# GSEB Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3

Question 1.
2x + 3y = sin x
Solution:
2x + 3y = sin x.
Differentiating w.r.t. x,
2x + 3$$\frac { dy }{ dx }$$ = cos x. ∴ $$\frac { dy }{ dx }$$ = $$\frac { 1 }{ 3 }$$(cosx-2)
2x + 3y = sin y.

Question 2.
2x + 3 y = sin y
Solution:
Differentiating w.r.t. x,
2 + 3$$\frac { dy }{ dx }$$ = cos y $$\frac { dy }{ dx }$$ or (cos y – 3) $$\frac { dy }{ dx }$$ = 2
∴ $$\frac { dy }{ dx }$$ = $$\frac { 2 }{ cos y – 3 }$$.
ax + by² = cos y.

Question 3.
ax + by² = cos y
Solution:
Differentiating w.r.t. x,
a + 2by$$\frac { dy }{ dx }$$ = – sin y $$\frac { dy }{ dx }$$
or (2b + sin y) $$\frac { dy }{ dx }$$ = – a
∴ $$\frac { dy }{ dx }$$ = – $$\frac { a }{ 2b + sin y }$$.
xy + y² = tan x + y.

Question 4.
xy + y² = tan x + y
Solution:
Differentiating w.r.t. x,
(1.y + x $$\frac { dy }{ dx }$$) = sec² x + $$\frac { dy }{ dx }$$
or (x + 2y – 1) $$\frac { dy }{ dx }$$ = sec² x – y
∴ $$\frac { dy }{ dx }$$ = $$\frac{\sec ^{2} x-y}{x+2 y-1}$$.

Question 5.
x² + xy + y² = 100
Solution:
x² + xy + y² = 100
Differentiating w.r.t. x,
2x + (1.y + x. $$\frac { dy }{ dx }$$) + 2y $$\frac { dy }{ dx }$$ = 0
or (x + 2y) $$\frac { dy }{ dx }$$ = – 2x – y
∴ $$\frac { dy }{ dx }$$ = – $$\frac{ 2x+y }{ x+2y }$$.

Question 6.
x³ + x²y + xy² + y³ = 81
Solution:
x³ + x²y + xy² + y³ = 81
Differentiating w.r.t. x,

Question 7.
sin² y + cos xy = π
Solution:
sin² y + cos xy = π
Differentiating w.r.t. x,

Question 8.
sin² x + cos² y = 1
Solution:
sin² x + cos² y = 1
Differentiating w.r.t. x,
2 sin x $$\frac { d }{ dx }$$ (sin x) + 2 cosy $$\frac { d }{ dx }$$ (cos y) = 0
or 2 sin x cos x + 2 cos y(- sin y) $$\frac { dy }{ dx }$$ = 0
or 2sin x cos x – 2cos y sin y $$\frac { dy }{ dx }$$ = 0
∴ $$\frac { dy }{ dx }$$ = $$\frac { 2 sin x cos x}{ 2sin y cos y }$$ = $$\frac { sin 2x }{ sin 2y }$$.

Question 9.
y = sin-1$$\left(\frac{2 x}{1+x^{2}}\right)$$
Solution:
y = sin-1$$\left(\frac{2 x}{1+x^{2}}\right)$$. put x = tanθ
So, y = sin-1$$\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$$ = sin-1(sin 2θ)
= 2θ = 2 tan-1x.

Question 10.
y = tan-1$$\frac { 2x }{ 2 }$$$$\frac{-1}{\sqrt{3}}$$ < x < $$\frac{1}{\sqrt{3}}$$
Solution:
y = tan-1$$\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$$. put x = tanθ
So, y = tan-1$$\left(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right)$$ = tan-1(tan 3θ)
= 3θ = 3 tan-1x.
∴ $$\frac { dy }{ dx }$$ = $$\frac{3}{1+x^{2}}$$

Question 11.
y = cos-1$$\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$$, 0 < x < 1
Solution:
y = cos-1$$\left(\frac{1-x^{2}}{1+x^{2}}\right)$$. put x = tanθ
So, y = cos-1$$\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)$$ = cos-1(cos 2θ)
= 2θ = 2 tan-1x.
∴ $$\frac { dy }{ dx }$$ = $$\frac{2}{1+x^{2}}$$

Question 12.
y = sin-1$$\left(\frac{1-x^{2}}{1+x^{2}}\right)$$, 0 < x < 1
Solution:

Question 13.
y = cos-1$$\left(\frac{2 x}{1+x^{2}}\right)$$, – 1 < x < 1
Solution:

Question 14.
y = sin-1$$\left(2 x \sqrt{1-x^{2}}\right)$$, – $$\frac{-1}{\sqrt{2}}$$ < x < $$\frac{1}{\sqrt{2}}$$
Solution:
y = sin-1$$\left(2 x \sqrt{1-x^{2}}\right)$$Put x = sin θ.
y = sin-1(2sinθ$$\left.\sqrt{1-\sin ^{2} \theta}\right)$$)
= sin-1(2 sin θ cos θ)
= sin-1(sin 2θ) = 2θ = 2 -1x.
∴ $$\frac { dy }{ dx }$$ = $$\frac{2}{\sqrt{1-x^{2}}}$$

Question 15.
y = sec-1$$\left(\frac{1}{2 x^{2}-1}\right)$$, 0 < x < $$\frac{1}{\sqrt{2}}$$
Solution:
y = sec-1$$\left(\frac{1}{2 x^{2}-1}\right)$$Put x = cos θ.
y = sec-1$$\left(\frac{1}{2 \cos ^{2} \theta-1}\right)$$ = sec-1$$\frac{1}{cosθ}$$
= sec-1(sec 2θ) = θ = 2 cos-1x.
= sec-1(sec 2θ) = 2θ = 2cos-1x.
∴ $$\frac { dy }{ dx }$$ = $$\frac{- 2}{\sqrt{1-x^{2}}}$$