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Class 12 NCERT Solutions- Mathematics Part I – Chapter 6 Application of Derivatives -Exercise 6.3 | Set 1
  • Last Updated : 06 Apr, 2021

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 y = \frac{x-1}{x-2} , x ≠ 2 at x = 10.

Solution:

Given curve: y=\frac{x-1}{x-2}

y=\frac{x-2+1}{x-2}=1+\frac{1}{x-2}

On differentiating w.r.t x, we get

\frac{dy}{dx}=0-\frac{1}{(x-2)^2}

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



\frac{dy}{dx}=\frac{-1}{(10-2)^2}=\frac{-1}{64}

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

\frac{dy}{dx}=\frac{d(x^3-x+1)}{dx}=3x^2-1

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

\frac{dy}{dx}=3(2)^2-1=11

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 \frac{-dx}{dy}

\frac{-dx}{dy}=-\frac{\frac{dx}{dθ}}{\frac{dy}{dθ}}       -(1)

\frac{dx}{d\theta} =\frac{d(a\cos^3θ) }{dθ}  = a.3.cos2 θ.(-sin θ)

\frac{dy}{dθ}=\frac{d(a\sin^3θ)}{dθ}  = a.3sin2 θ.cos θ

\frac{-dx}{dθ}=\frac{a.3.\cos^2θ.(-\sinθ)}{a.3.\sin^2θ.\cosθ}         -(using eq(1))

\frac{-dx}{dy}=\frac{\cosθ}{\sinθ}=\cotθ

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

\frac{-dx}{dy}=\cot(\frac{π}{4})=1

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 \frac{-dx}{dy}

\frac{-dx}{dy}=\frac{\frac{-dx}{dθ}}{\frac{dy}{dθ}}            -(1)

\frac{dx}{dθ}=\frac{d(1-a\sinθ)}{dθ}=-a\cosθ

\frac{dy}{dθ}=\frac{d(b\cos^2θ)}{dθ}=-2b\cosθ\sinθ

\frac{-dx}{dy}=\frac{-a\cosθ}{-2b\cosθ\sinθ}          -(using eq(1))

\frac{-dx}{dy}=\frac{-a}{2b\sinθ}

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

\frac{-dx}{dy}=\frac{-a}{2b\sin\frac{π}{2}}=\frac{-a}{2b}

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 = \frac{y_2-y_1}{x_2-x_1}=\frac{4-0}{4-2}=\frac{4}{2}=2

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 = \frac{1}{x-1} [Tex]  [/Tex], x ≠ 1.

Solution:

Given curve: y = \frac{1}{x-1}

\frac{dy}{dx}=\frac{-1}{(x-1)^2}                    -(1)

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

dy/dx = -1                  -(1)

From 1 & 2, we get

-1 = \frac{-1}{(x-1)^2}

(x – 1)2 = 1

x = 1 ± 1

x1 = 2 & x2 = 0

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

y_1=\frac{1}{x-1}=\frac{1}{2-1}=1

y_2=\frac{1}{x_2-1}=\frac{1}{0-1}=-1

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 y = \frac{1}{x-3} , 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 y=\frac{1}{x^2-2x+3} .

Solution:

Given curve,

y=\frac{1}{x^2-2x+3}     

On differentiating w.r.t x, we get

\frac{dy}{dx}=\frac{-1}{(x^2-2x+3)}.(2x-2)          -(chain rule)

Given slope = 0 = dy/dx  

So,\frac{dy}{dx}=\frac{-2(x-1)}{(x^2-2x+3)^2}=0

x – 1 = 0

x = 1

For x = 1, y = \frac{1}{1^2-2(1)+3}=\frac{1}{2}

So the equation of tangent from point slope from is 

y – y1 = m(x – x1)

y-\frac{1}{2}    = 0(x – 1)

2y – 1 = 0

Question 13. Find points on the curve \frac{x^2}{9}+\frac{y^2}{16}=1  at which the tangents are 

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

Solution:

Given curve:

\frac{x^2}{9}+\frac{y^2}{16}=1            -(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,

\frac{2x}{9}+\frac{2y}{16}.\frac{dy}{dx}=0    

\frac{dy}{dx}=\frac{-16x}{9y}

Now slope = 0, so

\frac{-16x}{9y}=0

= x1 = 0

For x1 = 0,   

\frac{0^2}{9}+\frac{y_1^2}{16}=1

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

\frac{2x}{9}.\frac{dx}{dy}+\frac{2y}{16}=0    

\frac{dx}{dy}=0,\frac{-9y}{16x}=0

y2 = 0

For y2 = 0, \frac{x_2^2}{9}+\frac{0}{16}=1

x22 = 9

x2 = ±3

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

Chapter 6 Application of Derivatives -Exercise 6.3 | Set 2

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