Let $$f: R\rightarrow R$$ be a twice differentiable function such that $$f''(x) > 0$$ for all $$x\in R$$ and f'(a-1)=0, where a is a real number. Let g(x)= $$f(\tan^{2}x- 2\tan x+a)$$, $$0 < x < \frac{\pi}{2}$$.
Consider the following two statements :
(I) $$\text{g is increasing in } \left(0, \frac{\pi}{4} \right)$$
(II) $$\text{g is deceasing in } \left( \frac{\pi}{4} , \frac{\pi}{2} \right)$$
Then,

$$f: \mathbb{R} \to \mathbb{R}$$ twice differentiable with $$f''(x) > 0$$ and $$f'(a-1) = 0$$. $$g(x) = f(\tan^2 x - 2\tan x + a)$$, $$0 < x < \pi/2$$.

Find $$g'(x)$$:

Let $$h(x) = \tan^2 x - 2\tan x + a = (\tan x - 1)^2 + (a-1)$$.

$$g'(x) = f'(h(x)) \cdot h'(x)$$

$$h'(x) = 2\tan x \sec^2 x - 2\sec^2 x = 2\sec^2 x(\tan x - 1)$$

Analyze the sign of $$g'(x)$$:

Since $$f''(x) > 0$$, $$f'$$ is strictly increasing. $$f'(a-1) = 0$$.

$$f'(h(x)) > 0$$ when $$h(x) > a-1$$, i.e., $$(\tan x - 1)^2 > 0$$, which is true for all $$x \neq \pi/4$$.

At $$x = \pi/4$$: $$h(\pi/4) = 0 + a - 1 = a - 1$$, so $$f'(h(\pi/4)) = f'(a-1) = 0$$.

For $$x \neq \pi/4$$: $$f'(h(x)) > 0$$.

$$h'(x) = 2\sec^2 x(\tan x - 1)$$: negative for $$x \in (0, \pi/4)$$, positive for $$x \in (\pi/4, \pi/2)$$.

Therefore:

(I) In $$(0, \pi/4)$$: $$g'(x) = f'(h(x)) \cdot h'(x) = (+)(−) < 0$$. g is DECREASING, not increasing. Statement I is FALSE.

(II) In $$(\pi/4, \pi/2)$$: $$g'(x) = (+)(+) > 0$$. g is INCREASING, not decreasing. Statement II is FALSE.

Neither statement is true.

The correct answer is Option B: Neither (I) nor (II) is True.