NCERT Class 12 Mathematics — Application of Derivatives, Exercise 6.2. All 19 questions solved.

One theorem powers the whole exercise. If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then on that interval

$$f'(x) > 0 \implies f \text{ is increasing}, \qquad f'(x) < 0 \implies f \text{ is decreasing}$$

So every question reduces to: differentiate, factorise $f’$, and read off where each factor is positive or negative.

Key insight. Once $f’$ is factorised, mark its zeros on a number line and test the sign in each resulting interval. You only need one test point per interval, because a continuous $f’$ cannot change sign without passing through zero.

Watch the even powers. In question 6(e), $f'(x) = 6(x+1)^2(x-3)^2(x-1)$: the squared factors are never negative, so they do not change the sign at all — only $(x-1)$ does. A student who marks $-1$ and $3$ as sign changes gets the answer wrong in both directions.

Prove or show the following.

Question 1

Show that the function given by $f(x) = 3x + 17$ is increasing on $\mathbf{R}$.

Solution. $f'(x) = 3 > 0$ for every real $x$. A positive derivative everywhere means $f$ is increasing on the whole of $\mathbf{R}$.

Question 2

Show that the function given by $f(x) = e^{2x}$ is increasing on $\mathbf{R}$.

Solution. $f'(x) = 2e^{2x}$. The exponential function is strictly positive for every real argument, so $f'(x) > 0$ throughout $\mathbf{R}$ and $f$ is increasing.

Question 3

Show that the function given by $f(x) = \sin x$ is (a) increasing in $\left(0, \tfrac{\pi}{2}\right)$ (b) decreasing in $\left(\tfrac{\pi}{2}, \pi\right)$ (c) neither increasing nor decreasing in $(0, \pi)$.

Solution. $f'(x) = \cos x$.

(a) On $\left(0, \tfrac{\pi}{2}\right)$ the cosine is positive, so $\sin x$ is increasing.

(b) On $\left(\tfrac{\pi}{2}, \pi\right)$ the cosine is negative, so $\sin x$ is decreasing.

(c) On $(0, \pi)$ the cosine takes both signs — positive before $\tfrac{\pi}{2}$ and negative after. A function that increases on part of an interval and decreases on the rest is neither increasing nor decreasing on the whole of it.

Question 4

Find the intervals in which the function $f$ given by $f(x) = 2x^2 – 3x$ is (a) increasing (b) decreasing.

Solution. $f'(x) = 4x – 3$, which is zero at $x = \tfrac34$ and changes sign there.

Question 5

Find the intervals in which the function $f$ given by $f(x) = 2x^3 – 3x^2 – 36x + 7$ is (a) increasing (b) decreasing.

Solution.

$$f'(x) = 6x^2 – 6x – 36 = 6\left(x^2 – x – 6\right) = 6(x – 3)(x + 2)$$

A product of two linear factors is positive when both have the same sign — that is, outside the roots — and negative between them.

Question 6

Find the intervals in which the following functions are strictly increasing or decreasing:

(a) $x^2 + 2x – 5$   (b) $10 – 6x – 2x^2$   (c) $-2x^3 – 9x^2 – 12x + 1$   (d) $6 – 9x – x^2$   (e) $(x+1)^3(x-3)^3$

Solution.

(a) $f'(x) = 2x + 2 = 2(x + 1)$, zero at $x = -1$.

Decreasing for $x < -1$, increasing for $x > -1$.

(b) $f'(x) = -6 – 4x$, zero at $x = -\tfrac32$. The negative leading coefficient reverses the usual order:

Increasing for $x < -\tfrac32$, decreasing for $x > -\tfrac32$.

(c) $f'(x) = -6x^2 – 18x – 12 = -6(x+1)(x+2)$. The overall minus sign means $f’$ is positive exactly where $(x+1)(x+2)$ is negative:

Increasing on $(-2, -1)$, decreasing for $x < -2$ and for $x > -1$.

(d) $f'(x) = -9 – 2x$, zero at $x = -\tfrac92$.

Increasing for $x < -\tfrac92$, decreasing for $x > -\tfrac92$.

(e) Differentiate as a product and then factor out everything common:

$$f'(x) = 3(x+1)^2(x-3)^3 + 3(x+1)^3(x-3)^2 = 3(x+1)^2(x-3)^2\big[(x-3) + (x+1)\big]$$

$$= 6(x+1)^2(x-3)^2(x-1)$$

Both squared factors are non-negative, so the sign of $f’$ is the sign of $(x – 1)$ alone. The function is therefore increasing to the right of $1$ and decreasing to its left, with $f’$ vanishing (but not changing sign) at $x = -1$ and $x = 3$.

Increasing on $(1, 3)$ and $(3, \infty)$; decreasing on $(-\infty, -1)$ and $(-1, 1)$.

Question 7

Show that $y = \log(1+x) – \dfrac{2x}{2+x}$, $x > -1$, is an increasing function of $x$ throughout its domain.

Solution. Differentiate, using the quotient rule on the second term:

$$\frac{dy}{dx} = \frac{1}{1+x} – \frac{(2+x)(2) – 2x(1)}{(2+x)^2} = \frac{1}{1+x} – \frac{4}{(2+x)^2}$$

Put over a common denominator:

$$= \frac{(2+x)^2 – 4(1+x)}{(1+x)(2+x)^2} = \frac{4 + 4x + x^2 – 4 – 4x}{(1+x)(2+x)^2} = \frac{x^2}{(1+x)(2+x)^2}$$

For $x > -1$ both $(1+x)$ and $(2+x)^2$ are positive, and $x^2 \ge 0$. So $\tfrac{dy}{dx} \ge 0$ throughout the domain, with equality only at the single point $x = 0$ — which does not stop the function increasing.

Question 8

Find the values of $x$ for which $y = [x(x-2)]^2$ is an increasing function.

Solution. Write $u = x^2 – 2x$ so that $y = u^2$, and use the chain rule:

$$\frac{dy}{dx} = 2\left(x^2 – 2x\right)(2x – 2) = 4x(x-2)(x-1)$$

Three simple roots at $0$, $1$, $2$. Testing the sign in each interval — a cubic with positive leading coefficient alternates sign across simple roots:

Interval	Sign of $\tfrac{dy}{dx}$
$x < 0$	negative
$0 < x < 1$	positive
$1 < x < 2$	negative
$x > 2$	positive

Question 9

Prove that $y = \dfrac{4\sin\theta}{2 + \cos\theta} – \theta$ is an increasing function of $\theta$ in $\left[0, \tfrac{\pi}{2}\right]$.

Solution. Using the quotient rule on the first term:

$$\frac{dy}{d\theta} = \frac{(2+\cos\theta)(4\cos\theta) – 4\sin\theta(-\sin\theta)}{(2+\cos\theta)^2} – 1$$

$$= \frac{8\cos\theta + 4\cos^2\theta + 4\sin^2\theta}{(2+\cos\theta)^2} – 1 = \frac{8\cos\theta + 4}{(2+\cos\theta)^2} – 1$$

Combining over the common denominator:

$$= \frac{8\cos\theta + 4 – \left(4 + 4\cos\theta + \cos^2\theta\right)}{(2+\cos\theta)^2} = \frac{4\cos\theta – \cos^2\theta}{(2+\cos\theta)^2} = \frac{\cos\theta(4 – \cos\theta)}{(2+\cos\theta)^2}$$

On $\left[0, \tfrac{\pi}{2}\right]$ we have $\cos\theta \ge 0$, and $4 – \cos\theta \ge 3 > 0$, and the denominator is positive. So $\tfrac{dy}{d\theta} \ge 0$ throughout and $y$ is increasing.

Question 10

Prove that the logarithmic function is increasing on $(0, \infty)$.

Solution. For $f(x) = \log x$ we have $f'(x) = \tfrac1x$, which is positive for every $x > 0$. Hence $f$ is increasing on $(0, \infty)$.

Question 11

Prove that the function $f$ given by $f(x) = x^2 – x + 1$ is neither strictly increasing nor decreasing on $(-1, 1)$.

Solution. $f'(x) = 2x – 1$, which is zero at $x = \tfrac12$ — a point inside $(-1, 1)$. On $\left(-1, \tfrac12\right)$ the derivative is negative and on $\left(\tfrac12, 1\right)$ it is positive.

Since $f’$ takes both signs on the interval, $f$ decreases on part of it and increases on the rest, so it is neither strictly increasing nor strictly decreasing there.

Question 12

Which of the following functions are decreasing on $\left(0, \tfrac{\pi}{2}\right)$?

Solution. Differentiate each and check the sign on the whole interval.

• (A) $\tfrac{d}{dx}\cos x = -\sin x < 0$ throughout. Decreasing.
• (B) $\tfrac{d}{dx}\cos 2x = -2\sin 2x$. As $x$ runs over $\left(0, \tfrac{\pi}{2}\right)$, $2x$ runs over $(0, \pi)$, where the sine is positive. So the derivative is negative throughout. Decreasing.
• (C) $\tfrac{d}{dx}\cos 3x = -3\sin 3x$. Here $3x$ runs over $\left(0, \tfrac{3\pi}{2}\right)$, and $\sin 3x$ changes sign at $3x = \pi$, that is at $x = \tfrac{\pi}{3}$. Not decreasing on the whole interval.
• (D) $\tfrac{d}{dx}\tan x = \sec^2 x > 0$. Increasing, not decreasing.

Two options are correct — the question says “functions”, plural, and the key gives both.

Question 13

On which of the following intervals is the function $f$ given by $f(x) = x^{100} + \sin x – 1$ decreasing?

Solution. $f'(x) = 100x^{99} + \cos x$.

On $(0, 1)$ and $\left(0, \tfrac{\pi}{2}\right)$ both terms are positive, so $f’$ is positive and $f$ increases. On $\left(\tfrac{\pi}{2}, \pi\right)$ the cosine is negative but never below $-1$, while $100x^{99}$ at $x > \tfrac{\pi}{2} > 1$ is already greater than $100$. The sum stays firmly positive.

So $f$ is increasing on all three intervals.

Question 14

For what values of $a$ is the function $f$ given by $f(x) = x^2 + ax + 1$ increasing on $[1, 2]$?

Solution. $f'(x) = 2x + a$. Requiring $f'(x) > 0$ at every point of $[1, 2]$ gives $a > -2x$ for all $x$ in the interval. The right-hand side is largest at $x = 1$, where it equals $-2$, so the condition is

$$a > -2$$

A note on the endpoint. At exactly $a = -2$ the function is $f(x) = x^2 – 2x + 1 = (x-1)^2$, whose derivative is $0$ at $x = 1$ and strictly positive on $(1, 2]$. That function is still strictly increasing on $[1, 2]$, so $a \ge -2$ is the sharper condition. NCERT prints $a > -2$ because it demands $f’ > 0$ at every point including the left endpoint. Either can be justified; if you write $a \ge -2$, say why.

Question 15

Let $\mathrm{I}$ be any interval disjoint from $[-1, 1]$. Prove that the function $f$ given by $f(x) = x + \dfrac1x$ is increasing on $\mathrm{I}$.

Solution.

$$f'(x) = 1 – \frac{1}{x^2} = \frac{x^2 – 1}{x^2}$$

The denominator is positive wherever $f$ is defined. The numerator is positive exactly when $x^2 > 1$, that is when $|x| > 1$.

An interval disjoint from $[-1, 1]$ consists entirely of points with $|x| > 1$, so $f’ > 0$ throughout it and $f$ is increasing there. (The restriction is essential: on $(-1, 1)$ the derivative is negative.)

Question 16

Prove that the function $f$ given by $f(x) = \log\sin x$ is increasing on $\left(0, \tfrac{\pi}{2}\right)$ and decreasing on $\left(\tfrac{\pi}{2}, \pi\right)$.

Solution. By the chain rule,

$$f'(x) = \frac{\cos x}{\sin x} = \cot x$$

On $(0, \pi)$ the sine is positive, so $\log\sin x$ is defined and the sign of $f’$ is the sign of $\cos x$: positive on $\left(0, \tfrac{\pi}{2}\right)$ and negative on $\left(\tfrac{\pi}{2}, \pi\right)$.

Question 17

Prove that the function $f$ given by $f(x) = \log|\cos x|$ is decreasing on $\left(0, \tfrac{\pi}{2}\right)$ and increasing on $\left(\tfrac{3\pi}{2}, 2\pi\right)$.

Solution. Differentiating $\log|\cos x|$ gives

$$f'(x) = \frac{-\sin x}{\cos x} = -\tan x$$

On $\left(0, \tfrac{\pi}{2}\right)$ the tangent is positive, so $f’ < 0$ and $f$ decreases. On $\left(\tfrac{3\pi}{2}, 2\pi\right)$ — the fourth quadrant — the tangent is negative, so $f’ > 0$ and $f$ increases.

The modulus is what allows $\log$ to be taken where $\cos x < 0$; it does not affect the derivative, because $\tfrac{d}{dx}\log|u| = \tfrac{u’}{u}$ regardless of the sign of $u$.

Question 18

Prove that the function given by $f(x) = x^3 – 3x^2 + 3x – 100$ is increasing in $\mathbf{R}$.

Solution.

$$f'(x) = 3x^2 – 6x + 3 = 3\left(x^2 – 2x + 1\right) = 3(x-1)^2$$

A perfect square is never negative, so $f’ \ge 0$ for every real $x$, vanishing only at the single point $x = 1$. A derivative that is non-negative everywhere and zero at isolated points still gives an increasing function.

Question 19

The interval in which $y = x^2 e^{-x}$ is increasing is

Solution. By the product rule,

$$\frac{dy}{dx} = 2xe^{-x} – x^2e^{-x} = xe^{-x}(2 – x)$$

The factor $e^{-x}$ is always positive, so the sign is that of $x(2-x)$ — a downward parabola, positive exactly between its roots $0$ and $2$.

Common mistakes

• Treating every zero of $f’$ as a sign change. Even-order factors such as $(x+1)^2$ in question 6(e) and $(x-1)^2$ in question 18 touch zero without crossing it.
• Forgetting the sign of the leading coefficient. In question 6(b) and 6(c) a negative coefficient reverses which side is increasing.
• Requiring $f’ > 0$ strictly everywhere. Questions 7, 18 and 9 all have derivatives that vanish at isolated points, and the functions are still increasing. The requirement is that $f’$ never becomes negative.
• Answering a “which interval” question from a single test point. In question 12(C) the derivative is negative near $x = 0$ and positive near $x = \tfrac{\pi}{2}$; one test point would give the wrong verdict.
• Not checking the domain. In question 15, $f(x) = x + \tfrac1x$ is undefined at $x = 0$, which is why the interval must avoid $[-1, 1]$ rather than merely $\{0\}$.

Practise next

• Exercise 6.3 — where $f’ = 0$ locates maxima and minima rather than interval boundaries.
• Miscellaneous Exercise on Chapter 6 — questions 3 and 4 there are harder versions of question 6 here.