NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2

Given:
$f(x) = 3x + 17$

Formula Used:
$f'(x) > 0$ for strictly increasing functions.

Step 1: Differentiate the given function with respect to $x$.
$$f'(x) = \frac{d}{dx}(3x + 17)$$
$$f'(x) = 3$$

Step 2: Analyze the derivative's sign across the domain $\mathbb{R}$.
Since $3 > 0$ for all real numbers $x$, $f'(x)$ is always positive.

Final Answer:
Since $f'(x) > 0$ for all $x \in \mathbb{R}$, $f(x) = 3x + 17$ is strictly increasing on $\mathbb{R}$.

Given:
$f(x) = e^{2x}$

Formula Used:
$f'(x) > 0$ for strictly increasing functions.

Step 1: Find the first derivative of the function.
$$f'(x) = \frac{d}{dx}(e^{2x}) = 2e^{2x}$$

Step 2: Evaluate the sign of the derivative.
For any real number $x$, the exponential function $e^{2x}$ is always positive. Therefore, $2e^{2x} > 0$ for all $x \in \mathbb{R}$.

Final Answer:
Since $f'(x) > 0$ for all $x \in \mathbb{R}$, the function $f(x) = e^{2x}$ is strictly increasing on $\mathbb{R}$.

Given:
$f(x) = \sin x$

Formula Used:
$f'(x) > 0$ (Strictly Increasing), $f'(x) < 0$ (Strictly Decreasing)

Step 1: Find the derivative of $f(x)$.
$$f'(x) = \cos x$$

Step 2: Analyze the intervals.
(a) In the interval $(0, \frac{\pi}{2})$ (first quadrant), $\cos x > 0$. Thus, $f'(x) > 0$.
(b) In the interval $(\frac{\pi}{2}, \pi)$ (second quadrant), $\cos x < 0$. Thus, $f'(x) < 0$.
(c) Over the entire interval $(0, \pi)$, $f'(x)$ changes sign from positive to negative.

Final Answer:
(a) Strictly increasing in $(0, \frac{\pi}{2})$.
(b) Strictly decreasing in $(\frac{\pi}{2}, \pi)$.
(c) Neither increasing nor decreasing in $(0, \pi)$.

Given:
$f(x) = 2x^2 - 3x$

Formula Used:
Set $f'(x) = 0$ to find critical points.

Step 1: Find the derivative and equate to zero.
$$f'(x) = 4x - 3$$
$$4x - 3 = 0 \implies x = \frac{3}{4}$$

Step 2: Check intervals $(-\infty, \frac{3}{4})$ and $(\frac{3}{4}, \infty)$.
For $x < \frac{3}{4}$ (e.g., $x = 0$): $f'(0) = -3 < 0$ (decreasing).
For $x > \frac{3}{4}$ (e.g., $x = 1$): $f'(1) = 1 > 0$ (increasing).

Final Answer:
(a) Strictly increasing in $(\frac{3}{4}, \infty)$
(b) Strictly decreasing in $(-\infty, \frac{3}{4})$

Given:
$f(x) = 2x^3 - 3x^2 - 36x + 7$

Step 1: Differentiate and factorize.
$$f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6)$$
$$f'(x) = 6(x - 3)(x + 2)$$

Step 2: Find critical points and test intervals.
Critical points are $x = 3, -2$. The intervals are $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$.
For $x \in (-\infty, -2)$: $f'(x) > 0$
For $x \in (-2, 3)$: $f'(x) < 0$
For $x \in (3, \infty)$: $f'(x) > 0$

Final Answer:
(a) Strictly increasing in $(-\infty, -2) \cup (3, \infty)$
(b) Strictly decreasing in $(-2, 3)$

Step 1: Differentiate each function and find roots.

(a) $f'(x) = 2x + 2 = 2(x + 1)$. Root: $x = -1$

(b) $f'(x) = -6 - 4x = -2(3 + 2x)$. Root: $x = -\frac{3}{2}$

(c) $f'(x) = -6x^2 - 18x - 12 = -6(x^2 + 3x + 2) = -6(x+1)(x+2)$. Roots: $x = -1, -2$

(d) $f'(x) = -9 - 2x$. Root: $x = -\frac{9}{2}$

(e) $f'(x) = 3(x+1)^2(x-3)^3 + 3(x-3)^2(x+1)^3$
$f'(x) = 3(x+1)^2(x-3)^2[x - 3 + x + 1]$
$f'(x) = 6(x+1)^2(x-3)^2(x-1)$. Root affecting sign: $x = 1$.

Final Answer:

• (a) Inc: $(-1, \infty)$, Dec: $(-\infty, -1)$
• (b) Inc: $(-\infty, -\frac{3}{2})$, Dec: $(-\frac{3}{2}, \infty)$
• (c) Inc: $(-2, -1)$, Dec: $(-\infty, -2) \cup (-1, \infty)$
• (d) Inc: $(-\infty, -\frac{9}{2})$, Dec: $(-\frac{9}{2}, \infty)$
• (e) Inc: $(1, \infty)$, Dec: $(-\infty, 1)$

Given: $y = \log(1+x) - \frac{2x}{2+x}$

Step 1: Differentiate $y$ with respect to $x$.
$$\frac{dy}{dx} = \frac{1}{1+x} - \frac{(2+x)(2) - (2x)(1)}{(2+x)^2}$$
$$\frac{dy}{dx} = \frac{1}{1+x} - \frac{4}{(2+x)^2}$$

Step 2: Simplify the derivative to check its sign.
$$\frac{dy}{dx} = \frac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2}$$
$$\frac{dy}{dx} = \frac{4 + x^2 + 4x - 4 - 4x}{(1+x)(2+x)^2}$$
$$\frac{dy}{dx} = \frac{x^2}{(1+x)(2+x)^2}$$

Since $x^2 \ge 0$ and for $x > -1$, $(1+x) > 0$ and $(2+x)^2 > 0$. Thus, $\frac{dy}{dx} \ge 0$.

Final Answer:
Since $\frac{dy}{dx} \ge 0$ for all $x > -1$, the function is increasing throughout its domain.

Given: $y = [x(x-2)]^2 = (x^2 - 2x)^2$

Step 1: Differentiate $y$.
$$\frac{dy}{dx} = 2(x^2 - 2x) \cdot (2x - 2)$$
$$\frac{dy}{dx} = 4x(x - 2)(x - 1)$$

Step 2: Find critical points and test intervals.
Roots are $x = 0, 1, 2$. Intervals: $(-\infty, 0)$, $(0, 1)$, $(1, 2)$, $(2, \infty)$.
Testing signs:
For $x \in (0, 1)$, $\frac{dy}{dx} = (+)(-)(-) > 0$.
For $x \in (2, \infty)$, $\frac{dy}{dx} = (+)(+)(+) > 0$.

Final Answer:
The function is increasing in the intervals $(0, 1) \cup (2, \infty)$.

Step 1: Differentiate $y$ with respect to $\theta$.
$$\frac{dy}{d\theta} = \frac{(2+\cos\theta)(4\cos\theta) - (4\sin\theta)(-\sin\theta)}{(2+\cos\theta)^2} - 1$$
$$\frac{dy}{d\theta} = \frac{8\cos\theta + 4\cos^2\theta + 4\sin^2\theta}{(2+\cos\theta)^2} - 1$$

Step 2: Simplify using $\sin^2\theta + \cos^2\theta = 1$.
$$\frac{dy}{d\theta} = \frac{8\cos\theta + 4}{(2+\cos\theta)^2} - 1$$
$$\frac{dy}{d\theta} = \frac{8\cos\theta + 4 - (4 + \cos^2\theta + 4\cos\theta)}{(2+\cos\theta)^2}$$
$$\frac{dy}{d\theta} = \frac{4\cos\theta - \cos^2\theta}{(2+\cos\theta)^2} = \frac{\cos\theta(4 - \cos\theta)}{(2+\cos\theta)^2}$$

In $[0, \frac{\pi}{2}]$, $\cos\theta \ge 0$ and $(4-\cos\theta) > 0$. Hence, $\frac{dy}{d\theta} \ge 0$.

Final Answer:
Since $\frac{dy}{d\theta} \ge 0$, $y$ is an increasing function in $[0, \frac{\pi}{2}]$.

Given: $f(x) = \log x$

Step 1: Differentiate the logarithmic function.
$$f'(x) = \frac{1}{x}$$

Step 2: Analyze the derivative in the given domain.
For any value in the interval $(0, \infty)$, the value of $x$ is strictly positive. Therefore, $\frac{1}{x} > 0$.

Final Answer:
Since $f'(x) > 0$ for all $x \in (0, \infty)$, the logarithmic function is strictly increasing on $(0, \infty)$.

Step 1: Differentiate and find the critical point.
$$f'(x) = 2x - 1$$
$$2x - 1 = 0 \implies x = \frac{1}{2}$$

Step 2: Evaluate intervals within $(-1, 1)$.
For $x \in (-1, \frac{1}{2})$, $f'(x) < 0$ (function is decreasing).
For $x \in (\frac{1}{2}, 1)$, $f'(x) > 0$ (function is increasing).

Final Answer:
Because $f'(x)$ changes sign in $(-1, 1)$, $f(x)$ is neither strictly increasing nor strictly decreasing on $(-1, 1)$.

Step 1: Differentiate each option.

(A) $f'(x) = -\sin x$. In $(0, \frac{\pi}{2})$, $\sin x > 0$, so $-\sin x < 0$. (Strictly Decreasing)

(B) $f'(x) = -2\sin 2x$. For $x \in (0, \frac{\pi}{2})$, $2x \in (0, \pi)$. $\sin 2x > 0$, so $-2\sin 2x < 0$. (Strictly Decreasing)

(C) $f'(x) = -3\sin 3x$. For $x \in (0, \frac{\pi}{2})$, $3x \in (0, \frac{3\pi}{2})$. Here $\sin 3x$ changes sign. (Not strictly decreasing)

(D) $f'(x) = \sec^2 x$. Always positive. (Strictly Increasing)

Final Answer:
Functions (A) $\cos x$ and (B) $\cos 2x$ are strictly decreasing on $(0, \frac{\pi}{2})$.

Step 1: Find the derivative.
$$f'(x) = 100x^{99} + \cos x$$

Step 2: Check each interval.
(A) In $(0, 1)$, $100x^{99} > 0$ and $\cos x > 0 \implies f'(x) > 0$.
(B) In $(\frac{\pi}{2}, \pi)$, $x > 1.57$ so $100x^{99} \gg 1$. Though $\cos x \in [-1, 0]$, $100x^{99} + \cos x > 0$.
(C) In $(0, \frac{\pi}{2})$, both terms are positive $\implies f'(x) > 0$.

Final Answer:
(D) None of these, because the function is strictly increasing in all the given intervals.

Step 1: Differentiate $f(x)$.
$$f'(x) = 2x + a$$

Step 2: Apply the increasing condition on the interval $(1, 2)$.
We need $2x + a \ge 0$ for all $x \in (1, 2)$.
Since $y = 2x + a$ is an increasing linear function, its minimum value on $[1, 2]$ occurs at $x = 1$.
So, $2(1) + a \ge 0 \implies 2 + a \ge 0 \implies a \ge -2$.

Final Answer:
The least value of $a$ is $-2$.

Given: $f(x) = x + \frac{1}{x}$ and domain disjoint from $[-1, 1]$. This means $x < -1$ or $x > 1$.

Step 1: Differentiate the function.
$$f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2}$$

Step 2: Check the condition $x < -1$ or $x > 1$.
For any $x$ such that $|x| > 1$, we have $x^2 > 1$.
Thus, $x^2 - 1 > 0$, making $f'(x) = \frac{\text{positive}}{\text{positive}} > 0$.

Final Answer:
Since $f'(x) > 0$ for all $x \notin [-1, 1]$, the function is strictly increasing on $I$.

Step 1: Differentiate $f(x)$.
$$f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x$$

Step 2: Evaluate the sign of $\cot x$.
In $(0, \frac{\pi}{2})$ (first quadrant), $\cot x > 0 \implies f'(x) > 0$.
In $(\frac{\pi}{2}, \pi)$ (second quadrant), $\cot x < 0 \implies f'(x) < 0$.

Final Answer:
Since $f'(x) > 0$ on $(0, \frac{\pi}{2})$ and $f'(x) < 0$ on $(\frac{\pi}{2}, \pi)$, the given properties are proven.

Step 1: Differentiate the function.
$$f'(x) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$$

Step 2: Check quadrants.
In $(0, \frac{\pi}{2})$ (quadrant I), $\tan x > 0 \implies -\tan x < 0$ (strictly decreasing).
In $(\frac{3\pi}{2}, 2\pi)$ (quadrant IV), $\tan x < 0 \implies -\tan x > 0$ (strictly increasing).

Final Answer:
The conditions $f'(x) < 0$ and $f'(x) > 0$ respectively prove the theorem for the given intervals.

Step 1: Find the derivative of $f(x)$.
$$f'(x) = 3x^2 - 6x + 3$$

Step 2: Factorize the derivative.
$$f'(x) = 3(x^2 - 2x + 1) = 3(x - 1)^2$$
Since $(x - 1)^2 \ge 0$ for all $x \in \mathbb{R}$, $f'(x) \ge 0$.

Final Answer:
Since $f'(x) \ge 0$ for all real numbers $x$, the function is increasing in $\mathbb{R}$.

Step 1: Find $y'$ and set to $0$.
$$y' = 2x e^{-x} + x^2(-e^{-x}) = e^{-x}(2x - x^2) = x e^{-x}(2 - x)$$
$$y' = 0 \implies x = 0, x = 2$$

Step 2: Analyze intervals.
Since $e^{-x}$ is always positive, the sign depends on $x(2 - x)$.
For $x \in (0, 2)$ (e.g., $x = 1$): $y' = (+)(+) > 0$.
For $x > 2$: $y' < 0$.
For $x < 0$: $y' < 0$.

Final Answer:
(D) The interval in which the function is increasing is $(0, 2)$.