# Class 12 NCERT Solutions- Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.3

• Last Updated : 09 Mar, 2021

### Question 1. 2x + 3y = sin x

Solution:

On differentiating both sides w.r.t. x, we get

2 + 3 = cos x = cos x – 2 = (cosx – 2)/3

### Question 2. 2x + 3y = sin y

Solution:

On differentiating both sides w.r.t. x, we get

2 + 3 = cos y (cosy – 3) = 2 ### Question 3. ax + by2 = cos y

Solution:

On differentiating both sides w.r.t. x, we get

a + b * 2y( ) = -sin y * (2by + siny) = -a ### Question 4. xy + y2 = tan x + y

Solution:

On differentiating both sides w.r.t. x, we get

(x * + y) + 2y = sec2x + (x + 2y – 1) = sec2x – y ### Question 5. x2 + xy + y2 = 100

Solution:

On differentiating both sides w.r.t. x, we get

2x + (x + y) + 2y = 0

(x + 2y) * = -(2x + y) ### Question 6. x3 + x2y + xy2 + y3 = 81

Solution:

Differentiate both sides w.r.t. x

3x2+(x2 + y * 2x) + (x * 2y * + y2) + 3y2 = 0

(x2 + 2xy + 3y2) = -(3x2 + 2xy + y2) ### Question 7. Sin2y + cos xy = π

Solution:

Differentiate both sides w.r.t. x

2 sin y * (sin⁡y) – sin(xy) * xy = 0

2sin y * cosy – sin(xy)(x * + y) = 0

(2sin cos y – sin (xy) – x)) = y(xy)  ### Question 8. sin2 x + cos2 y = 1

Solution:

2 sin x * (sin ⁡x) + 2 cos y * (cos ⁡y) = 0

2 sin x * cos x + 2 cos y*(-sin y) * = 0

2 sin x * cos x – 2 cos x – 2 cos y sin y * = 0

Sin(2x) – sin(2y) – = 0 ### Question 9. y = sin-1(\frac{2x}{(1 + x2)}

Solution:

Put x = tanθ

θ = tan-1x y = sin-1(sin 2θ)

y = 2θ

y = 2tan-1x         -(1)

On differentiating eq(1), we get ### Question 10. , -1/√3 < x < 1/√3

Solution:

Put x = tanθ

θ = tan-1x

y = y = tan-1(tan 3θ)

y = 3θ

y = 3tan-1x          -(1)

On differentiating eq(1), we get ### Question 11. , 0 < x < 1

Solution:

Put x = tanθ

θ  = tan-1 x

y = y = cos-1(cos 2θ)

y = 2θ

y = 2tan-1x         -(1)

On differentiating eq(1), we get ### Question 12. , 0 < x < 1

Solution:

Put x = tanθ

θ = tan-1x y = sin-1(cos 2θ)

y = sin-1(sin (π/2 – 2θ))

y = π/2 – 2θ

y = π/2 – 2 tan-1x ### Question 13. , -1 < x < 1

Solution:

Put x = tanθ

θ = tan-1x

y = cos^{-1}( )

y = cos-1(sin 2θ)

y = cos-1(cos (π/2 – 2θ))

y = π/2 – 2θ

y = π/2 – 2tan-1x ### Question 14. , -1/√2 < x < 1/√2

Solution:

Put x = sinθ

θ =  sin-1 x

y = sin-1(2sinθ√(1 – sin2θ))

y = sin-1(sin 2θ) = 2θ

y = 2sin-1x ### Question 15. , 0 < x < 1/√2

Solution:

Put x = tanθ y = sec-1(1/cos2θ))

y = sec-1(sec2θ) = 2θ

y = 2cos-1x = My Personal Notes arrow_drop_up