x dx
Solution:
Let I = ∫cos√x dx
Put √x = t
∴ x = t2
∴ dx = 2t dt
∴ I = ∫(cos t) 2t dt
= ∫2t cos t dt
= 2t ∫cos t dt – ∫[\(\frac{d}{d t}\)(2t) ∫cos t dt]dt
= 2t sin t – ∫2 sin t dt
= 2t sin t + 2 cos t + c
= 2[√x sin√x + cos√x] + c.

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3

Question 16.
∫sin θ . log(cos θ) dθ
Solution:
Let I = ∫sin θ . log (cos θ) dθ
= ∫log(cos θ) . sin θ dθ
Put cos θ = t
∴ -sin θ dθ = dt
∴ sin θ dθ = -dt
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q16
= -t log t + t + c
= -cos θ . log(cos θ) + cos θ + c
= -cos θ [log(cos θ) – 1] + c.

Question 17.
∫x cos3 x dx
Solution:
cos 3x = 4 cos3 x – 3 cos x
∴ cos3 x + 3 cos x = 4cos3x
∴ cos3 x = \(\frac{1}{4}\) cos 3x + \(\frac{3}{4}\) cos x
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q17

Question 18.
\(\int \frac{\sin (\log x)^{2}}{x} \cdot \log x d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q18

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3

Question 19.
\(\int \frac{\log x}{x} d x\)
Solution:
Let I = \(\int \frac{\log x}{x} d x\)
Put log x = t
\(\frac{1}{x}\) dx = dt
∴ I = ∫t dt
= \(\frac{1}{2}\) t2 + c
= \(\frac{1}{2}\) (log x)2 + c

Question 20.
∫x sin 2x cos 5x dx.
Solution:
Let I = ∫x sin 2x cos 5x dx
sin 2x cos 5x = \(\frac{1}{2}\)[2 sin 2x cos 5x]
= \(\frac{1}{2}\) [sin(2x + 5x) + sin(2x – 5x)]
= \(\frac{1}{2}\) [sin 7x – sin 3x]
∴ ∫sin 2x cos 5x dx = \(\frac{1}{2}\) [∫sin 7x dx – ∫sin 3x dx]
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q20

Question 21.
\(\int \cos (\sqrt[3]{x}) d x\)
Solution:
Let I = \(\int \cos (\sqrt[3]{x}) d x\)
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 I Q21

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

Question 1.
e2x sin 3x
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q1
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q1.1

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3

Question 2.
e-x cos 2x
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q2
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q2.1

Question 3.
sin(log x)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q3

Question 4.
\(\sqrt{5 x^{2}+3}\)
Solution:
Let I = \(\sqrt{5 x^{2}+3}\) dx
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q4

Question 5.
\(x^{2} \sqrt{a^{2}-x^{6}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q5

Question 6.
\(\sqrt{(x-3)(7-x)}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q6

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3

Question 7.
\(\sqrt{4^{x}\left(4^{x}+4\right)}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q7

Question 8.
(x + 1) \(\sqrt{2 x^{2}+3}\)
Solution:
Let I = ∫(x + 1) \(\sqrt{2 x^{2}+3}\) dx
Let x + 1 = A[\(\frac{d}{d x}\)(2x2 + 3)] + B
= A(4x) + B
= 4Ax + B
Comparing the coefficients of x and constant term on both the sides, we get
4A = 1, B = 1
∴ A = \(\frac{1}{4}\), B = 1
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q8
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q8.1

Question 9.
\(x \sqrt{5-4 x-x^{2}}\)
Solution:
Let I = ∫\(x \sqrt{5-4 x-x^{2}}\) dx
Let x = A[\(\frac{d}{d x}\)(5 – 4x – x2)] + B
= A[-4 – 2x] + B
= -2Ax + (B – 4A)
Comparing the coefficients of x and the constant term on both sides, we get
-2A = 1, B – 4A = 0
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q9
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q9.1

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3

Question 10.
\(\sec ^{2} x \sqrt{\tan ^{2} x+\tan x-7}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q10
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q10.1

Question 11.
\(\sqrt{x^{2}+2 x+5}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q11

Question 12.
\(\sqrt{2 x^{2}+3 x+4}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q12
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 II Q12.1

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

Question 1.
[2 + cot x – cosec2 x] ex
Solution:
Let I = ∫ex [2 + cot x – cosec2 x] dx
Put f(x) = 2 + cot x
∴ f'(x) = \(\frac{d}{d x}\)(2 + cot x)
= \(\frac{d}{d x}\)(2) + \(\frac{d}{d x}\)(cot x)
= 0 – cosec2 x
= -cosec2 x
∴ I = ∫ex [f(x) + f'(x)] dx
= ex f(x) + c
= ex (2 + cot x) + c.

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3

Question 2.
\(\left(\frac{1+\sin x}{1+\cos x}\right) e^{x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 III Q2

Question 3.
\(e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\)
Solution:
Let I = ∫\(e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\)
Let f(x) = \(\frac{1}{x}\), f'(x) = \(-\frac{1}{x^{2}}\)
∴ I = ∫ex [f(x) + f'(x)] dx
= ex f(x) + c
= ex . \(\frac{1}{x}\) + c

Question 4.
\(\left[\frac{x}{(x+1)^{2}}\right] e^{x}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 III Q4

Question 5.
\(\frac{e^{x}}{x}\) . [x(log x)2 + 2 log x]
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 III Q5

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3

Question 6.
\(e^{5 x}\left[\frac{5 x \log x+1}{x}\right]\)
Solution:
Let I = ∫\(e^{5 x}\left[\frac{5 x \log x+1}{x}\right]\)
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 III Q6

Question 7.
\(e^{\sin ^{-1} x}\left[\frac{x+\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}\right]\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 III Q7

Question 8.
log(1 + x)(1+x)
Solution :
Let I = ∫log(1 + x)(1+x) dx
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3 III Q8

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.3

Question 9.
cosec (log x)[1 – cot(log x)]
Solution:
Let I = ∫cosec (log x)[1 – cot(log x)] dx
Put log x = t
x = et
dx = et dt
I = ∫cosec t (1 – cot t). et dt
= ∫et [cosec t – cosec t cot t] dt
= ∫et [cosec t + \(\frac{d}{d t}\) (cosec t)] dt
= et cosec t + c ….. [∵ et [f(t) +f'(t) ] dt = et f(t) + c ]
= x . cosec(log x) + c.

Class 12 Maharashtra State Board Maths Solution 

Indefinite Integration Class 12 Maths 2 Exercise 3.2(C) Solutions Maharashtra Board

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

12th Maths Part 2 Indefinite Integration Exercise 3.2(C) Questions And Answers Maharashtra Board

I. Evaluate:

Question 1.
\(\int \frac{3 x+4}{x^{2}+6 x+5} d x\)
Solution:
Let I = \(\int \frac{3 x+4}{x^{2}+6 x+5} d x\)
Let 3x + 4 = A[\(\frac{d}{d x}\)(x2 + 6x + 5)] + B
= A(2x + B) + B
∴ 3x + 4 = 2Ax + (6A + B)
Comparing the coefficient of x and constant on both sides, we get
2A = 3 and 6A + B = 4
∴ A = \(\frac{3}{2}\) and 6(\(\frac{3}{2}\)) + B = 4
∴ B = -5
3x + 4 = \(\frac{3}{2}\) (2x + 6) – 5
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q1

Question 2.
\(\int \frac{2 x+1}{x^{2}+4 x-5} d x\)
Solution:
Let I = \(\int \frac{2 x+1}{x^{2}+4 x-5} d x\)
Let 2x + 1 = A[\(\frac{d}{d x}\)(x2 + 4x – 5)] + B
2x + 1 = A(2x + 1) + B
∴ 2x + 1 = 2Ax + (4A + B)
Comparing the coefficient of x and constant on both sides, we get
4A = 2 and 4A + B = 4
∴ A = \(\frac{3}{2}\) and 6(\(\frac{3}{2}\)) + B = 4
∴ B = -5
∴ 2x + 1 = \(\frac{3}{2}\)(2x + 1) – 5
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q2
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q2.1

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

Question 3.
\(\int \frac{2 x+3}{2 x^{2}+3 x-1} d x\)
Solution:
Let I = \(\int \frac{2 x+3}{2 x^{2}+3 x-1} d x\)
Let 2x+ 3 = A[\(\frac{d}{d x}\)(2x2 + 3x – 1)] + B
2x + 1 = A(4x + 3) + B
∴ 2x + 1 = 4Ax + (3A + B)
Comparing the coefficient of x and constant on both sides, we get
4A = 2 and 3A + B = 3
∴ A = \(\frac{1}{2}\) and 3(\(\frac{1}{2}\)) + B = 3
∴ B = \(\frac{3}{2}\)
∴ 2x + 3 = \(\frac{1}{2}\)(4x + 3) + \(\frac{3}{2}\)
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q3
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q3.1

Question 4.
\(\int \frac{3 x+4}{\sqrt{2 x^{2}+2 x+1}} d x\)
Solution:
Let I = \(\int \frac{3 x+4}{\sqrt{2 x^{2}+2 x+1}} d x\)
Let 3x + 4 = A[\(\frac{d}{d x}\)(2x2 + 2x + 1)] + B
∴ 3x + 4 = A (4x + 2) + B
∴ 3x + 4 = 4Ax + (2A + B)
Comparing the coefficient of x and the constant on both the sides, we get
4A = 3 and 2A + B = 4
∴ A = \(\frac{3}{4}\) and 2(\(\frac{3}{4}\)) + B = 4
∴ B = \(\frac{5}{2}\)
∴ 3x + 4 = \(\frac{3}{4}\) (4x + 2) + \(\frac{5}{2}\)
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q4
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q4.1

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

Question 5.
\(\int \frac{7 x+3}{\sqrt{3+2 x-x^{2}}} d x\)
Solution:
Let I = \(\int \frac{7 x+3}{\sqrt{3+2 x-x^{2}}} d x\)
Let 7x + 3 = A[\(\frac{d}{d x}\)(3 + 2x – x2)] + B
7x + 3 = A(2 – 2x) + B
∴ 7x + 3 = -2Ax + (2A + B)
Comparing the coefficient of x and constant on both the sides, we get
-2A = 7 and 2A + B = 3
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q5
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q5.1

Question 6.
\(\int \sqrt{\frac{x-7}{x-9}} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q6
Comparing the coefficients of x and constant term on both sides, we get
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q6.1
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q6.2

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

Question 7.
\(\int \sqrt{\frac{9-x}{x}} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q7
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q7.1
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q7.2

Question 8.
\(\int \frac{3 \cos x}{4 \sin ^{2} x+4 \sin x-1} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q8
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q8.1

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

Question 9.
\(\int \sqrt{\frac{e^{3 x}-e^{2 x}}{e^{x}+1}} d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q9
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q9.1
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(C) Q9.2

Class 12 Maharashtra State Board Maths Solution 

Indefinite Integration Class 12 Maths 2 Exercise 3.2(B) Solutions Maharashtra Board

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

12th Maths Part 2 Indefinite Integration Exercise 3.2(B) Questions And Answers Maharashtra Board

I. Evaluate the following:

Question 1.
\(\int \frac{1}{4 x^{2}-3} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q1
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q1.1

Question 2.
\(\int \frac{1}{25-9 x^{2}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q2

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

Question 3.
\(\int \frac{1}{7+2 x^{2}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q3

Question 4.
\(\int \frac{1}{\sqrt{3 x^{2}+8}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q4
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q4.1

Question 5.
\(\int \frac{1}{\sqrt{11-4 x^{2}}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q5

Question 6.
\(\int \frac{1}{\sqrt{2 x^{2}-5}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q6
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q6.1

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

Question 7.
\(\int \sqrt{\frac{9+x}{9-x}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q7

Question 8.
\(\int \sqrt{\frac{2+x}{2-x}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q8
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q8.1

Question 9.
\(\int \sqrt{\frac{10+x}{10-x}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q9
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q9.1

Question 10.
\(\int \frac{1}{x^{2}+8 x+12} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q10

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

Question 11.
\(\int \frac{1}{1+x-x^{2}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q11

Question 12.
\(\int \frac{1}{4 x^{2}-20 x+17} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q12
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q12.1

Question 13.
\(\int \frac{1}{5-4 x-3 x^{2}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q13
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q13.1

Question 14.
\(\int \frac{1}{\sqrt{3 x^{2}+5 x+7}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q14
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q14.1

Question 15.
\(\int \frac{1}{\sqrt{x^{2}+8 x-20}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q15

Question 16.
\(\int \frac{1}{\sqrt{8-3 x+2 x^{2}}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q16

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

Question 17.
\(\int \frac{1}{\sqrt{(x-3)(x+2)}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q17
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q17.1

Question 18.
\(\int \frac{1}{4+3 \cos ^{2} x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q18

Question 19.
\(\int \frac{1}{\cos 2 x+3 \sin ^{2} x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q19
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q19.1

Question 20.
\(\int \frac{\sin x}{\sin 3 x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) I Q20

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

Question 1.
\(\int \frac{1}{3+2 \sin x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q1
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q1.1

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

Question 2.
\(\int \frac{1}{4-5 \cos x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q2
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q2.1

Question 3.
\(\int \frac{1}{2+\cos x-\sin x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q3
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q3.1

Question 4.
\(\int \frac{1}{3+2 \sin x-\cos x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q4

Question 5.
\(\int \frac{1}{3-2 \cos 2 x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q5
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q5.1

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

Question 6.
\(\int \frac{1}{2 \sin 2 x-3} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q6
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q6.1

Question 7.
\(\int \frac{1}{3+2 \sin 2 x+4 \cos 2 x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q7
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q7.1

Question 8.
\(\int \frac{1}{\cos x-\sin x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q8

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

Question 9.
\(\int \frac{1}{\cos x-\sqrt{3} \sin x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.2(B) II Q9

Class 12 Maharashtra State Board Maths Solution 

Indefinite Integration Class 12 Maths 2 Exercise 3.2(A) Solutions Maharashtra Board

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

12th Maths Part 2 Indefinite Integration Exercise 3.2(A) Questions And Answers Maharashtra Board

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Class 12 Maharashtra State Board Maths Solution 

Indefinite Integration Class 12 Maths 2 Exercise 3.1 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Indefinite Integration Ex 3.1 Questions and Answers.

12th Maths Part 2 Indefinite Integration Exercise 3.1 Questions And Answers Maharashtra Board

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

(i) x3 + x2 – x + 1
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 I (i)

(ii) \(x^{2}\left(1-\frac{2}{x}\right)^{2}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 I (ii)

(iii) \(3 \sec ^{2} x-\frac{4}{x}+\frac{1}{x \sqrt{x}}-7\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 I (iii)

(iv) \(2 x^{3}-5 x+\frac{3}{x}+\frac{4}{x^{5}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 I (iv)

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1

(v) \(\frac{3 x^{3}-2 x+5}{x \sqrt{x}}\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 I (v)
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 I (v).1

II. Evaluate:

(i) ∫tan2 x . dx
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 II (i)

(ii) \(\int \frac{\sin 2 x}{\cos x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 II (ii)

(iii) \(\int \frac{\sin x}{\cos ^{2} x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 II (iii)

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1

(iv) \(\int \frac{\cos 2 x}{\sin ^{2} x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 II (iv)

(v) \(\int \frac{\cos 2 x}{\sin ^{2} x \cdot \cos ^{2} x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 II (v)
= -cot x – tan x + c

(vi) \(\int \frac{\sin x}{1+\sin x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 II (vi)

(vii) \(\int \frac{\tan x}{\sec x+\tan x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 II (vii)

(viii) \(\int \sqrt{1+\sin 2 x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 II (viii)

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1

(ix) \(\int \sqrt{1-\cos 2 x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 II (ix)

(x) ∫sin 4x cos 3x dx
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 II (x)

III. Evaluate:

(i) \(\int \frac{x}{x+2} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 III (i)

(ii) \(\int \frac{4 x+3}{2 x+1} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 III (ii)

(iii) \(\int \frac{5 x+2}{3 x-4} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 III (iii)

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1

(iv) \(\int \frac{x-2}{\sqrt{x+5}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 III (iv)

(v) \(\int \frac{2 x-7}{\sqrt{4 x-1}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 III (v)
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 III (v).1

(vi) \(\int \frac{\sin 4 x}{\cos 2 x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 III (vi)

(vii) \(\int \sqrt{1+\sin 5 x} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 III (vii)

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1

(viii) ∫cos2 x . dx
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 III (viii)

(ix) \(\int \frac{2}{\sqrt{x}-\sqrt{x+3}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 III (ix)

(x) \(\int \frac{3}{\sqrt{7 x-2}-\sqrt{7 x-5}} \cdot d x\)
Solution:
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 III (x)

Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1

IV.

Question 1.
If f'(x) = x – \(\frac{3}{x^{3}}\), f(1) = \(\frac{11}{2}\), find f(x).
Solution:
By the definition of integral,
Maharashtra Board 12th Maths Solutions Chapter 3 Indefinite Integration Ex 3.1 IV

Class 12 Maharashtra State Board Maths Solution 

Applications of Derivatives Class 12 Maths 2 Miscellaneous Exercise 2 Solutions Maharashtra Board

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

12th Maths Part 2 Applications of Derivatives Miscellaneous Exercise 2 Questions And Answers Maharashtra Board

I. Choose the correct option from the given alternatives:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

II. Solve the following:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Class 12 Maharashtra State Board Maths Solution 

Applications of Derivatives Class 12 Maths 2 Exercise 2.4 Solutions Maharashtra Board

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

12th Maths Part 2 Applications of Derivatives Exercise 2.4 Questions And Answers Maharashtra Board

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Class 12 Maharashtra State Board Maths Solution 

Applications of Derivatives Class 12 Maths 2 Exercise 2.3 Solutions Maharashtra Board

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

12th Maths Part 2 Applications of Derivatives Exercise 2.3 Questions And Answers Maharashtra Board

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Class 12 Maharashtra State Board Maths Solution 

Applications of Derivatives Class 12 Maths 2 Exercise 2.2 Solutions Maharashtra Board

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

12th Maths Part 2 Applications of Derivatives Exercise 2.2 Questions And Answers Maharashtra Board

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Class 12 Maharashtra State Board Maths Solution 

Applications of Derivatives Class 12 Maths 2 Exercise 2.1 Solutions Maharashtra Board

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

12th Maths Part 2 Applications of Derivatives Exercise 2.1 Questions And Answers Maharashtra Board

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Class 12 Maharashtra State Board Maths Solution