Maxima and Minima for CUET Maths

Q: What is the maximum value of a function?

The maximum value of f is a value f(c) that no other value of f exceeds: f(c) ≥ f(x) for every x in the domain. The minimum value is a value f(d) with f(d) ≤ f(x) for every x. A function may have both, one, or neither.

Q: What is a local maximum?

A function has a local maximum at c when f(c) is the largest value among the points near c, on some open interval around it; a local minimum is the same with "smallest". "Local" matters: a local maximum is a peak, not necessarily the highest point of the whole graph, and a local minimum value can even be larger than a local maximum value.

Q: What is a critical point?

A critical point is a point c of the domain where f'(c) = 0 or f is not differentiable. Every local maximum or minimum of a differentiable function is at a critical point, because the tangent is flat at a peak or a trough. But not every critical point is a peak or a trough.

Q: How does the second derivative test work?

At a critical point c with f'(c) = 0, look at f''(c). If it is negative, the graph bends downwards like the top of a hill, and c is a local maximum; if it is positive, the graph bends upwards like a bowl, and c is a local minimum; if it is 0, the test decides nothing, and the first derivative test must be used.

About 10 min to learn · 80 questions

Maxima and minima are part of Section A of CUET Maths, the section every candidate answers, whether you take Mathematics or Applied Mathematics. They turn derivatives into answers: the highest point, the least cost, the largest profit or area. Learn each rule here, from critical points and the first and second derivative tests to the greatest and least values on an interval and word problems, with worked examples, time-saving shortcuts and the traps CUET sets, then practise on free questions in CUET's own formats, each fully solved.

Syllabus area: Section A1

On this page · 10 sections

01Maximum and minimum values
02Local maxima and minima
03The second derivative test
04Greatest and least values on a closed interval
05Applied problems
06How CUET asks this chapter
07Exam shortcuts
08Common mistakes
09Quick revision
10Check your understanding

Maximum and minimum values

The ball we have been following, at height $h = 20t - 5t^2$ metres, rises and then falls. At some moment it is higher than at any other: that height is the maximum value of $h$ . This chapter finds such values, first by looking, then with derivatives.

What is the maximum value of a function?

The maximum value of $f$ is a value $f(c)$ that no other value of $f$ exceeds: $f(c) \ge f(x)$ for every $x$ in the domain. The minimum value is a value $f(d)$ with $f(d) \le f(x)$ for every $x$ . A function may have both, one, or neither.

Many questions can be answered without a derivative, because a square is never negative and $\sin$ and $\cos$ stay between $-1$ and 1. So $(x + 1)^2 - 4$ has minimum value $-4$ , at $x = -1$ , and no maximum; and $x^3$ has neither.

Example 1 · By inspection

Find the maximum and minimum values of $f(x) = 4 - 3\sin 2x$ .

$\sin 2x$ runs between $-1$ and 1, so $-3\sin 2x$ runs between $-3$ and 3.
Adding 4, $f(x)$ runs between 1 and 7.

Answer

Maximum value 7, when $\sin 2x = -1$ ; minimum value 1, when $\sin 2x = 1$ .

Watch out

Read what is asked: the value is $f(c)$ , the point is $c$ . Options often include both.

Local maxima and minima

What is a local maximum?

A function has a local maximum at $c$ when $f(c)$ is the largest value among the points near $c$ , on some open interval around it; a local minimum is the same with "smallest". "Local" matters: a local maximum is a peak, not necessarily the highest point of the whole graph, and a local minimum value can even be larger than a local maximum value.

What is a critical point?

A critical point is a point $c$ of the domain where $f'(c) = 0$ or $f$ is not differentiable. Every local maximum or minimum of a differentiable function is at a critical point, because the tangent is flat at a peak or a trough. But not every critical point is a peak or a trough.

The first derivative test

At a critical point $c$ : if $f'$ changes from $+$ to $-$ , $c$ is a local maximum; if from $-$ to $+$ , a local minimum; if it does not change sign, neither.

Example 2 · Two critical points

Find the local maxima and minima of $f(x) = x^3 - 3x^2 + 4$ .

$f'(x) = 3x^2 - 6x = 3x(x - 2)$ , so the critical points are $x = 0$ and $x = 2$ .
$f'$ is positive before 0, negative between 0 and 2, and positive after 2.

Answer

A local maximum at $x = 0$ , value $f(0) = 4$ ; a local minimum at $x = 2$ , value $f(2) = 0$ .

Watch out

$f(x) = (x - 2)^3$ has $f'(2) = 0$ , yet $f'(x) = 3(x - 2)^2$ is positive on both sides of 2. So $x = 2$ is a critical point that is neither a maximum nor a minimum, a point of inflection. A squared factor in $f'$ never changes sign.

The second derivative test

How does the second derivative test work?

At a critical point $c$ with $f'(c) = 0$ , look at $f''(c)$ . If it is negative, the graph bends downwards like the top of a hill, and $c$ is a local maximum; if it is positive, the graph bends upwards like a bowl, and $c$ is a local minimum; if it is 0, the test decides nothing, and the first derivative test must be used.

For the ball, $h'(t) = 20 - 10t = 0$ at $t = 2$ , and $h''(t) = -10 < 0$ : a maximum. Its greatest height is $h(2) = 40 - 20 = 20$ metres.

Example 3 · Values from the second derivative test

Find the local maximum and minimum values of $f(x) = 2x^3 - 3x^2 - 12x + 5$ .

$f'(x) = 6x^2 - 6x - 12 = 6(x - 2)(x + 1)$ , which is 0 at $x = -1$ and $x = 2$ .
$f''(x) = 12x - 6$ : $f''(-1) = -18 < 0$ , a maximum; $f''(2) = 18 > 0$ , a minimum.
$f(-1) = -2 - 3 + 12 + 5 = 12$ and $f(2) = 16 - 12 - 24 + 5 = -15$ .

Answer

Local maximum value 12, at $x = -1$ ; local minimum value $-15$ , at $x = 2$ .

Watch out

$f''(c) = 0$ does not mean "no extremum". $f(x) = x^6$ has $f''(0) = 0$ , but $x^6 \ge 0$ , so 0 is a minimum. When the second derivative test fails, go back to the sign of $f'$ .

Greatest and least values on a closed interval

A function that is continuous on a closed interval $[a, b]$ always has a greatest and a least value there, its absolute maximum and minimum. They can only occur at a critical point inside the interval or at one of the two ends.

How do you find the absolute maximum on a closed interval?

Find the critical points that lie inside the interval, then work out $f$ at each of them and at both end points. The largest of these numbers is the absolute maximum and the smallest is the absolute minimum; no second derivative is needed.

Example 4 · An end point wins

Find the greatest and least values of $f(x) = x^3 - 27x$ on $[-4, 7]$ .

$f'(x) = 3x^2 - 27 = 3(x - 3)(x + 3)$ , so the critical points are $x = -3$ and $x = 3$ , both inside.
$f(-4) = -64 + 108 = 44$ , $f(-3) = -27 + 81 = 54$ , $f(3) = 27 - 81 = -54$ and $f(7) = 343 - 189 = 154$ .

Answer

The greatest value is 154, at the end $x = 7$ ; the least value is $-54$ , at $x = 3$ .

The local maximum at $x = -3$ , value 54, is not the greatest value here: the end point beats it. That is why the ends are always checked.

Applied problems

How do you solve a maximum or minimum word problem?

Name the quantity to be made largest or smallest and write it in terms of the measurements. Use the condition the problem gives to reduce it to one variable, then find the critical point and confirm with the first or second derivative test. Answer exactly what is asked: a length, an area or a number of units.

Example 5 · The cheapest box

An open box with a square base must hold 500 cm³. What size uses the least material?

With base side $x$ and height $h$ : $x^2h = 500$ , so $h = \frac{500}{x^2}$ .
The material is the base plus four sides: $S = x^2 + 4xh = x^2 + \frac{2000}{x}$ .
$S' = 2x - \frac{2000}{x^2} = 0$ gives $x^3 = 1000$ , $x = 10$ ; $S'' = 2 + \frac{4000}{x^3} > 0$ , a minimum.

Answer

Base 10 cm by 10 cm and height $h = \frac{500}{100} = 5$ cm, using $S = 100 + 200 = 300$ cm².

Key idea

Three results come up so often that they are worth knowing. If two numbers have a fixed sum, their product is largest when they are equal. Of all rectangles with a given perimeter, the square has the largest area. A closed cylinder of given volume uses the least metal when its height equals its diameter.

How CUET asks this chapter

By inspection: the maximum or minimum value of a square plus a constant, or of $a\sin kx + b$ .
Local extrema: the critical points, the point of a local maximum or minimum, or its value, sometimes as a match-the-list question.
The tests: what $f'(c) = 0$ with a given $f''(c)$ says, or a statement question on critical points and extrema.
Closed intervals: the greatest or least value on $[a, b]$ , where the end points often win.
Word problems: largest product or area, least surface or cost, largest profit or revenue.

Exam shortcuts

Maximum of $a\sin kx + b$ is $\lvert a \rvert + b$ ; minimum is $b - \lvert a \rvert$ .
$x + \frac{a}{x}$ for $x > 0$ has its minimum $2\sqrt{a}$ at $x = \sqrt{a}$ .
Fixed sum $S$ : largest product $\frac{S^2}{4}$ ; least sum of squares $\frac{S^2}{2}$ .
Perimeter $P$ : largest rectangle is a square of area $\frac{P^2}{16}$ .
Profit or revenue $-x^2 + bx + c$ : best at $x = \frac{b}{2}$ .
On a closed interval, compare the critical values and the end values; no second derivative needed.

Common mistakes

Giving the point $c$ when the value $f(c)$ is asked, or the other way round.
Calling every critical point a maximum or minimum.
Treating $f''(c) = 0$ as "no extremum".
Forgetting to check the end points of a closed interval.
In a word problem, not using the condition to reduce to one variable, or answering with the wrong quantity.

Quick revision

Values

Maximum: $f(c) \ge f(x)$ for all $x$ ; a square $\ge 0$ , $-1 \le \sin \le 1$
The value is $f(c)$ ; the point is $c$

Local extrema

Critical point: $f'(c) = 0$ or not differentiable
$f'$ : $+$ to $-$ maximum, $-$ to $+$ minimum, no change neither
$f''(c) < 0$ maximum, $> 0$ minimum, $= 0$ test fails

Absolute and applied

Closed interval: compare critical points and both ends
Word problem: one variable, differentiate, check, answer what is asked

Differentiation has an inverse, and it gives the area under a curve: indefinite integrals.

Next · 09Quick revision

Check your understanding

80 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

Maximum and minimum valuesQ1 of 80Q1

Maximum and minimum values

13 questions

Question 1

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What is the minimum value of $f(x) = (2x - 3)^2 + 5$ ?

0

Correct answerYour answer

\frac{3}{2}

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5

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14

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AnswerC · $5$

A square is never negative, so $(2x - 3)^2 \ge 0$ and $f(x) \ge 5$ .
The square is 0 at $x = \frac{3}{2}$ , where $f = 5$ : the minimum value is 5. (There is no maximum: $f$ grows without limit.)

Worth noting

$\frac{3}{2}$ is where the minimum occurs; the question asks for the value.

Question 2

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What is the maximum value of $f(x) = 2\sin 3x + 5$ ?

5

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7

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8

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10

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AnswerB · $7$

$\sin 3x$ lies between $-1$ and 1, whatever the 3 inside, so $2\sin 3x$ lies between $-2$ and 2.
So $f$ lies between 3 and 7, and its maximum value is 7.

Worth noting

8 multiplies by the 3 inside the sine; that changes how fast it oscillates, not how high.

Question 3

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What is the maximum value of $f(x) = 7 - (x + 2)^2$ ?

-2

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7

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8

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9

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AnswerB · $7$

$(x + 2)^2 \ge 0$ , so $f(x) \le 7$ .
At $x = -2$ the square is 0 and $f = 7$ , so the maximum value is 7.

Worth noting

$-2$ is the point of the maximum, not its value.

Question 4

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What is the minimum value of $f(x) = 2x^2 - 8x + 11$ ?

-3

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2

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3

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11

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AnswerC · $3$

Complete the square: $2x^2 - 8x + 11 = 2(x - 2)^2 + 3$ .
$2(x - 2)^2 \ge 0$ , and it is 0 at $x = 2$ , so the minimum value is 3.

Worth noting

2 is where the minimum occurs; the question asks for the value.

Question 5

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What is the maximum value of $f(x) = 5 - 4\cos x$ ?

1

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4

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5

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9

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AnswerD · $9$

$-1 \le \cos x \le 1$ , so $-4\cos x$ runs from $-4$ to 4.
The maximum, at $\cos x = -1$ , is $5 + 4 = 9$ .

Worth noting

1 is the minimum, at $\cos x = 1$ : the minus sign turns the cosine's peak into $f$ 's lowest point.

Question 6

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What is the maximum value of $f(x) = 3 - \lvert x + 1 \rvert$ ?

-1

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3

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4

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5

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AnswerB · $3$

$\lvert x + 1 \rvert \ge 0$ , so $f(x) \le 3$ .
At $x = -1$ , $f = 3$ : the maximum value is 3. There is no minimum.

Worth noting

$-1$ is the point of the maximum, not its value.

Question 7

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What is the minimum value of $f(x) = 9x^2 + 6x + 4$ ?

- \frac{1}{3}

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1

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2

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3

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AnswerD · $3$

Complete the square: $9x^2 + 6x + 4 = (3x + 1)^2 + 3$ .
The square is least, 0, at $x = -\frac{1}{3}$ , so the minimum value is 3.

Worth noting

4 is $f(0)$ , but the parabola dips lower, to 3, just left of 0.

Question 8

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What is the least value of $f(x) = \lvert \sin 2x + 3 \rvert$ ?

0

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1

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2

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3

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AnswerC · $2$

$\sin 2x + 3$ lies between 2 and 4, so it is always positive and the absolute value changes nothing.
The least value is 2, when $\sin 2x = -1$ .

Worth noting

0 assumes an absolute value can always reach 0; here the inside never does.

Question 9

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Which is true of $f(x) = x^3 + 2$ on $\mathbb{R}$ ?

It has neither a maximum nor a minimum valueCorrect answerYour answerIts minimum value is 2Correct answerYour answerIts maximum value is 2Correct answerYour answerIt has a minimum value but no maximum valueCorrect answerYour answer

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AnswerA · It has neither a maximum nor a minimum value

As $x$ grows, $x^3$ grows without limit, and as $x$ falls, $x^3$ falls without limit.
So $x^3 + 2$ takes values above and below any number: there is no maximum and no minimum.

Worth noting

2 is $f(0)$ ; $f(-1) = 1$ is already smaller and $f(1) = 3$ larger.

Question 10

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Read the two statements and choose the correct option.

I. $f(x) = \lvert x - 1 \rvert + 2$ has minimum value 2.

II. $f(x) = \lvert x - 1 \rvert + 2$ has maximum value 3.

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerA · Only I

$\lvert x - 1 \rvert \ge 0$ , with 0 at $x = 1$ , so the minimum value is 2: I is true.
$\lvert x - 1 \rvert$ grows without limit, so there is no maximum: II is false.

Question 11

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What is the maximum value of $f(x) = 3\sin x + 4\cos x$ ?

1

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4

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5

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7

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AnswerC · $5$

$a\sin x + b\cos x$ can be written $\sqrt{a^{2} + b^{2}}\,\sin(x + \alpha)$ , so it runs from $-\sqrt{a^{2} + b^{2}}$ to $\sqrt{a^{2} + b^{2}}$ .
$\sqrt{9 + 16} = 5$ .

Worth noting

7 adds the two peaks, but $\sin x$ and $\cos x$ never reach 1 at the same $x$ .

Question 12

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Read the two statements and choose the correct option.

I. $f(x) = \sin x + 2$ has maximum value 3 and minimum value 1.

II. $f(x) = \frac{1}{x^{2} + 1}$ has maximum value 1 and no minimum value.

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerC · Both I and II

I: $\sin x$ runs from $-1$ to 1, so $f$ runs from 1 to 3. True.
II: $x^{2} + 1 \ge 1$ , so $f \le 1$ , reached at $x = 0$ ; $f$ gets close to 0 but never reaches it. True.

Question 13

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Which is true of $f(x) = e^{x}$ on $\mathbb{R}$ ?

Its minimum value is 0Correct answerYour answerIts minimum value is 1Correct answerYour answerIt has a maximum value but no minimum valueCorrect answerYour answerIt has neither a maximum nor a minimum valueCorrect answerYour answer

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AnswerD · It has neither a maximum nor a minimum value

$e^{x}$ grows without limit, so there is no maximum.
It comes as close to 0 as we like but never reaches it, since $e^{x} > 0$ ; so no value is the least.

Worth noting

0 is a bound that is never reached, so it is not a minimum value; 1 is only $f(0)$ .

Local extrema and the first derivative test

16 questions

Question 14

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What are the critical points of $f(x) = x^{3} - 3 x^{2} - 9 x + 5$ ?

x = -3, 1

Correct answerYour answer

x = -1

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x = 3

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x = -1, 3

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AnswerD · $x = -1, 3$

Critical points are where $f'(x) = 0$ : $f'(x) = 3 \left(x - 3\right) \left(x + 1\right)$ .
So $x = -1, 3$ .

Worth noting

Reading the roots off the factors with the signs flipped is the usual slip.

Question 15

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At which point does $f(x) = x^{3} - 12 x + 4$ have a local minimum?

-12

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-2

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2

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3

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AnswerC · $2$

$f'(x) = 3 \left(x - 2\right) \left(x + 2\right)$ .
$f'$ changes from negative to positive at $x = 2$ , so $f$ falls then rises there: a local minimum.

Worth noting

At $x = -2$ the sign changes the other way: that is the local maximum.

Question 16

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What are the critical points of $f(x) = x + \frac{1}{x}$ ?

x = -1

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x = 1

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x = 0, 2

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x = -1, 1

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AnswerD · $x = -1, 1$

Critical points are where $f'(x) = 0$ : $f'(x) = \frac{\left(x - 1\right) \left(x + 1\right)}{x^{2}}$ .
So $x = -1, 1$ .

Worth noting

$x = 0$ is not in the domain, so it is not a critical point here.

Question 17

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$f'$ changes sign from negative to positive as $x$ passes through 2. What does $f$ have at $x = 2$ ?

A local maximumCorrect answerYour answerA local minimumCorrect answerYour answerA point of inflexionCorrect answerYour answerNeither a maximum nor a minimumCorrect answerYour answer

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AnswerB · A local minimum

Negative $f'$ before 2 means $f$ falls; positive after means it rises.
Falling then rising: a local minimum at $x = 2$ .

Worth noting

A local maximum is the other change, from positive to negative.

Question 18

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A function has $f'(x) = (x - 1)(x - 3)^2$ . Where does it have local extrema?

A local minimum at

x = 1

and a local maximum at

x = 3

Correct answerYour answerA local maximum at

x = 1

onlyCorrect answerYour answerA local minimum at

x = 3

onlyCorrect answerYour answerA local minimum at

x = 1

onlyCorrect answerYour answer

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AnswerD · A local minimum at $x = 1$ only

$(x - 3)^2 \ge 0$ , so the sign of $f'$ is the sign of $x - 1$ : negative before 1, positive after.
So $x = 1$ is a local minimum. At $x = 3$ , $f'$ is positive on both sides: no extremum there.

Worth noting

A squared factor makes $f' = 0$ without a change of sign, so $x = 3$ is a critical point but not an extremum.

Question 19

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Does $f(x) = x^3$ have a local extremum at $x = 0$ ?

No:

f'

does not change sign at 0, so it is a point of inflectionCorrect answerYour answerYes, a local minimum, because

f'(0) = 0

Correct answerYour answerYes, a local maximum, because

f''(0) = 0

Correct answerYour answerYes, both a maximum and a minimumCorrect answerYour answer

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AnswerA · No: $f'$ does not change sign at 0, so it is a point of inflection

$f'(x) = 3x^2$ is positive on both sides of 0.
$f'(0) = 0$ makes 0 a critical point, but with no change of sign there is no maximum or minimum.

Question 20

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Read the two statements and choose the correct option.

I. Every local maximum of a differentiable function is a critical point.

II. Every critical point of a differentiable function is a local maximum or a local minimum.

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerA · Only I

At a local maximum of a differentiable function $f' = 0$ , so I is true.
$x^3$ has the critical point 0 that is neither, so II is false.

Question 21

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What are the critical points of $f(x) = 2 x^{3} - 9 x^{2} + 12 x + 3$ ?

x = -2, -1

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x = 1, 2

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x = 1

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x = 2

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AnswerB · $x = 1, 2$

Critical points are where $f'(x) = 0$ : $f'(x) = 6 \left(x - 2\right) \left(x - 1\right)$ .
So $x = 1, 2$ .

Worth noting

Reading the roots off the factors with the signs flipped is the usual slip.

Question 22

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At which point does $f(x) = - x^{3} + 3 x^{2} + 9 x - 2$ have a local maximum?

-3

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-1

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1

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3

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AnswerD · $3$

$f'(x) = - 3 \left(x - 3\right) \left(x + 1\right)$ .
$f'$ changes from positive to negative at $x = 3$ , so $f$ rises then falls there: a local maximum.

Worth noting

At $x = -1$ the sign changes the other way: that is the local minimum.

Question 23

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At which point does $f(x) = 2 x^{3} + 3 x^{2} - 12 x + 1$ have a local maximum?

-2

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1

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2

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21

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AnswerA · $-2$

$f'(x) = 6 \left(x - 1\right) \left(x + 2\right)$ .
$f'$ changes from positive to negative at $x = -2$ , so $f$ rises then falls there: a local maximum.

Worth noting

At $x = 1$ the sign changes the other way: that is the local minimum.

Question 24

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What are the critical points of $f(x) = x^{4} - 4 x^{3}$ ?

x = -3, 0

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x = 0, 3

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x = 0

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x = 3

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AnswerB · $x = 0, 3$

Critical points are where $f'(x) = 0$ : $f'(x) = 4 x^{2} \left(x - 3\right)$ .
So $x = 0, 3$ .

Worth noting

A repeated factor such as $x^{2}$ still gives a critical point, even though $f'$ does not change sign there.

Question 25

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At which point does $f(x) = x^{3} + 3 x^{2} - 9 x + 2$ have a local minimum?

-3

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-1

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1

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3

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AnswerC · $1$

$f'(x) = 3 \left(x - 1\right) \left(x + 3\right)$ .
$f'$ changes from negative to positive at $x = 1$ , so $f$ falls then rises there: a local minimum.

Worth noting

At $x = -3$ the sign changes from positive to negative: that is the local maximum.

Question 26

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At which points does $f(x) = x^{4} - 8 x^{2} + 3$ have local minima?

x = 0

onlyCorrect answerYour answer

x = -2

and

x = 2

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x = -2

x = 0

and

x = 2

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x = 2

onlyCorrect answerYour answer

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AnswerB · $x = -2$ and $x = 2$

$f'(x) = 4x^3 - 16x = 4x(x - 2)(x + 2)$ , which is 0 at $-2$ , 0 and 2.
The sign of $f'$ runs $-, +, -, +$ across these points: it turns from negative to positive at $-2$ and 2 (minima) and from positive to negative at 0 (a maximum).

Worth noting

$x = 0$ is a critical point too, but there $f$ rises and then falls: a local maximum.

Question 27

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Which of the following statements are true?

I. At a local maximum, $f'$ changes sign from positive to negative.

II. A function can have a local minimum value greater than one of its local maximum values.

III. A local maximum value is always the greatest value of the function.

IV. If $f$ has a local minimum at $c$ and $f'(c)$ exists, then $f'(c) = 0$ .

All four: I, II, III and IVCorrect answerYour answerI and IV onlyCorrect answerYour answerI, II and IV onlyCorrect answerYour answerI, III and IV onlyCorrect answerYour answer

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AnswerC · I, II and IV only

I is the first derivative test, and IV holds because the tangent is flat at a peak or trough of a differentiable function. For II: $x + \frac{1}{x}$ has a local maximum value $-2$ at $x = -1$ and a local minimum value 2 at $x = 1$ .
III is false: 'local' means only among nearby points. $x^3 - 3x$ has a local maximum value 2 at $x = -1$ , yet $f(3) = 18$ .

Question 28

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A function has $f'(x) = (x + 1)(x - 2)^2(x - 4)$ . Where does it have a local minimum?

x = 4

onlyCorrect answerYour answerAt

x = -1

onlyCorrect answerYour answerAt

x = 2

and

x = 4

Correct answerYour answerAt

x = -1

and

x = 4

Correct answerYour answer

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AnswerA · At $x = 4$ only

Signs of $f'$ : positive for $x < -1$ , negative on $(-1, 2)$ and on $(2, 4)$ , positive for $x > 4$ .
Negative to positive only at $x = 4$ : the local minimum. At $x = -1$ it goes positive to negative (a local maximum); at $x = 2$ it does not change.

Worth noting

The squared factor $(x - 2)^2$ makes $f'(2) = 0$ without a change of sign, so $x = 2$ is not an extremum.

Question 29

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Which of the following statements are true?

I. If $f'$ does not change sign at a critical point $c$ , then $c$ is not a local extremum.

II. Every critical point is a local maximum or a local minimum.

III. $f(x) = \lvert x \rvert$ has a local minimum at 0, though $f'(0)$ does not exist.

IV. A local maximum value can be smaller than a local minimum value of the same function.

I and III onlyCorrect answerYour answerAll four: I, II, III and IVCorrect answerYour answerIII and IV onlyCorrect answerYour answerI, III and IV onlyCorrect answerYour answer

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AnswerD · I, III and IV only

I is the first derivative test's third case, and III holds because $\lvert x \rvert \ge 0 = f(0)$ near 0.
IV: $x + \frac{1}{x}$ has local maximum value $-2$ and local minimum value 2. II is false: $x^{3}$ has a critical point at 0 that is neither.

Worth noting

'Local' means compared with nearby points only, which is why IV can happen.

The second derivative test

18 questions

Question 30

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What is the local maximum value of $f(x) = x^{3} - 6 x^{2} + 9 x + 1$ ?

-5

Correct answerYour answer

1

Correct answerYour answer

5

Correct answerYour answer

6

Correct answerYour answer

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AnswerC · $5$

$f'(x) = 3 \left(x - 3\right) \left(x - 1\right) = 0$ at $x = 1$ and $x = 3$ .
$f''(x) = 6 x - 12$ : $f''(1) = -6$ , so $x = 1$ is a local maximum.
The value there is $f(1) = 5$ .

Worth noting

$1$ is where it happens, not the value; and $1$ is the local minimum.

Question 31

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For a twice-differentiable $f$ , $f'(2) = 0$ and $f''(2) = -3$ . What does the second derivative test say about $x = 2$ ?

A local minimumCorrect answerYour answerA local maximumCorrect answerYour answerNothing: the test failsCorrect answerYour answerA point where

f

is not definedCorrect answerYour answer

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AnswerB · A local maximum

$f'' < 0$ at a critical point means a local maximum, $f'' > 0$ a local minimum.
Here $f''(2) = -3$ , so it is a local maximum.

Worth noting

A negative second derivative means the graph bends downwards, like the top of a hill.

Question 32

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What is the local minimum value of $f(x) = x^{3} - 3 x + 2$ ?

-1

Correct answerYour answer

0

Correct answerYour answer

1

Correct answerYour answer

4

Correct answerYour answer

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AnswerB · $0$

$f'(x) = 3 \left(x - 1\right) \left(x + 1\right) = 0$ at $x = -1$ and $x = 1$ .
$f''(x) = 6 x$ : $f''(1) = 6$ , so $x = 1$ is a local minimum.
The value there is $f(1) = 0$ .

Worth noting

$1$ is where it happens, not the value; and $4$ is the local maximum.

Question 33

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What is the local minimum value of $f(x) = x + \frac{9}{x}$ for $x > 0$ ?

3

Correct answerYour answer

6

Correct answerYour answer

9

Correct answerYour answer

10

Correct answerYour answer

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AnswerB · $6$

$f'(x) = 1 - \frac{9}{x^2} = 0$ at $x = 3$ , and $f''(x) = \frac{18}{x^3} > 0$ there: a local minimum.
The value is $f(3) = 3 + 3 = 6$ .

Worth noting

3 is where the minimum occurs; the value is $3 + \frac{9}{3}$ .

Question 34

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$f(x) = x^3 - 3x^2 - 9x + 4$ . Match each item in List I with its value in List II.

List I	List II
A. Point of local maximum	I. $9$
B. Point of local minimum	II. $-1$
C. Local maximum value	III. $-23$
D. Local minimum value	IV. $3$

A-II, B-IV, C-I, D-IIICorrect answerYour answerA-IV, B-II, C-I, D-IIICorrect answerYour answerA-II, B-IV, C-III, D-ICorrect answerYour answerA-IV, B-II, C-III, D-ICorrect answerYour answer

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AnswerA · A-II, B-IV, C-I, D-III

$f'(x) = 3(x - 3)(x + 1)$ and $f''(x) = 6x - 6$ : $f''(-1) = -12 < 0$ (maximum) and $f''(3) = 12 > 0$ (minimum).
$f(-1) = -1 - 3 + 9 + 4 = 9$ and $f(3) = 27 - 27 - 27 + 4 = -23$ .

Question 35

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For a twice-differentiable $f$ , $f'(4) = 0$ and $f''(4) = 0$ . What does the second derivative test say about $x = 4$ ?

A local maximumCorrect answerYour answerA local minimumCorrect answerYour answerA point where

f

is not definedCorrect answerYour answerNothing: the test failsCorrect answerYour answer

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AnswerD · Nothing: the test fails

$f'' < 0$ at a critical point means a local maximum, $f'' > 0$ a local minimum.
When $f'' = 0$ the test decides nothing: use the first derivative test instead.

Question 36

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What is the local maximum value of $f(x) = x + \frac{1}{x}$ ?

-2

Correct answerYour answer

-1

Correct answerYour answer

1

Correct answerYour answer

2

Correct answerYour answer

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AnswerA · $-2$

$f'(x) = 1 - \frac{1}{x^{2}} = 0$ at $x = \pm 1$ , and $f''(x) = \frac{2}{x^{3}}$ .
$f''(-1) = -2 < 0$ : a local maximum, with value $f(-1) = -2$ . (At $x = 1$ , the local minimum value is 2.)

Worth noting

2 is the local minimum value: here the local maximum is the smaller of the two, since the graph has two separate branches.

Question 37

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What is the least value of $f(x) = x^{2} + \frac{16}{x^{2}}$ , $x \ne 0$ ?

2

Correct answerYour answer

4

Correct answerYour answer

8

Correct answerYour answer

17

Correct answerYour answer

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AnswerC · $8$

$f'(x) = 2x - \frac{32}{x^{3}} = 0$ gives $x^{4} = 16$ , so $x = \pm 2$ ; $f''(x) = 2 + \frac{96}{x^{4}} > 0$ , a minimum.
$f(\pm 2) = 4 + 4 = 8$ .

Worth noting

2 is where the minimum occurs, not the value; 17 is $f(\pm 1)$ .

Question 38

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What is the local maximum value of $f(x) = x^{3} - 12 x + 5$ ?

-21

Correct answerYour answer

-11

Correct answerYour answer

-2

Correct answerYour answer

21

Correct answerYour answer

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AnswerD · $21$

$f'(x) = 3 \left(x - 2\right) \left(x + 2\right) = 0$ at $x = -2$ and $x = 2$ .
$f''(x) = 6 x$ : $f''(-2) = -12$ , so $x = -2$ is a local maximum.
The value there is $f(-2) = 21$ .

Worth noting

$-2$ is where it happens, not the value; and $-11$ is the local minimum.

Question 39

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What is the local minimum value of $f(x) = x^{3} + 3 x^{2} - 45 x + 4$ ?

-77

Correct answerYour answer

-76

Correct answerYour answer

3

Correct answerYour answer

179

Correct answerYour answer

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AnswerA · $-77$

$f'(x) = 3 \left(x - 3\right) \left(x + 5\right) = 0$ at $x = -5$ and $x = 3$ .
$f''(x) = 6 x + 6$ : $f''(3) = 24$ , so $x = 3$ is a local minimum.
The value there is $f(3) = -77$ .

Worth noting

$3$ is where it happens, not the value; and $179$ is the local maximum.

Question 40

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Assertion (A): $f(x) = x^4$ has a local minimum at $x = 0$ .

Reason (R): $f''(0) > 0$ for $f(x) = x^4$ .

Both A and R are true, and R is the correct explanation of A.Correct answerYour answerBoth A and R are true, but R is not the correct explanation of A.Correct answerYour answerA is true, but R is false.Correct answerYour answerA is false, but R is true.Correct answerYour answer

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AnswerC · A is true, but R is false.

$x^4 \ge 0 = f(0)$ for every $x$ , so 0 is a minimum: A is true.
$f''(x) = 12x^2$ , so $f''(0) = 0$ and R is false. The second derivative test fails here, but the first derivative test ( $4x^3$ goes from negative to positive) still finds the minimum.

Question 41

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What is the maximum value of $f(x) = xe^{-x}$ ?

e^{-1}

Correct answerYour answer

1

Correct answerYour answer

e^{-1} + 1

Correct answerYour answer

e

Correct answerYour answer

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AnswerA · $e^{-1}$

$f'(x) = e^{-x} - xe^{-x} = (1 - x)e^{-x}$ , which is 0 at $x = 1$ .
$f''(x) = (x - 2)e^{-x}$ , and $f''(1) = -\frac{1}{e} < 0$ : a maximum, with value $f(1) = \frac{1}{e}$ .

Worth noting

1 is where the maximum occurs, not its value.

Question 42

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$f(x) = ax^2 + bx$ has a local maximum value 8 at $x = 2$ . What is $b$ ?

-8

Correct answerYour answer

-2

Correct answerYour answer

4

Correct answerYour answer

8

Correct answerYour answer

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AnswerD · $8$

$f'(2) = 4a + b = 0$ and $f(2) = 4a + 2b = 8$ .
From the first, $b = -4a$ ; then $4a - 8a = 8$ gives $a = -2$ and $b = 8$ .

Worth noting

$-2$ is $a$ , negative as a maximum needs.

Question 43

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What is the local maximum value of $f(x) = - 2 x^{3} + 9 x^{2} - 12 x + 7$ ?

-3

Correct answerYour answer

1

Correct answerYour answer

2

Correct answerYour answer

3

Correct answerYour answer

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AnswerD · $3$

$f'(x) = - 6 \left(x - 2\right) \left(x - 1\right) = 0$ at $x = 1$ and $x = 2$ .
$f''(x) = 18 - 12 x$ : $f''(2) = -6$ , so $x = 2$ is a local maximum.
The value there is $f(2) = 3$ .

Worth noting

$2$ is where it happens, not the value; and $2$ is the local minimum.

Question 44

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At which point in $(0, \pi)$ does $f(x) = \sin 2x - x$ have a local maximum?

\frac{\pi}{6}

Correct answerYour answer

\frac{\pi}{4}

Correct answerYour answer

\frac{\pi}{3}

Correct answerYour answer

\frac{5 \pi}{6}

Correct answerYour answer

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AnswerA · $\frac{\pi}{6}$

$f'(x) = 2\cos 2x - 1 = 0$ when $\cos 2x = \frac{1}{2}$ : $2x = \frac{\pi}{3}$ or $\frac{5\pi}{3}$ , so $x = \frac{\pi}{6}$ or $\frac{5\pi}{6}$ .
$f''(x) = -4\sin 2x$ : at $\frac{\pi}{6}$ it is $-2\sqrt{3} < 0$ , a local maximum; at $\frac{5\pi}{6}$ it is $2\sqrt{3} > 0$ , a local minimum.

Worth noting

$\frac{5\pi}{6}$ is the other critical point, where $f'' > 0$ : the local minimum. $\frac{\pi}{4}$ is where $\sin 2x$ alone peaks.

Question 45

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$f(x) = x^{3} + ax^{2} + bx$ has local extrema at $x = 1$ and $x = 3$ . What are $a$ and $b$ ?

a = 6,\ b = 9

Correct answerYour answer

a = -6,\ b = 9

Correct answerYour answer

a = -4,\ b = 3

Correct answerYour answer

a = -6,\ b = -9

Correct answerYour answer

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AnswerB · $a = -6,\ b = 9$

$f'(x) = 3x^{2} + 2ax + b$ must be 0 at both points: $3 + 2a + b = 0$ and $27 + 6a + b = 0$ .
Subtracting: $24 + 4a = 0$ , so $a = -6$ and $b = 9$ . Check: $f'(x) = 3(x - 1)(x - 3)$ .

Worth noting

$a = -4,\ b = 3$ reads $f'$ as $x^{2} - 4x + 3$ , forgetting the factor 3 from differentiating $x^{3}$ .

Question 46

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What is the maximum value of $f(x) = \frac{\log x}{x}$ for $x > 0$ ?

-1 + e^{-1}

Correct answerYour answer

0

Correct answerYour answer

e^{-1}

Correct answerYour answer

e

Correct answerYour answer

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AnswerC · $e^{-1}$

$f'(x) = \frac{1 - \log x}{x^{2}} = 0$ at $\log x = 1$ , that is $x = e$ .
$f'$ is positive before $e$ and negative after, so the maximum value is $f(e) = \frac{1}{e}$ .

Worth noting

$e$ is where the maximum occurs; the value is $\frac{\log e}{e} = \frac{1}{e}$ .

Question 47

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What is the local maximum value of $f(x) = x^{2}e^{-x}$ ?

0

Correct answerYour answer

\frac{4}{e^{2}}

Correct answerYour answer

\frac{2}{e}

Correct answerYour answer

2

Correct answerYour answer

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AnswerB · $\frac{4}{e^{2}}$

$f'(x) = 2xe^{-x} - x^{2}e^{-x} = x(2 - x)e^{-x}$ , zero at $x = 0$ and $x = 2$ .
$f'$ goes from positive to negative at 2, so the local maximum value is $f(2) = 4e^{-2} = \frac{4}{e^{2}}$ .

Worth noting

0 is the local minimum value, at $x = 0$ ; 2 is the point of the maximum.

Greatest and least values on an interval

15 questions

Question 48

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What is the greatest value of $f(x) = x^{3} - 3 x$ on $[0, 2]$ ?

-2

Correct answerYour answer

0

Correct answerYour answer

2

Correct answerYour answer

3

Correct answerYour answer

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AnswerC · $2$

Critical points inside the interval: $f'(x) = 3 \left(x - 1\right) \left(x + 1\right) = 0$ at $x = 1$ .
Compare $f$ there and at the two ends: $f(0) = 0$ , $f(2) = 2$ , $f(1) = -2$ .
The greatest value is $2$ .

Worth noting

A local extremum is not always the greatest or least value: the ends of the interval must be checked too.

Question 49

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What is the least value of $\sin 2x$ on $[0, \pi]$ ?

-1

Correct answerYour answer

0

Correct answerYour answer

1

Correct answerYour answer

2

Correct answerYour answer

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AnswerA · $-1$

$f'(x) = 2 \cos 2x = 0$ inside the interval at $x = \frac{\pi}{4}, \frac{3 \pi}{4}$ .
Values: $f(0) = 0$ , $f(\pi) = 0$ , $f(\frac{\pi}{4}) = 1$ , $f(\frac{3 \pi}{4}) = -1$ ; the least value is $-1$ .

Question 50

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On $[1, 4]$ , where does $f(x) = x^2 - 6x + 10$ take its greatest value?

x = 3

, the critical pointCorrect answerYour answerAt

x = 1

, an end pointCorrect answerYour answerAt

x = 4

, an end pointCorrect answerYour answerIt has no greatest value on

[1, 4]

Correct answerYour answer

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AnswerB · At $x = 1$ , an end point

$f'(x) = 2x - 6 = 0$ at $x = 3$ , which is a minimum: $f(3) = 1$ .
The ends give $f(1) = 5$ and $f(4) = 2$ , so the greatest value, 5, is at $x = 1$ .

Worth noting

The only critical point is the least value here; the greatest value of a continuous function on a closed interval can sit at an end.

Question 51

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Assertion (A): On $[-2, 2]$ , the greatest value of $f(x) = x^3 - 3x$ is 2.

Reason (R): On a closed interval, a local maximum value is always the greatest value.

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AnswerC · A is true, but R is false.

$f'(x) = 3(x - 1)(x + 1)$ ; comparing $f(-2) = -2$ , $f(-1) = 2$ , $f(1) = -2$ and $f(2) = 2$ , the greatest value is 2: A is true.
R is false: on $[-2, 3]$ the same local maximum value 2 is beaten by the end value $f(3) = 18$ .

Question 52

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What is the greatest value of $f(x) = x^{3} - 3 x$ on $[-2, 3]$ ?

-2

Correct answerYour answer

2

Correct answerYour answer

17

Correct answerYour answer

18

Correct answerYour answer

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AnswerD · $18$

Critical points inside the interval: $f'(x) = 3 \left(x - 1\right) \left(x + 1\right) = 0$ at $x = -1, 1$ .
Compare $f$ there and at the two ends: $f(-2) = -2$ , $f(3) = 18$ , $f(-1) = 2$ , $f(1) = -2$ .
The greatest value is $18$ .

Worth noting

A local extremum is not always the greatest or least value: the ends of the interval must be checked too.

Question 53

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What is the greatest value of $\sin x + \cos x$ on $[0, \frac{\pi}{2}]$ ?

- \sqrt{2}

Correct answerYour answer

\frac{\sqrt{2}}{2}

Correct answerYour answer

1

Correct answerYour answer

\sqrt{2}

Correct answerYour answer

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AnswerD · $\sqrt{2}$

$f'(x) = - \sin x + \cos x = 0$ inside the interval at $x = \frac{\pi}{4}$ .
Values: $f(0) = 1$ , $f(\frac{\pi}{2}) = 1$ , $f(\frac{\pi}{4}) = \sqrt{2}$ ; the greatest value is $\sqrt{2}$ .

Question 54

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What is the least value of $f(x) = 2 x^{3} - 9 x^{2} + 12 x + 1$ on $[0, 3]$ ?

0

Correct answerYour answer

1

Correct answerYour answer

6

Correct answerYour answer

10

Correct answerYour answer

Show the explanationHide the explanation

AnswerB · $1$

Critical points inside the interval: $f'(x) = 6 \left(x - 2\right) \left(x - 1\right) = 0$ at $x = 1, 2$ .
Compare $f$ there and at the two ends: $f(0) = 1$ , $f(3) = 10$ , $f(1) = 6$ , $f(2) = 5$ .
The least value is $1$ .

Worth noting

A local extremum is not always the greatest or least value: the ends of the interval must be checked too.

Question 55

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What is the least value of $f(x) = x + \frac{4}{x}$ on $[1, 4]$ ?

2

Correct answerYour answer

3

Correct answerYour answer

4

Correct answerYour answer

5

Correct answerYour answer

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AnswerC · $4$

Critical points inside the interval: $f'(x) = \frac{\left(x - 2\right) \left(x + 2\right)}{x^{2}} = 0$ at $x = 2$ .
Compare $f$ there and at the two ends: $f(1) = 5$ , $f(4) = 5$ , $f(2) = 4$ .
The least value is $4$ .

Worth noting

A local extremum is not always the greatest or least value: the ends of the interval must be checked too.

Question 56

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Read the two statements and choose the correct option.

I. On $[0, 1]$ , $f(x) = x^{2}$ has greatest value 1.

II. On $(0, 1)$ , $f(x) = x^{2}$ has a greatest value.

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerA · Only I

I: $x^{2}$ rises on $[0, 1]$ , so its greatest value is $f(1) = 1$ . True.
II: on $(0, 1)$ , $x^{2}$ comes as close to 1 as we like but never reaches it, so no value is the greatest. False.

Worth noting

A continuous function is sure to reach a greatest and a least value only on a closed interval.

Question 57

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What is the greatest value of $f(x) = x^{3} - 6 x^{2} + 9 x + 2$ on $[0, 5]$ ?

2

Correct answerYour answer

6

Correct answerYour answer

20

Correct answerYour answer

22

Correct answerYour answer

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AnswerD · $22$

Critical points inside the interval: $f'(x) = 3 \left(x - 3\right) \left(x - 1\right) = 0$ at $x = 1, 3$ .
Compare $f$ there and at the two ends: $f(0) = 2$ , $f(5) = 22$ , $f(1) = 6$ , $f(3) = 2$ .
The greatest value is $22$ , at an end of the interval.

Worth noting

$6$ is the local maximum value; the end of the interval goes higher.

Question 58

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What is the greatest value of $f(x) = x^{4} - 2 x^{2} + 5$ on $[-2, 1]$ ?

4

Correct answerYour answer

5

Correct answerYour answer

11

Correct answerYour answer

13

Correct answerYour answer

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AnswerD · $13$

Critical points inside the interval: $f'(x) = 4 x \left(x - 1\right) \left(x + 1\right) = 0$ at $x = -1, 0$ .
Compare $f$ there and at the two ends: $f(-2) = 13$ , $f(1) = 4$ , $f(-1) = 4$ , $f(0) = 5$ .
The greatest value is $13$ , at an end of the interval.

Worth noting

$5$ is the local maximum value; the end of the interval goes higher.

Question 59

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What is the least value of $f(x) = - x^{3} + 12 x$ on $[-3, 3]$ ?

-16

Correct answerYour answer

-15

Correct answerYour answer

9

Correct answerYour answer

16

Correct answerYour answer

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AnswerA · $-16$

Critical points inside the interval: $f'(x) = - 3 \left(x - 2\right) \left(x + 2\right) = 0$ at $x = -2, 2$ .
Compare $f$ there and at the two ends: $f(-3) = -9$ , $f(3) = 9$ , $f(-2) = -16$ , $f(2) = 16$ .
The least value is $-16$ .

Worth noting

A local extremum is not always the greatest or least value: the ends of the interval must be checked too.

Question 60

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What is the least value of $f(x) = x - 2\log x$ on $[1, e]$ ?

2 - 2 \log 2

Correct answerYour answer

-2 + e

Correct answerYour answer

1

Correct answerYour answer

2

Correct answerYour answer

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AnswerA · $2 - 2 \log 2$

$f'(x) = 1 - \frac{2}{x} = 0$ at $x = 2$ , which lies in $[1, e]$ .
$f(1) = 1$ , $f(2) = 2 - 2\log 2 \approx 0.61$ , $f(e) = e - 2 \approx 0.72$ . The least is $2 - 2\log 2$ .

Worth noting

$e - 2$ is the value at the right end, which is not the lowest point.

Question 61

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On $[-2, 2]$ , at which points does $f(x) = 2x^{3} - 6x$ take its least value?

x = 1

onlyCorrect answerYour answerAt

x = -2

and

x = 1

Correct answerYour answerAt

x = -2

onlyCorrect answerYour answerAt

x = -1

and

x = 2

Correct answerYour answer

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AnswerB · At $x = -2$ and $x = 1$

$f'(x) = 6x^{2} - 6 = 0$ at $x = \pm 1$ .
$f(-2) = -4$ , $f(-1) = 4$ , $f(1) = -4$ , $f(2) = 4$ : the least value $-4$ comes at both $x = -2$ and $x = 1$ .

Worth noting

Stopping at the local minimum, $x = 1$ , misses the end $x = -2$ , where $f$ is just as low.

Question 62

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What is the greatest value of $f(x) = 2\sin x + \sin 2x$ on $[0, 2\pi]$ ?

\sqrt{3}

Correct answerYour answer

2

Correct answerYour answer

\frac{3 \sqrt{3}}{2}

Correct answerYour answer

3

Correct answerYour answer

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AnswerC · $\frac{3 \sqrt{3}}{2}$

$f'(x) = 2\cos x + 2\cos 2x = 0$ gives $2\cos^{2} x + \cos x - 1 = 0$ : $\cos x = \frac{1}{2}$ or $-1$ , so $x = \frac{\pi}{3}, \pi, \frac{5\pi}{3}$ .
$f\left(\frac{\pi}{3}\right) = \sqrt{3} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$ , $f(\pi) = 0$ , $f\left(\frac{5\pi}{3}\right) = -\frac{3\sqrt{3}}{2}$ , and 0 at both ends.

Worth noting

3 adds the two peaks, 2 and 1, but $\sin x$ and $\sin 2x$ do not peak at the same $x$ .

Applied problems

18 questions

Question 63

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Two positive numbers add up to 36. What is the largest possible value of their product?

18

Correct answerYour answer

36

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324

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648

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AnswerC · $324$

Let the numbers be $x$ and $36 - x$ ; the product is $P = 36x - x^2$ .
$P' = 36 - 2x = 0$ at $x = 18$ , and $P'' = -2 < 0$ : a maximum, $18 \times 18 = 324$ .

Worth noting

18 is each number at the maximum, not the product.

Question 64

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A rectangle has perimeter 40 cm. What is its largest possible area, in cm²?

100

Correct answerYour answer

101

Correct answerYour answer

200

Correct answerYour answer

400

Correct answerYour answer

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AnswerA · $100$

Sides $x$ and $20 - x$ ; area $A = x(20 - x)$ .
$A' = 20 - 2x = 0$ at $x = 10$ : the rectangle is a square, area $10^2 = 100$ .

Worth noting

400 uses the whole perimeter, 40, as the sum of two sides; two sides add up to half of it.

Question 65

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When the price is ₹ $(100 - x)$ per unit, a shop sells $x$ units. How many units give the largest revenue?

25

Correct answerYour answer

50

Correct answerYour answer

100

Correct answerYour answer

2{,}500

Correct answerYour answer

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AnswerB · $50$

Revenue $R(x) = x(100 - x) = 100x - x^2$ .
$R'(x) = 100 - 2x = 0$ at $x = 50$ , and $R'' < 0$ : a maximum.

Worth noting

2,500 is the largest revenue itself, in rupees.

Question 66

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Two positive numbers have product 64. What is the least possible value of their sum?

8

Correct answerYour answer

16

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32

Correct answerYour answer

65

Correct answerYour answer

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AnswerB · $16$

With the numbers $x$ and $\frac{64}{x}$ , the sum is $S = x + \frac{64}{x}$ , and $S' = 1 - \frac{64}{x^{2}} = 0$ at $x = 8$ .
$S'' = \frac{128}{x^{3}} > 0$ : a minimum, with both numbers 8 and sum 16.

Worth noting

8 is each number, not their sum.

Question 67

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A rectangular garden must have area 49 m². What is the least perimeter, in m, it can have?

7

Correct answerYour answer

14

Correct answerYour answer

28

Correct answerYour answer

50

Correct answerYour answer

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AnswerC · $28$

With sides $x$ and $\frac{49}{x}$ , $P = 2\left(x + \frac{49}{x}\right)$ and $P' = 2\left(1 - \frac{49}{x^{2}}\right) = 0$ at $x = 7$ .
The garden is a 7 m square, with perimeter 28 m.

Worth noting

14 is half the perimeter, the sum of two adjacent sides.

Question 68

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Two positive numbers add up to 16. What is the least possible value of the sum of their squares?

64

Correct answerYour answer

128

Correct answerYour answer

144

Correct answerYour answer

256

Correct answerYour answer

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AnswerB · $128$

$Q = x^2 + (16 - x)^2$ , so $Q' = 2x - 2(16 - x) = 0$ at $x = 8$ .
$Q'' = 4 > 0$ , a minimum: $Q = 8^2 + 8^2 = 128$ .

Worth noting

64 is the largest product, $8 \times 8$ , a different question.

Question 69

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What is the largest area of a rectangle that fits inside a circle of radius 5 cm, in cm²?

25

Correct answerYour answer

50

Correct answerYour answer

75

Correct answerYour answer

100

Correct answerYour answer

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AnswerB · $50$

The rectangle's diagonal is the diameter, 10. With sides $x$ and $y$ , $x^2 + y^2 = 100$ and the area is $A = x\sqrt{100 - x^2}$ .
$A^2 = 100x^2 - x^4$ is largest when $200x - 4x^3 = 0$ , that is $x^2 = 50$ : then $y^2 = 50$ too, a square, and $A = 50$ .

Worth noting

100 is the square on the diameter; the best rectangle, a square with diagonal 10, has half that area.

Question 70

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A farmer has 100 m of fencing for a rectangular plot beside a straight wall, so only three sides need fencing. What is the largest area, in m², he can enclose?

625

Correct answerYour answer

\frac{10000}{9}

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1{,}250

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2{,}500

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AnswerC · $1{,}250$

With the side parallel to the wall $y$ and the other two $x$ : $2x + y = 100$ , area $A = x(100 - 2x)$ .
$A' = 100 - 4x = 0$ at $x = 25$ , so $y = 50$ and $A = 1250$ .

Worth noting

625 fences all four sides, as if there were no wall.

Question 71

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A firm's profit, in thousands of rupees, from selling $x$ units is $P(x) = -x^2 + 60x - 500$ . What is the largest profit?

30

Correct answerYour answer

370

Correct answerYour answer

399

Correct answerYour answer

400

Correct answerYour answer

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AnswerD · $400$

$P'(x) = -2x + 60 = 0$ at $x = 30$ ; $P'' = -2 < 0$ , a maximum.
$P(30) = -900 + 1800 - 500 = 400$ thousand rupees.

Worth noting

30 is the number of units that earns it, not the profit.

Question 72

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Read the two statements and choose the correct option.

I. Of all rectangles with a given perimeter, the square has the largest area.

II. Of all closed cylinders with a given volume, the one with the least surface has its height equal to its radius.

Only ICorrect answerYour answerOnly IICorrect answerYour answerBoth I and IICorrect answerYour answerNeither I nor IICorrect answerYour answer

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AnswerA · Only I

With perimeter $2s$ , the area $x(s - x)$ is largest at $x = \frac{s}{2}$ : equal sides, a square. I is true.
For the cylinder the best height is the diameter, $h = 2r$ , not the radius, so II is false.

Question 73

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What is the largest area, in cm², of a right-angled triangle with hypotenuse 10 cm?

\frac{25}{2}

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24

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25

Correct answerYour answer

50

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AnswerC · $25$

With legs $a$ and $b$ , $a^{2} + b^{2} = 100$ and the area is $\frac{1}{2}ab = \frac{1}{2}a\sqrt{100 - a^{2}}$ .
Its square, $\frac{1}{4}(100a^{2} - a^{4})$ , is largest when $a^{2} = 50$ : then $b^{2} = 50$ too, and the area is $\frac{1}{2} \times 50 = 25$ .

Worth noting

24 is the 6-8-10 triangle; the isosceles right triangle beats it.

Question 74

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120 m of fencing must enclose a rectangular field and also divide it into two parts with one fence parallel to a side. What is the largest area, in m², that can be enclosed?

450

Correct answerYour answer

600

Correct answerYour answer

720

Correct answerYour answer

900

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AnswerB · $600$

With $x$ the side that is fenced three times (two sides and the divider) and $y$ the other: $3x + 2y = 120$ , so $A = xy = \frac{x(120 - 3x)}{2}$ .
$A' = \frac{120 - 6x}{2} = 0$ at $x = 20$ , so $y = 30$ and $A = 600$ .

Worth noting

900 forgets the dividing fence: a 30 m square uses all 120 m on the outside alone.

Question 75

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A firm sells $x$ units at ₹ $(60 - 2x)$ each, and making them costs ₹ $(12x + 50)$ . How many units give the largest profit?

12

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13

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15

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24

Correct answerYour answer

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AnswerA · $12$

Profit $P = (60 - 2x)x - (12x + 50) = 48x - 2x^{2} - 50$ .
$P' = 48 - 4x = 0$ at $x = 12$ , and $P'' = -4 < 0$ : a maximum.

Worth noting

15 maximises revenue alone, $(60 - 2x)x$ ; the cost per unit pulls the best output lower.

Question 76

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Positive numbers $x$ and $y$ satisfy $x + y = 40$ . For which value of $y$ is $xy^3$ largest?

10

Correct answerYour answer

20

Correct answerYour answer

29

Correct answerYour answer

30

Correct answerYour answer

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AnswerD · $30$

With $x = 40 - y$ , $P = (40 - y)y^3$ , so $P' = 3y^2(40 - y) - y^3 = y^2(120 - 4y)$ .
$P' = 0$ for positive $y$ at $y = 30$ ; $P'$ goes from positive to negative there, so it is a maximum, with $x = 10$ .

Worth noting

10 is the value of $x$ ; the power 3 on $y$ pulls the larger share to $y$ .

Question 77

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Equal squares are cut from the corners of a square sheet of side 24 cm and the sides folded up to make an open box. What side of square, in cm, gives the largest volume?

4

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6

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8

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12

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AnswerA · $4$

Volume $V = x(24 - 2x)^2$ for $0 < x < 12$ .
$V' = (24 - 2x)(24 - 6x) = 0$ gives $x = 4$ (the other root, 12, leaves no sheet).
At $x = 4$ cm the volume is $4 \times 16^2 = 1024$ cm³, the largest.

Worth noting

12 cm is the other root of $V' = 0$ ; it cuts the sheet away completely.

Question 78

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A closed cylindrical tin must hold $250\pi$ cm³. What radius, in cm, uses the least metal?

5

Correct answerYour answer

10

Correct answerYour answer

5 \sqrt{10}

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125

Correct answerYour answer

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AnswerA · $5$

$\pi r^2 h = 250\pi$ , so $h = \frac{250}{r^2}$ , and the surface is $S = 2\pi r^2 + \frac{500\pi}{r}$ .
$S' = 4\pi r - \frac{500\pi}{r^2} = 0$ gives $r^3 = 125$ , $r = 5$ .
Then $h = 10$ : the best tin has its height equal to its diameter.

Worth noting

10 cm is the height at the best shape, not the radius.

Question 79

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A closed cylinder has total surface area $24\pi$ cm². What is its largest possible volume, in cm³?

8 \pi

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12 \pi

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-1 + 16 \pi

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16 \pi

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AnswerD · $16 \pi$

$2\pi r^{2} + 2\pi rh = 24\pi$ gives $h = \frac{12 - r^{2}}{r}$ , so $V = \pi r^{2}h = \pi(12r - r^{3})$ .
$V' = \pi(12 - 3r^{2}) = 0$ at $r = 2$ , and $V'' = -6\pi r < 0$ : then $h = 4$ and $V = \pi \times 4 \times 4 = 16\pi$ .

Worth noting

$8\pi$ uses $h = 2$ , the radius, instead of $h = 2r = 4$ : at the best shape the height equals the diameter.

Question 80

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What is the shortest distance from the point $(0, 2)$ to the parabola $y = x^{2}$ ?

\frac{\sqrt{3}}{2}

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\frac{\sqrt{7}}{2}

Correct answerYour answer

\frac{7}{4}

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2

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AnswerB · $\frac{\sqrt{7}}{2}$

A point on the parabola is $(x, x^{2})$ ; the square of its distance from $(0, 2)$ is $D = x^{2} + (x^{2} - 2)^{2} = x^{4} - 3x^{2} + 4$ .
$D' = 4x^{3} - 6x = 0$ gives $x^{2} = \frac{3}{2}$ , where $D = \frac{9}{4} - \frac{9}{2} + 4 = \frac{7}{4}$ ; the distance is $\frac{\sqrt{7}}{2}$ .

Worth noting

2 is the distance to the vertex $(0, 0)$ , which is not the nearest point; $\frac{7}{4}$ is the squared distance.

NTA's Mathematics / Applied Mathematics syllabus has three sections: Section A1, which every candidate answers, then Section B1 (Mathematics) or Section B2 (Applied Mathematics), of which a candidate answers one. Each section lists units and topics but no chapters, so the chapters here are ours. Where both B sections list a topic that Section A also lists, its questions sit once, in Section A.