RBSE Solutions for Class 12 Maths Chapter 8 Application of Derivatives is part of RBSE Solutions for Class 12 Maths. Here we have given Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives.

## Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives Ex 8.3

Question 1.

For curve y = x^{3} – x find slope of tangent at x = 2.

Solution:

Question 2.

For curve y = \(\frac { x-1 }{ x-2 } \), x ≠ 2 find slope of tangent at x = 10.

Solution:

Question 3.

Find all points where slope of tangent to the curve

Solution:

Question 4.

Find all equations of lines which are tangent to the curve y + \(\frac { 2 }{ x-3 } \) = 0 and slope of those is 2.

Solution:

Putting x = 4 in eq. (i)

y = \(\frac { -2 }{ 4-3 } \) = -2

Then, point = (4, – 2)

Again putting, x = 2 in eq. (i)

y = \(\frac { -2 }{ 2-3 } \) = 2

Then, point = (2, 2)

Now, tangent at ponit (4, – 2),

y – (- 2) = 2 (x – 4)

⇒ y + 2 = 2x – 8

⇒ 2x – y – 10 = 0

Again, tangent at point (2, 2),

y – 2 = 2 (x – 2)

⇒ y – 2 = 2x – 4

⇒ 2x – y – 2 = 0

Hence, required equations are 2x – y – 10 = 0 and 2x – y – 2 = 0.

Question 5.

For curve

find those points where tangent is

(i) Parallel to x-axis

(ii) Parallel to y-axis.

Solution:

(i) When tangent is parallel to x-axis then,

Hence, tangents are parallel to x-axis at points (0, ± 5).

(ii) When tangent is parallel to y-axis then,

Hence, tangents are parallel to 7-axis at points (± 2, 0)

Question 6.

For curve x – a sin^{3} t,y = b cos^{3} t find equation n of tangent at t = \(\frac { \pi }{ 2 } \)

Solution:

So, at t = \(\frac { \pi }{ 2 } \) or at (a, 0), equation of tangent of given curve

y – 0 = 0 (x – a)

or y = 0

Question 7.

For curves = sin^{2} x, find equation of normal at

Solution:

Question 8.

Find equation of tangent and normal following curves, at given points.

Solution:

(a) Given curve, y = x^{2} + 4x + 1 …..(i)

Putting x = 3 in equation (i),

y = (3)^{2} + 4 (3) + 1

y = 22

So, point of contact = (3, 22)

DifF. (i) w.r.t. x,

⇒ 10y – 220 = – x + 3

⇒ x + 10y = 223

So, equation of normal x + 10y = 223

(b) Given curve, y^{2} = 4ax …..(i)

Putting at x = a in equation (i)

y^{2} = 4a (a)

⇒ y^{2} = 4a^{2}

⇒ y = ± 2a

Hence, points of contact are (a, 2a) and (a, – 2a).

DifF. (i) w.r.t. x,

Hence, at (a, 2a) equation of tangent is x – y + a = 0 and equation of normal is x + y – 3a = 0

⇒ m^{2} (my – 2d) = – (m^{2}x – a)

⇒m^{3}y – 2am^{2} = – m^{2}x + a

⇒ m^{2}x + m^{3} y = a (2m^{2} + 1)

Hence, equation of normal

m^{2}x + m^{3}y = a (2m^{2} + 1)

⇒ ay tan θ – ab tan^{2} θ = bx sec θ – ab sec^{2} θ

⇒ bx sec θ – ay tan θ = ab(sec^{2} θ – tan^{2} θ)

Equation of tangent

y – 2at = 1 (x – at^{2})

At t = 1

y – 2a = x – a

⇒ x – y + a = 0

Hence, equation of tangent x – y + a = 0

Again, equation of normal

y – 2at = – 1 (x – a t^{2})

At t = 1

y – 2a = – (x – a)

⇒ x + y – 3a = 0

Hence, equation of normal x + y – 3a = 0

(g) Given,

x = θ + sin θ, y – 1 – cos θ

Diff. w.r.t. θ,

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