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