Class 12 NCERT Solutions- Mathematics Part I – Chapter 6 Application of Derivatives -Exercise 6.3 | Set 1

Question 1. Find the slope of the tangent to the curve y = 3x⁴ – 4x at x = 4.

Solution:

Given curve: y = 3x⁴ – 4x

On differentiating w.r.t x, we get

dy/dx = 12x³ – 4

Now, we find the slope of the tangent to the given curve at x = 4 is

= dy/dx = 12(4)³ – 4 = 764

Hence, the slope is 764

Question 2. Find the slope of the tangent to the curve , x ≠ 2 at x = 10.

Solution:

Given curve: 

On differentiating w.r.t x, we get

Now, we find the slope of the tangent to the given curve at x = 10 is

Hence, the slope is -1/64

Question 3. Find the slope of the tangent to curve y = x³ – x + 1 at the point whose x-coordinate is 2. 

Solution:

Given curve: y = x³ – x + 1

On differentiating w.r.t x, we get

Now, we find the slope of the tangent to the given curve at x = 2 is

Hence, the slope is 11

Question 4. Find the slope of the tangent to the curve y = x³ –3x + 2 at the point whose x-coordinate is 3.

Solution:

Given curve: y = x³ – 3x + 2

On differentiating w.r.t x, we get

dy/dx = 3x² – 3

Now, we find the slope of the tangent to the given curve at x = 3 is

dy/dx = 3(3)² – 3 = 24

Hence, the slope is 24

Question 5. Find the slope of the normal to the curve x = acos³θ, y = asin³θ at θ = π/4.

Solution:

Given curve: x = acos³θ = f(θ)

y = asin³θ = g(θ)

To find slope of the normal of the curve at θ = π/4

Now, slope of the normal is 

    -(1)

 = a.3.cos² θ.(-sin θ)

 = a.3sin² θ.cos θ

      -(using eq(1))

Now, we find the slope of the tangent to the given curve at θ = π/4 is

The slope of normal of the parametric curve

Hence, the slope is 1

Question 6. Find the slope of the normal to the curve x = 1 – asinθ, y = bcos²θ at θ = π/2.

Solution:

Given curve: x = 1 – a sinθ

y = b cos2θ

Now, slope of normal is 

          -(1)

        -(using eq(1))

Now, we find the slope of the tangent to the given curve at θ = π/2 is

Hence, the slope is -a/2b.

Question 7. Find points at which the tangent to the curve y = x³ – 3x² – 9x + 7 is parallel to the x-axis.

Solution:

Given curve: y = x³ – 3x² – 9x + 7

On differentiating w.r.t x, we get

dy/dx = 3x² – 6x – 9

For tangent to be parallel to x-axis, slope is 0. So dy/dx = 0.

3x² – 6x – 9 = 0

3(x² – 2x – 3) = 0

3(x² + x – 3x – 3) = 0

3(x(x + 1) – 3(x + 1)) = 0

3(x + 1)(x – 3) = 0

x = -1 or x = 3

For x = -1, y = (-1)³ – 3(-1)² – 9(-1) + 7

x = -1, y = -1 – 3 + 9 + 7 = 12

Hence, the first point is (-1, 12)

Question 8. Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).

Solution:

Given curve: y = (x – 2)²

On differentiating w.r.t x, we get

dy/dx = 2(x – 2)                 -(1)

Given that, the tangent is parallel to the chord joining the points (2, 0) & (4, 4)

Slope of the chord = 

Now equality dy/dx = slope of chord

2(x – 2) = 2

x – 2 = 1

x = 3

y = (x – 2)²

y = (3 – 2)² = 1

Hence, the point on the curve y = (x – 2)² is (3, 1)

Question 9. Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11. 

Solution:

Given curve: y = x³ – 11x + 5

Given tangent: y = x – 11

From the given tangent, we can find out the slope comparing y = x – 11 with y = mx + c, we get

Slope(m) = 1

Now y = x³ – 11x + 5

dy/dx = 3x² – 11                  -(1)

dy/dx = slope = 1

So, from eq(1), we get  

3x² – 11 = 1

3x² = 12

x² = 4

x = ±2

If x = +2, y = 2³ – 11(2) + 5 = -9

If x = -2, y = (-2)³ – 11(-2) + 5 = 19

The points must lie on the tangent as well.

Only (2,-9) is satisfying the tangent equation.

So the point on the curve whose tangent is y = x – 11 is (2,-9). 

Question 10. Find the equation of all lines having slope –1 that are tangents to the curve y = [Tex]  [/Tex], x ≠ 1.

Solution:

Given curve: y = 

                  -(1)

Now given slope = -1 & we know that dy/dx = slope, so

dy/dx = -1                  -(1)

From 1 & 2, we get

-1 = 

(x – 1)² = 1

x = 1 ± 1

x₁ = 2 & x₂ = 0

Now corresponding to these x₁ & x₂ we need to find out y₁ & y₂ 

The points are (2, 1) & (0, -1)

Now equations slope is -1

Using point slope form the first tangent equation is 

(y – y₁) = m(x – x₁)

y – 1 = -1(x – 2)

= x + y = 3

Using point slope from the second tangent equation is 

(y – y₂) = m(x – x₂)

y – (-1) = -1(x – 0)

= x + y + 1 = 0 

Question 11. Find the equation of all lines having slope 2 which are tangents to the curve , x ≠ 3.

Solution:

Given curve: y = 1/(x – 3) 

dy/dx = -1/(x – 3)² = slope               -(dy/dx is slope)

Now given slope is 2, so

dy/dx = -1/(x – 3)² = 2

(x – 3)² = -1/2              -(1) (not possible)

Now because there is no real value of x which can satisfy 1, therefore no such tangent exists on the curve y = 1/x – 3 whose is 2.

Question 12. Find the equations of lines having slope 0 which are tangent to the curve .

Solution:

Given curve,

On differentiating w.r.t x, we get

      -(chain rule)

Given slope = 0 = dy/dx  

So,

x – 1 = 0

x = 1

For x = 1, y = 

So the equation of tangent from point slope from is 

y – y₁ = m(x – x₁)

= 0(x – 1)

2y – 1 = 0

Question 13. Find points on the curve  at which the tangents are 

(i) Parallel to x-axis            (ii)Parallel to y-axis

Solution:

Given curve:

         -(1)

(i) If tangent is parallel to x-axis then it means slope is 0 or dy/dx = 0

On differentiating both sides of equations (1) we get,

Now slope = 0, so

= x₁ = 0

For x₁ = 0,   

y₁² = 16

y₁ = ±4

The coordinates are (0, 4) & (0, -4)

(ii) Now, if tangent is parallel to y-axis to the dy/dx or slope is not defined or dy/dx = 0

On differentiating equations(1) with respect to y, we get

y₂ = 0

For y₂ = 0, 

x₂² = 9

x₂ = ±3

Hence, the coordinates are (3, 0) & (-3, 0)

Chapter 6 Application of Derivatives -Exercise 6.3 | Set 2