Class 12-NCERT Solutions-Chapter-06-Application of Derivatives-Ex 6.2

Solution:

Given \( f(x) = 3x + 17 \).

Differentiating w.r.t \( x \):

\( f'(x) = 3 \).

Since \( 3 > 0 \) for all \( x \in \mathbb{R} \), \( f'(x) > 0 \).

Therefore, \( f \) is strictly increasing on \( \mathbb{R} \).

Solution:

Given \( f(x) = e^{2x} \).

\( f'(x) = 2e^{2x} \).

We know that the exponential function \( e^{2x} > 0 \) for all \( x \in \mathbb{R} \).

Therefore, \( 2e^{2x} > 0 \Rightarrow f'(x) > 0 \) for all \( x \in \mathbb{R} \).

Hence, \( f \) is increasing on \( \mathbb{R} \).

Solution:

Given \( f(x) = \sin x \Rightarrow f'(x) = \cos x \).

(a) In interval \( (0, \frac{\pi}{2}) \) (1st Quadrant), \( \cos x > 0 \). Thus, \( f'(x) > 0 \). Increasing.

(b) In interval \( (\frac{\pi}{2}, \pi) \) (2nd Quadrant), \( \cos x < 0 \). Thus, \( f'(x) < 0 \). Decreasing.

(c) In interval \( (0, \pi) \), the derivative is positive in one part and negative in the other. Thus, it is neither strictly increasing nor strictly decreasing.

Solution:

\( f(x) = 2x^2 - 3x \Rightarrow f'(x) = 4x - 3 \).

For critical point, put \( f'(x) = 0 \Rightarrow 4x - 3 = 0 \Rightarrow x = \frac{3}{4} \).

This divides the number line into \( (-\infty, \frac{3}{4}) \) and \( (\frac{3}{4}, \infty) \).

(a) Increasing: \( f'(x) > 0 \Rightarrow 4x - 3 > 0 \Rightarrow x > \frac{3}{4} \). Interval: \( (\frac{3}{4}, \infty) \).

(b) Decreasing: \( f'(x) < 0 \Rightarrow 4x - 3 < 0 \Rightarrow x < \frac{3}{4} \). Interval: \( (-\infty, \frac{3}{4}) \).

Solution:

\( f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6) = 6(x - 3)(x + 2) \).

Critical points: \( x = 3, x = -2 \).

Intervals: \( (-\infty, -2), (-2, 3), (3, \infty) \).

(a) Increasing: \( f'(x) > 0 \) when \( x \in (-\infty, -2) \cup (3, \infty) \).

(b) Decreasing: \( f'(x) < 0 \) when \( x \in (-2, 3) \).

Solution:

(a) \( f(x) = x^2 + 2x - 5 \): \( f'(x) = 2x + 2 \). CP: \( x = -1 \).
Increasing: \( (-1, \infty) \), Decreasing: \( (-\infty, -1) \).

(b) \( f(x) = 10 - 6x - 2x^2 \): \( f'(x) = -6 - 4x \). CP: \( x = -3/2 \).
Increasing: \( (-\infty, -3/2) \), Decreasing: \( (-3/2, \infty) \).

(c) \( f(x) = -2x^3 - 9x^2 - 12x + 1 \): \( f'(x) = -6(x^2 + 3x + 2) = -6(x+1)(x+2) \).
Increasing: \( (-2, -1) \), Decreasing: \( (-\infty, -2) \cup (-1, \infty) \).

(d) \( f(x) = 6 - 9x - x^2 \): \( f'(x) = -9 - 2x \). CP: \( x = -9/2 \).
Increasing: \( (-\infty, -9/2) \), Decreasing: \( (-9/2, \infty) \).

(e) \( f(x) = (x+1)^3 (x-3)^3 \):
\( f'(x) = 3(x+1)^2(x-3)^3 + 3(x+1)^3(x-3)^2 \)
\( = 3(x+1)^2(x-3)^2 [x-3 + x+1] = 6(x+1)^2(x-3)^2(x-1) \).
Since squares are always positive, sign depends on \( (x-1) \).
Increasing: \( (1, \infty) \), Decreasing: \( (-\infty, 1) \).

Solution:

\( \frac{dy}{dx} = \frac{1}{1+x} - \frac{(2+x)(2) - 2x(1)}{(2+x)^2} \)

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

\( = \frac{4 + x^2 + 4x - 4 - 4x}{(1+x)(2+x)^2} = \frac{x^2}{(1+x)(2+x)^2} \).

Since \( x^2 \ge 0 \) and \( (2+x)^2 > 0 \), and given \( x > -1 \Rightarrow 1+x > 0 \), the derivative \( \frac{dy}{dx} \ge 0 \).

Hence, it is an increasing function.

Solution:

\( y = x^2(x-2)^2 \).
\( \frac{dy}{dx} = 2x(x-2)^2 + x^2 \cdot 2(x-2) = 2x(x-2)[x-2 + x] \).
\( = 2x(x-2)(2x-2) = 4x(x-1)(x-2) \).

Critical points: 0, 1, 2.
Intervals: \( (-\infty, 0), (0, 1), (1, 2), (2, \infty) \).

Check signs:
\( (-\infty, 0) \): (-)(-)(-) = - (Decreasing)
\( (0, 1) \): (+)(-)(-) = + (Increasing)
\( (1, 2) \): (+)(+)(-) = - (Decreasing)
\( (2, \infty) \): (+)(+)(+) = + (Increasing)

Increasing in \( (0, 1) \cup (2, \infty) \).

Solution:

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

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

\( = \frac{8\cos\theta + 4 - (4 + \cos^2\theta + 4\cos\theta)}{(2+\cos\theta)^2} = \frac{4\cos\theta - \cos^2\theta}{(2+\cos\theta)^2} \).

\( \frac{dy}{d\theta} = \frac{\cos\theta(4 - \cos\theta)}{(2+\cos\theta)^2} \).

In \( [0, \frac{\pi}{2}] \), \( \cos\theta \ge 0 \) and \( 4 - \cos\theta > 0 \) (since max cos is 1).
Thus, \( \frac{dy}{d\theta} \ge 0 \). Function is increasing.

Solution:

Let \( f(x) = \log x \).

\( f'(x) = \frac{1}{x} \).

For \( x \in (0, \infty) \), \( \frac{1}{x} > 0 \).

Therefore, logarithmic function is strictly increasing on \( (0, \infty) \).

Solution:

\( f'(x) = 2x - 1 \).

Critical point: \( 2x - 1 = 0 \Rightarrow x = 1/2 \).

Interval \( (-1, 1) \) splits into \( (-1, 1/2) \) and \( (1/2, 1) \).

In \( (-1, 1/2) \), \( f'(x) < 0 \) (Decreasing).

In \( (1/2, 1) \), \( f'(x) > 0 \) (Increasing).

Since it is decreasing in one part and increasing in the other, it is neither strictly increasing nor decreasing on the whole interval.

Solution:

(A) \( f(x) = \cos x \Rightarrow f'(x) = -\sin x \). In \( (0, \pi/2) \), \( \sin x > 0 \), so \( f'(x) < 0 \). Decreasing.

(B) \( f(x) = \cos 2x \Rightarrow f'(x) = -2\sin 2x \). In \( (0, \pi/2) \), \( 2x \in (0, \pi) \), sin is positive. So \( f'(x) < 0 \). Decreasing.

(C) \( f(x) = \cos 3x \Rightarrow f'(x) = -3\sin 3x \). In \( (0, \pi/2) \), \( 3x \in (0, 3\pi/2) \). Sin changes sign at \( \pi \). Neither strictly decreasing.

(D) \( f(x) = \tan x \Rightarrow f'(x) = \sec^2 x > 0 \). Increasing.

Answer: (A) and (B).

Answer: (D)

Solution:

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

(A) On \( (0, 1) \): \( x > 0 \Rightarrow 100x^{99} > 0 \). \( \cos x > 0 \) (since 1 radian approx 57 deg). \( f'(x) > 0 \). Increasing.

(B) On \( (\pi/2, \pi) \): \( x > 1 \), so \( 100x^{99} > 100 \). \( \cos x \in (-1, 0) \). Sum is positive. Increasing.

(C) On \( (0, \pi/2) \): Both terms positive. Increasing.

Since it is increasing in all given intervals, it is decreasing in None of these.

Solution:

\( f'(x) = 2x + a \).

For \( f(x) \) to be increasing on \( [1, 2] \), we need \( f'(x) \ge 0 \) for all \( x \in [1, 2] \).

The minimum value of \( 2x + a \) on \( [1, 2] \) occurs at \( x = 1 \).

\( 2(1) + a \ge 0 \Rightarrow 2 + a \ge 0 \Rightarrow a \ge -2 \).

So, the least value of \( a \) is -2.

Solution:

\( f'(x) = 1 - \frac{1}{x^2} = \frac{x^2 - 1}{x^2} \).

Since \( I \) is disjoint from \( [-1, 1] \), it means for all \( x \in I \), \( x < -1 \) or \( x > 1 \).

This implies \( |x| > 1 \Rightarrow x^2 > 1 \Rightarrow x^2 - 1 > 0 \).

Also \( x^2 > 0 \). Thus \( f'(x) > 0 \) for all \( x \in I \).

Hence, \( f \) is increasing on \( I \).

Solution:

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

In \( (0, \frac{\pi}{2}) \), \( \cot x > 0 \). Increasing.

In \( (\frac{\pi}{2}, \pi) \), \( \cot x < 0 \). Decreasing.

Solution:

For \( x \in (0, \frac{\pi}{2}) \), \( \cos x > 0 \), so \( f(x) = \log(\cos x) \).
\( f'(x) = \frac{1}{\cos x}(-\sin x) = -\tan x \).
Since tan x > 0 in 1st quad, \( f'(x) < 0 \). Decreasing.

For \( x \in (\frac{3\pi}{2}, 2\pi) \), \( \cos x > 0 \), so \( f(x) = \log(\cos x) \).
\( f'(x) = -\tan x \).
In 4th quad, tan x < 0, so \( -\tan x > 0 \). Increasing.

Solution:

\( f'(x) = 3x^2 - 6x + 3 \).

\( = 3(x^2 - 2x + 1) = 3(x-1)^2 \).

Since \( (x-1)^2 \ge 0 \) for all real \( x \), \( f'(x) \ge 0 \).

Therefore, \( f \) is increasing on \( \mathbb{R} \).

Answer: (D)

Solution:

\( \frac{dy}{dx} = x^2(-e^{-x}) + e^{-x}(2x) = e^{-x}(2x - x^2) = x e^{-x}(2 - x) \).

For increasing, \( \frac{dy}{dx} > 0 \).

Since \( e^{-x} > 0 \), we need \( x(2-x) > 0 \).

This implies \( x(x-2) < 0 \).

This holds for \( 0 < x < 2 \).

Interval: \( (0, 2) \).