# 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 = 3x^{4} – 4x at x = 4.

**Solution:**

Given curve: y = 3x

^{4 }– 4xOn differentiating w.r.t x, we get

dy/dx = 12x

^{3 }– 4Now, we find the slope of the tangent to the given curve at x = 4 is

= dy/dx = 12(4)

^{3 }– 4 = 764Hence, 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 = x^{3} – x + 1 at the point whose x-coordinate is 2.

**Solution:**

Given curve: y = x

^{3 }– x + 1On 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 = x^{3} –3x + 2 at the point whose x-coordinate is 3.

**Solution:**

Given curve: y = x

^{3 }– 3x + 2On differentiating w.r.t x, we get

dy/dx = 3x

^{2 }– 3Now, we find the slope of the tangent to the given curve at x = 3 is

dy/dx = 3(3)

^{2 }– 3 = 24Hence, the slope is 24

### Question 5. Find the slope of the normal to the curve x = acos^{3}θ, y = asin^{3}θ at θ = π/4.

**Solution:**

Given curve: x = acos

^{3}θ = f(θ)y = asin

^{3}θ = g(θ)To find slope of the normal of the curve at θ = π/4

Now, slope of the normal is

-(1)

= a.3.cos

^{2}θ.(-sin θ)= a.3sin

^{2}θ.cos θ-(using eq(1))

Now, we find the slope of the tangent to the given curve at θ = π/4 isThe 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 = bcos^{2}θ 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 isHence, the slope is -a/2b.

### Question 7. Find points at which the tangent to the curve y = x^{3} – 3x^{2} – 9x + 7 is parallel to the x-axis.

**Solution:**

Given curve: y = x

^{3}– 3x^{2}– 9x + 7On differentiating w.r.t x, we get

dy/dx = 3x

^{2 }– 6x – 9For tangent to be parallel to x-axis, slope is 0. So dy/dx = 0.

3x

^{2 }– 6x – 9 = 03(x

^{2 }– 2x – 3) = 03(x

^{2 }+ x – 3x – 3) = 03(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) + 7x = -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 }= 1Hence, the point on the curve y = (x – 2)

^{2}is (3, 1)

### Question 9. Find the point on the curve y = x^{3} – 11x + 5 at which the tangent is y = x – 11.

**Solution:**

Given curve: y = x

^{3 }– 11x + 5Given 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 = x

^{3 }– 11x + 5dy/dx = 3x

^{2 }– 11 -(1)dy/dx = slope = 1

So, from eq(1), we get

3x

^{2 }– 11 = 13x

^{2 }= 12x

^{2 }= 4x = ±2

If x = +2, y = 2

^{3 }– 11(2) + 5 = -9If x = -2, y = (-2)

^{3 }– 11(-2) + 5 = 19The 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 }= 1x = 1 ± 1

x

_{1 }= 2 & x_{2 }= 0Now corresponding to these x

_{1}& x_{2 }we need to find out y_{1 }& y_{2 }The points are (2, 1) & (0, -1)

Now equations slope is -1

Using point slope form the first tangent equation is

(y – y

_{1}) = m(x – x_{1})y – 1 = -1(x – 2)

= x + y = 3

Using point slope from the second tangent equation is

(y – y

_{2}) = m(x – x_{2})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 – y

_{1 }= m(x – x_{1})= 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 = 0On differentiating both sides of equations (1) we get,

Now slope = 0, so

= x

_{1 }= 0For x

_{1 }= 0,y

_{1}^{2 }= 16y

_{1 }= ±4The 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 = 0On differentiating equations(1) with respect to y, we get

y

_{2}= 0For y

_{2 }= 0,x

_{2}^{2 }= 9x

_{2 }= ±3Hence, the coordinates are (3, 0) & (-3, 0)