ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives MCQs

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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
(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 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 = 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 – 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 = 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 × 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 = 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
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) = 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, 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) = 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 = – 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 \(\sqrt{ab}\)