Question ID: #672
Let $f:R\rightarrow R$ be a twice differentiable function such that $f^{\prime\prime}(x) \gt 0$ for all $x\in R$ and $f^{\prime}(a-1)=0$, where $a$ is a real number.
Let $g(x)=f(\tan^{2}x-2\tan x+a), \quad 0 \lt x \lt \frac{\pi}{2}$.
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,
- (1) Neither (I) nor (II) is True
- (2) Only (II) is True
- (3) Only (I) is True
- (4) Both (I) and (II) are True
Solution:
$$g(x) = f((\tan x – 1)^2 + a – 1)$$
$$g'(x) = f'((\tan x – 1)^2 + a – 1) \cdot \frac{d}{dx}((\tan x – 1)^2 + a – 1)$$
$$g'(x) = f'((\tan x – 1)^2 + a – 1) \cdot 2(\tan x – 1)\sec^2 x$$
Given $f”(x) \gt 0 \Rightarrow f'(x)$ is strictly increasing.
Also $f'(a-1) = 0$.
Since $(\tan x – 1)^2 \ge 0 \Rightarrow (\tan x – 1)^2 + a – 1 \ge a – 1$.
$\Rightarrow f'((\tan x – 1)^2 + a – 1) \ge f'(a-1) = 0$.
For $x \ne \frac{\pi}{4}$, $f'(…) \gt 0$ and $\sec^2 x \gt 0$.
$\therefore$ Sign of $g'(x)$ depends on $(\tan x – 1)$.
**Case 1:** $x \in \left(0, \frac{\pi}{4}\right)$
$\tan x \lt 1 \Rightarrow \tan x – 1 \lt 0$
$\Rightarrow g'(x) \lt 0 \Rightarrow g(x)$ is Decreasing. (Statement I is False)
**Case 2:** $x \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$
$\tan x \gt 1 \Rightarrow \tan x – 1 \gt 0$
$\Rightarrow g'(x) \gt 0 \Rightarrow g(x)$ is Increasing. (Statement II is False)
Ans. (1)
Was this solution helpful?
YesNo