These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.3

Question 1.

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

Solution:

y = 3x^{4} – 4x

Differentiating w.r.t. x,

∴ \(\frac { dy }{ dx }\) = 12x^{3} – 4

∴ Req. slope = \({ \left( \frac { dy }{ dx } \right) }_{ x=4 }\)

= 12(4^{3}) – 4 = 764.

Question 2.

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

Solution:

y = \(\frac { x-1 }{ x-2 }\), x ≠ 2

Differentiating w.r.t. x,

Question 3.

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

Solution:

The curve is y = x^{3} – x + 1

Differentiating w.r.t. x,

\(\frac { dy }{ dx }\) = 3x² – 1

∴ slope of tangent = \(\frac { dy }{ dx }\)

= 3x² – 1

= 3 x 2² – 1

= 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:

The curve is y = x^{3} – 3x + 2

\(\frac { dy }{ dx }\) = 3x² – 3

∴ slope of tangent = \(\frac { dy }{ dx }\)

= 3x² – 3

= 3 x 3² – 3

= 24

Question 5.

Find the slope of the normal to the curve x = a cos^{3} θ, y = a sin^{3} θ at θ = \(\frac { \pi }{ 4 } \).

Solution:

y = a sin^{3}θ and x = a cos^{3}θ

Differentiating w.r.t. x,

Question 6.

Find the slope of the normal to the curve x = 1 – a sin θ, y = b cos² θ at θ = \(\frac { \pi }{ 2 } \)

Solution:

x = 1 – a sin θ and y = b cos² θ

Differentiating w.r.t. x,

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:

y = x^{3} – 3x^{2} – 9x + 7

Differentiating w.r.t. x,

\(\frac { dy }{ dx }\) = 3(x – 3)(x + 1)

Tangent is parallel to x-axis if the slope of tangent = 0

or \(\frac { dy }{ dx }=0\)

⇒ 3(x + 3)(x + 1) = 0

⇒ x = – 1, 3

when x = – 1, y = 12 & When x = 3, y = – 20

Hence the tangent to the given curve are parallel to x-axis at the points (- 1, – 12), (3, – 20)

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:

The equation of the curve is y = (x – 2)²

Differentiating w.r.t x

\(\frac { dy }{ dx }\) = 2(x – 2 )

The point A and B are (2,0) and (4,4) respectively.

Slope of AB = \(\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }\)

= \(\frac { 4-0 }{ 4-2 } =\frac { 4 }{ 2 } \) = 2 … (i)

Slope of the tangent = 2 (x – 2) ….(ii)

from (i) & (ii) 2 (x – 2) = 2

∴ x – 2 = 1 or x = 3

when x = 3,y = (3 – 2)² = 1

∴ The tangent is parallel to the chord AB at (3, 1)

Question 9.

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

Solution:

Here, y = x^{3} – 11x + 5

⇒ \(\frac { dy }{ dx }\) = 3x² – 11

The slope of tangent line y = x – 11 is 1

∴ 3x² – 11 = 1

⇒ 3x² = 12

⇒ x² = 4, x = ±2

When x = 2, y = – 9 & when x = -2,y = -13

But (-2, -13) does not lie on the curve

∴ y = x – 11 is the tangent at (2, -9)

Question 10.

Find the equation of all lines having slope -1 that are tangents to the curve y = \(\frac { 1 }{ x-1 }\), x ≠ 1

Solution:

y = \(\frac { 1 }{ x-1 }\), x ≠ 1

Differentiating w.r.t x

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

Since the tangent have slope – 1,

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

⇒ (x – 1)² = 1 ⇒ x – 1 = ± 1

⇒ x = 2 or x = 0

When x = 0, y = \(\frac { 1 }{ 0 – 1 }\) = – 1

When x = 2, y = \(\frac { 1 }{ 2 – 1 }\) = 1

∴ The required points are (0, – 1) and (2, 1)

Equation of the tangent at (0, – 1) is

y – 1 = – 1 (x – 0) ⇒ x + y + 1 = 0

Equation of the tangent at (2, 1) is

y – 1 = – 1(x – 2) ⇒ x + y – 3 = 0

Question 11.

Find the equation of ail lines having slope 2 which are tangents to the curve y = \(\frac { 1 }{ x-3 }\), x ≠ 3.

Solution:

Here y = \(\frac { 1 }{ x-3 }\)

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

∵ slope of tangent = 2

\(\frac { -1 }{ { (x-3) }^{ 2 } } =2\Rightarrow { (x-3) }^{ 2 }=-\frac { 1 }{ 2 } \)

Which is not possible as (x – 3)² > 0

Thus, no tangent to y = \(\frac { 1 }{ x-3 }\)

Hence there is no tangent to the curve with slope 2.

Question 12.

Find the equations of all lines having slope 0 which are tangent to the curve y = \(\frac { 1 }{ { x }^{ 2 }-2x+3 }\)

Solution:

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

Differentiating w.r.t x

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

Since the slope of the tangent is zero,

\(\frac { dy }{ dx }\) = 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:

The equation of the curve is \(\frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 16 }\) = 1 … (i)

Differentiating both sides w.r.t x

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

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

i. Since the tangent is parallel to x axis,

\(\frac { dy }{ dx }\) = 0 ⇒ \(\frac{-16x}{9y}\)

Hence x = 0

When x = 0, we get \(\frac{0}{9}\) + \(\frac { { y }^{ 2 } }{ 16 }\) = 1

⇒ y² = 16 ⇒ y = ±4

∴ (0,4) and (0, – 4) are the points at which the tangents are parallel to the x axis

ii. Since the tangent is parallel to y axis, its slope is not defined, then the normal is parallel to x-axis whose slope is zero.

∴ (3, 0) and (- 3, 0) are the points at which the tangents are parallel to they axis.

Question 14.

Find the equations of the tangent and normal to the given curves at the indicated points:

(i) y = x^{4} – 6x^{3} + 13x^{2} – 10x + 5 at (0,5)

(ii) y = x^{4} – 6x^{3} + 13x^{2} – 10x + 5 at (1,3)

(iii) y = x^{3} at (1, 1)

(iv) y = x^{2} at (0,0)

(v) x = cos t, y = sin t at t = \(\frac { \pi }{ 4 } \)

Solution:

(i) y = x^{4} – 6x^{3} + 13x^{2} – 10x + 5

Differentiating w.r.t x

Equation of the tangent at (0, 5) is

(y- 5) = – 10(x- 0)

i.e., 10x+y-5 = 0

Equation of the normal at (0, 5) is

(y – 5) = – 10(x – 0)

i.e., x – 10y + 50 = 0

(ii) y = x^{4} – 6x^{3} + 13x^{2} – 10x + 5

Differentiating w.r.t. x,

\(\frac { dy }{ dx }\) = 4x^{3} – 18x^{2} + 26x – 10

\(\frac { dy }{ dx }\)_{(1, 3)} = 4(1) – 18(1) + 26(1) – 10 = 2

At (1, 3), the slope of the tangent = 2

At (1, 3), the slope of the normal = \(\frac { – 1 }{ 2 }\)

Equation of the tangent at (1, 3) is y – 3 = 2( x – 1)

i.e., y = 2x + 1

Equation of the normal at (1, 3) is

y – 3 = y(x – 1)

i.e., x + 2y – 7 = 0

(iii) y = x^{3}

Differentiating w.r.t. x,

\(\frac { dy }{ dx }\) = 3x^{2}

At(1, 1), \(\frac { dy }{ dx }\) = 3 x 1² = 3

At (1, 1), the slope of the tangent = 3

Equation of the tangent at(1, 1) is

y – 1 = 3(x – 1)

i.e., y = 3x – 2

At (1, 1), the slope o fthe normal = \(\frac { -1 }{ 3 }\)

Equation of the normal at (1, 1) is

y – 1 = \(\frac { -1 }{ 3 }\) (x – 1)

i.e., x + 3y – 4 = 0

(iv) y = x^{2}

Differentiating w.r.t. x,

\(\frac { dy }{ dx }\) = 2x

[ \(\frac { dy }{ dx }\) ]_{(0, 0)} = 0

At (0, 0), the slope of the tangent = 0

Equation of the tangent at (0, 0) is

y – 0 = 0(x – 0)

i.e., y = 0 or the x axis

Since the tangent is the x axis, the normal is parallel to y axis.

∴ The equation of the normal at (0, 0) is x = 0, since the normal passes through (0,0)

(v) y = sin t and x = cos t

Differentiating w.r.t. t, we get \(\frac { dy }{ dt }\) = cos t

Question 15.

Find the equation of the tangent line to the curve y = x^{2} – 2x + 7 which is

(i) parallel to the line 2x – y + 9 = 0

(ii) perpendicular to the line 5y – 15x = 13.

Solution:

y = x^{2} – 2x + 7

Differentiating w.r.t. x,

\(\frac { dy }{ dx }\) = 2x – 2

\(\frac { dy }{ dx }\) = 2(x – 1) … (1)

(i) Slope of the line 2x – y + 9 = 0 is 2

Since the tangent to the curve is parallel to the line 2x – y + 9 = 0, we get

\(\frac { dy }{ dx }\) = 2

From (1), we get 2(x – 1) = 2 ⇒ x = 2

When x = 2, y = 2² – 2 x 2 + 7 = 7

∴ At (2, 7) the tangent is parallel to 2x – y + 9 = 0

The equation of the tangent to the given curve at (2, 7) is y – 7 = 2(x – 2)

i.e., 2x – y + 3 = 0 or y – 2x – 3 = 0

Slope of the line – 15x = 13 is 3

(ii) Since the tangent to the curve is perpendicular to the line 5y – 15x = 13,

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

From (1), we get 2(x – 1) = \(\frac { -1 }{ 3 }\)

The tangent is perpendicular to the line 5y – 15x = 13

The equation of the tangent to the given

Question 16.

Show that the tangents to the curve y = 7x^{3} + 11 at the points where x = 2 and x = – 2 are parallel.

Solution:

Here, y = 7x^{3} + 11

⇒ x \(\frac { dy }{ dx }\) = 21 x²

Now m_{1} = slope at x = 2 is

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

= 21 x 2² = 84

and m_{2} = slope at x = -2 is

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

= 21 x (- 2)² = 84

Hence, m_{1} = m_{2} Thus, the tangents to the given curve at the points where x = 2 and x = – 2 are parallel

Question 17.

Find the points on the curve y = x³ at which the slope of the tangent is equal to the y-coordinate of the point

Solution:

y = x³

Differentiating w.r.t. x,

\(\frac { dy }{ dx }\) = 3x²

The slope of the tangent = 3x²

Since slope of the tangent = y coordinate,

3x² = y or 3x² = x³

= 3x² – x³ = 0 = x²(3 – x) = 0

⇒ x = 3 or x = 0

When x = 0, y = (0)³ = O

When x = 3, y= (3)³ = 27

∴ (0, 0) and (3, 27) are the points at which the slope of the tangent is equal to the y coordinate of the point.

Question 18.

For the curve y = 4x^{3} – 2x^{5}, find all the points at which the tangent passes through the origin.

Solution:

Let, (h, k) be the point at which the tangent passes through the origin.

y = 4x³ – 2x^{5}

Differentiating w.r.t. x,

\(\frac { dy }{ dx }\) = 12x² – 10x^{4}

At (h, k) the slope of the tangent = \(\frac { dy }{ dx }\)_{(h,k)} = 12h² – 10h^{4}

Equation of the tangent at (h, k) is

y – k = (12h² – 10h^{4}) (x – h)

Since the tangent passes through the origin,

(0 – k) = (12h² – 10h^{4}) (0 – h)

⇒ k = 12h³ – 10h^{5} … (1)

(h, k) is a point on the curve y = 4x³ – 2x^{5}

∴ k = 4h³ – 2h^{5} … (2)

From (1) & (2) we get

12h³ – 10h^{5} = 4h³ – 2h^{5}

⇒ 8h³ – 8h^{5} = 0

⇒ 8h³ (1 – h²) = 0

⇒ 8h³ (1 – h) (1 +h) = 0

⇒ h = 0 or h= 1 or h = – 1

When h = 0, k = 0

When h = 1, k = 2

When h = – 1, k = – 2

∴ (0, 0), (1, 2) and (- 1, – 2) are the points at which the tangent passes through the origin.

Question 19.

Find the points on the curve x^{2} + y^{2} – 2x – 3 = 0 at which the tangents are parallel to the x-axis.

Solution:

Here, x^{2} + y^{2} – 2x – 3 = 0

⇒ \(\frac { dy }{ dx } =\frac { 1-x }{ y }\)

Tangent is parallel to x-axis, if \(\frac { dy }{ dx }\) = 0 i.e.

if 1 – x = 0

⇒ x = 1

Putting x = 1 in (i)

⇒ y = ±2

Hence, the required points are (1,2), (1, -2) i.e. (1, ±2).

Question 20.

Find the equation of the normal at the point (am², am³ ) for the curve ay² = x³ .

Solution:

ay² = x³

Differentiating w.r.t. x,

Question 21.

Find the equation of the normal’s to the curve y = x^{3} + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

Solution:

y = x^{3} + 2x + 6

Differentiating w.r.t. x,

\(\frac { dy }{ dx }\) = 3x² + 2

Slope of the normal = \(\frac{-1}{3 x^{2}+2}\)

Slope of the line x + 14y + 4 = 0 is \(\frac { -1 }{ 14 }\)

Since the normal is parallel to x + 14y + 4 = 0

\(\frac{-1}{3 x^{2}+2}\) = \(\frac { -1 }{ 14 }\) ⇒ 3x² + 2 = 14

⇒ 3x² = 12 ⇒ x² = 4 ⇒ x = ± 2

When x = 2, y = 18 and when x = – 2, y = – 6

∴ At (2, 18) and (- 2, – 6), the normals are parallel to x + 14y + 4 = 0.

Slope of the normal = \(\frac { -1 }{ 14 }\)

Equation of the normal at (2, 18) is

y – 8 = \(\frac { -1 }{ 14 }\)(x – 2)

i.e., x + 14y – 254 = 0

Equation of the normal at ( -2, -6) is

y + 6 = \(\frac { -1 }{ 14 }\) (x + 2) i.e., x + 14y + 86 = 0

Question 22.

Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at²,2at).

Solution:

y² = 4ax

Differentiating w.r.t. x,

2y\(\frac { dy }{ dx }\) = 4a

Equation of the tangent at (at², 2at) is

y – 2at = \(\frac { 1 }{ t }\)(x – at²)

yt – 2at² = x – at²

x – yt + at² = 0 or ty = x + at²

Slope of the normal at (at², 2at) is – t.

Equation of the normal at (at², 2at) is

y – 2at = – t(x – at²)

xt + y = at³ + 2at

y = – tx + 2at + at³

Question 23.

Prove that the curves x = y² and xy = k cut at right angles if 8k² = 1.

Solution:

The curve are

x = y² … (1)

xy = k … (2)

Let P be the point of intersection

Substituting x = y² in (2),

Since the two curves cut at right angles at P, the product of their slopes at P is – 1.

Taking the cubes we get k² = \(\frac { 1 }{ 8 }\) or 8k² = I

Question 24.

Find the equations of the tangent and normal to the hyperbola \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } }\) = 1 at the point (x_{0}, y_{0}).

Solution:

Question 25.

Find the equation of the tangent to the curve y = \(\sqrt { 3x – 2 } \) which is parallel to the line 4x – 2y + 5 = 0.

Solution:

y = \(\sqrt { 3x – 2 } \)

Differentiating w.r.t. x,

\(\frac { dy }{ dx }\) = \(\frac{3}{2 \sqrt{3 x-2}}\)

Slope of the tangent = \(\frac{3}{2 \sqrt{3 x-2}}\)

Slope of the line 4x – 2y + 5 = 0 is 2

Since the tangent is parallel to the line dy

4x – 2y + 5 = 0, we get \(\frac { dy }{ dx }\) = 2

∴ At(\(\frac { 41 }{ 48 }\), \(\frac { 3 }{ 4 }\)), the tangent is parallel to the line 4x – 2y + 5 = 0

Equation of the tangent at (\(\frac { 41 }{ 48 }\), \(\frac { 3 }{ 4 }\)) is

y – \(\frac { 3 }{ 4 }\) = 2(x – \(\frac { 41 }{ 48 }\)

24y – 18 = 48x – 41

i.e., 48x – 24y = 23

Question 26.

The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is

(a) 3

(b) \(\frac { 1 }{ 3 }\)

(c) -3

(d) \(-\frac { 1 }{ 3 }\)

Solution:

(d) ∵ y = 2x² + 3sinx

∴ \(\frac { dy }{ dx }\) = 4x+3cosx

at

x = 0, \(\frac { dy }{ dx }\) = 3

∴ slope = 3

⇒ slope of normal is = \(\frac { 1 }{ 3 }\)

Question 27.

The line y = x + 1 is a tangent to the curve y² = 4x at the point

(a) (1,2)

(b) (2,1)

(c) (1,-2)

(d) (-1,2)

Solution:

(a) The curve is y² = 4x,

∴ \(\frac { dy }{ dx } =\frac { 4 }{ 2y } =\frac { 2 }{ y } \)

Slope of the given line y = x + 1 is 1

∴ \(\frac { 2 }{ y }\) = 1

y = 2 Putting y = 2 in

⇒ y² = 4x

⇒ 2² = 4x

⇒ x = 1

∴ The Point of contact is (1, 2)