Students can cross-reference their work with Understanding ISC Mathematics Class 12 Solutions Chapter 7 Applications of Derivatives MCQs to ensure accuracy.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives MCQ’s

Choose the correct answer from the given four options in questions (1 to 47):

Question 1.

If the radius of a circle is increasing at the rate of 2 cm / sec, then the area of the circle when its radius is 20 cm is increasing at the rate of

(a) 80 π m^{2} / sec

(b) 80 m^{2} / sec

(c) 80 π cm^{2} /sec

(d) 80 cm^{2} /sec

Solution:

(c) 80 π cm^{2} /sec

Let r be the radius of circle at any time t

Then A = area of circle = πr²

∴ \(\frac{d A}{d t}\) = 2 cm / sec

Given \(\frac{d r}{d t}\) = 2 cm/sec ;

r = 20 cm

∴ \(\frac{d A}{d t}\) = 2π × 20 × 2 = 80π cm^{2} / sec

Question 2.

The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which its area increases, when side is 10 cm, is

(a) 10 cm^{2} / sec

(b) 10√3 cm^{2} /sec

(c) \(\frac{10}{3}\) cm^{2} / sec

(d) √3 cm^{2} / sec

Solution:

(b) 10√3 cm^{2} /sec

Let x be the length of each side of equilateral triangle at any time t.

Then A = area of equilateral triangle = \(\frac{\sqrt{3}}{4}\) x^{2}

∴ \(\frac{d \mathrm{~A}}{d t}=\frac{\sqrt{3}}{4} \times 2 x \frac{d x}{d t}\)

= \(\frac{\sqrt{3}}{2} x \frac{d x}{d t}\)

Given \(\frac{d x}{d t}\) = 2 cm / sec ;

x = 10 cm

∴ \(\frac{d A}{d t}\) = \(\frac{\sqrt{3}}{2}\) × 10 × 2 = 10√3 cm^{2} / sec.

Question 3.

A spherical ice ball is melting at the rate of 100 π cm^{3} / min. The rate at which its radius is decreasing, when its radius is 15 cm, is

(a) \(\frac{1}{9}\) cm / min

(b) \(\frac{1}{9 \ pi}\) cm / min

(c) \(\frac{1}{18}\) cm / min

(d) \(\frac{1}{36}\) cm / min

Solution:

(a) \(\frac{1}{9}\) cm / min

Let r be the radius of spherical ice ball

Then \(\frac{d V}{d t}\) = – 100π cm^{3} / min.

⇒ \(\frac{d}{d t}\) (\(\frac{4}{3}\) πr^{3}) = – 100π

⇒ \(\frac{4}{3}\) π × 3r^{2} \(\frac{d r}{d t}\) = – 100π

⇒ 4π × 15^{2} . \(\frac{d r}{d t}\) = – 100π [given r = 15]

⇒ \(\frac{d r}{d t}\) = – \(\frac{1}{9}\) cm / min.

Question 4.

The radius of a cylinder is increasing at the rate of 3 cm/sec and its height is decreasing at the rate of 4 cm/sec. The rate of change of its volume when radius is 4 cm and height is 6 cm, is

(a) 64 cm^{3} / sec

(b) 144 cm^{3}/ sec

(c) 80 cm^{3}/ sec

(d) 80 π cm^{3} / sec

Solution:

(d) 80 π cm^{3} / sec

Let r be the radius and h be the height of cylinder at any time.

Then V = volume of cylinder = πr²h

∴ \(\frac{d V}{d t}=\pi\left[r^2 \frac{d h}{d t}+2 r h \frac{d r}{d t}\right]\)

Given \(\frac{d r}{d t}\) = 3 cm/sec ;

\(\frac{d h}{d t}\) = – 4 cm/sec ;

r = 4 cm

and h = 6 cm

Thus \(\frac{d V}{d t}\) = π [4^{2} × (- 4) + 2 × 4 × 6 × 3]

= π [- 64 + 144]

= 80 π cm^{3} / sec

Question 5.

A point on the curve y^{2} = 18x at which the ordinate increases at twice the rate of abscissa is

(a) (2, 4)

(b) (2, – 4)

(c) \(\left(-\frac{9}{8}, \frac{9}{2}\right)\)

(d) \(\left(\frac{9}{8}, \frac{9}{2}\right)\)

Solution:

(d) \(\left(\frac{9}{8}, \frac{9}{2}\right)\)

Given eqn. of curve y^{2} = 18x

⇒ 2y \(\frac{d y}{d x}\) = 18

⇒ \(\frac{d y}{d x}=\frac{9}{y}\)

⇒ 2y \(\frac{d y}{d t}\) = 18 \(\frac{d x}{d t}\)

⇒ \(\frac{d y}{d t}=\frac{9}{y} \frac{d x}{d t}\) ……..(1)

Also \(\frac{d y}{d t}\) = 2 \(\frac{d x}{d t}\)

from (1) and (2) ; we have

⇒ 2 \(\frac{d x}{d t}\) = \(\frac{9}{y}\) \(\frac{d x}{d t}\)

⇒ y = \(\frac{9}{2}\) \(

∴ from given curve, we have,

[latex]\left(\frac{9}{2}\right)^2\) = 18x

⇒ x = \(\frac{81}{4 \times 18}=\frac{9}{8}\)

Thus required point on given curve be \(\left(\frac{9}{8}, \frac{9}{2}\right)\).

Question 6.

A ladder, 5 metres long standing on a floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when the lower end of the ladder is 2 metres away from the wall is

(a) \(\frac{1}{10}\) radian / sec

(b) \(\frac{1}{20}\) radian/sec

(c) 5 radian/sec

(d) 10 radian/sec

Solution:

(b) \(\frac{1}{20}\) radian/sec

Let the bottom of the ladder be at a distance x m from the wall and the top be at a height h from the ground.

Then x^{2} + y^{2} = 25 ………..(1)

and tan θ = \(\frac{y}{x}\) …………(2)

Diff. eqn. (1) both sides w.r.t. t, we have

2x \(\frac{d x}{d t}\) + 2y \(\frac{d y}{d t}\) = 0

Given \(\frac{d y}{d t}\) = – 10 cm / sec = – \(\frac{1}{10}\) m / sec

⇒ 2x \(\frac{d x}{d t}\) – \(\frac{2 y}{10}\) = 0

⇒ \(\frac{d x}{d t}=\frac{y}{10 x}\)

Diff. eqn. (2) both sides w.r.t. t, we have

Question 7.

The curve y = x^{1/5} has at (0, 0)

(a) a vertical tangent (parallel to y-axis)

(b) a horizonlal tangent (parallel to x-axis)

(c) an oblique tangent

(d) no tangent

Solution:

(a) a vertical tangent (parallel to y-axis)

Equation of given curve be,

y = x^{1/5}

∴ slope of tangent to given curve at point

P(0, 0) = (\(\frac{d y}{d x}\))_{(0, 0)} → ∞

Thus tangent is parallel to y-axis.

Question 8.

The equation of the tangent to the curve y = (4 – x^{2})^{2/3} at x = 2 is

(a) x = – 2

(b) x = 2

(c) y = 2

(d) y = – 2

Solution:

(b) x = 2

eqn. of given curve be y = (4 – x^{2})^{2/3}

When x = 2 ; y = 0

Thus any point on given curve be (2, 0)

∴ \(\frac{d y}{d x}\) = \(\frac{2}{3}\) (4 – x^{2})^{-1/3} (- 2x)

∴ slope of tangent to given curve at x = 2 = (\(\frac{d y}{d x}\))_{x = 2} = ∞

Thus required eqn. of tangent to given curve at (2, 0) be given by

y – 0 = \(\frac{\frac{1}{d x}}{d y}\) (x – 2)

⇒ x – 2 = 0

⇒ x = 2.

Question 9.

The equation of the tangent to the curve y = e^{2x} at (0, 1) is

(a) y + 1 = 2x

(b) 1 – y = 2x

(c) y – 1 = 2x

(d) none of these

Solution:

(c) y – 1 = 2x

eqn. of given curve be

y = e^{2x} …………(1)

∴ \(\frac{d y}{d x}\) = 2e^{2x}

Thus slope of tangent to curve (1) at (0, 1) = (\(\frac{d y}{d x}\))_{(0, 1)}

=2 e^{2 × 0} = 2

Thus eqn. of tangent to given curve at point (0, 1) be given by

y – 1 = 2(x – 0)

⇒ y – 1 = 2x.

Question 10.

The tangent to the curve y = e at the point (0, 1) meets x-axis at:

(a) (0, 1)

(b) (- \(\frac{1}{2}\), o)

(c) (2, 0)

(d) (0, 2)

Solution:

(b) (- \(\frac{1}{2}\), o)

Equation of given curve be y = e^{2x} ………….(1)

Differentiate equation (1) both sides w.r.t. x; we get

\(\frac{d y}{d x}\) = 2e^{2x}

∴ slope of tangent to given curve (1) at (0, 1) = (\(\frac{d y}{d x}\))_{(0, 1)}

= 2 × e^{0} = 2

Thus required equation of tangent to given curve (1) at (0, 1) is given by

y – 1 = 2 (x – 0)

⇒ y – 1 = 2x …………(2)

Equation (2) meets x-axis at y = 0

Hence the coordinates of required point are (- \(\frac{1}{2}\), o).

Question 11.

The tangent to the curve x^{2} = 2y at the point (1, \(\frac{1}{2}\)) makes with the x-axis an angle of

(a) 0

(b) \(\frac{\pi}{6}\)

(c) \(\frac{\pi}{4}\)

(d) \(\frac{\pi}{3}\)

Solution:

(c) \(\frac{\pi}{4}\)

Given eqn. of curve be x^{2} = 2y

∴ 2x = 2 \(\frac{d y}{d x}\)

⇒ \(\frac{d y}{d x}\) = x

∴ slope of tangent to given curve at (1, \(\frac{1}{2}\)) = (\(\frac{d}{d x}\))_{(1, \(\frac{1}{2}\))} = 1

Let θ be the angle made by tangent with the +ve direction of x-axis.

∴ slope of tangent to given curve = tan θ

∴ tan θ = 1

⇒ θ = \(\frac{\pi}{4}\)

Question 12.

The tangents to the curve x^{2} + y^{2} = 2 at points (1, 1) and (- 1, 1) are

(a) parallel

(b) at right angles

(c) neither parallel nor at right angles

(d) none of these

Solution:

eqn. of given curve be

x^{2} + y^{2} = 2 …………(1)

Diff. eqn. (1) w.r.t. x; we have

2x + 2y \(\frac{d y}{d x}\) = 0

⇒ \(\frac{d y}{d x}\) = – \(\frac{x}{y}\)

∴ m_{1} = slope of tangent to given curve at (1, 1) = (\(\frac{d y}{d x}\))_{(0, 1)} = – 1

m_{2} = slope of tangent to given curve at (- 1, 1) = (\(\frac{d y}{d x}\))_{(- 1, 1)} = 1

∴ m_{1} m_{2} = – 1 × 1 = – 1

Hence, tangents to given curve at points (1, 1) and (- 1, 1) are at right angles.

Question 13.

The point on the curve y^{2} = x, where tangent make an angle of with the x-axis, is

(a) \(\left(\frac{1}{2}, \frac{1}{4}\right)\)

(b) \(\left(\frac{1}{4}, \frac{1}{2}\right)\)

(c) (4, 2)

(d) (1, 1)

Solution:

(b) \(\left(\frac{1}{4}, \frac{1}{2}\right)\)

eqn. of given curve be

y^{2}= x …………….(1)

diff, both sides of eqn. (1) w.r.t. x;

2y \(\frac{d y}{d x}\) = 1

⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2 y}\)

∴ slope oftangentto curve (1) = \(\frac{d y}{d x}\) = \(\frac{1}{2 y}\)

Also, slope of tangent to given curve (1) = tan \(\frac{\pi}{4}\) = 1

∴ \(\frac{1}{2 y}\) = 1

⇒ y = \(\frac{1}{2}\)

∴ from (1) ;

x = \(\frac{1}{4}\)

Thus, required point on given curve be (\(\left(\frac{1}{4}, \frac{1}{2}\right)\))

Question 14.

The point on the curve y = 6x – x^{2} where the tangent is parallel to the line 4x – 2y – 1 = 0 is

(a) (2, 8)

(b) (8, 2)

(c) (6, 1)

(d) (4, 2)

Solution:

(a) (2, 8)

Let P (x_{1}, y_{1}) be any point on curve

y = 6x – x^{2} ……….(1)

∴ \(\frac{d y}{d x}\) = 6 – 2x

Thus slope of tangent to given curve at

P(x_{1}, y_{1}) = \(\left(\frac{d y}{d x}\right)_{\mathrm{P}\left(x_1, y_1\right)}\) = 6 – 2x_{1}

eqn.of given line be 4x – 2y – 1 = 0 ………….(2)

∴ slope of hne (2) = \(\frac{- 4}{- 2}\) = 2

Since it is given that tangent to given curve is parallel to line (2)

∴ 6 – 2x_{1} = 2

⇒ 2x_{1} = 4

⇒ x_{1} = 2

Also P (x_{1}, y_{1}) lies on eqn. (1) ;

y_{1} = 6x_{1} – x_{1}

= 6 × 2 + 2^{2} = 8

Hence coordinates of any point on given curve be (2, 8).

Question 15.

The points at which the tangents to the curve y = x^{3} – 12x + 18 are parallel to x-axis are :

(a) (2, – 2), (- 2, – 34)

(b) (2, 34), (- 2, 0)

(c) (0, 34), (- 2, 0)

(d) (2, 2), (- 2, 34)

Solution:

(d) (2, 2), (- 2, 34)

Equation of given curve be

y = x^{3} – 12x + 18 …………..(1)

Let the required points on curve (1) are (x_{1}, y_{1})

Differentiate equation (1) w.r.t. x; we get

\(\frac{d y}{d x}\) = 3x^{2} – 12

slope of tangent to given curve (1) at (x_{1}, y_{1})

= \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)

= 3x_{1}^{2} – 12

Also tangents are parallel to x – axis.

∴ \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\) = 0

[since slope of the tangent is 0]

3x_{1}^{2} – 12 = 0

⇒ x_{1} = ± 2

Also, the point (x_{1}, y_{1}) lies on curve (1)

y_{1}= x_{1} – 12x_{1} + 18 …………….(2)

when x_{1} = 2

∴ from (2) ; we have

y_{1} = 8 – 24 + 18 = 2

when x_{1} = 2

∴ from (2) ; we have

y_{1} = – 8 + 24 + 18 = 34

Thus the required points on given curve (1) are (2, 2) and (- 2, 34).

Question 16.

The point at which the tangent to the curve y = 2x^{2} – x + 1 is parallel to the line y = 3x + 9 is

(a) (2, 1)

(b) (1, 2)

(c) (3, 9)

(d) (- 2, 1)

Solution:

(b) (1, 2)

Let P (x_{1}, y_{1}) be any point on given curve

y = 2x^{2} – x + 1

∴ y_{1} = 2x_{1}^{2} – x_{1} + 1 ………..(1)

∴ slope of tangent to given curve at P (x_{1}, y_{1})

= \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}\)

= 4x_{1} – 1

Again eqn. of given line be 3x – y + 9 = 0 …………..(2)

∴ slope of line (2) = \(\frac{- 3}{- 1}\) = 3

Since tangent to curve (1) is parallel to line (2)

∴ 4x_{1} – 1 = 3

⇒ x_{1} = 1

∴ from (1);

y_{1} = 2 – 1 + 1 = 2

Thus, required point on tangent to given curve be (1, 1).

Question 17.

If the tangent to the curve x = t^{2} – 1, y = t^{2} – t is parallel to x-axis, then

(a) t = 0

(b) t = 2

(c) t = \(\frac{1}{2}\)

(d) t = – \(\frac{1}{2}\)

Solution:

(c) t = \(\frac{1}{2}\)

Given eqn. of curve be,

x = t^{2} – 1

and y = t^{2} – t

∴ \(\frac{d x}{d t}\) = 2t ;

\(\frac{d y}{d t}\) = 2t – 1

Thus \(\frac{d y}{d x}\) = \(\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)

= \(\frac{2 t-1}{2 t}\)

Since tangent to given curve is parallel to x-axis

∴ \(\frac{d y}{d x}\) = 0

⇒ \(\frac{2 t-1}{2 t}\) = 0

⇒ t = \(\frac{1}{2}\).

Question 18.

The tangent to the curve x = e^{t} cos t, y = e^{t} sin t at t = \(\frac{\pi}{4}\) makes with x-axis an angle

(a) 0

(b) \(\frac{\pi}{4}\)

(c) \(\frac{\pi}{3}\)

(d) \(\frac{\pi}{2}\)

Solution:

(d) \(\frac{\pi}{2}\)

Given parametric eqns. of curve be,

x = e^{t} cos t ……………..(1)

y = e^{t} sin t …………..(2)

Diff. eqns. (1) and (2) w.r.t. t; we have

\(\frac{d x}{d t}\) = e^{t} (- sin t) + cos t . e^{t}

and \(\frac{d y}{d t}\) = e^{t} cos t + sin t e^{t}

∴ \(\frac{d y}{d x}\) = \(\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)

= \(\frac{e^t(\cos t+\sin t)}{e^t(\cos t-\sin t)}\)

= \(\frac{\cos t+\sin t}{\cos t-\sin t}\)

∴ slope of tangent to given curve at t = \(\frac{\pi}{4}\)

= \(\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{4}}\)

= \(\frac{\cos \frac{\pi}{4}+\sin \frac{\pi}{4}}{\cos \frac{\pi}{4}-\sin \frac{\pi}{4}}\)

= \(\frac{\sqrt{2}}{0}\) → ∞

Let the tangent to given curve at t = \(\frac{\pi}{4}\) makes an angle θ with x – axis.

Then slope of tangent to given curve at t = \(\frac{\pi}{4}\) = tan θ

∴ tan θ = ∞

⇒ θ = \(\frac{\pi}{2}\).

Question 19.

The equation to the normal to the curve sin x at (0, 0) is

(a) x = 0

(b) y =

(c) x + y = 0

(d) x – y = 0

Solution:

(c) x + y = 0

Given eqn. of curve be y = sin x …………..(1)

∴ \(\frac{d y}{d x}\) = cos x

∴ \(\left(\frac{d y}{d x}\right)_{(0,0)}\) = cos 0 = 1

Thus required equation of normal at (0, 0) to given curve be

y – 0 = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{(0,0)}}\) (x – 0)

⇒ y = – 1 (x)

⇒ x + y = 0.

Question 20.

The slope of the normal to the curve X2 + 3y +ý = 5 at the poInt (1, 1) ¡s

(a) – \(\frac{2}{5}\)

(b) \(\frac{5}{2}\)

(c) \(\frac{2}{5}\)

(d) – \(\frac{5}{2}\)

Solution:

(b) \(\frac{5}{2}\)

eqn. of given curve be

x^{2} + 3y + y^{2} = 5 …………(1)

duff. eqn. (1) both sides w.r.t. X; we have

2x + 3 \(\frac{d y}{d x}\) + 2y \(\frac{d y}{d x}\) = 0

⇒ (2y + 3) \(\frac{d y}{d x}\) = – 2x

⇒ \(\frac{d y}{d x}\) = \(\frac{-2 x}{2 y+3}\)

∴ slope of normal to given curve at (1, 1) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{(1,1)}}\)

= \(\frac{-1}{\frac{-2 \times 1}{2 \times 1+3}}\)

= \(\frac{5}{2}\).

Question 21.

The point on the curve 9y^{2} = x^{3}, where the normal to the curve makes equal intercepts with the axes is

(a) (4, ± 8/3)

(b) (- 4, 8/3)

(c) (- 4, – 8/3)

(d) (8/3, 4)

Solution:

(a) (4, ± 8/3)

Given eqn. of curve be

9y^{2} = x^{3} ………….(1)

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

∴ 9y_{1}^{2} = x_{1}^{3} ………….(2)

Diff. eqn. (1) both sides w.r.t. x; we get

18 y = \(\frac{d y}{d x}\) = 3x^{2}

Since normal to curve makes an equal intercept with axes.

∴ slope of normal at (x_{1}, y_{1}) = ± 1

∴ \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}}\) = ± 1

⇒ \(\frac{-6 y_1}{x_1^2}\) = ± 1

⇒ \(\frac{6 y_1}{x_1^2}\) = ± 1

when \(\frac{6 y_1}{x_1^2}\) = 1

⇒ 6y_{1} = x_{1}^{2} ………….(3)

∴ from (2) ; we have

\(\frac{9 x_1^4}{36}\) = x_{1}^{3}

⇒ x_{1}^{3} (x_{1} – 4) = 0

⇒ x_{1} = 0, 4

When x_{1} = 0

∴ from (3) ;

y_{1} = 0

When x_{1} = 4

∴ from (2) ;

y_{1} = ± 8/3

When \(\frac{6 y_1}{x_1^2}\) = – 1

⇒ 6y_{1} = – x_{1}^{2} …………(4)

∴ from (2) ;

x_{1} = 0, 4

When x_{1} = 4

∴ from (4) ;

y_{1} = 0

When x_{1} = 0

∴ from (4) ; y_{1} = 0

Thus required points are (0. 0), (4, ± 8/3).

Question 22.

The equation of the normal to the curve 3x^{2} – y^{2} = 8 which is parallel to x + 3y = 8 is

(a) x – 3y = 8

(b) x – 3y + 8 = 0

(c) x + 3y ± 8 = 0

(d) x + 3y = 0

Solution:

Given eqn. of curve be 3x^{2} – y^{2} = 8 ………..(1)

Let (x_{1}, y_{1}) be point on curve (1) where normal to curve (1) || to given line

x+3y – 8 = 0

Diff. (1) w.r.t. x; we get

6x – 2y \(\frac{d y}{d x}\) = 0

⇒ \(\frac{d y}{d x}\) = \(\frac{d3 x}{y}\)

∴ \(\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=\frac{3 x_1}{y_1}\)

Since the normal to curve (1) || to line x + 3y – 8 = 0

∴ slope of normal at (x_{1}, y_{1}) = slope of line

x + 3y – 8 = 0

⇒ \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}}=-\frac{1}{3}\)

⇒ \(-\frac{y_1}{3 x_1}=-\frac{1}{3}\)

⇒ x_{1} = y_{1} …………….(2)

also point (x_{1}, y_{1}) lies on eqn. (1)

∴ 3x_{1}^{2} – y_{1}^{2} = 8

⇒ 3x_{1}^{2} – x_{1}^{2} = 8

⇒ x_{1}^{2} = 4

⇒ x_{1} = ± 2

from (2) ;

y_{1} = ± 2

Thus point of contact be (± 2, ± 2).

∴ required eqn. of normal at (2, 2) to given curve be

y – 2 = – (x – 2)

3y – 6 = – x + 2

x + 3y -8 = 0

and eqn. of normal at (- 2, – 2) to given curve be

y – 2 = – \(\frac{1}{3}\) (x – 2)

⇒ 3y – 6 = – x + 2

⇒ x + 3y – 8 = 0

and eqn. of normal at (- 2, – 2) to given curve be

y + 2 = – \(\frac{1}{3}\) (x + 2)

⇒ x + 3y + 8 = 0

Question 23.

The angle between the tangents to the curve y = x^{2} – 5x + 6 at the point (2, 0) and (3, 0) is

(a) \(\frac{\pi}{2}\)

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

(c) \(\frac{\pi}{4}\)

(d) \(\frac{\pi}{6}\)

Solution:

(a) \(\frac{\pi}{2}\)

eqn. of given curve be y = x^{2} – 5x + 6

∴ \(\frac{d y}{d x}\) = 2x – 5

m_{1} = slope of tangent to given curve at (2, 0)

= \(\left(\frac{d y}{d x}\right)_{(2,0)}\)

= 2 × 2 – 5 = – 1

and m_{2} = slope of tangent to given curve at (3, 0)

= \(\left(\frac{d y}{d x}\right)_{(2,0)}\)

= 2 × 3 – 5 = 1

Here m_{1}m_{2} = (- 1 ) × 1 = – 1

Thus both tangents to given curve at points (2, 0) and (3, 0) cut orthogonally

i.e. the angle between them is \(\frac{\pi}{2}\).

Question 24.

If the curve ay + x^{2} = 7 and x^{3} = y cut orthogonally at (1, 1), then a is equal to

(a) 1

(b) – 6

(c) 6

(d) 0

Solution:

(c) 6

Given eqns. of curve be,

ay + x^{2} = 7 ……….(1)

and x^{3} = y ………..(2)

Diff. (1) and (2) both sides w.r.t. x, we get

a \(\frac{d y}{d x}\) + 2x = 0

∴ \(\frac{d y}{d x}\) = – \(\frac{2 x}{a}\)

∴ m_{1} = \(\left(\frac{d y}{d x}\right)_{(1,1)}\)

= \(-\frac{2}{a}\)

and 3x^{2} = \(\frac{d y}{d x}\)

∴ m_{2} = \(\left(\frac{d y}{d x}\right)_{(1,1)}\) = 3

Since both cuves cut orthogonally.

∴ m_{1}m_{2} = – 1

⇒ – \(\frac{2}{3}\) × a = – 1

⇒ a = 6

Question 25.

The two curves x^{3} – 3xy^{2} + 2 = 0 and 3x^{2}y – y^{3} – 2 = 0 intersect at an angle of:

(a) \(\frac{\pi}{4}\)

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

(c) \(\frac{\pi}{2}\)

(d) \(\frac{\pi}{6}\)

Solution:

(c) \(\frac{\pi}{2}\)

Given eqns. of curves be,

x^{3} – 3xy^{2} = – 2 ……………(1)

and 3x^{2}y – y^{3} = 2 …………..(2)

From (1) and (2) ; we get

x^{3} – 3xy^{2} = – 3x^{2}y + y^{3}

⇒ x^{3} – 3xy^{2} + 3x^{2}y – y^{3} = 0

⇒ (x – y)^{3} = 0

⇒ x = y

putting x =y in eqn. (1); we get

x^{3} – 3x^{3} = – 2

⇒ x^{3} = 1

⇒ x = + 1

∴ from (3) ;

y = + 1

Hence the points of intersection of two curves be (1, 1).

Diff. (1) w.r.t. x; we get

3x^{2} – 3 [2xy \(\frac{d y}{d x}\) + y^{2}] = 0

⇒ \(\frac{d y}{d x}=\frac{x^2-y^2}{2 x y}\) ……..(3)

Diff. (2) w.r.t. x; we get

3 [x^{2} \(\frac{d y}{d x}\) + y . 2x] – 3y^{2} \(\frac{d y}{d x}\) = 0

\(\frac{d y}{d x}=-\frac{2 x y}{x^2-y^2}\) ………….(4)

at (1, 1) :

using (3);

m_{1} = \(\left(\frac{d y}{d x}\right)_{(1,1)}\) = 0

m_{2} = \(\left(\frac{d y}{d x}\right)_{(1,1)}\) = ∞

If θ be the angle of intersection between two curves.

Then tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)

= \(\left|\frac{m_2\left(\frac{m_1}{m_2}-1\right)}{m_2\left(\frac{1}{m_2}+m_1\right)}\right|\) = ∞

⇒ θ = \(\frac{\pi}{2}\)

Hence both curves intersect orthogonally at (1, 1).

Question 26.

The interval on which the function f(x) = 2x^{3} + 9x^{2} + 12x – 1 is decreasing in:

(a) [- 1, ∞)

(b) [- 2, – 1]

(c) (- ∞, – 2]

(d) [- 1, 1]

Solution:

(b) [- 2, – 1]

Given f (x) = 2x^{3} + 9x^{2} + 12x + 20

∴ f’(x) = 6x^{2} + 18x + 12

= 6 (x^{2} + 3x + 2)

= 6 (x + 1) (x + 2)

Now f’(x) > 0

⇒ (x + 1) (x + 2) > 0

⇒ x > – 1 or x < – 2

∴ f(x) is strictly increasing in (- ∞, – 2) ∪ (- 1, ∞).

Now f’(x) < 0

⇒ (x + 1) (x + 2) < 0

⇒ – 2 < x < – 1

[If (x – a) (x – b) < 0 and a < b ⇒ a < x < b]

∴ f(x) is strictly decreasing in [- 2, – 1].

[If f’(x) < 0 ∀ x ∈ (a, b)Thenf(x) is decreasg in [a, b]]

Question 27.

The interval in which the function f(x) = 2x^{3} + 3x^{2} – 12x + 1 is strictly increasing is

(a) [- 2, 1]

(b) (- ∞, – 2] ∪ [1, ∞)

(e) (- ∞, 1]

(f)(- ∞, – 1] ∪ [2, ∞)

Solution:

(b) (- ∞, – 2] ∪ [1, ∞)

Given f(x) = 2x^{3} + 3x^{2} – 12x + 1

∴ f’(x) = 6x^{2} + 6x – 12

= 6 (x^{2} + x – 2)

= 6 (x – 1) (x + 2)

Now f’(x) > 0 iff 6 (x – 1) (x + 2) > 0

iff x > 1 or x< – 2 [if (x – a) (x – b) > 0 and a > b ⇒ x > a or x < b]

iff x ∈ (- ∞, – 2) ∪ (1, ∞)

Thus f(x) is strictly increasing in (- ∞, – 2] ∪ [1, x).

Question 28.

The function f (x) = x^{2} e^{– x} is monotonic increasing when

(a) x ∈ R – [0, 2]

(b) 0 < x < 2

(c) 2 < x < ∞

(d) x < 0

Solution:

(b) 0 < x < 2

Given f(x) = x^{2} e^{– x}

∴ f'(x) = x^{2} e^{– x} (- 1) + e^{– x} 2x

= e^{– x} [- x^{2} + 2x]

= e^{– x} (2 – x) x

For f(x) is to be monotomically decreasing we must have

f’(x) ≥ 0

⇒ e^{– x} (2 – x) x ≥ 0 [where e^{– x} ≥ 0]

⇒ – (x – 2)x ≤ 0

⇒ (x – 2) x ≤ 0

⇒ 0 < x < 2

⇒ x ∈ (0, 2)

[if (x – a) (x – b)< 0 and a < b then a < x < b].

Question 29.

The function f(x) = tan x – x

(a) always increases

(b) always decreases

(c) never increases

(d) sometimes increases and decreases.

Solution:

(a) always increases

Given f(x) = tan x – x

∴ f'(x) = sec^{2} x – 1 ≥ 0

[sec^{2} x ≥ 1

⇒ sec^{2} x – 1 ≥ 0 ∀ x ∈ R]

Hence the function f(x) is always increases.

Question 30.

The function f(x) = x^{4} – 4x is strictly

(a) decreasing in [1, ∞)

(b) increasing in [1, ∞)

(c) Increasing in (- ∞, 1]

(d) increasing in [- 1, 1]

Solution:

Given f(x) = x^{4} – 4x

f’(x) = 4x^{3} – 4

= 4 (x^{3} – 1)

= 4 (x – 1) (x^{2} + x + 1)

= 4 (x – 1) (x^{2} + x + \(\frac{1}{4}\) + \(\frac{3}{4}\))

= 4 (x – 1) [(x + \(\frac{1}{2}\))^{2} + \(\frac{3}{4}\)]

Now f’ (x) > 0

iff 4 (x – 1) [(x + \(\frac{1}{2}\))^{2} + \(\frac{3}{4}\)] > 0

iff x – 1 > 0

[∵ (x + \(\frac{1}{2}\)) + \(\frac{3}{4}\) > 0 ∀ x ∈ R]

iff x > 1

iff x ∈ (1, ∞)

Thus f(x) is strictly increasing in [1, ∞)

Question 31.

The function f (x) = x^{2} – 2x is strictly decreasing in the interval

(a) (- ∞, 1]

(b) [1, ∞)

(c) [- 1, ∞)

(d) none of these

Solution:

(a) (- ∞, 1]

Given f(x) = x^{2} – 2x

∴ f’(x) = 2x – 2

Now f’(x) < 0 iff 2x – 2 < 0

iff x < 1 iff x ∈ (- ∞, 1)

Thus,f(x) is strictly decreasing in (- ∞, 1].

Question 32.

y = x (x – 3)^{2} decreases for the values of x given by:

(a) 1 < x < 3

(b) x < 0 (c) x > 0

(d) 0 < x < \(\frac{3}{2}\)

Solution:

(a) 1 < x < 3

Given y = x (x – 3)^{2} = f(x)

∴ \(\frac{d y}{d x}\) = (x – 3)^{2} + 2x (x – 3)

= (x – 3) (x – 3 + 2x)

= 3 (x – 3) (x – 1)

since f(x) is decreases if f’ (x) < 0

i.e. \(\frac{d y}{d x}\) < 0

⇒ (x – 3) (x – 1) < 0

⇒ 1 < x < 3

[if a < b and (x – a) (x- b) < 0 ⇒ a < x < b]

Question 33.

Which of the following functions is decreasing on (o, \(\frac{\pi}{2}\))

(a) sin 2x

(b) tan x

(c) cos x

(d) cos 3x

Solution:

When 0 < x < \(\frac{\pi}{2}\)

⇒ 0 < 2x < π For option (A) ; f (x) = sin 2x ∴ f’(x) = 2 cos 2x may be > 0 or < 0.

Hence f(x) may be increasing or decreasing.

For option (B) ;

f (x) = tan x

∴ f'(x) = sec^{2} x > 0 ∀ x ∈ (0)

f(x) is increasing on (o, \(\frac{\pi}{2}\)).

For option (C) ;

f (x) = cos x

∴ f’(x) = – sin x < 0 ∀ x ∈ (o, \(\frac{\pi}{2}\))

∴ f(x) is decreasing on (o, \(\frac{\pi}{2}\)).

For option (D) ;

f (x) = cos 3x

∴ f ‘(x) = 3 sin 3x

[∵ 0 < x < \(\frac{\pi}{2}\) ⇒ 0 < 3x < \(\frac{3 \pi}{2}\)]

it may be positive or negative.

Thus f(x) may be increasing or decreasing.

Question 34.

The function f (x) = x^{x}, x > 0, is increasing on the interval

(a) (0, e]

(b) (0, \(\frac{1}{e}\))

(c) [\(\frac{1}{e}\), ∞)

(d) none of these

Solution:

(c) [\(\frac{1}{e}\), ∞)

Given f(x) = x^{x}

∴ log f(x) = Iog x^{x} = x log x;

Diff, both sides w.r.t. x, we have

⇒ \(\frac{1}{f(x)}\) f’(x) = x × \(\frac{1}{x}\) + log x × 1

⇒ f’ (x) = x^{x} (1 + log x)

Now f’(x) ≥ 0 iff x^{x} (1 + log x) ≥ 0

iff 1 + log x ≥ 0 [∵ x^{x} > 0]

iff log x ≥ 1 iff x ≥ \(\frac{1}{e}\)

Thus f(x) is increasing in [\(\frac{1}{e}\), ∞).

Question 35.

The value of p so that the function f (x) = sin x – cos x – px + q decreases for all real values of x is given by

(a) p ≥ √2

(b) p < √2

(c) p ≥ 1

(d) p < 1

Solution:

(a) p ≥ √2

f’(x) = cos x + sin x – p

f(x) is decreasing iff f’cos ≤ 0

iff cosx + sinx – p ≤ 0

iff cos x + sin x ≤ p

iff √2 sin (x + \(\frac{\pi}{4}\)) ≤ p

Since |sin x| ≤ 1

∴ p ≥ √2

Question 36.

If x is real, x^{2} – 8x + 17 is

(a) – 1

(b) 0

(c) 1

(d) 2

Solution:

(c) 1

Let f(x) = x^{2} – 8x + 17

= x^{2} – 8x + 16 + 1

= (x – 4)^{2} + 1 ≥ 1 ∀ x ∈ R

Hence 1 be the minimum value of f (x) and is attains at x = 4.

Question 37.

The smallest value of the polynomial x^{3} – 18x^{2} + 96x in [0, 9] is

(a) 126

(b) 135

(c) 160

(d) 0

Solution:

(d) 0

Given f(x) = x^{3} – 18x^{2} + 96x

∴ f’(x) = 3x^{2} – 36x+96

= 3 (x^{2} – 12x + 32)

= 3 (x – 8) (x – 4)

For critical points, f’ (x) = 0

⇒ 3 (x – 8) (x – 4) = 0

⇒ x = 4, 8

Now we compute f (x) at critical points x = 4, 8 and at end points x = 0, 9.

Here, f(0) = 0;

f(4) = 4^{3} – 18 x 4^{2} + 96 x 4

= 64 – 288 + 384

= 160

f(8) = 8^{3} – 18 × 64 + 96 × 8 = 128

f(9) = 9^{3} – 18 × 81 + 96 × 9 = 135

∴ smallest value of f(x) = f(0) = 0.

Question 38.

The minimum value of (x^{2} + \(\frac{250}{x}\)) is

(a) 75

(b) 50

(c) 25

(d) 55

Solution:

(a) 75

Let f(x) = x^{2} + \(\frac{250}{x}\)

∴ f’(x) = 2x – \(\frac{250}{x^{2}}\)

For maxima/minima, f’ (x) = 0

⇒ 2x – \(\frac{250}{x^{2}}\) = 0

⇒ x^{3} = 125 = 5^{3}

[∵ x^{3} – 5^{3}

⇒ (x – 5) (x^{2} + 5x + 25) = 0

⇒ x = 5 and x^{2} + 5x + 25 = 0 does not gives any real values of x]

⇒ f”(x) = 2 + \(\frac{500}{x^{3}}\)

∴ f”(5) = 2 + \(\frac{500}{125}\)

= 6 > 0

Hence x = 5 is a point of minima and min.

value of f(x) = f(5) = 25 + \(\frac{250}{5}\) = 75

Question 39.

The function f(x) = \(\frac{x}{2}+\frac{2}{x}\) has a local minimum at

(a) x = 1

(b) x = 2

(c) x = – 2

(d) x = – 1

Solution:

(b) x = 2

Given f(x) = \(\frac{x}{2}+\frac{2}{x}\)

∴ f'(x) = \(\frac{1}{2}-\frac{2}{x^2}\)

For local maxima/minima, f'(x) = 0

⇒ \(\frac{1}{2}-\frac{2}{x^2}\) = 0

⇒ \(\frac{1}{2}=\frac{2}{x^2}\)

⇒ x^{2} = 4

⇒ x = ± 2

at x = 2;

f”(2) = \(\frac{4}{2^3}=\frac{1}{2}\) >0

Thus f(x) is minimise when x = 2.

Question 40.

Let the function f: R → R be defined by f(x) = 2 x+ cos x, then f(x)

(a) has a minimum at x = it

(b) has a maximum at x = 0

(c) is a decreasing function

(d) is an increasing function [NCERT Exemplar]

Solution:

(d) is an increasing function

Given f(x) = 2x + cos x

∴ f’(x) = 2 – sinx

Now – 1 ≤ sin x ≤ 1

⇒ 1 ≥ – sin x ≥ – 1

⇒ 3 ≥ 2 – sin x ≥ 1

⇒ 1 ≤ f’(x) ≤ 3

∴ f'(x) > 0

Thus f(x) is an increasing function.

Question 41.

At x = \(\frac{5 \pi}{6}\), f(x) = 2 sin 3x + 3 cos 3x is

(a) 0

(b) maximum

(c) minimum

(d) none of these

Solution:

(d) none of these

f(x) = 2 sin 3x + 3 cos 3x

f’(x) = 6 cos 3x – 9 sin 3x

f” (x) = – 18 sin 3x – 27 cos 3x

∴ f’(5 π / 6) = 6 cos \(\frac{5 \pi}{2}\) – 9 sin 5π /2

= 6 cos (2π+ π/2) – 9 sin (2π + π/2)

= 6 × 0 – 9 × 1

= – 9 ≠ 0

f(5 π / 6) = 2 sin 5 π / 2 + 3 cos 5 π / 2

= 2 × 1 + 3 × 0

=2 ≠ 0

Thus x = \(\frac{5 \pi}{2}\) is not an extreme point.

Question 42.

The f(x) = x^{x} has a stationary point at

(a) x = e

(b) x = \(\frac{1}{e}\)

(c) x = 1

(d) x = √e

Solution:

(b) x = \(\frac{1}{e}\)

Given f (x) = x^{x};

Taking logarithm on both sides ; we have

∴ log f(x) = x log x

Differentiate both sides wr.t. x; we have

\(\frac{1}{f(x)}\) f’(x) = x . \(\frac{1}{x}\) + log x . 1

⇒ f ‘(x) = f(x) [1 + log x]

= x^{x} (1 + log x)

For stationary point, we put f ‘(x) = 0

⇒ x^{x} (1 + log x) = 0

⇒ 1 + log x = 0 [∵ x^{x} > o]

⇒ log x = – 1

⇒ x = e^{-1}

= \(\frac{1}{e}\)

When x slightly < \(\frac{1}{e}\)

⇒ log x < – 1

⇒ 1 + log x < 0

⇒ f'(x) < 0 When x slightly > \(\frac{1}{e}\)

⇒ log x > – 1

⇒ 1 + log x > 0

⇒ f’(x) > 0

Thus f'(x) changes its sign from -ve to +ve.

∴ x = \(\frac{1}{e}\) be a point of minima.

Question 43.

The maximum value of \(\frac{\log x}{x}\) is

(a) e

(b) 2e

(c) \(\frac{1}{e}\)

(d) \(\frac{2}{e}\)

Solution:

(c) \(\frac{1}{e}\)

Given y = \(\frac{\log x}{x}\)

Thus y is maximise at x = e

∴ Maximum value of f(x) = f(e) = \(\frac{\log e}{e}=\frac{1}{e}\).

Question 44.

The minimum value of \(\frac{x}{\log _e x}\) is

(a) e

(b) 1/e

(c) 1

(d) none of these

Solution:

(a) e

Given f(x) = \(\frac{x}{\log _e x}\) is defined for x > 0

∴ f'(x) = \(\frac{\log x \cdot 1-x \times \frac{1}{x}}{(\log x)^2}\)

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

For maxima / minima,

f’ (x) = 0

⇒ log x = 1

⇒ log x = log e

⇒ x = e

when x slightly < e

⇒ log x< 1

⇒ log x – 1 < 0

⇒ f'(x) < 0 when x slightly > e

⇒ log x > log e

⇒ log x – 1 > 0

⇒ f'(x) > 0

Thusf’ (x) changes its sign from negative to positive when x increases through e.

Thus x = e is a point of local minima and local min. value of f(x) = f(e) = \(\frac{e}{log e}\) = e.

Question 45.

Maximum slope of the curve y = – x^{3} + 3x^{2} + 9x – 27 is:

(a) 0

(b) 12

(c) 16

(d) 32

Solution:

(b) 12

Given equation of curve be

y = – x^{3} + 3x^{2} + 9x – 27

∴ slope of curve m = – 3x^{2} + 6x + 9

we want to maximise m.

∴ \(\frac{d m}{d x}\) = – 6x + 6

for maximalminíma, we have \(\frac{d m}{d x}\) = 0

⇒ – 6x + 6 = 0

⇒ x = 1

and \(\frac{d^2 m}{d x^2}\) = – 6

Thus at x = 1;

\(\frac{d^2 m}{d x^2}\) = – 6 < 0

Hence m is maximise when x = 1.

∴ Max value of m = – 3 (1)^{2} + 6 + 9 = 12

Question 46.

If f(x) = Zx^{3} – 21x^{2} + 36x – 30, then

(a) f(x) has minimum at x = 1

(b) f(x) has maximum at x = 6

(c) f(x) has maximum at x = 1

(d) f(x) has no maxima or minima

Solution:

(c) f(x) has maximum at x = 1

Given f(x) = 2x^{3} – 21x^{2} + 36x – 30

f’(x) = 6x^{2} – 42x + 36

= 6 (x^{2} – 7x +6)

= 6 (x – 1) (x – 6)

For maxima / minima, f’ (x) = 0

⇒ 6 (x -1) (x – 6)O = 0

⇒ x = 1, 6

∴ f”(x) = 6 (2x – 7)

Now f”(1) = 6 (2 – 7) = – 30 < 0

∴ f(x) is maximise when x = 1

and f”(6) = 6 (12 – 7) = 30 > 0

Thus f(x) is minimise at x = 6.

Question 47.

The least value of the function f(x) = ax + \(\frac{b}{x}\) (x > 0, a > 0, b > 0) is

(a) \(\sqrt{ab}\)

(b) 2 \(\sqrt{ab}\)

(c) ab

(d) 2ab

Solution:

(b) 2 \(\sqrt{ab}\)

Given f(x) = ax + \(\frac{b}{x}\)

∴ f'(x) = a – \(\frac{b}{x^{2}}\)

For critical points ; f'(x) = 0

⇒ a – \(\frac{b}{x^{2}}\) = 0

⇒ x^{2} = \(\frac{b}{a}\)

⇒ x = \(\sqrt{\frac{b}{a}}\) [∵ x > 0]

[Since a > 0, b > 0

∴ \(\sqrt{\frac{b}{a}}\) > 0

∴ x ≠ \(\sqrt{\frac{b}{a}}\)]

∴ least value of f(x) = f (\(\sqrt{\frac{b}{a}}\))

= a \(\sqrt{\frac{b}{a}}\) + \(\frac{b}{\sqrt{\frac{b}{a}}}\)

= \(\sqrt{ab}\) + \(\sqrt{ab}\)

= 2 \(\sqrt{ab}\)