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 m2 / sec
(b) 80 m2 / sec
(c) 80 π cm2 /sec
(d) 80 cm2 /sec
Solution:
(c) 80 π cm2 /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 cm2 / 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 cm2 / sec
(b) 10√3 cm2 /sec
(c) $$\frac{10}{3}$$ cm2 / sec
(d) √3 cm2 / sec
Solution:
(b) 10√3 cm2 /sec

Let x be the length of each side of equilateral triangle at any time t.
Then A = area of equilateral triangle = $$\frac{\sqrt{3}}{4}$$ x2
∴ $$\frac{d \mathrm{~A}}{d t}=\frac{\sqrt{3}}{4} \times 2 x \frac{d x}{d t}$$
= $$\frac{\sqrt{3}}{2} x \frac{d x}{d t}$$
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 cm2 / sec.

Question 3.
A spherical ice ball is melting at the rate of 100 π cm3 / 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 cm3 / min.
⇒ $$\frac{d}{d t}$$ ($$\frac{4}{3}$$ πr3) = – 100π
⇒ $$\frac{4}{3}$$ π × 3r2 $$\frac{d r}{d t}$$ = – 100π
⇒ 4π × 152 . $$\frac{d r}{d t}$$ = – 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 cm3 / sec
(b) 144 cm3/ sec
(c) 80 cm3/ sec
(d) 80 π cm3 / sec
Solution:
(d) 80 π cm3 / 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 [42 × (- 4) + 2 × 4 × 6 × 3]
= π [- 64 + 144]
= 80 π cm3 / sec

Question 5.
A point on the curve y2 = 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 y2 = 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
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 x2 + y2 = 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 = x1/5 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 = x1/5
∴ 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 – x2)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 – x2)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 – x2)-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 = e2x 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 = e2x …………(1)
∴ $$\frac{d y}{d x}$$ = 2e2x
Thus slope of tangent to curve (1) at (0, 1) = ($$\frac{d y}{d x}$$)(0, 1)
=2 e2 × 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 = e2x ………….(1)
Differentiate equation (1) both sides w.r.t. x; we get
$$\frac{d y}{d x}$$ = 2e2x
∴ slope of tangent to given curve (1) at (0, 1) = ($$\frac{d y}{d x}$$)(0, 1)
= 2 × e0 = 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 x2 = 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 x2 = 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 x2 + y2 = 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
x2 + y2 = 2 …………(1)
Diff. eqn. (1) w.r.t. x; we have
2x + 2y $$\frac{d y}{d x}$$ = 0
⇒ $$\frac{d y}{d x}$$ = – $$\frac{x}{y}$$
∴ m1 = slope of tangent to given curve at (1, 1) = ($$\frac{d y}{d x}$$)(0, 1) = – 1
m2 = slope of tangent to given curve at (- 1, 1) = ($$\frac{d y}{d x}$$)(- 1, 1) = 1
∴ m1 m2 = – 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 y2 = 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
y2= 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 – x2 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 (x1, y1) be any point on curve
y = 6x – x2 ……….(1)
∴ $$\frac{d y}{d x}$$ = 6 – 2x
Thus slope of tangent to given curve at
P(x1, y1) = $$\left(\frac{d y}{d x}\right)_{\mathrm{P}\left(x_1, y_1\right)}$$ = 6 – 2x1
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 – 2x1 = 2
⇒ 2x1 = 4
⇒ x1 = 2
Also P (x1, y1) lies on eqn. (1) ;
y1 = 6x1 – x1
= 6 × 2 + 22 = 8
Hence coordinates of any point on given curve be (2, 8).

Question 15.
The points at which the tangents to the curve y = x3 – 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 = x3 – 12x + 18 …………..(1)
Let the required points on curve (1) are (x1, y1)
Differentiate equation (1) w.r.t. x; we get
$$\frac{d y}{d x}$$ = 3x2 – 12
slope of tangent to given curve (1) at (x1, y1)
= $$\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}$$
= 3x12 – 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]
3x12 – 12 = 0
⇒ x1 = ± 2
Also, the point (x1, y1) lies on curve (1)
y1= x1 – 12x1 + 18 …………….(2)
when x1 = 2
∴ from (2) ; we have
y1 = 8 – 24 + 18 = 2
when x1 = 2
∴ from (2) ; we have
y1 = – 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 = 2x2 – 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 (x1, y1) be any point on given curve
y = 2x2 – x + 1
∴ y1 = 2x12 – x1 + 1 ………..(1)
∴ slope of tangent to given curve at P (x1, y1)
= $$\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}$$
= 4x1 – 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)
∴ 4x1 – 1 = 3
⇒ x1 = 1
∴ from (1);
y1 = 2 – 1 + 1 = 2
Thus, required point on tangent to given curve be (1, 1).

Question 17.
If the tangent to the curve x = t2 – 1, y = t2 – 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 = t2 – 1
and y = t2 – 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 = et cos t, y = et 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 = et cos t ……………..(1)
y = et sin t …………..(2)
Diff. eqns. (1) and (2) w.r.t. t; we have
$$\frac{d x}{d t}$$ = et (- sin t) + cos t . et
and $$\frac{d y}{d t}$$ = et cos t + sin t et
∴ $$\frac{d y}{d x}$$ = $$\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$
= $$\frac{e^t(\cos t+\sin t)}{e^t(\cos t-\sin t)}$$
= $$\frac{\cos t+\sin t}{\cos t-\sin t}$$
∴ 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
x2 + 3y + y2 = 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 9y2 = x3, 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
9y2 = x3 ………….(1)
Let the required point on curve (1) be (x1, y1)
∴ 9y12 = x13 ………….(2)
Diff. eqn. (1) both sides w.r.t. x; we get
18 y = $$\frac{d y}{d x}$$ = 3x2
Since normal to curve makes an equal intercept with axes.
∴ slope of normal at (x1, y1) = ± 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
⇒ 6y1 = x12 ………….(3)
∴ from (2) ; we have
$$\frac{9 x_1^4}{36}$$ = x13
⇒ x13 (x1 – 4) = 0
⇒ x1 = 0, 4
When x1 = 0
∴ from (3) ;
y1 = 0
When x1 = 4
∴ from (2) ;
y1 = ± 8/3
When $$\frac{6 y_1}{x_1^2}$$ = – 1
⇒ 6y1 = – x12 …………(4)
∴ from (2) ;
x1 = 0, 4
When x1 = 4
∴ from (4) ;
y1 = 0
When x1 = 0
∴ from (4) ; y1 = 0
Thus required points are (0. 0), (4, ± 8/3).

Question 22.
The equation of the normal to the curve 3x2 – y2 = 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 3x2 – y2 = 8 ………..(1)
Let (x1, y1) 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 (x1, y1) = slope of line
x + 3y – 8 = 0
⇒ $$\frac{-1}{\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}}=-\frac{1}{3}$$
⇒ $$-\frac{y_1}{3 x_1}=-\frac{1}{3}$$
⇒ x1 = y1 …………….(2)
also point (x1, y1) lies on eqn. (1)
∴ 3x12 – y12 = 8
⇒ 3x12 – x12 = 8
⇒ x12 = 4
⇒ x1 = ± 2
from (2) ;
y1 = ± 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 = x2 – 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 = x2 – 5x + 6
∴ $$\frac{d y}{d x}$$ = 2x – 5
m1 = slope of tangent to given curve at (2, 0)
= $$\left(\frac{d y}{d x}\right)_{(2,0)}$$
= 2 × 2 – 5 = – 1
and m2 = slope of tangent to given curve at (3, 0)
= $$\left(\frac{d y}{d x}\right)_{(2,0)}$$
= 2 × 3 – 5 = 1
Here m1m2 = (- 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 + x2 = 7 and x3 = 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 + x2 = 7 ……….(1)
and x3 = y ………..(2)
Diff. (1) and (2) both sides w.r.t. x, we get
a $$\frac{d y}{d x}$$ + 2x = 0
∴ $$\frac{d y}{d x}$$ = – $$\frac{2 x}{a}$$
∴ m1 = $$\left(\frac{d y}{d x}\right)_{(1,1)}$$
= $$-\frac{2}{a}$$
and 3x2 = $$\frac{d y}{d x}$$
∴ m2 = $$\left(\frac{d y}{d x}\right)_{(1,1)}$$ = 3
Since both cuves cut orthogonally.
∴ m1m2 = – 1
⇒ – $$\frac{2}{3}$$ × a = – 1
⇒ a = 6

Question 25.
The two curves x3 – 3xy2 + 2 = 0 and 3x2y – y3 – 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,
x3 – 3xy2 = – 2 ……………(1)
and 3x2y – y3 = 2 …………..(2)
From (1) and (2) ; we get
x3 – 3xy2 = – 3x2y + y3
⇒ x3 – 3xy2 + 3x2y – y3 = 0
⇒ (x – y)3 = 0
⇒ x = y
putting x =y in eqn. (1); we get
x3 – 3x3 = – 2
⇒ x3 = 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
3x2 – 3 [2xy $$\frac{d y}{d x}$$ + y2] = 0
⇒ $$\frac{d y}{d x}=\frac{x^2-y^2}{2 x y}$$ ……..(3)
Diff. (2) w.r.t. x; we get
3 [x2 $$\frac{d y}{d x}$$ + y . 2x] – 3y2 $$\frac{d y}{d x}$$ = 0
$$\frac{d y}{d x}=-\frac{2 x y}{x^2-y^2}$$ ………….(4)

at (1, 1) :
using (3);
m1 = $$\left(\frac{d y}{d x}\right)_{(1,1)}$$ = 0
m2 = $$\left(\frac{d y}{d x}\right)_{(1,1)}$$ = ∞
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) = 2x3 + 9x2 + 12x – 1 is decreasing in:
(a) [- 1, ∞)
(b) [- 2, – 1]
(c) (- ∞, – 2]
(d) [- 1, 1]
Solution:
(b) [- 2, – 1]

Given f (x) = 2x3 + 9x2 + 12x + 20
∴ f’(x) = 6x2 + 18x + 12
= 6 (x2 + 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) = 2x3 + 3x2 – 12x + 1 is strictly increasing is
(a) [- 2, 1]
(b) (- ∞, – 2] ∪ [1, ∞)
(e) (- ∞, 1]
(f)(- ∞, – 1] ∪ [2, ∞)
Solution:
(b) (- ∞, – 2] ∪ [1, ∞)

Given f(x) = 2x3 + 3x2 – 12x + 1
∴ f’(x) = 6x2 + 6x – 12
= 6 (x2 + 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) = x2 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) = x2 e– x
∴ f'(x) = x2 e– x (- 1) + e– x 2x
= e– x [- x2 + 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) = sec2 x – 1 ≥ 0
[sec2 x ≥ 1
⇒ sec2 x – 1 ≥ 0 ∀ x ∈ R]
Hence the function f(x) is always increases.

Question 30.
The function f(x) = x4 – 4x is strictly
(a) decreasing in [1, ∞)
(b) increasing in [1, ∞)
(c) Increasing in (- ∞, 1]
(d) increasing in [- 1, 1]
Solution:

Given f(x) = x4 – 4x
f’(x) = 4x3 – 4
= 4 (x3 – 1)
= 4 (x – 1) (x2 + x + 1)
= 4 (x – 1) (x2 + 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) = x2 – 2x is strictly decreasing in the interval
(a) (- ∞, 1]
(b) [1, ∞)
(c) [- 1, ∞)
(d) none of these
Solution:
(a) (- ∞, 1]

Given f(x) = x2 – 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) = sec2 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) = xx, 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) = xx
∴ log f(x) = Iog xx = 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) = xx (1 + log x)
Now f’(x) ≥ 0 iff xx (1 + log x) ≥ 0
iff 1 + log x ≥ 0 [∵ xx > 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, x2 – 8x + 17 is
(a) – 1
(b) 0
(c) 1
(d) 2
Solution:
(c) 1

Let f(x) = x2 – 8x + 17
= x2 – 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 x3 – 18x2 + 96x in [0, 9] is
(a) 126
(b) 135
(c) 160
(d) 0
Solution:
(d) 0

Given f(x) = x3 – 18x2 + 96x
∴ f’(x) = 3x2 – 36x+96
= 3 (x2 – 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) = 43 – 18 x 42 + 96 x 4
= 64 – 288 + 384
= 160
f(8) = 83 – 18 × 64 + 96 × 8 = 128
f(9) = 93 – 18 × 81 + 96 × 9 = 135
∴ smallest value of f(x) = f(0) = 0.

Question 38.
The minimum value of (x2 + $$\frac{250}{x}$$) is
(a) 75
(b) 50
(c) 25
(d) 55
Solution:
(a) 75

Let f(x) = x2 + $$\frac{250}{x}$$
∴ f’(x) = 2x – $$\frac{250}{x^{2}}$$
For maxima/minima, f’ (x) = 0
⇒ 2x – $$\frac{250}{x^{2}}$$ = 0
⇒ x3 = 125 = 53
[∵ x3 – 53
⇒ (x – 5) (x2 + 5x + 25) = 0
⇒ x = 5 and x2 + 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
⇒ $$\frac{1}{2}=\frac{2}{x^2}$$
⇒ x2 = 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) = xx 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) = xx;
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]
= xx (1 + log x)
For stationary point, we put f ‘(x) = 0
⇒ xx (1 + log x) = 0
⇒ 1 + log x = 0 [∵ xx > 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 = – x3 + 3x2 + 9x – 27 is:
(a) 0
(b) 12
(c) 16
(d) 32
Solution:
(b) 12

Given equation of curve be
y = – x3 + 3x2 + 9x – 27
∴ slope of curve m = – 3x2 + 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) = Zx3 – 21x2 + 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) = 2x3 – 21x2 + 36x – 30
f’(x) = 6x2 – 42x + 36
= 6 (x2 – 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
⇒ x2 = $$\frac{b}{a}$$
⇒ 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 $$\sqrt{ab}$$