PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 1.
Using differentials, find the approximate value of each of the following:
Solution.
(a) \(\left(\frac{17}{81}\right)^{\frac{1}{4}}\)
Solution.
Consider y = \(x^{\frac{1}{4}}\).

Let x = \(\frac{16}{81}\) and ∆x = \(\frac{1}{81}\).

Then, ∆y = (x + ∆x)\(\frac{1}{4}\) – x\(\frac{1}{4}\)

= \(\left(\frac{17}{81}\right)^{\frac{1}{4}}-\left(\frac{16}{81}\right)^{\frac{1}{4}}\)

= \(\left(\frac{17}{81}\right)^{\frac{1}{4}}-\frac{2}{3}\)

∴ \(\left(\frac{17}{81}\right)^{\frac{1}{4}}\) = \(\frac{2}{3}\) + ∆y

Now, dy is approximately equal to ∆y and is given by

dy = \(\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x)\)

= \(\frac{1}{4\left(\frac{16}{81}\right)^{\frac{3}{4}}}\left(\frac{1}{81}\right)=\frac{27}{4 \times 8} \times \frac{1}{81}=\frac{1}{32 \times 3}=\frac{1}{96}\) = 0.010

Hence, the approximate value of \(\left(\frac{17}{81}\right)^{\frac{1}{4}}\) is \(\frac{2}{3}\) + 0.010
= 0.667 + 0.010 = 0.677.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

(b) Consider y = \(x^{-\frac{1}{5}}\).
Let x = 32 and ∆x = 1.
Then, ∆y = (x + ∆x)\(-\frac{1}{5}\) – x\(-\frac{1}{5}\)

= (33)\(-\frac{1}{5}\) – (32)\(-\frac{1}{5}\)

= (33)\(-\frac{1}{5}\) – \(\frac{1}{2}\)
∴ (33)\(-\frac{1}{5}\) = \(\frac{1}{2}\) + ∆y

Now, dy is approximately equal to ∆y and is given by
dy = \(\left(\frac{d y}{d x}\right) \Delta x=\frac{-1}{5(x)^{\frac{6}{5}}}(\Delta x)\) (As y = x\(-\frac{1}{5}\))

= \(\frac{-1}{5(2)^{6}}(1)=-\frac{1}{320}\) = – 0.003.

Hence, the approximate value of (33)\(-\frac{1}{5}\)
= \(\frac{1}{2}\) + (- 0.003) = 0.5 – 0.003 = 0.497.

Question 2.
Show that the function given by f(x) = \(\frac{\log x}{x}\) maximum at x = e.
Solution.
The given function is f(x) = \(\frac{\log x}{x}\)

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise 1

Therefore, by second derivative test, f is the maximum at x = e.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 3.
The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast s the area decreasing when two equal sides are equal to the base?
Solution.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise 2

Let ∆ ABC be isosceles, where BC is the base of fixed length b.
Let the length of the two equal sides of ∆ABC be a.
Draw AD ⊥ BC

Now, in ∆ADC, by applying the Pythagoras theorem, we have

AD = \(\sqrt{a^{2}-\frac{b^{2}}{4}}\)

:. Area of triangle (A) = \(\frac{1}{2} b \sqrt{a^{2}-\frac{b^{2}}{4}}\)

The rate of change of the area with respect to time (t) is given by

\(\frac{d A}{d t}=\frac{1}{2} b \cdot \frac{2 a}{2 \sqrt{a^{2}-\frac{b^{2}}{4}}} \cdot \frac{d a}{d t}=\frac{a b}{\sqrt{4 a^{2}-b^{2}}} \cdot \frac{d a}{d t}\)

It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second.

∴ \(\frac{d a}{d t}\) = – 3 cm/s (Negative sign shows for decreasing)

∴\(\frac{d A}{d t}=\frac{-3 b}{\sqrt{4 a^{2}-b^{2}}} \mathrm{~cm}^{2} / \mathrm{s}\) cm2 / s

When a = b, we have \(\frac{d A}{d t}=\frac{-3 b^{2}}{\sqrt{4 b^{2}-b^{2}}}=\frac{-3 b^{2}}{\sqrt{3 b^{2}}}\) = √3b cm2 / s
Hence, if the two equal sides are equal to the base, then the area of the triangle ¡s decreasing at the rate of √3 db cm2/s.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 4.
Find the equation of the normal to curve y2 = 4x at the point (1, 2).
Solution.
The equation of the given curve is y2 = 4x.
On differentiating w.r.t. x, we get
2y \(\frac{d y}{d x}\) = 4

⇒ \(\frac{d y}{d x}\) = \(\frac{4}{2 y}=\frac{2}{y}\)

Slope of the tangent at (1, 2) is \(\left(\frac{d y}{d x}\right)_{(1,2)}=\frac{2}{2}\) = 1
and slope of the normal at the point (1, 2) is = \(\frac{-1}{1}\) = – 1.
∴ Equation of the normal at (1, 2) is
y – 2 = – 1(x – 1)
⇒ y – 2 = – x + 1
⇒ x + y – 3 = 0.

Question 5.
Show that the normal at any point 0 to the curve x = a cos0 + a0 sin0, y = a sin0 – a0 cos0 is at a constant distance from the origin.
Solution.
The given curve is x = a cos θ + a θ sin θ, y = a sin θ – a θ cos θ
On differentiating w.r.t. θ, we get dx
\(\frac{d x}{d \theta}\) = – a sin θ + a sin θ + a θ cos θ
= a θ cos θ

and \(\frac{d y}{d \theta}\) = a cos θ – a cos θ + a θ sin θ
= a θ sin θ

Slope of the tangent at θ,

\(\frac{d y}{d x}=\frac{d y}{d \theta} \cdot \frac{d \theta}{d x}=\frac{a \theta \sin \theta}{a \theta \cos \theta}\) = tan θ

∴ Slope of the normal at θ is \(\frac{1}{\tan \theta}\) = – cot θ.

The equation of the normal at a given point (x, y) is given by
y – a sin θ + a θ cos θ = – \(\frac{1}{\tan \theta}\) (x – a cos θ – a θ sin θ)

⇒ y sin θ – a sin2 θ + a θ sin θ cos θ = – x cos θ + a cos2 θ + a θ sin θ cos θ
⇒ x cos θ + y sin θ – a (sin2 θ + cos2 θ) = 0
⇒ x cos θ + y sin θ – a = 0
Now, the perpendicular distance of the normal from the origin is \(\frac{|-a|}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}=\frac{|-a|}{\sqrt{1}}=|-a|\), which is independent of θ.
Hence, the perpendicular distance of the normal from the origin is constant.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 6.
Find the intervals in which the function f given by f(x) = \(\frac{4 \sin x-2 x-x \cos x}{2+\cos x}\) is
(i) strictly increasing
(ii) strictly decreasing
Solution.
Given, f(x) = \(\frac{4 \sin x-2 x-x \cos x}{2+\cos x}\)

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise 3

Now, f'(x) = 0
⇒ cos x = 0 or cos x = 4
But, cos x ≠ 4
∴ cos x = 0
⇒ x = \(\frac{\pi}{2}, \frac{3 \pi}{2}\)
Now, x = \(\frac{\pi}{2}\) and x = \(\frac{3 \pi}{2}\) divides (0, 2π) into three disjoint intervals i.e.,
(0, \(\frac{\pi}{2}\)), (\(\frac{\pi}{2}, \frac{3 \pi}{2}\)), and (\(\frac{3 \pi}{2}\), 2π)

In intervals (0, \(\frac{\pi}{2}\)) and (\(\frac{3 \pi}{2}\), 2π), f'(x) > 0.

Thus, f(x) is increasing for 0 < x < \(\frac{x}{2}\) and \(\frac{3 \pi}{2}\) < x < 2π.
In the interval \(\frac{\pi}{2}, \frac{3 \pi}{2}\), f'(x) < 0.
Thus, f(x) is decreasing for \(\frac{\pi}{2}\) < x < \(\frac{3 \pi}{2}\).

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 7.
Find the intervals in which the function f given by f(x) = x3 + \(\frac{1}{x^{3}}\), x ≠ 0 is
(i) increasing
(ii) decreasing
Solution.
f(x) = x3 + \(\frac{1}{x^{3}}\)

∴ f'(x) = 3x2 – \(\frac{3}{x^{4}}=\frac{3 x^{6}-3}{x^{4}}\)

Then, f’(x) = 0 = 3x6 – 3 = 0
⇒ x6 = 1
⇒ x = ± 1
Now, the points x = 1 and x = – 1 divide the real line into three disornt intervals i.e., (- ∞, – 1), (- 1, 1), and (1, ∞).
In intervals (- ∞, – 1) and (1, ∞) i.e., when x < – 1 and x > 1, f’(x) > 0.
Thus, when x < – 1 and x > 1, f is increasing.
In interval (- 1, 1) i.e., when – 1 < x < 1, f’(x) < 0.
Thus, when – 1 < x < 1, f is decreasing.

Question 8.
Find the maximum area of an isosceles triangle inscribed in the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\) = 1 with Its vertex at one end of the major
Solution.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise 4

A = area of isosceles ∆ APP’
= \(\frac{1}{2}\) . PP’ . AM

= \(\frac{1}{2}\) ab (2b sin θ) (a – a cos θ)

= ab (sin θ – \(\frac{1}{2}\) sin 2θ)

Differentiating w.r.t. θ, we get
\(\frac{d A}{d \theta}\) = ab (cos θ – cos 2θ)

For maxima and minima,
\(\frac{d A}{d \theta}\) = 0

∴ ab(cos θ – cos 2θ) = 0
⇒ cos 2θ = cos θ
⇒ 2θ = 2π – θ
⇒ θ = \(\frac{2 \pi}{3}\)

Now, \(\frac{d^{2} A}{d \theta^{2}}\) = ab(- sin θ + 2 sin 2θ)

At θ = \(\frac{2 \pi}{3}\),

\(\frac{d^{2} A}{d \theta^{2}}\) = ab \(\left(-\sin \frac{2 \pi}{3}+2 \sin \frac{4 \pi}{3}\right)\)

= ab \(\left[-\left(\frac{\sqrt{3}}{2}\right)+2\left(-\frac{\sqrt{3}}{2}\right)\right]\)

= ab \(\left(-\frac{\sqrt{3}}{2}-\frac{2 \sqrt{3}}{2}\right)=\frac{-3 \sqrt{3}}{2}\) ab < 0

⇒ A is maximum, when θ = \(\frac{2 \pi}{3}\) = 120°

Maximum value of A = ab(sin 120° – \(\frac{1}{2}\) sin 240°)
= ab \(\left[\frac{\sqrt{3}}{2}-\frac{1}{2}\left(-\frac{\sqrt{3}}{2}\right)\right]=\frac{3 \sqrt{3}}{4} a b\)

Thus, rnaximum area o the isosceles triangle is \(\frac{3 \sqrt{3}}{4} a b\) ab sq. unit.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 9.
A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth in 2 m and volume is 8 m3. If building of tank costs ₹ 70 per sq m for the base and ₹ 45 per sq metre for sides. What is the cost of least expensive tank?
Solution.
Let l, b and h represent the length, breadth and height of the tank, respectively.
Then, we have height (ft) = 2 m
∴ Volume of the tank = 8 m3
⇒ Volume of the tank = l × b × h
⇒ 8 = l × b × 2
⇒ lb = 4
⇒ b = \(\frac{4}{l}\)
Now, area of the base = lb = 4
Area of the 4 walls (A) = 2h (l + b)
∴ A = 4 (l + \(\frac{4}{l}\))
⇒ \(\frac{d A}{d l}=4\left(1-\frac{4}{l^{2}}\right)\)
However, the length cannot be negative.
Therefore, we have l = 4
∴ b = \(\frac{4}{l}=\frac{4}{2}\) = 2

Now, \(\frac{d^{2} A}{d l^{2}}=\frac{32}{l^{3}}\)

when l = 2, then \(\frac{d^{2} A}{d l^{2}}=\frac{32}{8}\) = 4 > 0

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise 5

Thus, by second derivative test, the area is the minimum when l = 2.
We have l = b = h = 2
∴ Cost of building the base = ₹ 70 × (lb)
= ₹ 70 (4) = ₹ 280
Cost of building the walls = ₹ 2h (l + b) × 45
= ₹ 90 (2) (2 + 2)
= ₹ 8 (90) = ₹ 720
Required total cost = ₹ 280 + 720 = ₹ 1000
Hence, the total cost of the tank will be ₹ 1000.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 10.
The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.
Solution.
Let x be the radius of the circle and y be the side of the square
Circumference of the circle = 2πx
Perimeter of square = 4y
Sum of perimeters of circle and square = 2πx + 4y = k ……………(i)
Area of circle = πx2
Area of square = y2
Sum of areas of circle and square = πx2 + y2 …………..(ii)
From Eq. (i),
y = \(\frac{k-2 \pi x}{4}\) ……………..(iii)

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise 6

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 11.
A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.
Solution.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise 7

Let x and y be the length and breadth of the rectangular window.
Then, radius of the semicircular opening = \(\frac{x}{2}\)
It is given that the perimeter of the window is 10 m.
∴ x + 2y + \(\frac{\pi x}{2}\) = 10

⇒ (x + \(\frac{\pi}{2}\)) + 2y = 10

⇒ 2y = 10 – x (1 + \(\frac{\pi}{2}\))

⇒ y = 5 – x (\(\frac{1}{2}+\frac{\pi}{4}\))

∴ Area of the window (A) is given by

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise 8

Therefore, by second derivative test, the area is the maximum when length x = \(\frac{20}{\pi+4}\) m.

y = \(5-\frac{20}{\pi+4}\left(\frac{2+\pi}{4}\right)=5-\frac{5(2+\pi)}{\pi+4}=\frac{10}{\pi+4} \mathrm{~m}\)

Hence, the required dimensions of the window to admit maximum light is given by length = \(\frac{20}{\pi+4}\) m and breadth = \(\frac{10}{\pi+4}\) m.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 12.
A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the maximum length of the hypotenuse is \(\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\).
Solution.
Let ∆ ABC be right angled at B.
Let AB = x and BC = y.
Let P be a point on the hypotenuse of the triangle such that P is at a distance of a and b from the sides AB and BC respectively.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise 9

Let ∠C = θ.
We have, AC = \(\sqrt{x^{2}+y^{2}}\)
Now, PC = b cosec θ and, AP = a sec θ
∴ AC = AP + PC
⇒ AC = b cosec θ + a sec θ ……………. (i)

∴ \(\frac{d(A C)}{d \theta}\) = – cosec θ cot θ + a sec θ tan θ

∴ \(\frac{d(A C)}{d \theta}\) = 0

⇒ a sec θ tan θ = cosec θ cot θ

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise 10

It can be clearly shown that \(\frac{d^{2}(A C)}{d \theta^{2}}\) < 0 when tan θ = \(\left(\frac{b}{a}\right)^{\frac{1}{3}}\).

Therefore, by second derivative test, the length of the hypotenuse is the tan θ = \(\left(\frac{b}{a}\right)^{\frac{1}{3}}\).

Now, when tan θ = \(\left(\frac{b}{a}\right)^{\frac{1}{3}}\), then we have

AC = \(\frac{b \sqrt{a^{\frac{2}{3}}}+b^{\frac{2}{3}}}{b^{\frac{1}{3}}}+\frac{a \sqrt{a^{\frac{2}{3}}}+b^{\frac{2}{3}}}{a^{\frac{1}{3}}}\) [Using Eqs. (i) and (ii)]

= \(\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}\left(b^{\frac{2}{3}}+a^{\frac{2}{3}}\right)=\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\)

Hence, the maximum length of the hypotenuse is \(\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\).

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 13.
Find the points at which the function f given by f(x) = (x – 2)4 (x + 1)3 has –
(i) local maxima
(ii) local minima
(iii) point of inflexion
Solution.
The given function is f(x) = (x – 2)4 (x + 1)3
∴ f'(x) = 4 (x – 2)3 (x + 1)3 + 3(x + 1)2 (x – 2)4
= (x – 2)3 (x +1)2 [4(x + 1) + 3(x – 2)]
= (x – 2)3 (x + 1)2 (7x – 2)
f'(x) = 0
⇒ x = – 1 and x = \(\frac{2}{7}\) or x = 2
Now, for values of x close to \(\frac{2}{7}\) and to the left of \(\frac{2}{7}\), f'(x) > 0.

Also, for values of x close to \(\frac{2}{7}\) and to the right of \(\frac{2}{7}\), f'(x) < 0.

Thus, x = \(\frac{2}{7}\) is the point of local maxima.
Now, for values of x close to 2 and to the left of 2, f'(x) < 0. Also, for values of x close to 2 and to the right of 2, f'(x) > 0.
Thus, x = 2 is the point of local minima.
Now, as the value of x varies through – 1, f (x) does not change its sign.
Thus, x = – 1 is the point of inflexion.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 14.
Find the absolute maximum and minimum values of the function f’given by f(x) = cos2 x + sin x, x ∈ [0, π].
Solution.
Given, f(x) = cos2 x + sin x
∴ f'(x) = 2 cos x (- sin x) + cos x = – 2 sin x cos x + cos x
Now, f’ (x) = 0
⇒ 2 sin x cos x = cos x
⇒ 2 sin x cos x – cos x = 0
⇒ cos x (2 sin x – 1) = 0
⇒ sin x = – or cos x = 0
⇒ x = \(\frac{\pi}{6}\) or \(\frac{\pi}{2}\) as x ∈ [0, π]
Now, evaluating the value of f at critical points x = \(\frac{\pi}{2}\) and x = \(\frac{\pi}{6}\) and at
the end points of the interval [0, π] (i.e., at x = 0 and x = π), we have

f(\(\frac{\pi}{6}\)) = cos2 \(\frac{\pi}{6}\) + sin \(\frac{\pi}{6}\)
= \(\left(\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{2}=\frac{5}{4}\)

f(0) = cos2 0 + sin 0
= 1 + 0 = 1

f(π) = cos2 π + sin π
= (- 1)2 + 0 = 1

f(\(\frac{\pi}{2}\)) = cos2 \(\frac{\pi}{2}\) + sin \(\frac{\pi}{2}\)
= 0 + 1 = 1

Hence, the absolute maximum value of f is \(\frac{5}{4}\) occurring at x = \(\frac{\pi}{6}\) and the absolute minimum value of f is 1 occurring at x = 0, \(\frac{\pi}{2}\) and π.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 15.
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r, is -.
Solution.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise 11

Let R be the radius and h be the height of cone.
∴OA = h – r
In ∆OAB, r2 = R2 + (h – r)2
⇒ r2 = R2 + h2 + r2 – 2rh
⇒ R2 = 2rh – h2
The volume V of the cone is given by V = \(\frac{1}{3}\) πR2h
= \(\frac{1}{3}\) πh (2rh – h2)

= \(\frac{1}{3}\) π (2rh2 – h3)

On differentiating w.r.t. h, we get
\(\frac{d V}{d h}\) = \(\frac{1}{3}\) π (4rh – 3h2)

For maximum or minimum put \(\frac{d V}{d h}\) = 0
⇒ 4rh = 3h2
⇒ 4r = 3h
∴ h = \(\frac{4 r}{3}\) (h ≠ 0)

Now, \(\frac{d^{2} V}{d h^{2}}\) = \(\frac{1}{3}\) π (4r – 6h)

At h = \(\frac{4 r}{3}\),

\(\left(\frac{d^{2} V}{d h^{2}}\right)_{h=\frac{4 r}{3}}\) = \(\frac{1}{3}\) π (4r – 6 × \(\frac{4 r}{3}\))

= \(\frac{pi}{3}\) (4r – 8r) = \(\frac{-4 r \pi}{3}\) < 0 ⇒ V is maximum when h = \(\frac{4 r}{3}\) Hence, volume of the cone is maximum when h = \(\frac{4 r}{3}\), which is the altitude of cone. Question 16. Let f be a function defined on [a, b] such that f'(x) > 0, for all x ∈ (a, b). Then prove that f is an increasing function on (a, b).
Solution.
Given, f'(x) > 0 on (a, b)
f is a differentiable function on (a, b).
Also, every differentiable function is continuous, therefore, f is continuous on (a, b).
Let x1, x2 ∈ [a, b] and x2 > x1; then by LMV theorem, there exist c ∈ [a, b] such that

f'(c) = \(\frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}\)
⇒ f(x2) – f(x1) = (x2 – x1) f'(c)
⇒ f(x2) – f(x1) > 0 as x2 > x1 and f'(x) > 0
⇒ f(x2) > f(x1)
For x1 < x2
⇒ f(x1) < f(x2)
Hence, f is an increasing function on (a, b).

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 17.
Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R, is . Also, find the maximum volume.
Solution.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise 12

Radius of the sphere = R
Let h be the height and x be the diameter of the base of the inscribed cylinder Then.
h2 + x2 = (2R)2
⇒ h2 + x2 = 4R2 …………(i)
Volume of the cylinder = π (radius)2 × height

⇒ V = π (\(\frac{x}{2}\))2 . h
= \(\frac{1}{4}\) πx2h
⇒ V = \(\frac{1}{4}\) πh(4R2 – h2) …………(ii)
[From Eq. (i), x2 = 4R2 – h2]

⇒ V = πR2h – \(\frac{1}{4}\) πh3
On differentiating w.r.t h, we get
\(\frac{d V}{d h}\) = πR2h – \(\frac{3}{4}\) πh2
= π (R2 – \(\frac{3}{4}\) h2)

Put \(\frac{d V}{d h}\) = 0
⇒ R2 = \(\frac{3}{4}\) h2
⇒ h = \(\frac{2 R}{\sqrt{3}}\)

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise 14

Thus, volume of the cylinder is maximum when h = \(\frac{2 R}{\sqrt{3}}\).

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 18.
Show that height of the cylinder of greatest volume which can be inscribed in a circular cone of height h and having
semi-vertical angle a is one-third that of one cone and the
greatest volume of cylinder Is ida3 tan2 a.
Solution.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise 15

Let VAR be the cone of height h, semi-vertical angle a and let x be the radius of the base of the cylinder A’ B’ DC which is inscribed in the cone VAB.
Then, OO’ is the height of the cylinder = VO – VO’ = h – x cot α
Volume of the cylinder,
V = πx2 (h – x cot α) ………………(i)
On differentiating w.r.t. x, we get
\(\frac{d V}{d x}\) = 2πrh – 3πx2 cot α
For maxima or minima put \(\frac{d V}{d x}=\)= 0
⇒ 2πxh – 3πx2 cot α = 0

⇒ x = \(\frac{2 h}{3}\) tan α (∵ x ≠ 0)

Now, \(\frac{d^{2} V}{d x^{2}}\) = 2πh – 6π x cot α

At x= \(\frac{2 h}{3}\) tan α,

\(\frac{d^{2} V}{d x^{2}}\) = π (2h – 4h) = – 2πh < 0

Now, OO’ = h – x cot α = h – \(\frac{2 h}{3}\) = \(\frac{h}{3}\)

∴ The maximum volume of the cylinder is V = π (\(\frac{2 h}{3}\) tan α)2 (h – \(\frac{2 h}{3}\))

= \(\frac{4}{27}\) πh3 tan2 α.

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Direction (19 – 24):
Choose the correct answer.

Question 19.
A cylindrical tank of radius 10 m is being filled with wheat at the rate of 314 cubic meter per hour. Then the depth of the wheat is increasing at the rate of
(A) 1 m/h
(B) 0.1 m/h
(C) 1.1 m/h
(D) 0.5 m/h
Solution.
Let r be the radius of the cylinder.
Then, volume (V) of the cylinder is given by,
V = π (radius)2 × height
= π (10)2 h (radius = 10 m)
= 100 πh
On differentiating w.r.t. t, we get
\(\frac{d V}{d t}\) = 100 π \(\frac{d h}{d t}\)
The tank is being filled with wheat at the rate of 314 cubic metres per hour.
\(\frac{d V}{d t}\) = 314 m3/h

Thus, we have
314 = 100π \(\frac{d h}{d t}\)

⇒ \(\frac{d h}{d t}\) = \(\frac{314}{100(3.14)}=\frac{314}{314}\) = 1

Hence, the depth of wheat is increasing at the rate of 1 m / h.
The correct answer is (A).

Question 20.
The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2, – 1) is
(A) \(\frac{22}{7}\)

(B) \(\frac{6}{7}\)

(C) \(\frac{7}{6}\)

(D) \(\frac{-6}{7}\)
Solution.
The given curve is x = t2 + 3t – 8 and y = 2t2 – 2t – 5
∴ \(\frac{d x}{d t}\) = 2t + 3 and

\(\frac{d y}{d t}\) =4t – 2

∴ \(\frac{d y}{d x}=\frac{d y}{d t} \cdot \frac{d t}{d x}=\frac{4 t-2}{2 t+3}\)

The given point is (2, – 1)
At x = 2, we have r2+3r-8=2
⇒ t2 + 3t – 10 = 0
⇒ (t – 2) (t + 5) = 0
⇒ t = 2 or t = – 5
At y = – 1, we have
⇒ 2t2 – 2t – 4 = 0
⇒ 2 (t2 – t – 2) = 0
⇒ t = 2 or t = – 1
The common value oft is 2.
Hence, the slope of the tangent to the given curve at point (2, – 1) is

\(\left[\frac{d y}{d x}\right]_{t=2}=\frac{4(2)-2}{2(2)+3}=\frac{8-2}{4+3}=\frac{6}{7}\)
The correct answer is (B).

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 21.
The line y = mx + 1 is a tangent to the curve y2 = 4# if the value of m is
(A) 1
(B) 2
(C) 3
(D) \(\frac{1}{2}\)
Solution.
The equation of the tangent to the given curve is y = mx + 1.
Now, substituting y = mx + 1 in y2 = 4x, we get
⇒ (mx + 1)2 = 4x
⇒ m2x2 +1 + 2mx – 4x = 0
⇒ m2x2 + x (2m – 4) + 1 = 0 …………….(i)
Since a tangent touches the curve at one point, the roots of equation (i) must be equal.
Therefore, we have
Discriminant = 0
⇒ (2m – 4)2 – 4(m2) (1) = 0
⇒ 4m2 + 16 – 16m – 4m2 = 0
⇒ 16 – 16m = 0
⇒ m = 1
Hence, the required value of m is 1.
The correct answer is A.

Question 22.
The normal at the point (1, 1) on the curve 2y + x2 = 3 is
(A) x + y = 0
(B) x – y = 0
(C) x + y + 1 = 0
(D) x – y = 0
Solution.
The equation of the given curve is 2y + x2 = 3

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

\(\frac{d y}{d x}\) = – x

Slope of the normal to the given curve at point (1, 1) is \(\frac{-1}{\left[\frac{d y}{d x}\right]_{(1,1)}}=\frac{-1}{-1}\) = 1
Hence, the equation of the normal to the given curve at (1, 1) is given as
⇒ y – 1 = 1 (x – 1)
⇒ y – 1 = x – 1
⇒ x – y = 0
The correct answer is (B).

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 23.
The normal to the curve x2 = 4y passing (1, 2) is
(A) x + y = 3
(B) x – y = 3
(C) x + y = 1
(D) x – y = 1
Solution.
The equation of the given curve is x = 4y
Differentiating w.r.t x, we get
2x = 4 \(\frac{d y}{d x}\)

∴ \(\frac{d y}{d x}=\frac{2 x}{4}=\frac{x}{2}\) = slope of the tangent
∴Slope of the normal = (- 1) / slope of tangent
∴ Equation of normal at (x1, y1) is
y – y1 = (x – x1) ……………..(i)
It passes through (1, 2), therefore
2 – y1 = – \(\frac{2}{x_{1}}\) (1 – x1)
⇒ 2x1 – x1y1 = – 2 + 2x1
x1y1 = 2
or y1 = \(\frac{2}{x_{1}}\) …………(ii)
The point (x1, y1) lies on x2 = 4y
x12 = 4y1 ……………(iii)
From Eqs. (i) and (iii), we get
x12 = 4 . \(\frac{2}{x_{1}}\)
∴ x13 = 8
x1 = 2
From Eq. (iii), 4 = 4y1
∴ y1 = 1
Putting these values in Eq. (i), we get
y – 1 = – \(\frac{2}{24}\) (x – 2)
y – 1 = – x + 2
⇒ x + y = 3
The correct answer is (A).

PSEB 12th Class Maths Solutions Chapter 6 Application of Derivatives Miscellaneous Exercise

Question 24.
The points on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with the axes are
(A) (4, ± \(\frac{8}{3}\))

(B) (4, – \(\frac{8}{3}\))

(C) (4, ± \(\frac{3}{8}\))

(D)(± 4, \(\frac{3}{8}\))
Solution.
The equation of the given curve is 9y2 = x3
Differentiating w.r.t. x, we get
18 y \(\text { dy }\) = 3x2

⇒ \(\frac{d y}{d x}=\frac{x^{2}}{6 y}\)

Let P(x1, y1) be the point where normal is drawn
Slope of tangent = \(\frac{x_{1}^{2}}{6 y_{1}}\)

∴ Slope of normal = \(\frac{6 y_{1}}{x_{1}^{2}}\)

Normal make equal intercepts on the curve
∴ Its slope = ± 1
– \(\frac{6 y_{1}}{x_{1}^{2}}\) = ± 1
⇒ 6y1 = ± x12 ……………..(i)
(x1, y1)lies on the curve 9y2 = x3
⇒ 9y12 = x13 …………..(ii)
Taking +ve sign, eliminating y1 from Eq. (1) and Eq. (ii),
9 \(\left(\frac{x_{1}^{2}}{6}\right)^{2}\) = x13
⇒ \(\frac{9 x_{1}^{4}}{36}\) = x13
⇒ x1 = 4
From Eq. (i),
y1 = ± \(\frac{x_{1}^{2}}{6}=\pm \frac{16}{6}\)[Putting x = 4]
= ± \(\frac{8}{3}\)
∴ The point P is (4, ± \(\frac{8}{3}\))
The correct answer is (A).

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