Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3

Question 1.

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

Solution:

The curve is y = 3r^{4} – 4x.

âˆ´ \(\frac { dy }{ dx }\) = 12xÂ³ – 4.

âˆ´ Slope of the tangent at x = 4 is

(\(\frac { dy }{ dx }\))_{x=4} = 12 x 4Â³ – 4 = 764.

Question 2.

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

Solution:

Question 3.

Find the slope of the tangent to the curve y = xÂ³ – x + 1 at the point whose x-coordinate is 2.

Solution:

The curve is y = xÂ³ – x + 1.

âˆ´ \(\frac { dy }{ dx }\) = 3xÂ² – 4.

âˆ´ Slope of the tangent at x = 2

(\(\frac { dy }{ dx }\))_{x=2} = 3 x 2Â² – 1 = 11.

Question 4.

Find the slope of the tangent to the curve y = xÂ³ – 3x + 2 at the point whose x-coordinate is 3.

Solution:

The curve is y = xÂ³ – 3x + 2

âˆ´ \(\frac { dy }{ dx }\) = 3xÂ² – 3.

âˆ´ Slope of the tangent at x = 3

(\(\frac { dy }{ dx }\))_{x=3} = 3 . 3Â² – 3 = 24.

Question 5.

Find the slope of the normal to the curve x = a cosÂ³Î¸, y = a sinÂ³Î¸ at Î¸ = \(\frac { Ï€ }{ 4 }\).

Solution:

Equation of the curve is x = a cosÂ³Î¸, y = a sinÂ³Î¸.

Differentiating w.r.t. Î¸, we get

\(\frac { dx }{ dÎ¸ }\) = – 3a cosÂ² Î¸, \(\frac { dy }{ dÎ¸ }\) = 3a sinÂ²Î¸ cosÎ¸

Hence, slope of the normal = – \(\frac { 1 }{ m }\) = \(\frac { -1 }{ -1 }\) = 1.

Question 6.

Find the slope of the normal to the curve x = 1 – a sin Î¸, y = b cosÂ² Î¸ at Î¸ = \(\frac { Ï€ }{ 2 }\).

Solution:

The equation of curve is x = 1 – a sin Î¸, y = b cosÂ²Î¸.

Question 7.

Find points at which the tangent to the curve y = xÂ³ – 3xÂ² – 9x + 7 is parallel to the x-axis.

Solution:

Equation of the curve is y = xÂ³ – 3xÂ² – 9x + 7.

Differentiating w.r.t. x, we get

\(\frac { dy }{ dx }\) = 3xÂ² – 6x – 9 = 3(xÂ² – 2x – 3)

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

â‡’ 3(x – 3)(x + 1) = 0

âˆ´ x = – 1, 3.

When x = – 1, y = (- 1)Â³ – 3(- 1)Â² – 9(- 1) + 7

= – 1 – 3 + 9 + 7 = 12.

When x = 3, y = 3Â³ – 3.3Â² – 9 . 3 + 7

= 27 – 27 – 27 + 7 = – 20.

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

Question 8.

Find a point on the curve y = (x – 2)Â² 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)Â²

Differentiating w.r.t. x, we get

\(\frac { dy }{ dx }\) = 2(x – 2)

The points A and B are (2, 0) and (4, 4) respectively :

âˆ´ Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) = \(\frac { 4-0 }{ 4-2 }\)

= \(\frac { 4 }{ 2 }\) = 2

Slope of the tangent = 2(x – 2)

Equating (1) and (2), we get

2(x – 2) = 2

âˆ´ x – 2 = 1 or x = 3.

When x = 3, y = (3 – 2)Â² = 1.

Hence, the tangent is parallel to the chord AB at (3, 1).

Alternatively : By MVT,

Hence, the result.

Question 9.

Find the point on the curve y = xÂ³ – 11x + 5 at which the tangent is y = x – 11.

Solution:

Equation of the curve is

y = xÂ³ – 11x + 5.

Differentiating w.r.t. x, we get

\(\frac { dy }{ dx }\) = 3xÂ² – 11 … (1)

Also, the slope of the tangent y = x – 11

= coeff. of x = 1 … (2)

Equating (1) and (2), we get

3xÂ² – 11 = 1.

âˆ´ 3xÂ² = 12.

â‡’ xÂ² = 4, x = Â± 2.

When x = 2, y = x – 11 = 2 – 11 = – 9.

When x = -2, y = x – 11 = – 2 – 11 = – 13.

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

âˆ´ y = x – 11 is the tangent at (2, – 9).

Question 10.

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

Solution:

Equation of the curve is y = \(\frac { 1 }{ x-1 }\)

Differentiating w.r.t. x,

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

Slope of the tangent = – 1

âˆ´ – \(\frac{1}{(x-1)^{2}}\) = – 1.

âˆ´ (x – 1)Â² = 1.

âˆ´ x – 1 = Â± 1 â‡’ x = 2 or 0.

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

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

The points where the tangents to the given curve have the slope -1 are (2,1) and (0,- 1).

âˆ´ Equation of the tangent at (2, 1) is

y – 1 = – 1 x (x – 2) = – x + 2

or x + y = 3.

Equation of tangent at (0, – 1) is

y + 1 = – 1 x (x – 0)

â‡’ x + y + 1 = 0.

Thus, the required lines are x + y- 3 = 0 and x + y + 1 = 0.

Question 11.

Find the equations of all lines having slope 2 and that are tangents to the curve y = \(\frac { 1 }{ x-3 }\)

Solution:

The equation of the given curve is y = \(\frac { 1 }{ x-3 }\)

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

But slope of tangent = 2

âˆ´ \(\frac{-1}{(x-3)^{2}}\) = 2 â‡’ (x-3)Â² = – \(\frac { 1 }{ 2 }\),

Which is not possible as (x – 3)Â² > 0.

Thus, no tangent to the curve y = \(\frac { 1 }{ x-3 }\) has slope 2.

Question 12.

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

Solution:

Let the tangent be at the point (x_{1}, y_{1}) to the curve

The tangent at {xv y,), i.e., (1, \(\frac { 1 }{ 2 }\)) having slope 0 is

y – \(\frac { 1 }{ 2 }\) = 0(x – 1) â‡’ y – \(\frac { 1 }{ 2 }\) = 0 â‡’ y = \(\frac { 1 }{ 2 }\)

Question 13.

Find the 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

Differentiating w.r.t. x,

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

âˆ´ \(\frac { dy }{ dx }\) = – \(\frac { 16 }{ 9 }\) . \(\frac { x }{ y }\).

(i) If the tangent is parallel to x-axis, \(\frac { dy }{ dx }\) = 0.

âˆ´ – \(\frac { 16 }{ 9 }\) . \(\frac { x }{ y }\) = 0 â‡’ x = 0.

Putting x = 0 in (1), we get

\(\frac{y^{2}}{16}\) = 1. âˆ´ y = Â± 4.

Tangents are parallel to x-axis at (0, 4) and (0, – 4).

(ii) When tangents are parallel to y-axis, then denominator of \(\frac { dy }{ dx }\) is zero.

â‡’ y = 0.

Putting y = 0 in (1), \(\frac{x^{2}}{9}\) = 1 â‡’ x = Â± 3.

âˆ´ Tangents are parallel to y-axis at (3, 0) and (- 3, 0).

Question 14.

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

(i) y = x^{4} – 6xÂ³ + 13xÂ² – 10x + 5 at (0, 5).

(ii) y = x^{4} – 6xÂ³ + 13xÂ² – 10x + 5 at (1, 3).

(iii) y = xÂ³ at (1,1).

(iv) y = xÂ² at (0, 0

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

Solution:

(i) We have : y = x^{4} – 6xÂ³ + 13xÂ² – 10x + 5

\(\frac { dy }{ dx }\) = 4xÂ³ – 18xÂ² + 26x – 10.

Putting x = 0, \(\frac { dy }{ dx }\) at (0, 5) = – 10.

Slope of tangent at (0, 5) = – 10.

Thus, the equation of tangent at P(0, 5) is

y – y_{1} = \(\frac { dy }{ dx }\)_{at P} (x – x_{1})

â‡’ y = 5 = – 10(x – 0)

â‡’ y + 10x – 5 = 0.

Further, the equation of normal at P(0, 5) is

(x – x_{1}) = \(\frac { dy }{ dx }\)_{at P} (y – y_{1}) = 0

â‡’ (x – 0) + (-10)(y – 5) = 0

â‡’ y – 10y + 50 = 0.

(ii) We have : y = x^{4} – 6xÂ³ + 13xÂ² – 10x + 5

\(\frac { dy }{ dx }\) = 4xÂ³ – 18xÂ² + 26x – 10.

Putting x = 1, \(\frac { dy }{ dx }\) = 4 – 18 + 26 – 10 = 2.

âˆ´ Slope of tangent at (1, 3) is 2.

âˆ´ Equation of tangent at (1, 3) is

y – 3 = 2(x – 1) â‡’ y – 3 = 2x – 2

â‡’ y = 2x + 1.

Further, the equation of normal is

(x – 1) + 2 . (y – 3) = 0 â‡’ x – 1 + 2y – 6 = 0.

â‡’ x + 2y – 7 = 0.

(iii) We have : y = xÂ³ at (1,1).

âˆ´ \(\frac { dy }{ dx }\) = 3xÂ².

Now, \(\frac { dy }{ dx }\) at (1, 1) = 3(1)Â² = 3.

i.e., Slope of tangent at (1, 1) is 3.

Equation of tangent at (1, 1) is

y – 1 = 3(x – 1) â‡’ y – 1 = 3x – 3

â‡’ y = 3x – 2.

Further, the equation of normal at (1, 1) is

(x – 1) + 3(y – 1) = 0 â‡’ x – 1 + 3y – 3 = 0.

â‡’ x + 3y – 4 = 0.

(iv) We have : y = xÂ²

âˆ´ \(\frac { dy }{ dx }\) = 2x.

Now, \(\frac { dy }{ dx }\) at (0, 0) is 0.

i.e., Slope of tangent at (0, 0) is 0.

âˆ´ Equation of tangent at (0, 0) is

y – 0 = 0(x – 0) â‡’ y = 0.

Further, the equation of normal at (0, 0) is

(x – 0) + 0(y – 0) = 0 â‡’ x = 0.

(v) Equation of the curve are x = cos t … (1) and y = sin t … (2)

At t = \(\frac { Ï€ }{ 4 }\), x = cos \(\frac { Ï€ }{ 4 }\) = \(\frac{1}{\sqrt{2}}\)

and y = sin \(\frac { Ï€ }{ 4 }\) = \(\frac{1}{\sqrt{2}}\)

âˆ´ t = \(\frac { Ï€ }{ 4 }\) â‡’ the point is (\(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\))

From (1) and (2), we have :

Now, slope of the normal at (\(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\)) = – \(\frac { 1 }{ -1 }\) = 1.

Equation of the normal is

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

â‡’ y = x.

Question 15.

Find the equation of the tangent line to the curve y = xÂ² – 2x + 7 which is

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

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

Solution:

Equation of the curve is y = xÂ² – 2x + 7 … (1)

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

(a) Slope of the line 2x-y + 9 = 0 is 2.

â‡’ Slope of tangent = \(\frac { dy }{ dx }\) = 2(x – 1) = 2.

â‡’ x – 1 = 1 ox x = 2.

Putting x = 2 in (1), we get

y = 2Â² – 2 . 2 + 7 = 7.

âˆ´ Tangent parallel to 2x – y + 9 = 0 at (2, 7) is

y – 7 = 2(x – 2)

or 2x – y + 3 = 0.

(b) Tangent is perpendicular to the line 5y – 15x = 13.

â‡’ Slope of tangent x slope of line = – 1,

Slope of the line = \(\frac { 15 }{ 5 }\) = 3.

Slope of tangent = 2(x – 1).

âˆ´ 2(x – 1) x 3 = – 1

âˆ´ x – 1 = \(\frac { – 1 }{ 6 }\) or x = 1 – \(\frac { 1 }{ 6 }\) = \(\frac { 5 }{ 6 }\).

Putting x = \(\frac { 5 }{ 6 }\) in (1), we get

Question 16.

Show that the tangents to the curve y = 7xÂ³ + 11 at the points, where x = 2 and x = – 2, are parallel.

Solution:

The equation of the curve is y = 7xÂ³+ 11. … (1)

Clearly, 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 = xÂ³ at which the slope of the tangent is equal to the y – coordinate of the point.

Solution:

Let P(x_{1}, y_{1}) be the required point. The given curve is

y = xÂ³

â‡’ \(\frac { dy }{ dx }\) = 3xÂ².

â‡’ \(\frac { dy }{ dx }\)(x_{1}, y_{1}) = 3xÂ²_{1}.

Since the slope of the tangent at (x_{1}, y_{1}) = y_{1}, therefore

3xÂ²_{1} = y_{1} … (2)

Also, (x_{1}, y_{1}) lies on (1). So, y_{1} = xÂ³_{1} … (3)

From (2) and (3), we have :

3xÂ²_{1} = xÂ³_{1} â‡’ xÂ²_{1}(3 – x_{1}) = 0.

â‡’ x_{1} = 0 or x_{1} = 3.

When x_{1} = 0, y_{1} = (0)Â³ = 0.

When x_{1} = 3, y_{1} = (3)Â³ = 27.

âˆ´ The required points are (0, 0) and (3, 27).

Question 18.

For the curve y = 4xÂ³ – 2x^{5}, find all the points at which the tangents pass through the origin.

Solution:

Let (x_{1}, y_{1}) be the required point on the given curve

y = 4xÂ³ – 2x^{5}

Then, y_{1} = 4xÂ³_{1} – 2x^{5}_{1} … (1)

Differentiating y = 4xÂ³- 2×5 w.r.t. x, we get

When x_{1} = 0, then from (2), y_{1} = 0

When x_{1} = 1, then from (2), y_{1} = 4 . 1Â³ – 2 . 1^{5} = 4 – 2 = 2

When x_{1} = – 1, then from (2), y_{1} = 4.(-1)Â³ – 2.(-1)^{5} = – 4 + 2 = – 2

Hence, the required points are (0, 0), (1, 2) and (- 1, – 2).

Question 19.

Find the points on the curve xÂ² + yÂ² – 2x = 3 at which tangents are parallel to the x-axis.

Solution:

The curve is

xÂ² + yÂ² – 2x – 3 = 0 … (1)

Differentiating w.r.t. x, we have :

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

â‡’ 2y \(\frac { dy }{ dx }\) = 2 – 2x

â‡’ \(\frac { dy }{ dx }\) = \(\frac { 1 – x }{ y }\)

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

i.e., if 1 – x = 0 â‡’ x=1.

Putting x = 1 in (1), we get

1 + yÂ² – 2 – 3 = 0 â‡’ yÂ² = 4 â‡’ y = Â± 2.

Hence, the required points are (1,2) and (1, – 2), i.e., (1, Â± 2).

Question 20.

Find the equation of the normal at the point (amÂ², amÂ³) for the curve ayÂ² = xÂ³.

Solution:

Equation of the curve is ayÂ² = xÂ³.

Differentiating w.r.t. x, we have :

â‡’ Slope of the tangent at (amÂ², amÂ³) is \(\frac { 3m }{ 2 }\).

and so slope of the normal at (amÂ², amÂ³) is \(\frac { -2 }{ 3m }\)

âˆ´ Equation of the normal is

y – amÂ³ = – \(\frac { 2 }{ 3m }\) (x – amÂ²)

â‡’ 3my – 3am^{4} = – 2x + 2amÂ²

â‡’ 2x + 3my – amÂ²(2 + 3mÂ²) = 0.

Question 21.

Find the equation of the normal to the curve y = xÂ³ + 2x + 6 which is parallel to the line x + 14y + 4 = 0.

Solution:

Let the required normal be drawn at the point (x_{1}, y_{1}). The equation of the given curve is

y = xÂ³ + 2x + 6 … (1)

Differentiating both sides w.r.t x, we get

\(\frac { dy }{ dx }\) = 3xÂ² + 2

â‡’ \(\frac { dy }{ dx }\)(x_{1}, y_{1}) = 3xÂ²_{1} + 2.

Since the normal at (xt, y\) is parallel to the line x + 14y + 4 = 0. Therefore, slope of the normal at (x, y,)

= Slope of the line x + 14y + 4 = 0

Thus, the co-ordinates of the points are (2, 18) and (- 2, – 6).

The equation of the normal at (2, 18) is

y – 18 = – \(\frac { 1 }{ 14 }\) (x – 2)

â‡’ 14y – 252 = – x + 2

â‡’ x + 14y – 254 = 0.

The equation of the normal at (- 2, – 6) is

y + 6 = – \(\frac { 1 }{ 14 }\)(x + 2)

â‡’ 14y + 84 = – x – 2

â‡’ x + 14y + 86 = 0.

Question 22.

Find the equations of the tangent and normal to the parabola yÂ² = 4ax at (atÂ², 2at).

Solution:

The equation of the given curve is yÂ² = 4ax. … (1)

Differentiating (1) w.r.t. x, we get

Question 23.

Prove that the curves x = yÂ² and xy = k cut at right angles, if 8kÂ² = 1.

Solution:

The given curves are :

x = yÂ² … (1)

and xy = k … (2)

Putting x = yÂ² in (2), we get

yÂ³ = k â‡’ y = k \(\frac { 1 }{ 3 }\).

From (1), x = yÂ² = \(\left(k^{\frac{1}{3}}\right)^{2}\) = k\(k^{\frac{2}{3}}\)

Thus, the point of intersection is (\(k^{\frac{2}{3}}\), \(k^{\frac{1}{3}}\))

Differentiating (1) w.r.t. x, we have :

Now the curves cut at right angles, if the product of slopes of tangents to two curves at (\(k^{\frac{2}{3}}\)), \(k^{\frac{1}{3}}\)) is – 1.

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:

Let the point of contact of the tangent line parallel to the given line be P(x_{1}, y_{1}).

The equation of the curve is y = \(\sqrt{3x-2}\)

Differentiating both sides w.r.t. x, we get

Since the tangent at (xx, is parallel to the line 4x – 2y + 5 = 0, therefore

So, the point of contact is (\(\frac { 41 }{ 8 }\), \(\frac { 3 }{ 4 }\)).

Hence, the equation of the required tangent is

Question 26.

The slope of the normal to the curve y = 2xÂ³ + 3 sin x at x = 0 is

(A) 3

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

(C) – 3

(D) – \(\frac { 1 }{ 3 }\)

Answer:

The curve is y

= 2xÂ² + 3 sin .t.

âˆ´ \(\frac { dy }{ dx }\) = 4x + 3 cos x

At x = 0, \(\frac { dy }{ dx }\) = 4 x 0 + 3 cos 0 = 3. âˆ´ Slope = 3.

Slope of the normal is – \(\frac { 1 }{ 3 }\).

Hence, Part (D) is the correct answer.

Question 27.

The line y = x + 1 is a tangent to the curve yÂ² = 4x at the point

(A) (1, 2)

(B) (2, 1)

(C) (1, – 2)

(D) (- 1, 2)

Solution:

The curve is yÂ² = 4x. Therefore, 2y \(\frac { dy }{ dx }\) = 4.

âˆ´ \(\frac { dy }{ dx }\) = \(\frac { 4 }{ 2y }\) = \(\frac { 2 }{ y }\).

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

âˆ´ \(\frac { 2 }{ y }\) âˆ´ y = 2.

Putting y = 2 in yÂ² = 4x, we get

2Â² = 4 â‡’ x = 1.

Point of contact is (1, 2).

âˆ´ Correct Answer is (A).