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# NCERT Solutions Of Chapter 6 Exercise 6.3

NCERT solutions of chapter 6 exercise 6.3 is going to help all the maths students of class 12,in this exercise 6.3 of the maths chapter class 12 -Application of Derivatives you will learn solutions of the questions related to the slope of the curve. All questions are explained by an expert of the subject as per the CBSE norms. NCERT Solutions of the exercise 6.3 Application of Derivatives will give an idea to all students about the concept and the way of solving the questions of exercise 6.3 of chapter 6 which is required to solve all the questions of Application of Derivatives in the Class 12 CBSE Board exam.

You can also study

Exercise 6.1 – Application of Derivatives

Exercise 6.2 – Application of Derivatives

Exercise 6.3 – Application of Derivatives

Exercise 6.4- Application of Derivatives

## NCERT Solutions Of Chapter 6 Exercise 6.3

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Q1.Find the slope of the tangent to the curve $y=3x^{^{4}}-4x$ at x = 4.

Ans. The slope of the tangent to the curve $y=3x^{^{4}}-4x$  is

$\frac{dy}{dx}=12x^{3}-4$

If x = 4, the slope =

$12(4)^{3}-4= 768-4= 764$

Q2.Find the slope of the tangent to the curve  $y=\frac{x-1}{x-2}, x\neq 2\, at\, x = 10.$

Ans. We are given the equation of the curve

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

The slope of the given curve is the differentiation of y with respect to x as follows

$\frac{dy}{dx}= \frac{d}{dx}\left ( \frac{x-1}{x-2} \right )$

$\frac{dy}{dx}= \frac{\left ( x-2 \right )\frac{d}{dx}\left ( x-1 \right )-\left ( x-1 \right )\frac{d}{dx}\left ( x-2 \right )}{\left ( x-2 \right )^{2}}$

$=\frac{x-2-x+1}{\left ( x-2 \right )^{2}}$

$=-\frac{1}{\left ( x-2 \right )^{2}}$

If x = 10, the slope =

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

Q3.Find the slope of the tangent to the curve y = x³ – x + 1 at the point whose x-coordinate is 2.

Ans. The slope of the tangent to the curve

$y=x^{3}-x + 1$  is

y = x³ – x + 1

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

= 3x²- 1

Slope of the curve at x = 2

3 × 2²- 1

12 -1 = 11

Q4.Find the slope of the tangent to the curve y = x3 – 3x + 2 at the point whose x-coordinate is 3

Ans. We are given the curve y = x3 – 3x + 2

y = x3 – 3x + 2

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

= 3x² – 3

The slope of the curve at x = 3

3 × 3² – 3 = 27 – 3 = 24

$\left [ \frac{dy}{dx} \right ]_{x=3}=24$

Q5.Find slope of the normal to the curve x = a cos³θ, y = a sin³θ at θ = π/4.

Ans. We are given

x = a cos³θ, y = a sin³θ

$\frac{dx}{d\Theta }=-3a\: cos^{2}\Theta.sin\Theta ,\frac{dy}{d\Theta } =3a\: sin^{2}\Theta .cos\Theta$

$\frac{dy/d\Theta }{dx/d\Theta }=\frac{3a\: sin^{2}\Theta .cos\Theta }{-3a\: cos^{2}\Theta .sin\Theta }$

$\frac{dy}{dx}=\frac{-sin\Theta }{cos\Theta }$

$\frac{dy}{dx}=-tan\Theta$

$\left [ \frac{dy}{dx} \right ]_{\Theta =\frac{\pi }{4}}=-tan\frac{\pi }{4}=-1$

The slope of the curve is = -1

Let the slope of the normal to the curve = m

We know the relation

the slope of the curve × m = -1

-1 × m = -1

m = 1

Q6. Find the slope of the normal to the curve x = 1 – a sinθ, y = b cos”θ at θ = π/2.

Ans. We are given the curve

x = 1 – a sinθ,        y = b cos”θ

$\frac{dx}{d\Theta }=\frac{d}{d\Theta }\left ( 1-a sin\Theta \right ),\: \: \frac{dy}{d\Theta }=\frac{d}{d\Theta }\left ( b\: cos^{2}\Theta \right )$

$\frac{dx}{d\Theta }= a\: cos\Theta ,\: \: \frac{dy}{d\Theta }=-2b\: cos\Theta .sin\Theta$

$\frac{dy/d\Theta }{dx/d\Theta }=\frac{-2b\: cos\Theta .sin\Theta }{-a\: cos\Theta }$

$\frac{dy}{dx}=\frac{2b}{a}sin\Theta$

The slope of the given curve is

$=\frac{2b}{a}sin\Theta$

The slope of the curve at θ = π/2

$\left [ \frac{dy}{dx} \right ]_{\Theta =\frac{\pi }{2}}=\frac{2b}{a}sin\frac{\pi }{2}=\frac{2b}{a}$

So, slope of the curve, m = 2b/a

Let slope of the normal to the curve is = M

We have

$m\times M= -1$

$\frac{2b}{a}\times M= -1$

$M= -\frac{a}{2b}$

Hence normal to the given curve is = -a/2b

### NCERT Solutions Of Chapter 6 Exercise 6.3

Q7. Find points at which the tangent to the curve y = x3 – 3x2 – 9x + 7 is parallel to the x-axis.

Ans. The given curve is

y = x3 – 3x2 – 9x + 7

Slope of the tangent dy/dx is evaluated as follows

$\frac{dy}{dx}=\frac{d}{dx}\left ( x^{3}-3x^{2} -9x+7\right )$

$\frac{dy}{dx}=3x^{2}-6x-9$

Since the tangent to the curve is parallel x-axis

The equation of x-axis is

y = 0

$\therefore \frac{dy}{dx}=0$

It is given to us that   the tangent to the curve is parallel to each other

So, 3x² – 6x – 9 = 0

x² – 2x – 3 = 0

x² – 3x + x – 3 = 0

x( x – 3) + ( x – 3) = 0

( x – 3)(x + 1) = 0

x = 3, -1

Hence at x = 3, -1 the tangent to the curve is parallel to x-axis

Q8.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).

Ans.

Calculating the slope of tagent to the given curve, y = (x – 2)²

y = (x – 2)²

$\frac{dy}{dx}=\frac{d}{dx}\left ( x -2 \right )^{2}$

$\frac{dy}{dx}=2\left ( x-2 \right )$

Slope of the chord joining the points (2,0) and (4,4) = m

$m=\frac{4-0}{4-2}=2$

We are given that the tangent is parallel to the given chord of the curve, so their slopes must be equal

2(x – 2) = 2

x = 3

Putting this value of x in the equation of the curve

y = (3 – 2)² = 1

Therefore the point on the given curve is (3,1) in which tangent is parallel to the given chord.

Q9. Find the point on the curve y = x³ – 11x + 5 at which the tangent is  y = x -11.

Ans. The given curve is

y = x³ – 11x + 5

Finding its slope as following

$\frac{dy}{dx}=\frac{d}{dx}\left ( x^{3}-11x +5 \right )$

= 3x² – 11

The slope of the given tangent y = x -11 is calculated as follows

$\frac{dy}{dx}=1$

The slope of the curve = Slope of the tangent

3x² – 11 = 1

x = ±2

If x = 2, then y = 2 – 11 = -9, if x = -2, then y= -2 -11 = -13

(-2,-13) does’t satisfy the equation of the curve so it does’t lie on the curve

The point (2, -9) satisfy the equation of the curve so the given tangent on the curve is at (2, -9)

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