Comprehensive solutions for Class 12 Maths (NCERT) Exercise 6.2. Learn how to determine the intervals of increasing and decreasing functions using the first derivative test ($f'(x) > 0$ or $f'(x) < 0$). Includes proofs for trigonometric, logarithmic, and polynomial functions.

Rbse Solutions Class 12 Maths Chapter 6 Exercise 6.2 | Increasing and Decreasing functions

This exercise applies the derivative, $f'(x)$, to determine the intervals of increasing and decreasing behavior of a function $f(x)$.

A function $f(x)$ is strictly increasing on an interval if $f'(x) > 0$ for all $x$ in that interval.
A function $f(x)$ is strictly decreasing on an interval if $f'(x) < 0$ for all $x$ in that interval.
A function $f(x)$ is increasing if $f'(x) \ge 0$.
A function $f(x)$ is decreasing if $f'(x) \le 0$.

1. Show that $f(x) = 3x + 17$ is increasing on R

Find the derivative: $f'(x) = 3$.
Analyze the sign: Since $f'(x) = 3 > 0$ for all $x \in \mathbb{R}$, the function is strictly increasing on $\mathbb{R}$.

2. Show that $f(x) = e^{2x}$ is increasing on R

Find the derivative: $f'(x) = 2e^{2x}$.
Analyze the sign: Since $e^{2x} > 0$ for all $x \in \mathbb{R}$ and $2$ is positive, $f'(x) = 2e^{2x} > 0$ for all $x \in \mathbb{R}$. The function is strictly increasing on $\mathbb{R}$.

3. Show the behavior of $f(x) = \sin x$

Find the derivative: $f'(x) = \cos x$.

(a) Increasing in $\left(0, \frac{\pi}{2}\right)$: In the first quadrant, $\cos x > 0$. Since $f'(x) = \cos x > 0$, $f(x)$ is strictly increasing in $\left(0, \frac{\pi}{2}\right)$.

(b) Decreasing in $\left(\frac{\pi}{2}, \pi\right)$: In the second quadrant, $\cos x < 0$. Since $f'(x) = \cos x < 0$, $f(x)$ is strictly decreasing in $\left(\frac{\pi}{2}, \pi\right)$.

(c) Neither increasing nor decreasing in $(0, \pi)$: Since $f(x)$ is increasing on $\left(0, \frac{\pi}{2}\right)$ and decreasing on $\left(\frac{\pi}{2}, \pi\right)$, it is neither strictly increasing nor strictly decreasing on the entire interval $(0, \pi)$.

4. Find intervals for $f(x) = 2x^2 – 3x$

Find the derivative: $f'(x) = 4x – 3$.
Find the critical point: Set $f'(x) = 0 \implies 4x – 3 = 0 \implies x = \frac{3}{4}$. The critical point divides $\mathbb{R}$ into $\left(-\infty, \frac{3}{4}\right)$ and $\left(\frac{3}{4}, \infty\right)$.

(a) Increasing interval ($f'(x) > 0$):

$$4x – 3 > 0 \implies 4x > 3 \implies x > \frac{3}{4}$$

The function is increasing on $\mathbf{\left[\frac{3}{4}, \infty\right)}$.

(b) Decreasing interval ($f'(x) < 0$):

$$4x – 3 < 0 \implies 4x < 3 \implies x < \frac{3}{4}$$

The function is decreasing on $\mathbf{\left(-\infty, \frac{3}{4}\right]}$.

5. Find intervals for $f(x) = 2x^3 – 3x^2 – 36x + 7$

Find the derivative: $f'(x) = 6x^2 – 6x – 36 = 6(x^2 – x – 6)$.
Find the critical points: Set $f'(x) = 0$.$$6(x^2 – x – 6) = 0 \implies 6(x – 3)(x + 2) = 0$$The critical points are $\mathbf{x = 3}$ and $\mathbf{x = -2}$. These points divide $\mathbb{R}$ into three intervals: $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$.
Test the intervals:

Interval	Test Value (x)	f′(x)=6(x−3)(x+2)	Sign of f′(x)	Monotonicity
$(-\infty, -2)$	$-3$	$6(-)(-)$	Positive (+)	Increasing
$(-2, 3)$	$0$	$6(-)(+)$	Negative (-)	Decreasing
$(3, \infty)$	$4$	$6(+)(+)$	Positive (+)	Increasing

(a) Increasing intervals: $\mathbf{(-\infty, -2]}$ and $\mathbf{[3, \infty)}$.

(b) Decreasing interval: $\mathbf{[-2, 3]}$.

🔄 Solutions for Exercises 6-11

6. Find intervals where the following functions are strictly increasing or decreasing

(a) $f(x) = x^2 + 2x – 5$

$f'(x) = 2x + 2$. Critical point: $2x + 2 = 0 \implies x = -1$.
Strictly Increasing: $2x + 2 > 0 \implies x > -1$. Interval: $\mathbf{(-1, \infty)}$.
Strictly Decreasing: $2x + 2 < 0 \implies x < -1$. Interval: $\mathbf{(-\infty, -1)}$.

(b) $f(x) = 10 – 6x – 2x^2$

$f'(x) = -6 – 4x$. Critical point: $-6 – 4x = 0 \implies 4x = -6 \implies x = -\frac{3}{2}$.
Strictly Increasing: $-6 – 4x > 0 \implies -4x > 6 \implies x < -\frac{6}{4} \implies x < -\frac{3}{2}$. Interval: $\mathbf{\left(-\infty, -\frac{3}{2}\right)}$.
Strictly Decreasing: $-6 – 4x < 0 \implies x > -\frac{3}{2}$. Interval: $\mathbf{\left(-\frac{3}{2}, \infty\right)}$.

$f'(x) = -6x^2 – 18x – 12 = -6(x^2 + 3x + 2) = -6(x + 1)(x + 2)$.
Critical points: $x = -1$ and $x = -2$. Intervals: $(-\infty, -2)$, $(-2, -1)$, $(-1, \infty)$.
Test Intervals:
- $(-\infty, -2)$: $x=-3$, $f'(-3) = -6(-)(-)$ which is Negative. Strictly Decreasing
- $(-2, -1)$: $x=-1.5$, $f'(-1.5) = -6(+)(-)$ which is Positive. Strictly Increasing
- $(-1, \infty)$: $x=0$, $f'(0) = -6(+)(+)$ which is Negative. Strictly Decreasing
Strictly Increasing: $\mathbf{(-2, -1)}$.
Strictly Decreasing: $\mathbf{(-\infty, -2)}$ and $\mathbf{(-1, \infty)}$.

(d) $f(x) = 6 – 9x – x^2$

$f'(x) = -9 – 2x$. Critical point: $-9 – 2x = 0 \implies 2x = -9 \implies x = -\frac{9}{2}$.
Strictly Increasing: $-9 – 2x > 0 \implies -2x > 9 \implies x < -\frac{9}{2}$. Interval: $\mathbf{\left(-\infty, -\frac{9}{2}\right)}$.
Strictly Decreasing: $-9 – 2x < 0 \implies x > -\frac{9}{2}$. Interval: $\mathbf{\left(-\frac{9}{2}, \infty\right)}$.

(e) $f(x) = (x + 1)^3 (x – 3)^3 = [(x+1)(x-3)]^3 = (x^2 – 2x – 3)^3$

$f'(x) = 3(x^2 – 2x – 3)^2 \cdot \frac{d}{dx}(x^2 – 2x – 3)$
$f'(x) = 3(x^2 – 2x – 3)^2 \cdot (2x – 2) = 6(x – 1)(x^2 – 2x – 3)^2$
Critical points: $2x – 2 = 0 \implies x = 1$. Also, $x^2 – 2x – 3 = 0 \implies (x-3)(x+1) = 0 \implies x=3, x=-1$.
$f'(x)$ is zero at $x=-1, 1, 3$. The factor $(x^2 – 2x – 3)^2$ is non-negative everywhere except at $x=-1, 3$ where it is zero. For the behavior, we only need to look at the sign of $6(x-1)$.
Strictly Increasing: $f'(x) > 0 \implies 6(x – 1)(x^2 – 2x – 3)^2 > 0$. This requires $x – 1 > 0$ and $x \ne -1, 3$. Interval: $\mathbf{(1, 3) \cup (3, \infty)}$ or simply $\mathbf{(1, \infty) \setminus \{3\}}$.
Strictly Decreasing: $f'(x) < 0 \implies 6(x – 1)(x^2 – 2x – 3)^2 < 0$. This requires $x – 1 < 0$ and $x \ne -1$. Interval: $\mathbf{(-\infty, -1) \cup (-1, 1)}$ or simply $\mathbf{(-\infty, 1) \setminus \{-1\}}$.

7. Show that $y = \log(1+x) – \frac{2x}{2+x}, x > -1$, is an increasing function

Find the derivative:$$\frac{dy}{dx} = \frac{d}{dx}(\log(1+x)) – \frac{d}{dx}\left(\frac{2x}{2+x}\right)$$$$\frac{dy}{dx} = \frac{1}{1+x} – \left[\frac{(2+x)(2) – 2x(1)}{(2+x)^2}\right]$$ (Using Quotient Rule)$$\frac{dy}{dx} = \frac{1}{1+x} – \left[\frac{4 + 2x – 2x}{(2+x)^2}\right] = \frac{1}{1+x} – \frac{4}{(2+x)^2}$$
Combine the terms:$$\frac{dy}{dx} = \frac{(2+x)^2 – 4(1+x)}{(1+x)(2+x)^2} = \frac{(4 + 4x + x^2) – (4 + 4x)}{(1+x)(2+x)^2}$$$$\frac{dy}{dx} = \frac{x^2}{(1+x)(2+x)^2}$$
Analyze the sign:

Domain: $x > -1$.
$x^2 \ge 0$.
$1+x > 0$ (since $x > -1$).
$(2+x)^2 > 0$ (since $x \ne -2$, and $x > -1$ avoids this).
Since the numerator $x^2 \ge 0$ and the denominator is strictly positive, $\frac{dy}{dx} \ge 0$ for all $x > -1$.
The derivative is zero only at $x=0$, which is a single point, not an interval.
Therefore, the function is increasing on its domain $(-1, \infty)$.

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

Find the derivative:$$y = (x^2 – 2x)^2$$$$\frac{dy}{dx} = 2(x^2 – 2x) \cdot \frac{d}{dx}(x^2 – 2x) = 2(x^2 – 2x)(2x – 2)$$$$\frac{dy}{dx} = 4x(x – 2)(x – 1)$$
Find the critical points: $4x(x – 2)(x – 1) = 0 \implies x = 0, 1, 2$. These divide $\mathbb{R}$ into $(-\infty, 0)$, $(0, 1)$, $(1, 2)$, and $(2, \infty)$.
Test the intervals for increasing behavior ($\frac{dy}{dx} > 0$):

Interval	Test Value (x)	dxdy=4x(x−2)(x−1)	Sign of dxdy	Monotonicity
$(-\infty, 0)$	$-1$	$(-) (-)(-)$	Negative (-)	Decreasing
$(0, 1)$	$0.5$	$(+) (-)(-)$	Positive (+)	Increasing
$(1, 2)$	$1.5$	$(+) (-)(+)$	Negative (-)	Decreasing
$(2, \infty)$	$3$	$(+) (+)(+)$	Positive (+)	Increasing

The function is increasing on $\mathbf{[0, 1]}$ and $\mathbf{[2, \infty)}$.

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

Find the derivative ($\frac{dy}{d\theta}$) using the Quotient Rule:$$\frac{dy}{d\theta} = \frac{(2 + \cos \theta) \frac{d}{d\theta}(4 \sin \theta) – 4 \sin \theta \frac{d}{d\theta}(2 + \cos \theta)}{(2 + \cos \theta)^2}$$$$\frac{dy}{d\theta} = \frac{(2 + \cos \theta) (4 \cos \theta) – 4 \sin \theta (-\sin \theta)}{(2 + \cos \theta)^2}$$$$\frac{dy}{d\theta} = \frac{8 \cos \theta + 4 \cos^2 \theta + 4 \sin^2 \theta}{(2 + \cos \theta)^2}$$$$\frac{dy}{d\theta} = \frac{8 \cos \theta + 4(\cos^2 \theta + \sin^2 \theta)}{(2 + \cos \theta)^2} = \frac{8 \cos \theta + 4}{(2 + \cos \theta)^2}$$
Analyze the sign in $\left[0, \frac{\pi}{2}\right]$:

In $\left[0, \frac{\pi}{2}\right]$, we know $0 \le \cos \theta \le 1$.
Numerator: $8 \cos \theta + 4$. Since $\cos \theta \ge 0$, $8 \cos \theta + 4 \ge 4 > 0$.
Denominator: $(2 + \cos \theta)^2$. Since $\cos \theta \ge 0$, the denominator is $\ge (2+0)^2 = 4 > 0$.
Since the numerator is strictly positive and the denominator is strictly positive, $\frac{dy}{d\theta} > 0$ for all $\theta \in \left[0, \frac{\pi}{2}\right]$ (The derivative is never zero on the interior of the interval).
Therefore, the function is strictly increasing in $\left[0, \frac{\pi}{2}\right]$.

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

Define the function: $f(x) = \log x$. Domain is $(0, \infty)$.
Find the derivative: $f'(x) = \frac{1}{x}$.
Analyze the sign: For all $x \in (0, \infty)$, $x$ is a positive number.

Therefore, $f'(x) = \frac{1}{x} > 0$.
Since $f'(x) > 0$ on $(0, \infty)$, the logarithmic function is strictly increasing on its domain.

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

Find the derivative: $f'(x) = 2x – 1$.
Find the critical point: $f'(x) = 0 \implies 2x – 1 = 0 \implies x = \frac{1}{2}$.
Check intervals within $(-1, 1)$: The critical point $x=\frac{1}{2}$ is in the interval $(-1, 1)$, dividing it into $\left(-1, \frac{1}{2}\right)$ and $\left(\frac{1}{2}, 1\right)$.

On $\left(-1, \frac{1}{2}\right)$: Test $x=0$. $f'(0) = -1 < 0$. The function is decreasing.
On $\left(\frac{1}{2}, 1\right)$: Test $x=0.6$. $f'(0.6) = 2(0.6) – 1 = 0.2 > 0$. The function is increasing.Since the function changes from decreasing to increasing within the interval $(-1, 1)$, it is neither strictly increasing nor strictly decreasing on $(-1, 1)$.

📊 Solutions for Exercises 12-19

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

The interval $\left(0, \frac{\pi}{2}\right)$ is the first quadrant. A function is decreasing if $f'(x) < 0$.

(A) $f(x) = \cos x$. $f'(x) = -\sin x$. In $\left(0, \frac{\pi}{2}\right)$, $\sin x > 0$, so $f'(x) < 0$. Decreasing.
(B) $f(x) = \cos 2x$. $f'(x) = -2 \sin 2x$. In $\left(0, \frac{\pi}{2}\right)$, $2x \in (0, \pi)$, where $\sin 2x > 0$. Thus $f'(x) < 0$. Decreasing.
(C) $f(x) = \cos 3x$. $f'(x) = -3 \sin 3x$. In $\left(0, \frac{\pi}{2}\right)$, $3x \in \left(0, \frac{3\pi}{2}\right)$. $\sin 3x$ is positive in $\left(0, \frac{\pi}{3}\right)$ and negative in $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$. Thus $f'(x)$ changes sign. Neither.
(D) $f(x) = \tan x$. $f'(x) = \sec^2 x$. In $\left(0, \frac{\pi}{2}\right)$, $\sec^2 x > 0$. Increasing.

The functions decreasing on $\left(0, \frac{\pi}{2}\right)$ are (A) $\cos x$ and (B) $\cos 2x$.

13. On which interval is $f(x) = x^{100} + \sin x – 1$ decreasing?

Find the derivative: $f'(x) = 100x^{99} + \cos x$.
Analyze the intervals:
- (A) $(0, 1)$: $x^{99} > 0$, $\cos x \in (\cos 1, 1)$. Both terms are positive. $f'(x) > 0$. Increasing.
- (B) $\left(\frac{\pi}{2}, \pi\right)$: $x^{99} > 0$, $\cos x \in (-1, 0)$. $x^{99}$ is significantly large ($\approx 1.5^{99}$), so $100x^{99}$ is much greater than 1. $f'(x) > 0$. Increasing.
- (C) $\left(0, \frac{\pi}{2}\right)$: $x^{99} > 0$, $\cos x \in (0, 1)$. Both terms are positive. $f'(x) > 0$. Increasing.

Since $100x^{99}$ is overwhelmingly positive on all given intervals, and $\cos x$ is generally small, $f'(x)$ is positive in all these intervals.

The function is never decreasing in any of the given intervals.

The correct answer is (D) None of these.

14. For what values of $a$ is $f(x) = x^2 + ax + 1$ increasing on $[1, 2]$?

Condition for increasing: $f'(x) \ge 0$ for all $x \in [1, 2]$.
Find the derivative: $f'(x) = 2x + a$.
Apply the condition: We require $2x + a \ge 0$ for all $x \in [1, 2]$.Since $2x$ is an increasing function, its minimum value on $[1, 2]$ occurs at the left endpoint, $x=1$.$$2x + a \ge 0 \implies 2(1) + a \ge 0$$$$2 + a \ge 0 \implies a \ge -2$$The function is increasing on $[1, 2]$ when $\mathbf{a \ge -2}$.

15. Prove that $f(x) = x + \frac{1}{x}$ is increasing on $I$ (where $I$ is disjoint from $[-1, 1]$)

Find the derivative: $f'(x) = 1 – \frac{1}{x^2} = \frac{x^2 – 1}{x^2}$.
Analyze the sign: We want to show $f'(x) > 0$.$$f'(x) > 0 \implies \frac{x^2 – 1}{x^2} > 0$$Since $x \in I$ is disjoint from $[-1, 1]$, $x$ cannot be between $-1$ and $1$ (and $x \ne 0$).This means: $|x| > 1$ (i.e., $x > 1$ or $x < -1$).If $|x| > 1$, then $x^2 > 1$, so $\mathbf{x^2 – 1 > 0}$.The denominator $x^2$ is always positive.Since the numerator is positive and the denominator is positive, $f'(x) > 0$ on $I$.Thus, $f(x)$ is strictly increasing on $I$.

16. Prove the behavior of $f(x) = \log (\sin x)$

Find the derivative: $f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$.

Increasing on $\left(0, \frac{\pi}{2}\right)$: In the first quadrant, $\cot x > 0$. Since $f'(x) > 0$, the function is strictly increasing.
Decreasing on $\left(\frac{\pi}{2}, \pi\right)$: In the second quadrant, $\cot x < 0$. Since $f'(x) < 0$, the function is strictly decreasing.

17. Prove the behavior of $f(x) = \log |\cos x|$

Find the derivative: $f'(x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$. (Assuming $|\cos x| = \cos x$ or using Chain Rule for $\log|u|$ to get $u’/u$)

Decreasing on $\left(0, \frac{\pi}{2}\right)$: In the first quadrant, $\tan x > 0$, so $f'(x) = -\tan x < 0$. The function is strictly decreasing.
Increasing on $\left(\frac{3\pi}{2}, 2\pi\right)$: In the fourth quadrant, $\tan x < 0$. $f'(x) = -\tan x > 0$. The function is strictly increasing.

18. Prove that $f(x) = x^3 – 3x^2 + 3x – 100$ is increasing in R

Find the derivative: $f'(x) = 3x^2 – 6x + 3 = 3(x^2 – 2x + 1) = 3(x – 1)^2$.
Analyze the sign: Since $(x – 1)^2 \ge 0$ for all $x \in \mathbb{R}$, $f'(x) = 3(x – 1)^2 \ge 0$ for all $x \in \mathbb{R}$.

The derivative is zero only at $x=1$, which is an isolated point.
Therefore, the function is increasing on $\mathbb{R}$.

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

Find the derivative (using the Product Rule):$$\frac{dy}{dx} = \frac{d}{dx}(x^2) e^{-x} + x^2 \frac{d}{dx}(e^{-x})$$$$\frac{dy}{dx} = 2x e^{-x} + x^2 (-e^{-x}) = x e^{-x} (2 – x)$$
Condition for increasing: $\frac{dy}{dx} > 0$.$$x e^{-x} (2 – x) > 0$$Since $e^{-x}$ is always positive, the sign of the expression depends on $x(2 – x)$.$$x(2 – x) > 0 \implies x > 0 \text{ and } 2 – x > 0 \text{ OR } x < 0 \text{ and } 2 – x < 0$$

Case 1: $x > 0$ and $x < 2$. This gives the interval $\mathbf{(0, 2)}$.
Case 2: $x < 0$ and $x > 2$. This is impossible.

The function is increasing on $\mathbf{(0, 2)}$.

The correct answer is (D) $(0, 2)$.

Ex 6.1

Ex 6.2

Ex 6.3

Ch 6 Misc.