Class 12 Maths NCERT Solutions Chapter 6 Ex 6.2: Increasing and Decreasing Functions

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

f'(x) = 2e^{2x} > 0 for all x, since e^{2x} > 0 always.

f'(x) = \cos x.

(a) On \left(0, \tfrac{\pi}{2}\right): \cos x > 0 — increasing.

(b) On \left(\tfrac{\pi}{2}, \pi\right): \cos x < 0 — decreasing.

(c) f' changes sign on (0,\pi), so f is neither throughout.

f'(x) = 4x - 3 = 0 \Rightarrow x = \tfrac{3}{4}.

For x < \tfrac{3}{4}: decreasing. For x > \tfrac{3}{4}: increasing.

f'(x) = 6x^2 - 6x - 36 = 6(x-3)(x+2) = 0 \Rightarrow x = -2, 3.

Test each interval: (-\infty,-2) gives 6 > 0 (increasing); (-2,3) gives -6 < 0 (decreasing); (3,\infty) gives 6 > 0 (increasing).

f'(x) = 2x+2 = 2(x+1) = 0 \Rightarrow x=-1.

f'(x) = -6-4x = -2(2x+3) = 0 \Rightarrow x=-\tfrac{3}{2}.

f'(x) = -6x^2-18x-12 = -6(x+1)(x+2) = 0 \Rightarrow x=-2,-1.

Test each interval: negative on (-\infty,-2), positive on (-2,-1), negative on (-1,\infty).

f'(x) = -9-2x = 0 \Rightarrow x=-\tfrac{9}{2}.

Product rule: f'(x) = 3(x+1)^2(x-3)^3 + 3(x+1)^3(x-3)^2.

Factoring: f'(x) = 3(x+1)^2(x-3)^2\big[(x-3)+(x+1)\big] = 6(x+1)^2(x-3)^2(x-1).

Since (x+1)^2(x-3)^2 \geq 0 always, the sign of f'(x) follows (x-1).

Quotient rule: \dfrac{dy}{dx} = \dfrac{1}{1+x} - \dfrac{4}{(2+x)^2}.

Combine over a common denominator: \dfrac{dy}{dx} = \dfrac{(2+x)^2 - 4(1+x)}{(1+x)(2+x)^2} = \dfrac{x^2}{(1+x)(2+x)^2}.

For x > -1: (1+x) > 0, (2+x)^2 > 0, x^2 \geq 0.

Let u = x^2-2x, so y=u^2 \Rightarrow \dfrac{dy}{dx} = 2u(2x-2) = 4x(x-2)(x-1) = 0 \Rightarrow x = 0, 1, 2.

Sign of x(x-2)(x-1): negative on x<0, positive on 0<x<1, negative on 1<x<2, positive on x>2.

Quotient rule (using \cos^2\theta+\sin^2\theta=1): \dfrac{dy}{d\theta} = \dfrac{8\cos\theta+4}{(2+\cos\theta)^2} - 1.

Combine and expand: \dfrac{dy}{d\theta} = \dfrac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^2}.

On \left[0,\tfrac{\pi}{2}\right]: \cos\theta \geq 0 and 4-\cos\theta > 0 (since \cos\theta \leq 1).

Let f(x) = \log x, x > 0. Then f'(x) = \dfrac{1}{x} > 0 always, since x > 0.

f'(x) = 2x-1 = 0 \Rightarrow x=\tfrac{1}{2}, inside (-1,1).

On \left(-1,\tfrac{1}{2}\right): decreasing. On \left(\tfrac{1}{2},1\right): increasing.

(A) \cos x: derivative -\sin x < 0 throughout — decreasing.

(B) \cos 2x: derivative -2\sin 2x; 2x ranges over (0,\pi) where \sin 2x > 0 — decreasing.

(C) \cos 3x: 3x passes \pi at x=\tfrac{\pi}{3}, so \sin 3x changes sign — not decreasing throughout.

(D) \tan x: derivative \sec^2 x > 0 always — increasing, not decreasing.

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

On all three intervals x>0, so 100x^{99} dominates even where \cos x turns negative — f'(x) > 0 throughout, so f is increasing, not decreasing, on each.

f'(x) = 2x+a; need f'(x) \geq 0 throughout [1,2].

f'(x) is increasing in x, so its minimum on [1,2] is f'(1) = 2+a. Requiring 2+a \geq 0 gives a \geq -2.

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

I is disjoint from [-1,1], so every x \in I has x^2 > 1, making x^2-1>0.

f'(x) = \dfrac{\cos x}{\sin x} = \cot x.

On \left(0,\tfrac{\pi}{2}\right): \cot x > 0 — increasing. On \left(\tfrac{\pi}{2},\pi\right): \cot x < 0 — decreasing.

f'(x) = -\tan x.

On \left(0,\tfrac{\pi}{2}\right): \tan x > 0 \Rightarrow f'(x) < 0 — decreasing. On \left(\tfrac{3\pi}{2},2\pi\right): \tan x < 0 \Rightarrow f'(x)>0 — increasing.

f'(x) = 3x^2-6x+3 = 3(x-1)^2 \geq 0 for all x, touching zero only at x=1.

y' = 2xe^{-x} - x^2e^{-x} = e^{-x}\,x(2-x).

Since e^{-x}>0 always, y' takes the sign of x(2-x), positive exactly when 0<x<2.