Application of Derivatives — Questions and Answers
Welcome to HSLC Guru. This lesson gives complete, step-by-step answers to every question in ASSEB Class 12 Mathematics Chapter 6, Application of Derivatives (অৱকলজৰ প্ৰয়োগ). It covers Exercise 6.1, Exercise 6.2, Exercise 6.3 and the Miscellaneous Exercise on Chapter 6, with each answer fully worked out so that you can follow the reasoning and prepare confidently for your examination.
Summary
The derivative $\frac{dy}{dx}$ measures the rate of change of $y$ with respect to $x$. If two quantities $x$ and $y$ both vary with time $t$, then by the Chain Rule $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt}$ (provided $\frac{dx}{dt} \neq 0$), and $\frac{dy}{dx}$ is positive when $y$ increases as $x$ increases and negative when $y$ decreases. This idea is used for related-rate problems, marginal cost ($\frac{dC}{dx}$) and marginal revenue ($\frac{dR}{dx}$).
A function $f$ is increasing on an interval where $f^{\prime}(x) \geq 0$ and decreasing where $f^{\prime}(x) \leq 0$. A point $c$ where $f^{\prime}(c) = 0$ or $f$ is not differentiable is a critical point. By the First Derivative Test, $c$ is a local maximum if $f^{\prime}$ changes from $+$ to $-$, a local minimum if it changes from $-$ to $+$, and a point of inflexion if it does not change sign. By the Second Derivative Test, $c$ is a local maximum if $f^{\prime}(c) = 0$ and $f^{\prime\prime}(c) < 0$, and a local minimum if $f^{\prime}(c) = 0$ and $f^{\prime\prime}(c) > 0$.
To find the absolute maximum and minimum of a continuous function on a closed interval $[a, b]$, evaluate $f$ at every critical point inside the interval and at both end points, then pick the greatest and least values. These tools solve applied optimisation problems — largest boxes, cheapest tanks, shortest distances, and inscribed cones and cylinders of maximum volume.
Summary: ASSEB Class 12 Mathematics Chapter 6 Application of Derivatives explains rate of change of quantities, marginal cost and revenue, increasing and decreasing functions, the first and second derivative tests for local maxima and minima, and absolute maxima and minima on closed intervals, with fully worked answers to Exercise 6.1, Exercise 6.2, Exercise 6.3 and the Miscellaneous Exercise for Assam Board (ASSEB) Class 12 students.
Textbook Questions and Answers
Exercise 6.1
1. Find the rate of change of the area of a circle with respect to its radius $r$ when (a) $r = 3$ cm (b) $r = 4$ cm.
Answer: The area of a circle of radius $r$ is $A = \pi r^2$, so the rate of change of area with respect to the radius is $\frac{dA}{dr} = 2\pi r$.
(a) When $r = 3$ cm, $\frac{dA}{dr} = 2\pi(3) = 6\pi$ cm²/cm. (b) When $r = 4$ cm, $\frac{dA}{dr} = 2\pi(4) = 8\pi$ cm²/cm.
2. The volume of a cube is increasing at the rate of $8$ cm³/s. How fast is the surface area increasing when the length of an edge is $12$ cm?
Answer: Let the edge be $x$. Then $V = x^3$ and $S = 6x^2$. Given $\frac{dV}{dt} = 8$. Differentiating $V = x^3$: $\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$, so $\frac{dx}{dt} = \frac{8}{3x^2}$.
Now $\frac{dS}{dt} = 12x \frac{dx}{dt} = 12x \cdot \frac{8}{3x^2} = \frac{32}{x}$. At $x = 12$ cm, $\frac{dS}{dt} = \frac{32}{12} = \frac{8}{3}$ cm²/s.
3. The radius of a circle is increasing uniformly at the rate of $3$ cm/s. Find the rate at which the area of the circle is increasing when the radius is $10$ cm.
Answer: $A = \pi r^2$, so $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$. Given $\frac{dr}{dt} = 3$ cm/s. At $r = 10$ cm, $\frac{dA}{dt} = 2\pi(10)(3) = 60\pi$ cm²/s.
4. An edge of a variable cube is increasing at the rate of $3$ cm/s. How fast is the volume of the cube increasing when the edge is $10$ cm long?
Answer: $V = x^3$, so $\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$. Given $\frac{dx}{dt} = 3$ cm/s. At $x = 10$ cm, $\frac{dV}{dt} = 3(10)^2(3) = 900$ cm³/s.
5. A stone is dropped into a quiet lake and waves move in circles at the speed of $5$ cm/s. At the instant when the radius of the circular wave is $8$ cm, how fast is the enclosed area increasing?
Answer: $A = \pi r^2$, so $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$. Given $\frac{dr}{dt} = 5$ cm/s. At $r = 8$ cm, $\frac{dA}{dt} = 2\pi(8)(5) = 80\pi$ cm²/s.
6. The radius of a circle is increasing at the rate of $0.7$ cm/s. What is the rate of increase of its circumference?
Answer: Circumference $C = 2\pi r$, so $\frac{dC}{dt} = 2\pi \frac{dr}{dt} = 2\pi(0.7) = 1.4\pi$ cm/s.
7. The length $x$ of a rectangle is decreasing at the rate of $5$ cm/minute and the width $y$ is increasing at the rate of $4$ cm/minute. When $x = 8$ cm and $y = 6$ cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.
Answer: Since $x$ is decreasing and $y$ increasing, $\frac{dx}{dt} = -5$ cm/min and $\frac{dy}{dt} = 4$ cm/min.
(a) Perimeter $P = 2(x + y)$, so $\frac{dP}{dt} = 2\left(\frac{dx}{dt} + \frac{dy}{dt}\right) = 2(-5 + 4) = -2$ cm/min. The perimeter is decreasing at $2$ cm/min.
(b) Area $A = xy$, so $\frac{dA}{dt} = y\frac{dx}{dt} + x\frac{dy}{dt} = 6(-5) + 8(4) = -30 + 32 = 2$ cm²/min. The area is increasing at $2$ cm²/min.
8. A balloon, which always remains spherical on inflation, is being inflated by pumping in $900$ cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is $15$ cm.
Answer: $V = \frac{4}{3}\pi r^3$, so $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. Given $\frac{dV}{dt} = 900$ cm³/s, we get $\frac{dr}{dt} = \frac{900}{4\pi r^2}$.
At $r = 15$ cm, $\frac{dr}{dt} = \frac{900}{4\pi(225)} = \frac{900}{900\pi} = \frac{1}{\pi}$ cm/s.
9. A balloon, which always remains spherical, has a variable radius. Find the rate at which its volume is increasing with the radius when the radius is $10$ cm.
Answer: $V = \frac{4}{3}\pi r^3$, so $\frac{dV}{dr} = 4\pi r^2$. At $r = 10$ cm, $\frac{dV}{dr} = 4\pi(10)^2 = 400\pi$ cm³/cm.
10. A ladder $5$ m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of $2$ cm/s. How fast is its height on the wall decreasing when the foot of the ladder is $4$ m away from the wall?
Answer: Let $x$ be the distance of the foot from the wall and $y$ the height on the wall. Then $x^2 + y^2 = 25$. Differentiating with respect to $t$: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$, so $\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}$.
Given $\frac{dx}{dt} = 2$ cm/s. When $x = 4$ m, $y = \sqrt{25 – 16} = 3$ m. So $\frac{dy}{dt} = -\frac{4}{3}(2) = -\frac{8}{3}$ cm/s. The height on the wall is decreasing at $\frac{8}{3}$ cm/s.
11. A particle moves along the curve $6y = x^3 + 2$. Find the points on the curve at which the $y$-coordinate is changing $8$ times as fast as the $x$-coordinate.
Answer: Differentiating $6y = x^3 + 2$ with respect to $t$: $6\frac{dy}{dt} = 3x^2\frac{dx}{dt}$. We are given $\frac{dy}{dt} = 8\frac{dx}{dt}$, so $6(8) = 3x^2$, giving $x^2 = 16$, i.e. $x = \pm 4$.
When $x = 4$: $y = \frac{4^3 + 2}{6} = \frac{66}{6} = 11$. When $x = -4$: $y = \frac{(-4)^3 + 2}{6} = \frac{-62}{6} = -\frac{31}{3}$. The required points are $(4, 11)$ and $\left(-4, -\frac{31}{3}\right)$.
12. The radius of an air bubble is increasing at the rate of $\frac{1}{2}$ cm/s. At what rate is the volume of the bubble increasing when the radius is $1$ cm?
Answer: A bubble is a sphere, so $V = \frac{4}{3}\pi r^3$ and $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. Given $\frac{dr}{dt} = \frac{1}{2}$ cm/s. At $r = 1$ cm, $\frac{dV}{dt} = 4\pi(1)^2 \cdot \frac{1}{2} = 2\pi$ cm³/s.
13. A balloon, which always remains spherical, has a variable diameter $\frac{3}{2}(2x + 1)$. Find the rate of change of its volume with respect to $x$.
Answer: Radius $r = \frac{1}{2} \cdot \frac{3}{2}(2x + 1) = \frac{3}{4}(2x + 1)$. Volume $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \cdot \frac{27}{64}(2x + 1)^3 = \frac{9}{16}\pi(2x + 1)^3$.
Therefore $\frac{dV}{dx} = \frac{9}{16}\pi \cdot 3(2x + 1)^2 \cdot 2 = \frac{27}{8}\pi(2x + 1)^2$.
14. Sand is pouring from a pipe at the rate of $12$ cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is $4$ cm?
Answer: Given $h = \frac{1}{6}r$, so $r = 6h$. Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(6h)^2 h = 12\pi h^3$.
Then $\frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt}$. Given $\frac{dV}{dt} = 12$ cm³/s, so $\frac{dh}{dt} = \frac{12}{36\pi h^2} = \frac{1}{3\pi h^2}$. At $h = 4$ cm, $\frac{dh}{dt} = \frac{1}{3\pi(16)} = \frac{1}{48\pi}$ cm/s.
15. The total cost $C(x)$ in Rupees associated with the production of $x$ units of an item is given by $C(x) = 0.007x^3 – 0.003x^2 + 15x + 4000$. Find the marginal cost when $17$ units are produced.
Answer: Marginal cost $MC = \frac{dC}{dx} = 0.021x^2 – 0.006x + 15$. At $x = 17$: $MC = 0.021(289) – 0.006(17) + 15 = 6.069 – 0.102 + 15 = 20.967$. The marginal cost is ₹$20.967$ (nearly ₹$20.97$).
16. The total revenue in Rupees received from the sale of $x$ units of a product is given by $R(x) = 13x^2 + 26x + 15$. Find the marginal revenue when $x = 7$.
Answer: Marginal revenue $MR = \frac{dR}{dx} = 26x + 26$. At $x = 7$: $MR = 26(7) + 26 = 182 + 26 = 208$. The marginal revenue is ₹$208$.
17. The rate of change of the area of a circle with respect to its radius $r$ at $r = 6$ cm is
(A) $10\pi$ (B) $12\pi$ (C) $8\pi$ (D) $11\pi$
Answer: (B) $12\pi$. Since $A = \pi r^2$, $\frac{dA}{dr} = 2\pi r$, and at $r = 6$ this equals $2\pi(6) = 12\pi$.
18. The total revenue in Rupees received from the sale of $x$ units of a product is given by $R(x) = 3x^2 + 36x + 5$. The marginal revenue, when $x = 15$, is
(A) $116$ (B) $96$ (C) $90$ (D) $126$
Answer: (D) $126$. Marginal revenue $MR = \frac{dR}{dx} = 6x + 36$, and at $x = 15$ this equals $6(15) + 36 = 90 + 36 = 126$.
Exercise 6.2
1. Show that the function given by $f(x) = 3x + 17$ is increasing on $\mathbb{R}$.
Answer: $f^{\prime}(x) = 3 > 0$ for every $x \in \mathbb{R}$. Since the derivative is positive throughout, $f$ is (strictly) increasing on $\mathbb{R}$.
2. Show that the function given by $f(x) = e^{2x}$ is increasing on $\mathbb{R}$.
Answer: $f^{\prime}(x) = 2e^{2x}$. Since $e^{2x} > 0$ for all $x$, we have $f^{\prime}(x) > 0$ throughout $\mathbb{R}$, so $f$ is increasing on $\mathbb{R}$.
3. Show that the function given by $f(x) = \sin x$ is (a) increasing in $\left(0, \frac{\pi}{2}\right)$ (b) decreasing in $\left(\frac{\pi}{2}, \pi\right)$ (c) neither increasing nor decreasing in $(0, \pi)$.
Answer: $f^{\prime}(x) = \cos x$.
(a) For $x \in \left(0, \frac{\pi}{2}\right)$, $\cos x > 0$, so $f^{\prime}(x) > 0$ and $f$ is increasing there. (b) For $x \in \left(\frac{\pi}{2}, \pi\right)$, $\cos x < 0$, so $f^{\prime}(x) < 0$ and $f$ is decreasing there. (c) On $(0, \pi)$ the derivative $\cos x$ is positive on part of the interval and negative on another part, so $f$ is neither increasing nor decreasing on the whole of $(0, \pi)$.
4. Find the intervals in which the function $f$ given by $f(x) = 2x^2 – 3x$ is (a) increasing (b) decreasing.
Answer: $f^{\prime}(x) = 4x – 3$. Setting $f^{\prime}(x) = 0$ gives $x = \frac{3}{4}$.
(a) For $x > \frac{3}{4}$, $f^{\prime}(x) > 0$, so $f$ is increasing on $\left(\frac{3}{4}, \infty\right)$. (b) For $x < \frac{3}{4}$, $f^{\prime}(x) < 0$, so $f$ is decreasing on $\left(-\infty, \frac{3}{4}\right)$.
5. Find the intervals in which the function $f$ given by $f(x) = 2x^3 – 3x^2 – 36x + 7$ is (a) increasing (b) decreasing.
Answer: $f^{\prime}(x) = 6x^2 – 6x – 36 = 6(x^2 – x – 6) = 6(x – 3)(x + 2)$. So $f^{\prime}(x) = 0$ at $x = -2$ and $x = 3$, which split the line into $(-\infty, -2)$, $(-2, 3)$ and $(3, \infty)$.
(a) On $(-\infty, -2)$ and $(3, \infty)$ both factors have the same sign, so $f^{\prime}(x) > 0$ and $f$ is increasing there. (b) On $(-2, 3)$ the factors have opposite signs, so $f^{\prime}(x) < 0$ and $f$ is decreasing there.
6. Find the intervals in which the following functions are strictly increasing or decreasing:
(a) $x^2 + 2x – 5$
Answer: $f^{\prime}(x) = 2x + 2 = 2(x + 1)$. So $f$ is strictly increasing on $(-1, \infty)$ and strictly decreasing on $(-\infty, -1)$.
(b) $10 – 6x – 2x^2$
Answer: $f^{\prime}(x) = -6 – 4x = -2(2x + 3)$. This is positive when $x < -\frac{3}{2}$ and negative when $x > -\frac{3}{2}$. So $f$ is strictly increasing on $\left(-\infty, -\frac{3}{2}\right)$ and strictly decreasing on $\left(-\frac{3}{2}, \infty\right)$.
(c) $-2x^3 – 9x^2 – 12x + 1$
Answer: $f^{\prime}(x) = -6x^2 – 18x – 12 = -6(x^2 + 3x + 2) = -6(x + 1)(x + 2)$. This is positive when $(x + 1)(x + 2) < 0$, i.e. for $-2 < x < -1$. So $f$ is strictly increasing on $(-2, -1)$ and strictly decreasing on $(-\infty, -2)$ and $(-1, \infty)$.
(d) $6 – 9x – x^2$
Answer: $f^{\prime}(x) = -9 – 2x$. This is positive when $x < -\frac{9}{2}$ and negative when $x > -\frac{9}{2}$. So $f$ is strictly increasing on $\left(-\infty, -\frac{9}{2}\right)$ and strictly decreasing on $\left(-\frac{9}{2}, \infty\right)$.
(e) $(x + 1)^3(x – 3)^3$
Answer: Using the product rule, $f^{\prime}(x) = 3(x + 1)^2(x – 3)^3 + 3(x + 1)^3(x – 3)^2 = 3(x + 1)^2(x – 3)^2\big[(x – 3) + (x + 1)\big] = 6(x + 1)^2(x – 3)^2(x – 1)$. Since $(x + 1)^2 \geq 0$ and $(x – 3)^2 \geq 0$, the sign of $f^{\prime}$ is the sign of $(x – 1)$. Hence $f$ is strictly increasing on $(1, \infty)$ and strictly decreasing on $(-\infty, 1)$.
7. Show that $y = \log(1 + x) – \frac{2x}{2 + x}$, $x > -1$, is an increasing function of $x$ throughout its domain.
Answer: Differentiate: $\frac{dy}{dx} = \frac{1}{1 + x} – \frac{2(2 + x) – 2x}{(2 + x)^2} = \frac{1}{1 + x} – \frac{4}{(2 + x)^2}$.
Combining over a common denominator: $\frac{dy}{dx} = \frac{(2 + x)^2 – 4(1 + x)}{(1 + x)(2 + x)^2} = \frac{4 + 4x + x^2 – 4 – 4x}{(1 + x)(2 + x)^2} = \frac{x^2}{(1 + x)(2 + x)^2}$.
For $x > -1$, $(1 + x) > 0$, $(2 + x)^2 > 0$ and $x^2 \geq 0$, so $\frac{dy}{dx} \geq 0$ throughout the domain. Hence $y$ is an increasing function of $x$.
8. Find the values of $x$ for which $y = [x(x – 2)]^2$ is an increasing function.
Answer: $y = (x^2 – 2x)^2$, so $\frac{dy}{dx} = 2(x^2 – 2x)(2x – 2) = 4x(x – 2)(x – 1)$. The zeros are $x = 0, 1, 2$. Testing signs, $\frac{dy}{dx} > 0$ on $(0, 1)$ and on $(2, \infty)$. Hence $y$ is increasing for $0 < x < 1$ and $x > 2$ (i.e. on $(0, 1) \cup (2, \infty)$).
9. Prove that $y = \frac{4\sin\theta}{2 + \cos\theta} – \theta$ is an increasing function of $\theta$ in $\left[0, \frac{\pi}{2}\right]$.
Answer: $\frac{dy}{d\theta} = \frac{4\cos\theta(2 + \cos\theta) – 4\sin\theta(-\sin\theta)}{(2 + \cos\theta)^2} – 1 = \frac{8\cos\theta + 4\cos^2\theta + 4\sin^2\theta}{(2 + \cos\theta)^2} – 1 = \frac{8\cos\theta + 4}{(2 + \cos\theta)^2} – 1$.
Combining: $\frac{dy}{d\theta} = \frac{8\cos\theta + 4 – (2 + \cos\theta)^2}{(2 + \cos\theta)^2} = \frac{8\cos\theta + 4 – 4 – 4\cos\theta – \cos^2\theta}{(2 + \cos\theta)^2} = \frac{4\cos\theta – \cos^2\theta}{(2 + \cos\theta)^2} = \frac{\cos\theta(4 – \cos\theta)}{(2 + \cos\theta)^2}$.
For $\theta \in \left[0, \frac{\pi}{2}\right]$, $\cos\theta \geq 0$ and $4 – \cos\theta > 0$, so $\frac{dy}{d\theta} \geq 0$. Hence $y$ is an increasing function of $\theta$ on $\left[0, \frac{\pi}{2}\right]$.
10. Prove that the logarithmic function is increasing on $(0, \infty)$.
Answer: Let $f(x) = \log x$, $x > 0$. Then $f^{\prime}(x) = \frac{1}{x}$. For every $x > 0$, $\frac{1}{x} > 0$, so $f^{\prime}(x) > 0$. Hence $\log x$ is increasing on $(0, \infty)$.
11. Prove that the function $f$ given by $f(x) = x^2 – x + 1$ is neither strictly increasing nor decreasing on $(-1, 1)$.
Answer: $f^{\prime}(x) = 2x – 1$, which is zero at $x = \frac{1}{2}$, a point inside $(-1, 1)$. For $x < \frac{1}{2}$, $f^{\prime}(x) < 0$ (decreasing), and for $x > \frac{1}{2}$, $f^{\prime}(x) > 0$ (increasing). Since the derivative changes sign inside the interval, $f$ is neither strictly increasing nor strictly decreasing on $(-1, 1)$.
12. Which of the following functions are decreasing on $\left(0, \frac{\pi}{2}\right)$?
(A) $\cos x$ (B) $\cos 2x$ (C) $\cos 3x$ (D) $\tan x$
Answer: (A) and (B). For $\cos x$, the derivative is $-\sin x < 0$ on $\left(0, \frac{\pi}{2}\right)$, so it is decreasing. For $\cos 2x$, the derivative is $-2\sin 2x$; since $2x \in (0, \pi)$, $\sin 2x > 0$, so it is decreasing. For $\cos 3x$, the derivative $-3\sin 3x$ changes sign at $x = \frac{\pi}{3}$ (where $3x = \pi$), so it is not decreasing throughout. For $\tan x$, the derivative $\sec^2 x > 0$, so it is increasing. Hence only (A) and (B) are decreasing on $\left(0, \frac{\pi}{2}\right)$.
13. On which of the following intervals is the function $f$ given by $f(x) = x^{100} + \sin x – 1$ decreasing?
(A) $(0, 1)$ (B) $\left(\frac{\pi}{2}, \pi\right)$ (C) $\left(0, \frac{\pi}{2}\right)$ (D) None of these
Answer: (D) None of these. Here $f^{\prime}(x) = 100x^{99} + \cos x$. On $(0, 1)$, $100x^{99} > 0$ and $\cos x > 0$, so $f^{\prime}(x) > 0$. On $\left(\frac{\pi}{2}, \pi\right)$, $x > 1$ so $100x^{99}$ is very large positive and dominates, giving $f^{\prime}(x) > 0$. On $\left(0, \frac{\pi}{2}\right)$, $f^{\prime}(x) > 0$ as well. The function is increasing on all the given intervals, so it is decreasing on none of them.
14. For what values of $a$ the function $f$ given by $f(x) = x^2 + ax + 1$ is increasing on $[1, 2]$?
Answer: $f^{\prime}(x) = 2x + a$. For $f$ to be increasing on $[1, 2]$ we need $f^{\prime}(x) \geq 0$ for all $x \in [1, 2]$. The smallest value of $2x + a$ on this interval is at $x = 1$, namely $2 + a$. So $2 + a \geq 0$, giving $a \geq -2$.
15. Let $I$ be any interval disjoint from $[-1, 1]$. Prove that the function $f$ given by $f(x) = x + \frac{1}{x}$ is increasing on $I$.
Answer: $f^{\prime}(x) = 1 – \frac{1}{x^2} = \frac{x^2 – 1}{x^2}$. If $I$ is disjoint from $[-1, 1]$, then $|x| > 1$ for every $x \in I$, so $x^2 > 1$ and hence $x^2 – 1 > 0$. Since $x^2 > 0$ too, $f^{\prime}(x) > 0$ on $I$. Therefore $f$ is increasing on $I$.
16. Prove that the function $f$ given by $f(x) = \log \sin x$ is increasing on $\left(0, \frac{\pi}{2}\right)$ and decreasing on $\left(\frac{\pi}{2}, \pi\right)$.
Answer: $f^{\prime}(x) = \frac{\cos x}{\sin x} = \cot x$. On $\left(0, \frac{\pi}{2}\right)$, $\cot x > 0$, so $f$ is increasing. On $\left(\frac{\pi}{2}, \pi\right)$, $\cot x < 0$, so $f$ is decreasing.
17. Prove that the function $f$ given by $f(x) = \log |\cos x|$ is decreasing on $\left(0, \frac{\pi}{2}\right)$ and increasing on $\left(\frac{3\pi}{2}, 2\pi\right)$.
Answer: $f^{\prime}(x) = \frac{-\sin x}{\cos x} = -\tan x$. On $\left(0, \frac{\pi}{2}\right)$, $\tan x > 0$, so $f^{\prime}(x) = -\tan x < 0$ and $f$ is decreasing. On $\left(\frac{3\pi}{2}, 2\pi\right)$ (fourth quadrant), $\sin x < 0$ and $\cos x > 0$, so $\tan x < 0$ and $f^{\prime}(x) = -\tan x > 0$; hence $f$ is increasing.
18. Prove that the function given by $f(x) = x^3 – 3x^2 + 3x – 100$ is increasing in $\mathbb{R}$.
Answer: $f^{\prime}(x) = 3x^2 – 6x + 3 = 3(x^2 – 2x + 1) = 3(x – 1)^2$. Since $(x – 1)^2 \geq 0$ for all $x$, $f^{\prime}(x) \geq 0$ throughout $\mathbb{R}$ (and is zero only at the single point $x = 1$). Hence $f$ is increasing on $\mathbb{R}$.
19. The interval in which $y = x^2 e^{-x}$ is increasing is
(A) $(-\infty, \infty)$ (B) $(-2, 0)$ (C) $(2, \infty)$ (D) $(0, 2)$
Answer: (D) $(0, 2)$. Here $\frac{dy}{dx} = 2x e^{-x} + x^2(-e^{-x}) = e^{-x}(2x – x^2) = e^{-x}\, x(2 – x)$. Since $e^{-x} > 0$, the sign of $\frac{dy}{dx}$ is that of $x(2 – x)$, which is positive for $0 < x < 2$. Hence $y$ is increasing on $(0, 2)$.
Exercise 6.3
1. Find the maximum and minimum values, if any, of the following functions given by:
(i) $f(x) = (2x – 1)^2 + 3$
Answer: Since $(2x – 1)^2 \geq 0$, the smallest value of $f$ is $0 + 3 = 3$, attained when $2x – 1 = 0$, i.e. $x = \frac{1}{2}$. As $x \to \pm\infty$, $f \to \infty$, so there is no maximum. Minimum value $= 3$; no maximum value.
(ii) $f(x) = 9x^2 + 12x + 2$
Answer: Completing the square, $f(x) = 9\left(x + \frac{2}{3}\right)^2 – 2$. Since the square is $\geq 0$, the minimum value is $-2$, attained at $x = -\frac{2}{3}$. There is no maximum value.
(iii) $f(x) = -(x – 1)^2 + 10$
Answer: Since $-(x – 1)^2 \leq 0$, the largest value of $f$ is $10$, attained when $x = 1$. As $x \to \pm\infty$, $f \to -\infty$, so there is no minimum. Maximum value $= 10$; no minimum value.
(iv) $g(x) = x^3 + 1$
Answer: As $x \to \infty$, $g \to \infty$ and as $x \to -\infty$, $g \to -\infty$. Hence $g$ has neither a maximum nor a minimum value.
2. Find the maximum and minimum values, if any, of the following functions given by:
(i) $f(x) = |x + 2| – 1$
Answer: $|x + 2| \geq 0$, so $f(x) \geq -1$, with equality when $x = -2$. Minimum value $= -1$ at $x = -2$; no maximum value (it grows without bound).
(ii) $g(x) = -|x + 1| + 3$
Answer: $-|x + 1| \leq 0$, so $g(x) \leq 3$, with equality when $x = -1$. Maximum value $= 3$ at $x = -1$; no minimum value.
(iii) $h(x) = \sin 2x + 5$
Answer: Since $-1 \leq \sin 2x \leq 1$, we have $4 \leq h(x) \leq 6$. Maximum value $= 6$ (when $\sin 2x = 1$); minimum value $= 4$ (when $\sin 2x = -1$).
(iv) $f(x) = |\sin 4x + 3|$
Answer: Since $\sin 4x + 3$ lies between $2$ and $4$, it is always positive, so $f(x) = \sin 4x + 3$. Thus $2 \leq f(x) \leq 4$. Maximum value $= 4$; minimum value $= 2$.
(v) $h(x) = x + 1$, $x \in (-1, 1)$
Answer: On the open interval $(-1, 1)$, $h$ is increasing and takes values between $0$ and $2$, but neither endpoint is included, so the value $2$ (or $0$) is never actually attained. Hence $h$ has neither a maximum nor a minimum value on $(-1, 1)$.
3. Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:
(i) $f(x) = x^2$
Answer: $f^{\prime}(x) = 2x = 0$ gives $x = 0$. $f^{\prime\prime}(x) = 2 > 0$, so $x = 0$ is a point of local minima. Local minimum value $= f(0) = 0$.
(ii) $g(x) = x^3 – 3x$
Answer: $g^{\prime}(x) = 3x^2 – 3 = 3(x – 1)(x + 1) = 0$ gives $x = 1, -1$. $g^{\prime\prime}(x) = 6x$. At $x = 1$, $g^{\prime\prime}(1) = 6 > 0$, so it is a local minimum with value $g(1) = 1 – 3 = -2$. At $x = -1$, $g^{\prime\prime}(-1) = -6 < 0$, so it is a local maximum with value $g(-1) = -1 + 3 = 2$.
(iii) $h(x) = \sin x + \cos x$, $0 < x < \frac{\pi}{2}$
Answer: $h^{\prime}(x) = \cos x – \sin x = 0$ gives $\tan x = 1$, i.e. $x = \frac{\pi}{4}$ (in the given interval). $h^{\prime\prime}(x) = -\sin x – \cos x$, and $h^{\prime\prime}\left(\frac{\pi}{4}\right) = -\sqrt{2} < 0$, so $x = \frac{\pi}{4}$ is a point of local maxima. Local maximum value $= \sin\frac{\pi}{4} + \cos\frac{\pi}{4} = \sqrt{2}$.
(iv) $f(x) = \sin x – \cos x$, $0 < x < 2\pi$
Answer: $f^{\prime}(x) = \cos x + \sin x = 0$ gives $\tan x = -1$, i.e. $x = \frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$. $f^{\prime\prime}(x) = -\sin x + \cos x$. At $x = \frac{3\pi}{4}$, $f^{\prime\prime} = -\sqrt{2} < 0$, a local maximum with value $f\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}$. At $x = \frac{7\pi}{4}$, $f^{\prime\prime} = \sqrt{2} > 0$, a local minimum with value $f\left(\frac{7\pi}{4}\right) = -\frac{\sqrt{2}}{2} – \frac{\sqrt{2}}{2} = -\sqrt{2}$.
(v) $f(x) = x^3 – 6x^2 + 9x + 15$
Answer: $f^{\prime}(x) = 3x^2 – 12x + 9 = 3(x – 1)(x – 3) = 0$ gives $x = 1, 3$. $f^{\prime\prime}(x) = 6x – 12$. At $x = 1$, $f^{\prime\prime}(1) = -6 < 0$, a local maximum with value $f(1) = 1 – 6 + 9 + 15 = 19$. At $x = 3$, $f^{\prime\prime}(3) = 6 > 0$, a local minimum with value $f(3) = 27 – 54 + 27 + 15 = 15$.
(vi) $g(x) = \frac{x}{2} + \frac{2}{x}$, $x > 0$
Answer: $g^{\prime}(x) = \frac{1}{2} – \frac{2}{x^2} = 0$ gives $x^2 = 4$, i.e. $x = 2$ (since $x > 0$). $g^{\prime\prime}(x) = \frac{4}{x^3} > 0$ at $x = 2$, so $x = 2$ is a point of local minima. Local minimum value $= \frac{2}{2} + \frac{2}{2} = 2$.
(vii) $g(x) = \frac{1}{x^2 + 2}$
Answer: $g^{\prime}(x) = -\frac{2x}{(x^2 + 2)^2} = 0$ gives $x = 0$. For $x < 0$, $g^{\prime}(x) > 0$; for $x > 0$, $g^{\prime}(x) < 0$. So $g^{\prime}$ changes from $+$ to $-$, and $x = 0$ is a point of local maxima. Local maximum value $= g(0) = \frac{1}{2}$.
(viii) $f(x) = x\sqrt{1 – x}$, $0 < x < 1$
Answer: $f^{\prime}(x) = \sqrt{1 – x} + x \cdot \frac{-1}{2\sqrt{1 – x}} = \frac{2(1 – x) – x}{2\sqrt{1 – x}} = \frac{2 – 3x}{2\sqrt{1 – x}}$. Setting $f^{\prime}(x) = 0$ gives $x = \frac{2}{3}$. For $x < \frac{2}{3}$, $f^{\prime}(x) > 0$; for $x > \frac{2}{3}$, $f^{\prime}(x) < 0$, so $x = \frac{2}{3}$ is a point of local maxima. Local maximum value $= \frac{2}{3}\sqrt{1 – \frac{2}{3}} = \frac{2}{3} \cdot \frac{1}{\sqrt{3}} = \frac{2\sqrt{3}}{9}$.
4. Prove that the following functions do not have maxima or minima:
(i) $f(x) = e^x$
Answer: $f^{\prime}(x) = e^x$, which is never zero (it is $> 0$ for all $x$). Since there is no critical point and $f$ is everywhere increasing, $f$ has neither a maximum nor a minimum.
(ii) $g(x) = \log x$
Answer: $g^{\prime}(x) = \frac{1}{x}$, which is never zero for $x > 0$. There is no critical point, and $g$ is everywhere increasing, so $g$ has neither a maximum nor a minimum.
(iii) $h(x) = x^3 + x^2 + x + 1$
Answer: $h^{\prime}(x) = 3x^2 + 2x + 1$. Its discriminant is $2^2 – 4(3)(1) = 4 – 12 = -8 < 0$, so $h^{\prime}(x) > 0$ for all $x$ (no real root). Hence there is no critical point and $h$ is strictly increasing, so it has neither a maximum nor a minimum.
5. Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
(i) $f(x) = x^3$, $x \in [-2, 2]$
Answer: $f^{\prime}(x) = 3x^2 = 0$ gives $x = 0$. Values: $f(-2) = -8$, $f(0) = 0$, $f(2) = 8$. Absolute maximum $= 8$ at $x = 2$; absolute minimum $= -8$ at $x = -2$.
(ii) $f(x) = \sin x + \cos x$, $x \in [0, \pi]$
Answer: $f^{\prime}(x) = \cos x – \sin x = 0$ gives $x = \frac{\pi}{4}$. Values: $f(0) = 1$, $f\left(\frac{\pi}{4}\right) = \sqrt{2}$, $f(\pi) = -1$. Absolute maximum $= \sqrt{2}$ at $x = \frac{\pi}{4}$; absolute minimum $= -1$ at $x = \pi$.
(iii) $f(x) = 4x – \frac{1}{2}x^2$, $x \in \left[-2, \frac{9}{2}\right]$
Answer: $f^{\prime}(x) = 4 – x = 0$ gives $x = 4$. Values: $f(-2) = -8 – 2 = -10$, $f(4) = 16 – 8 = 8$, $f\left(\frac{9}{2}\right) = 18 – \frac{81}{8} = \frac{63}{8} = 7.875$. Absolute maximum $= 8$ at $x = 4$; absolute minimum $= -10$ at $x = -2$.
(iv) $f(x) = (x – 1)^2 + 3$, $x \in [-3, 1]$
Answer: $f^{\prime}(x) = 2(x – 1) = 0$ gives $x = 1$ (an endpoint). Values: $f(-3) = 16 + 3 = 19$, $f(1) = 0 + 3 = 3$. Absolute maximum $= 19$ at $x = -3$; absolute minimum $= 3$ at $x = 1$.
6. Find the maximum profit that a company can make, if the profit function is given by $p(x) = 41 – 72x – 18x^2$.
Answer: $p^{\prime}(x) = -72 – 36x = 0$ gives $x = -2$. $p^{\prime\prime}(x) = -36 < 0$, so $x = -2$ gives a maximum. Maximum profit $= p(-2) = 41 – 72(-2) – 18(4) = 41 + 144 – 72 = 113$.
7. Find both the maximum value and the minimum value of $3x^4 – 8x^3 + 12x^2 – 48x + 25$ on the interval $[0, 3]$.
Answer: Let $f(x) = 3x^4 – 8x^3 + 12x^2 – 48x + 25$. Then $f^{\prime}(x) = 12x^3 – 24x^2 + 24x – 48 = 12(x^3 – 2x^2 + 2x – 4) = 12(x – 2)(x^2 + 2)$. Since $x^2 + 2 > 0$, the only critical point is $x = 2$.
Values: $f(0) = 25$, $f(2) = 48 – 64 + 48 – 96 + 25 = -39$, $f(3) = 243 – 216 + 108 – 144 + 25 = 16$. Maximum value $= 25$ at $x = 0$; minimum value $= -39$ at $x = 2$.
8. At what points in the interval $[0, 2\pi]$ does the function $\sin 2x$ attain its maximum value?
Answer: The maximum value of $\sin 2x$ is $1$, attained when $2x = \frac{\pi}{2} + 2k\pi$, i.e. $x = \frac{\pi}{4} + k\pi$. In $[0, 2\pi]$ this gives $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$. Hence $\sin 2x$ attains its maximum value at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.
9. What is the maximum value of the function $\sin x + \cos x$?
Answer: Let $f(x) = \sin x + \cos x$. Then $f^{\prime}(x) = \cos x – \sin x = 0$ gives $\tan x = 1$, e.g. $x = \frac{\pi}{4}$. $f^{\prime\prime}(x) = -\sin x – \cos x = -\sqrt{2} < 0$ at $x = \frac{\pi}{4}$, so it is a maximum. Maximum value $= \sin\frac{\pi}{4} + \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$.
10. Find the maximum value of $2x^3 – 24x + 107$ in the interval $[1, 3]$. Find the maximum value of the same function in $[-3, -1]$.
Answer: Let $f(x) = 2x^3 – 24x + 107$. Then $f^{\prime}(x) = 6x^2 – 24 = 6(x^2 – 4) = 0$ gives $x = \pm 2$.
On $[1, 3]$ the critical point is $x = 2$. Values: $f(1) = 85$, $f(2) = 75$, $f(3) = 89$. Maximum $= 89$ at $x = 3$.
On $[-3, -1]$ the critical point is $x = -2$. Values: $f(-3) = 125$, $f(-2) = 139$, $f(-1) = 129$. Maximum $= 139$ at $x = -2$.
11. It is given that at $x = 1$, the function $x^4 – 62x^2 + ax + 9$ attains its maximum value on the interval $[0, 2]$. Find the value of $a$.
Answer: Let $f(x) = x^4 – 62x^2 + ax + 9$, so $f^{\prime}(x) = 4x^3 – 124x + a$. Since the maximum on $[0, 2]$ occurs at the interior point $x = 1$, we must have $f^{\prime}(1) = 0$: $4 – 124 + a = 0$, giving $a = 120$.
12. Find the maximum and minimum values of $x + \sin 2x$ on $[0, 2\pi]$.
Answer: Let $f(x) = x + \sin 2x$. Then $f^{\prime}(x) = 1 + 2\cos 2x = 0$ gives $\cos 2x = -\frac{1}{2}$, so $2x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}$, i.e. $x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$.
Comparing values at these points and at the end points $x = 0$ and $x = 2\pi$: $f(0) = 0$ and $f(2\pi) = 2\pi$ (since $\sin 4\pi = 0$), while every interior critical value lies between these. The largest value is $2\pi$ (at $x = 2\pi$) and the smallest is $0$ (at $x = 0$). Hence maximum value $= 2\pi$ and minimum value $= 0$.
13. Find two numbers whose sum is $24$ and whose product is as large as possible.
Answer: Let the numbers be $x$ and $24 – x$. Product $P = x(24 – x) = 24x – x^2$. Then $P^{\prime} = 24 – 2x = 0$ gives $x = 12$, and $P^{\prime\prime} = -2 < 0$, so this is a maximum. The two numbers are $12$ and $12$, with maximum product $144$.
14. Find two positive numbers $x$ and $y$ such that $x + y = 60$ and $xy^3$ is maximum.
Answer: Substitute $x = 60 – y$, so $P = (60 – y)y^3 = 60y^3 – y^4$. Then $\frac{dP}{dy} = 180y^2 – 4y^3 = 4y^2(45 – y) = 0$ gives $y = 45$ (rejecting $y = 0$). $\frac{d^2P}{dy^2} = 360y – 12y^2 = 360(45) – 12(2025) < 0$, so it is a maximum. Then $x = 60 – 45 = 15$. The numbers are $x = 15$, $y = 45$.
15. Find two positive numbers $x$ and $y$ such that their sum is $35$ and the product $x^2 y^5$ is a maximum.
Answer: Substitute $x = 35 – y$, so $P = (35 – y)^2 y^5$. Then $\frac{dP}{dy} = 2(35 – y)(-1)y^5 + (35 – y)^2 \cdot 5y^4 = (35 – y)y^4\big[-2y + 5(35 – y)\big] = 7(35 – y)y^4(25 – y)$. Setting this to zero (with $x, y > 0$) gives $y = 25$, and then $x = 10$. Checking the sign of $\frac{dP}{dy}$ confirms a maximum. The numbers are $x = 10$, $y = 25$.
16. Find two positive numbers whose sum is $16$ and the sum of whose cubes is minimum.
Answer: Let the numbers be $x$ and $16 – x$. Then $S = x^3 + (16 – x)^3$. $S^{\prime} = 3x^2 – 3(16 – x)^2 = 0$ gives $x^2 = (16 – x)^2$, so $x = 16 – x$, i.e. $x = 8$. $S^{\prime\prime} = 6x + 6(16 – x) = 96 > 0$, so it is a minimum. The two numbers are $8$ and $8$.
17. A square piece of tin of side $18$ cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible?
Answer: If a square of side $x$ is cut from each corner, the box has height $x$ and a square base of side $(18 – 2x)$. Volume $V = x(18 – 2x)^2$.
$V^{\prime}(x) = (18 – 2x)^2 + x \cdot 2(18 – 2x)(-2) = (18 – 2x)\big[(18 – 2x) – 4x\big] = (18 – 2x)(18 – 6x)$. Setting $V^{\prime} = 0$ gives $x = 9$ (rejected, since the base would vanish) or $x = 3$. $V^{\prime\prime}(x) = 24x – 144$, and $V^{\prime\prime}(3) = -72 < 0$, so $x = 3$ gives the maximum. The side of the square to be cut off is $3$ cm.
18. A rectangular sheet of tin $45$ cm by $24$ cm is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?
Answer: Cutting a square of side $x$ from each corner gives a box of height $x$, length $(45 – 2x)$ and breadth $(24 – 2x)$. Volume $V = x(45 – 2x)(24 – 2x) = 4x^3 – 138x^2 + 1080x$.
$V^{\prime}(x) = 12x^2 – 276x + 1080 = 12(x^2 – 23x + 90) = 12(x – 5)(x – 18)$. So $x = 5$ or $x = 18$; but $x = 18$ makes $(24 – 2x)$ negative, so it is rejected. $V^{\prime\prime}(x) = 24x – 276$, and $V^{\prime\prime}(5) = -156 < 0$, so $x = 5$ gives the maximum. The side of the square to be cut off is $5$ cm.
19. Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Answer: Let the circle have radius $a$, and let an inscribed rectangle have sides $2x$ and $2y$. Its diagonal is a diameter, so $(2x)^2 + (2y)^2 = (2a)^2$, giving $x^2 + y^2 = a^2$.
Area $A = (2x)(2y) = 4xy = 4x\sqrt{a^2 – x^2}$. To maximise, consider $A^2 = 16x^2(a^2 – x^2)$. Let $Z = x^2(a^2 – x^2) = a^2x^2 – x^4$. Then $\frac{dZ}{dx} = 2a^2 x – 4x^3 = 2x(a^2 – 2x^2) = 0$ gives $x^2 = \frac{a^2}{2}$. Then $y^2 = a^2 – \frac{a^2}{2} = \frac{a^2}{2} = x^2$, so $x = y$. Since the sides $2x$ and $2y$ are equal, the rectangle of maximum area is a square.
20. Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.
Answer: Let the cylinder have radius $r$, height $h$ and fixed total surface area $S = 2\pi r^2 + 2\pi r h$. Then $h = \frac{S – 2\pi r^2}{2\pi r}$. Volume $V = \pi r^2 h = \pi r^2 \cdot \frac{S – 2\pi r^2}{2\pi r} = \frac{r}{2}(S – 2\pi r^2) = \frac{Sr}{2} – \pi r^3$.
$\frac{dV}{dr} = \frac{S}{2} – 3\pi r^2 = 0$ gives $S = 6\pi r^2$. $\frac{d^2V}{dr^2} = -6\pi r < 0$, so this is a maximum. Substituting $S = 6\pi r^2$ into $S = 2\pi r^2 + 2\pi r h$: $6\pi r^2 = 2\pi r^2 + 2\pi r h$, so $2\pi r h = 4\pi r^2$, giving $h = 2r$. Thus the height equals the diameter of the base.
21. Of all the closed cylindrical cans (right circular) of a given volume of $100$ cubic centimetres, find the dimensions of the can which has the minimum surface area.
Answer: Let the can have radius $r$ and height $h$, with $V = \pi r^2 h = 100$, so $h = \frac{100}{\pi r^2}$. Total surface area $S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \cdot \frac{100}{\pi r^2} = 2\pi r^2 + \frac{200}{r}$.
$\frac{dS}{dr} = 4\pi r – \frac{200}{r^2} = 0$ gives $4\pi r^3 = 200$, so $r^3 = \frac{50}{\pi}$, i.e. $r = \left(\frac{50}{\pi}\right)^{1/3}$ cm. $\frac{d^2S}{dr^2} = 4\pi + \frac{400}{r^3} > 0$, so this is a minimum. Also $h = \frac{100}{\pi r^2} = \frac{100}{\pi r^2}$; using $\pi r^3 = 50$ gives $\pi r^2 = \frac{50}{r}$, so $h = \frac{100 r}{50} = 2r$. Hence $r = \left(\frac{50}{\pi}\right)^{1/3}$ cm and $h = 2\left(\frac{50}{\pi}\right)^{1/3}$ cm.
22. A wire of length $28$ m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
Answer: Let a length $x$ (m) be bent into a square and the remaining $(28 – x)$ into a circle. The square has side $\frac{x}{4}$ and area $\frac{x^2}{16}$. The circle has circumference $28 – x = 2\pi r$, so $r = \frac{28 – x}{2\pi}$ and area $\pi r^2 = \frac{(28 – x)^2}{4\pi}$.
Combined area $A = \frac{x^2}{16} + \frac{(28 – x)^2}{4\pi}$. $\frac{dA}{dx} = \frac{x}{8} – \frac{28 – x}{2\pi} = 0$ gives $\pi x = 4(28 – x)$, so $x(\pi + 4) = 112$, i.e. $x = \frac{112}{\pi + 4}$. $\frac{d^2A}{dx^2} = \frac{1}{8} + \frac{1}{2\pi} > 0$, so it is a minimum. The pieces should be $\frac{112}{\pi + 4}$ m (for the square) and $28 – \frac{112}{\pi + 4} = \frac{28\pi}{\pi + 4}$ m (for the circle).
23. Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere.
Answer: Let the cone have base radius $\rho$ and height $H$. If the centre of the sphere is at distance $x$ from the base ($0 \leq x < R$), then $H = R + x$ and $\rho^2 = R^2 – x^2$. Volume $V = \frac{1}{3}\pi \rho^2 H = \frac{1}{3}\pi (R^2 – x^2)(R + x)$.
$\frac{dV}{dx} = \frac{\pi}{3}\big[-2x(R + x) + (R^2 – x^2)\big] = \frac{\pi}{3}(R^2 – 2Rx – 3x^2) = \frac{\pi}{3}(R – 3x)(R + x)$. Setting $\frac{dV}{dx} = 0$ gives $x = \frac{R}{3}$ (rejecting $x = -R$). $\frac{d^2V}{dx^2} = \frac{\pi}{3}(-2R – 6x) < 0$, so it is a maximum.
Then $H = R + \frac{R}{3} = \frac{4R}{3}$ and $\rho^2 = R^2 – \frac{R^2}{9} = \frac{8R^2}{9}$, so $V = \frac{1}{3}\pi \cdot \frac{8R^2}{9} \cdot \frac{4R}{3} = \frac{32\pi R^3}{81} = \frac{8}{27}\left(\frac{4}{3}\pi R^3\right)$. Hence the largest cone has volume $\frac{8}{27}$ of the volume of the sphere.
24. Show that the right circular cone of least curved surface and given volume has an altitude equal to $\sqrt{2}$ times the radius of the base.
Answer: Let the cone have base radius $r$, height $h$ and fixed volume $V = \frac{1}{3}\pi r^2 h$, so $h = \frac{3V}{\pi r^2}$. Curved surface $S = \pi r l = \pi r \sqrt{r^2 + h^2}$. It is convenient to minimise $S^2 = \pi^2 r^2 (r^2 + h^2) = \pi^2 r^4 + \pi^2 r^2 h^2$. Substituting $h^2 = \frac{9V^2}{\pi^2 r^4}$: $S^2 = \pi^2 r^4 + \frac{9V^2}{r^2}$.
$\frac{d(S^2)}{dr} = 4\pi^2 r^3 – \frac{18V^2}{r^3} = 0$ gives $4\pi^2 r^6 = 18V^2$, so $V^2 = \frac{2\pi^2 r^6}{9}$. But $V^2 = \frac{\pi^2 r^4 h^2}{9}$, so $\frac{\pi^2 r^4 h^2}{9} = \frac{2\pi^2 r^6}{9}$, giving $h^2 = 2r^2$, i.e. $h = \sqrt{2}\, r$. The second derivative is positive, confirming a minimum. Hence the altitude equals $\sqrt{2}$ times the radius.
25. Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan^{-1}\sqrt{2}$.
Answer: Let the slant height $l$ be fixed and let $\alpha$ be the semi-vertical angle. Then $r = l\sin\alpha$ and $h = l\cos\alpha$. Volume $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi l^3 \sin^2\alpha \cos\alpha$.
$\frac{dV}{d\alpha} = \frac{1}{3}\pi l^3\big[2\sin\alpha\cos^2\alpha – \sin^3\alpha\big] = \frac{1}{3}\pi l^3 \sin\alpha\,(2\cos^2\alpha – \sin^2\alpha)$. Setting this to zero (with $\sin\alpha \neq 0$) gives $2\cos^2\alpha = \sin^2\alpha$, i.e. $\tan^2\alpha = 2$, so $\tan\alpha = \sqrt{2}$. The sign of $\frac{dV}{d\alpha}$ changes from $+$ to $-$ here, so it is a maximum. Hence $\alpha = \tan^{-1}\sqrt{2}$.
26. Show that semi-vertical angle of right circular cone of given surface area and maximum volume is $\sin^{-1}\left(\frac{1}{3}\right)$.
Answer: Let the cone have radius $r$, slant height $l$ and fixed total surface area $S = \pi r^2 + \pi r l$, so $l = \frac{S – \pi r^2}{\pi r}$. With height $h = \sqrt{l^2 – r^2}$, the volume satisfies $V^2 = \frac{\pi^2 r^4}{9}(l^2 – r^2) = \frac{\pi^2 r^4}{9}\left[\left(\frac{S – \pi r^2}{\pi r}\right)^2 – r^2\right] = \frac{r^2}{9}\big[(S – \pi r^2)^2 – \pi^2 r^4\big] = \frac{r^2}{9}\big[S^2 – 2\pi r^2 S\big] = \frac{S r^2}{9}(S – 2\pi r^2)$.
Maximising $V^2 = \frac{S}{9}(S r^2 – 2\pi r^4)$: $\frac{d(V^2)}{dr} = \frac{S}{9}(2Sr – 8\pi r^3) = 0$ gives $S = 4\pi r^2$. Then $l = \frac{S – \pi r^2}{\pi r} = \frac{4\pi r^2 – \pi r^2}{\pi r} = 3r$, so $\sin\alpha = \frac{r}{l} = \frac{r}{3r} = \frac{1}{3}$. Hence the semi-vertical angle is $\sin^{-1}\left(\frac{1}{3}\right)$.
27. The point on the curve $x^2 = 2y$ which is nearest to the point $(0, 5)$ is
(A) $(2\sqrt{2}, 4)$ (B) $(2\sqrt{2}, 0)$ (C) $(0, 0)$ (D) $(2, 2)$
Answer: (A) $(2\sqrt{2}, 4)$. For a point $(x, y)$ on the curve, $x^2 = 2y$. The squared distance to $(0, 5)$ is $D^2 = x^2 + (y – 5)^2 = 2y + (y – 5)^2 = y^2 – 8y + 25$. Then $\frac{d(D^2)}{dy} = 2y – 8 = 0$ gives $y = 4$, and $x^2 = 2(4) = 8$, so $x = \pm 2\sqrt{2}$. The nearest point (matching the options) is $(2\sqrt{2}, 4)$.
28. For all real values of $x$, the minimum value of $\frac{1 – x + x^2}{1 + x + x^2}$ is
(A) $0$ (B) $1$ (C) $3$ (D) $\frac{1}{3}$
Answer: (D) $\frac{1}{3}$. Let $f(x) = \frac{x^2 – x + 1}{x^2 + x + 1}$. Then $f^{\prime}(x) = \frac{(2x – 1)(x^2 + x + 1) – (x^2 – x + 1)(2x + 1)}{(x^2 + x + 1)^2} = \frac{2(x^2 – 1)}{(x^2 + x + 1)^2}$. Setting $f^{\prime}(x) = 0$ gives $x = \pm 1$. At $x = 1$, $f = \frac{1}{3}$ (a minimum); at $x = -1$, $f = 3$ (a maximum). The minimum value is $\frac{1}{3}$.
29. The maximum value of $[x(x – 1) + 1]^{1/3}$, $0 \leq x \leq 1$, is
(A) $\left(\frac{1}{3}\right)^{1/3}$ (B) $\frac{1}{2}$ (C) $1$ (D) $0$
Answer: (C) $1$. Let $g(x) = x(x – 1) + 1 = x^2 – x + 1$. On $[0, 1]$, $g^{\prime}(x) = 2x – 1 = 0$ at $x = \frac{1}{2}$, where $g = \frac{3}{4}$ (the minimum). At the end points $g(0) = 1$ and $g(1) = 1$. So the maximum of $g$ on $[0, 1]$ is $1$, and hence the maximum of $g^{1/3}$ is $1^{1/3} = 1$.
Miscellaneous Exercise on Chapter 6
1. Show that the function given by $f(x) = \frac{\log x}{x}$ has maximum at $x = e$.
Answer: $f^{\prime}(x) = \frac{\frac{1}{x} \cdot x – \log x \cdot 1}{x^2} = \frac{1 – \log x}{x^2}$. Setting $f^{\prime}(x) = 0$ gives $\log x = 1$, i.e. $x = e$.
For $x < e$, $\log x < 1$ so $f^{\prime}(x) > 0$; for $x > e$, $\log x > 1$ so $f^{\prime}(x) < 0$. Since $f^{\prime}$ changes from positive to negative at $x = e$, by the first derivative test $f$ has a maximum at $x = e$.
2. The two equal sides of an isosceles triangle with fixed base $b$ are decreasing at the rate of $3$ cm per second. How fast is the area decreasing when the two equal sides are equal to the base?
Answer: Let each equal side be $a$. The height from the apex to the base is $\sqrt{a^2 – \frac{b^2}{4}}$, so the area is $A = \frac{1}{2}b\sqrt{a^2 – \frac{b^2}{4}}$.
$\frac{dA}{dt} = \frac{1}{2}b \cdot \frac{a}{\sqrt{a^2 – \frac{b^2}{4}}} \cdot \frac{da}{dt}$. Given $\frac{da}{dt} = -3$ cm/s. When $a = b$: $\sqrt{b^2 – \frac{b^2}{4}} = \frac{\sqrt{3}}{2}b$, so $\frac{dA}{dt} = \frac{1}{2}b \cdot \frac{b}{\frac{\sqrt{3}}{2}b} \cdot (-3) = \frac{1}{2}b \cdot \frac{2}{\sqrt{3}} \cdot (-3) = -\sqrt{3}\, b$. The area is decreasing at the rate of $\sqrt{3}\, b$ cm²/s.
3. Find the intervals in which the function $f$ given by $f(x) = \frac{4\sin x – 2x – x\cos x}{2 + \cos x}$ is (i) increasing (ii) decreasing.
Answer: Split the numerator: $f(x) = \frac{4\sin x – x(2 + \cos x)}{2 + \cos x} = \frac{4\sin x}{2 + \cos x} – x$. Differentiating, $f^{\prime}(x) = \frac{4\cos x(2 + \cos x) + 4\sin^2 x}{(2 + \cos x)^2} – 1 = \frac{8\cos x + 4}{(2 + \cos x)^2} – 1 = \frac{8\cos x + 4 – (2 + \cos x)^2}{(2 + \cos x)^2} = \frac{\cos x(4 – \cos x)}{(2 + \cos x)^2}$.
Since $4 – \cos x > 0$ and $(2 + \cos x)^2 > 0$, the sign of $f^{\prime}(x)$ is that of $\cos x$. On $[0, 2\pi]$, $\cos x > 0$ for $x \in \left(0, \frac{\pi}{2}\right) \cup \left(\frac{3\pi}{2}, 2\pi\right)$ and $\cos x < 0$ for $x \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$. Hence (i) $f$ is increasing on $\left(0, \frac{\pi}{2}\right)$ and $\left(\frac{3\pi}{2}, 2\pi\right)$, and (ii) decreasing on $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$.
4. Find the intervals in which the function $f$ given by $f(x) = x^3 + \frac{1}{x^3}$, $x \neq 0$, is (i) increasing (ii) decreasing.
Answer: $f^{\prime}(x) = 3x^2 – \frac{3}{x^4} = \frac{3(x^6 – 1)}{x^4} = \frac{3(x^2 – 1)(x^4 + x^2 + 1)}{x^4}$. Since $x^4 > 0$ and $x^4 + x^2 + 1 > 0$, the sign of $f^{\prime}(x)$ is that of $(x^2 – 1)$, which is positive when $|x| > 1$ and negative when $|x| < 1$.
Hence (i) $f$ is increasing on $(-\infty, -1)$ and $(1, \infty)$, and (ii) decreasing on $(-1, 0)$ and $(0, 1)$.
5. 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 axis.
Answer: Take the fixed vertex at $A = (a, 0)$. By symmetry the base is a vertical chord with end points $(x, y)$ and $(x, -y)$ on the ellipse, where $y = b\sqrt{1 – \frac{x^2}{a^2}}$. Base length $= 2y$ and height $= a – x$, so area $A(x) = \frac{1}{2}(2y)(a – x) = y(a – x)$.
Put $x = a\cos\theta$, $y = b\sin\theta$. Then area $= b\sin\theta(a – a\cos\theta) = ab\sin\theta(1 – \cos\theta)$. Differentiating $g(\theta) = \sin\theta(1 – \cos\theta)$: $g^{\prime}(\theta) = \cos\theta – \cos 2\theta = 0$ leads to $2\cos^2\theta – \cos\theta – 1 = 0$, i.e. $(2\cos\theta + 1)(\cos\theta – 1) = 0$, giving $\cos\theta = -\frac{1}{2}$, i.e. $\theta = \frac{2\pi}{3}$. Then $\sin\theta = \frac{\sqrt{3}}{2}$ and the maximum area is $ab \cdot \frac{\sqrt{3}}{2}\left(1 + \frac{1}{2}\right) = \frac{3\sqrt{3}}{4}ab$ square units.
6. A tank with rectangular base and rectangular sides, open at the top, is to be constructed so that its depth is $2$ m and volume is $8$ m³. If building of tank costs ₹$70$ per sq metre for the base and ₹$45$ per square metre for the sides, what is the cost of least expensive tank?
Answer: Let the base be $x$ by $y$ metres. Volume $= 2xy = 8$, so $xy = 4$, i.e. $y = \frac{4}{x}$. The base area is $xy = 4$ m², and the four sides have total area $2(2x) + 2(2y) = 4(x + y)$ m².
Cost $C = 70(4) + 45 \cdot 4(x + y) = 280 + 180\left(x + \frac{4}{x}\right)$. $\frac{dC}{dx} = 180\left(1 – \frac{4}{x^2}\right) = 0$ gives $x^2 = 4$, i.e. $x = 2$ (and $y = 2$). $\frac{d^2C}{dx^2} = 180 \cdot \frac{8}{x^3} > 0$, so it is a minimum. Least cost $= 280 + 180(2 + 2) = 280 + 720 = $ ₹$1000$.
7. 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.
Answer: Let the circle have radius $r$ and the square have side $s$, with $2\pi r + 4s = k$, so $s = \frac{k – 2\pi r}{4}$. Sum of areas $A = \pi r^2 + s^2$.
$\frac{dA}{dr} = 2\pi r + 2s \cdot \frac{ds}{dr} = 2\pi r + 2s\left(-\frac{2\pi}{4}\right) = 2\pi r – \pi s$. Setting $\frac{dA}{dr} = 0$ gives $2\pi r = \pi s$, i.e. $s = 2r$. $\frac{d^2A}{dr^2} = 2\pi + \frac{\pi^2}{2} > 0$, so this is a minimum. Hence the sum of the areas is least when the side of the square is double the radius of the circle.
8. 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.
Answer: Let the rectangle have width $2r$ (the diameter of the semicircle) and height $h$, so the semicircle has radius $r$. The perimeter is $2h + 2r + \pi r = 10$, giving $h = 5 – r – \frac{\pi r}{2}$.
Light admitted is proportional to the area $A = 2rh + \frac{1}{2}\pi r^2 = 2r\left(5 – r – \frac{\pi r}{2}\right) + \frac{1}{2}\pi r^2 = 10r – 2r^2 – \frac{1}{2}\pi r^2$. $\frac{dA}{dr} = 10 – 4r – \pi r = 0$ gives $r = \frac{10}{4 + \pi}$. $\frac{d^2A}{dr^2} = -4 – \pi < 0$, so this is a maximum. Then $h = 5 – r\left(1 + \frac{\pi}{2}\right) = \frac{10}{4 + \pi}$. So the width is $2r = \frac{20}{4 + \pi}$ m and the height is $h = \frac{10}{4 + \pi}$ m (with the semicircle radius $\frac{10}{4 + \pi}$ m).
9. A point on the hypotenuse of a triangle is at distance $a$ and $b$ from the sides of the triangle. Show that the minimum length of the hypotenuse is $\left(a^{2/3} + b^{2/3}\right)^{3/2}$.
Answer: Let the hypotenuse make an angle $\theta$ with the base. The point $P$ on the hypotenuse is at perpendicular distance $a$ from one side and $b$ from the other. Splitting the hypotenuse at the foot of the perpendiculars, its length is $L = \frac{a}{\sin\theta} + \frac{b}{\cos\theta}$.
$\frac{dL}{d\theta} = -\frac{a\cos\theta}{\sin^2\theta} + \frac{b\sin\theta}{\cos^2\theta} = 0$ gives $\frac{b\sin\theta}{\cos^2\theta} = \frac{a\cos\theta}{\sin^2\theta}$, so $\tan^3\theta = \frac{a}{b}$, i.e. $\tan\theta = \left(\frac{a}{b}\right)^{1/3}$.
Then $\sin\theta = \frac{a^{1/3}}{\sqrt{a^{2/3} + b^{2/3}}}$ and $\cos\theta = \frac{b^{1/3}}{\sqrt{a^{2/3} + b^{2/3}}}$. Substituting, $L = \frac{a}{\sin\theta} + \frac{b}{\cos\theta} = \left(a^{2/3} + b^{2/3}\right)^{1/2}\left(a^{2/3} + b^{2/3}\right) = \left(a^{2/3} + b^{2/3}\right)^{3/2}$. Since $\frac{d^2L}{d\theta^2} > 0$ here, this is the minimum length of the hypotenuse.
10. 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.
Answer: $f^{\prime}(x) = 4(x – 2)^3(x + 1)^3 + 3(x – 2)^4(x + 1)^2 = (x – 2)^3(x + 1)^2\big[4(x + 1) + 3(x – 2)\big] = (x – 2)^3(x + 1)^2(7x – 2)$. The critical points are $x = 2$, $x = -1$ and $x = \frac{2}{7}$.
Examining the sign of $f^{\prime}$: it changes from $+$ to $-$ at $x = \frac{2}{7}$ (local maxima) and from $-$ to $+$ at $x = 2$ (local minima). At $x = -1$ the factor $(x + 1)^2$ does not change sign, so $f^{\prime}$ does not change sign there — it is a point of inflexion. Hence (i) local maxima at $x = \frac{2}{7}$, (ii) local minima at $x = 2$, (iii) point of inflexion at $x = -1$.
11. Find the absolute maximum and minimum values of the function $f$ given by $f(x) = \cos^2 x + \sin x$, $x \in [0, \pi]$.
Answer: $f(x) = \cos^2 x + \sin x = 1 – \sin^2 x + \sin x$. $f^{\prime}(x) = -2\cos x\sin x + \cos x = \cos x(1 – 2\sin x) = 0$ gives $\cos x = 0$ (i.e. $x = \frac{\pi}{2}$) or $\sin x = \frac{1}{2}$ (i.e. $x = \frac{\pi}{6}, \frac{5\pi}{6}$).
Values: $f(0) = 1$, $f\left(\frac{\pi}{6}\right) = 1 – \frac{1}{4} + \frac{1}{2} = \frac{5}{4}$, $f\left(\frac{\pi}{2}\right) = 1 – 1 + 1 = 1$, $f\left(\frac{5\pi}{6}\right) = \frac{5}{4}$, $f(\pi) = 1$. Absolute maximum $= \frac{5}{4}$ (at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$); absolute minimum $= 1$ (at $x = 0$, $\frac{\pi}{2}$ and $\pi$).
12. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius $r$ is $\frac{4r}{3}$.
Answer: Let the cone have altitude $h$ and base radius $\rho$. If the centre of the sphere is at distance $(h – r)$ from the base, then $\rho^2 = r^2 – (h – r)^2 = 2rh – h^2$. Volume $V = \frac{1}{3}\pi \rho^2 h = \frac{1}{3}\pi(2rh – h^2)h = \frac{1}{3}\pi(2rh^2 – h^3)$.
$\frac{dV}{dh} = \frac{1}{3}\pi(4rh – 3h^2) = \frac{1}{3}\pi h(4r – 3h) = 0$ gives $h = \frac{4r}{3}$ (rejecting $h = 0$). $\frac{d^2V}{dh^2} = \frac{1}{3}\pi(4r – 6h)$, and at $h = \frac{4r}{3}$ this is $\frac{1}{3}\pi(4r – 8r) = -\frac{4\pi r}{3} < 0$, confirming a maximum. Hence the altitude of the cone of maximum volume is $\frac{4r}{3}$.
13. Let $f$ be a function defined on $[a, b]$ such that $f^{\prime}(x) > 0$ for all $x \in (a, b)$. Then prove that $f$ is an increasing function on $(a, b)$.
Answer: Take any $x_1, x_2 \in (a, b)$ with $x_1 < x_2$. Since $f$ is differentiable on $[a, b]$, it is continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$. By the Mean Value Theorem there exists $c \in (x_1, x_2)$ with $f(x_2) – f(x_1) = f^{\prime}(c)(x_2 – x_1)$. Here $f^{\prime}(c) > 0$ and $x_2 – x_1 > 0$, so $f(x_2) – f(x_1) > 0$, i.e. $f(x_1) < f(x_2)$. Since this holds for every such pair, $f$ is increasing on $(a, b)$.
14. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius $R$ is $\frac{2R}{\sqrt{3}}$. Also find the maximum volume.
Answer: Let the cylinder have radius $x$ and height $h$. Then $x^2 + \left(\frac{h}{2}\right)^2 = R^2$, so $x^2 = R^2 – \frac{h^2}{4}$. Volume $V = \pi x^2 h = \pi\left(R^2 – \frac{h^2}{4}\right)h = \pi R^2 h – \frac{\pi h^3}{4}$.
$\frac{dV}{dh} = \pi R^2 – \frac{3\pi h^2}{4} = 0$ gives $h^2 = \frac{4R^2}{3}$, i.e. $h = \frac{2R}{\sqrt{3}}$. $\frac{d^2V}{dh^2} = -\frac{3\pi h}{2} < 0$, confirming a maximum. The maximum volume is $V = \pi\left(R^2 – \frac{R^2}{3}\right)\frac{2R}{\sqrt{3}} = \pi \cdot \frac{2R^2}{3} \cdot \frac{2R}{\sqrt{3}} = \frac{4\pi R^3}{3\sqrt{3}} = \frac{4\sqrt{3}\,\pi R^3}{9}$ cubic units.
15. Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height $h$ and semi-vertical angle $\alpha$ is one-third that of the cone and the greatest volume of cylinder is $\frac{4}{27}\pi h^3 \tan^2\alpha$.
Answer: The cone has base radius $R = h\tan\alpha$. Inscribe a cylinder of radius $x$ and height $H$. By similar triangles, at height $H$ the remaining cone height is $h – H$ and $x = (h – H)\tan\alpha$. Volume $V = \pi x^2 H = \pi \tan^2\alpha\,(h – H)^2 H$.
$\frac{dV}{dH} = \pi\tan^2\alpha\big[(h – H)^2 + H \cdot 2(h – H)(-1)\big] = \pi\tan^2\alpha\,(h – H)(h – 3H) = 0$ gives $H = \frac{h}{3}$ (rejecting $H = h$). The second derivative is negative here, so it is a maximum; thus the cylinder height is one-third of the cone height.
Greatest volume $= \pi\tan^2\alpha\left(h – \frac{h}{3}\right)^2 \cdot \frac{h}{3} = \pi\tan^2\alpha \cdot \frac{4h^2}{9} \cdot \frac{h}{3} = \frac{4}{27}\pi h^3 \tan^2\alpha$.
16. A cylindrical tank of radius $10$ m is being filled with wheat at the rate of $314$ cubic metre 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
Answer: (A) $1$ m/h. The volume is $V = \pi r^2 d = \pi(10)^2 d = 100\pi d$, where $d$ is the depth. So $\frac{dV}{dt} = 100\pi \frac{dd}{dt}$. Given $\frac{dV}{dt} = 314$, we get $\frac{dd}{dt} = \frac{314}{100\pi} = \frac{314}{314} = 1$ m/h (taking $\pi \approx 3.14$).
Additional Questions and Answers
Multiple Choice Questions
1. If the radius of a circle increases at $2$ cm/s, the rate of increase of its area when the radius is $5$ cm is
(A) $10\pi$ cm²/s (B) $20\pi$ cm²/s (C) $5\pi$ cm²/s (D) $25\pi$ cm²/s
Answer: (B) $20\pi$ cm²/s. Since $\frac{dA}{dt} = 2\pi r \frac{dr}{dt} = 2\pi(5)(2) = 20\pi$.
2. The function $f(x) = x^2 – 2x + 5$ is increasing on
(A) $(-\infty, 1)$ (B) $(1, \infty)$ (C) $\mathbb{R}$ (D) $(-1, 1)$
Answer: (B) $(1, \infty)$. $f^{\prime}(x) = 2x – 2 = 2(x – 1) > 0$ when $x > 1$.
3. The slope of the tangent to the curve $y = x^3$ at the point $(1, 1)$ is
(A) $1$ (B) $2$ (C) $3$ (D) $0$
Answer: (C) $3$. The slope is $\frac{dy}{dx} = 3x^2$, and at $x = 1$ this equals $3$.
4. The function $f(x) = x^3$ has at $x = 0$
(A) a local maximum (B) a local minimum (C) a point of inflexion (D) an absolute maximum
Answer: (C) a point of inflexion. $f^{\prime}(x) = 3x^2$ is zero at $x = 0$ but does not change sign there, so $x = 0$ is a point of inflexion.
5. At a point of local maximum of a twice differentiable function $f$,
(A) $f^{\prime}(c) = 0$ and $f^{\prime\prime}(c) > 0$ (B) $f^{\prime}(c) = 0$ and $f^{\prime\prime}(c) < 0$ (C) $f^{\prime}(c) > 0$ (D) $f^{\prime\prime}(c) = 0$ always
Answer: (B) $f^{\prime}(c) = 0$ and $f^{\prime\prime}(c) < 0$. This is the second derivative test for a local maximum.
6. The maximum value of $\sin x \cos x$ is
(A) $1$ (B) $2$ (C) $\frac{1}{2}$ (D) $\frac{1}{4}$
Answer: (C) $\frac{1}{2}$. Since $\sin x \cos x = \frac{1}{2}\sin 2x$ and the maximum of $\sin 2x$ is $1$, the maximum value is $\frac{1}{2}$.
7. The absolute minimum value of $f(x) = x^2$ on $[-2, 3]$ is
(A) $4$ (B) $9$ (C) $0$ (D) $-2$
Answer: (C) $0$. The minimum of $x^2$ occurs at the interior critical point $x = 0$, giving $f(0) = 0$.
8. Marginal revenue is defined as
(A) $\frac{dC}{dx}$ (B) $\frac{dR}{dx}$ (C) $\frac{R}{x}$ (D) $R \cdot x$
Answer: (B) $\frac{dR}{dx}$. Marginal revenue is the instantaneous rate of change of total revenue with respect to the number of units sold.
Fill in the Blanks
1. A point $c$ in the domain of $f$ where $f^{\prime}(c) = 0$ or $f$ is not differentiable is called a ________ point.
Answer: critical
2. A function is increasing on an interval if its derivative is ________ there.
Answer: non-negative (i.e. $f^{\prime}(x) \geq 0$)
3. If $f^{\prime}(c) = 0$ and $f^{\prime\prime}(c) > 0$, then $x = c$ is a point of local ________.
Answer: minima
4. A point where $f^{\prime}$ vanishes but does not change sign is called a point of ________.
Answer: inflexion
5. The maximum value of $\sin x + \cos x$ is ________.
Answer: $\sqrt{2}$
True or False
1. Every critical point of a function is a point of local maximum or minimum.
Answer: False. A critical point may be a point of inflexion, for example $x = 0$ for $f(x) = x^3$.
2. If $f^{\prime}(x) > 0$ on an interval, then $f$ is increasing on that interval.
Answer: True. A positive derivative throughout an interval means the function is increasing there.
3. A continuous function on a closed interval always attains its absolute maximum and minimum.
Answer: True. This is guaranteed by the extreme value theorem for continuous functions on a closed interval.
4. The second derivative test always decides whether a critical point is a maximum or a minimum.
Answer: False. The test fails when $f^{\prime}(c) = 0$ and $f^{\prime\prime}(c) = 0$; then one must use the first derivative test.
5. The absolute maximum of a function on a closed interval can occur at an end point.
Answer: True. The absolute extremum may occur either at an interior critical point or at an end point of the interval.
Short Answer Questions
1. Find the intervals in which $f(x) = x^2 – 4x + 6$ is increasing and decreasing.
Answer: $f^{\prime}(x) = 2x – 4 = 0$ at $x = 2$. For $x < 2$, $f^{\prime}(x) < 0$ (decreasing); for $x > 2$, $f^{\prime}(x) > 0$ (increasing). So $f$ is decreasing on $(-\infty, 2)$ and increasing on $(2, \infty)$.
2. Find the local maximum and local minimum values of $f(x) = x^3 – 3x + 3$.
Answer: $f^{\prime}(x) = 3x^2 – 3 = 3(x – 1)(x + 1) = 0$ at $x = \pm 1$. $f^{\prime\prime}(x) = 6x$. At $x = 1$, $f^{\prime\prime} > 0$: local minimum $f(1) = 1 – 3 + 3 = 1$. At $x = -1$, $f^{\prime\prime} < 0$: local maximum $f(-1) = -1 + 3 + 3 = 5$.
3. The volume of a spherical balloon is increasing at $25$ cm³/s. Find the rate of change of its radius when the radius is $5$ cm.
Answer: $V = \frac{4}{3}\pi r^3$, so $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. Then $\frac{dr}{dt} = \frac{25}{4\pi r^2}$. At $r = 5$, $\frac{dr}{dt} = \frac{25}{4\pi(25)} = \frac{1}{4\pi}$ cm/s.
4. Find the absolute maximum and minimum values of $f(x) = 2x^3 – 3x^2 – 12x + 1$ on $[-2, 3]$.
Answer: $f^{\prime}(x) = 6x^2 – 6x – 12 = 6(x – 2)(x + 1) = 0$ at $x = 2, -1$. Values: $f(-2) = -16 – 12 + 24 + 1 = -3$, $f(-1) = -2 – 3 + 12 + 1 = 8$, $f(2) = 16 – 12 – 24 + 1 = -19$, $f(3) = 54 – 27 – 36 + 1 = -8$. Absolute maximum $= 8$ at $x = -1$; absolute minimum $= -19$ at $x = 2$.
Key Terms
| Term | Meaning |
|---|---|
| Rate of change | The derivative $\frac{dy}{dx}$, which measures how fast $y$ changes with respect to $x$. |
| Related rates | Problems in which several quantities vary with time and their rates are linked by the Chain Rule. |
| Marginal cost | The rate of change of total cost with output, $\frac{dC}{dx}$. |
| Marginal revenue | The rate of change of total revenue with the number of units sold, $\frac{dR}{dx}$. |
| Increasing function | A function for which $x_1 < x_2 \Rightarrow f(x_1) \leq f(x_2)$; equivalently $f^{\prime}(x) \geq 0$. |
| Decreasing function | A function for which $x_1 < x_2 \Rightarrow f(x_1) \geq f(x_2)$; equivalently $f^{\prime}(x) \leq 0$. |
| Critical point | A point $c$ where $f^{\prime}(c) = 0$ or $f$ is not differentiable. |
| Local maximum / minimum | A value that is the greatest / least among nearby values of the function. |
| First derivative test | Determines a local maximum, minimum or inflexion from the sign change of $f^{\prime}$ at a critical point. |
| Second derivative test | At $f^{\prime}(c) = 0$: local maximum if $f^{\prime\prime}(c) < 0$, local minimum if $f^{\prime\prime}(c) > 0$. |
| Point of inflexion | A critical point where $f^{\prime}$ does not change sign, so it is neither a maximum nor a minimum. |
| Absolute maximum / minimum | The greatest / least value of $f$ over its whole domain or a given closed interval. |