Application of Derivatives Part 3

Concept:If $f''(x) > 0$, then $f'(x)$ is strictly increasing for all $x \in \mathbb{R}$.The monotonicity of $g(x)$ is determined by the sign of its derivative $g'(x)$.Explanation:First rewrite $g(x) = f( \tan^2 x - 2 \tan x + a )$.Complete the square: $\tan^2 x - 2 \tan x = (\tan x - 1)^2 - 1$, so the inner argument is $(\tan x - 1)^2 + (a - 1)$.Thus $g(x) = f\big((\tan x - 1)^2 + a - 1\big)$.Differentiate using the chain rule:$g'(x) = f'\big((\tan x - 1)^2 + a - 1\big) \cdot 2(\tan x - 1) \cdot \sec^2 x$.Since $f''(x) > 0$, $f'$ is increasing. Given $f'(a-1)=0$, we have:If the argument of $f'$ is $> a-1$, then $f' > 0$; if argument $< a-1$, then $f' < 0$.Consider $x \in (0, \pi/4)$:$\tan x$ increases from $0$ to $1$, so $\tan x - 1$ is from $-1$ to $0$.Then $0 < (\tan x - 1)^2 < 1$.Hence the argument $(\tan x - 1)^2 + (a-1)$ lies in $(a-1, a)$.Since argument $> a-1$, $f'(\text{argument}) > f'(a-1) = 0$.In this interval, $\tan x - 1$ is negative, so $2(\tan x -1) < 0$, and $\sec^2 x > 0$.Thus $g'(x) = (+)(-)(+) = -$, so $g$ is decreasing on $(0, \pi/4)$.Now consider $x \in (\pi/4, \pi/2)$:$\tan x > 1$, so $\tan x - 1 > 0$. Thus $(\tan x - 1)^2 > 0$ and the argument is $> a$.Since argument $> a-1$, $f'(\text{argument}) > 0$.Now $2(\tan x -1) > 0$, $\sec^2 x > 0$, so $g'(x) = (+)(+)(+) = +$.Hence $g$ is increasing on $(\pi/4, \pi/2)$.Therefore statement (I) "$g$ is increasing in $(0,\pi/4)$" is false (it is decreasing).Statement (II) "$g$ is decreasing in $(\pi/4,\pi/2)$" is false (it is increasing).Thus neither statement is true.Answer:Both (I) and (II) are False.Correct option: B (Neither (I) nor (II) is True).