Application Of Derivatives Question 2
Question 2 - 2024 (27 Jan Shift 2)
Let $g(x)=3 f\left(\frac{x}{3}\right)+f(3-x)$ and $f^{\prime \prime}(x)>0$ for all $x \in(0,3)$. If $g$ is decreasing in $(0, \alpha)$ and increasing in $(\alpha, 3)$, then $8 \alpha$ is
(1) 24
(2) 0
(3) 18
(4) 20
Show Answer
Answer (3)
Solution
$g(x)=3 f\left(\frac{x}{3}\right)+f(3-x)$ and $f^{\prime \prime}(x)>0 \forall x \in(0,3) \Rightarrow f^{\prime}(x)$ is increasing function
$g^{\prime}(x)=3 \times \frac{1}{3} \cdot f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x)$
$=f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x)$
If $g$ is decreasing in $(0, \alpha)$
$g^{\prime}(x)<0$
$f^{\prime}\left(\frac{x}{3}\right)+f^{\prime}(3-x)<0$
$f^{\prime}\left(\frac{x}{3}\right)<f^{\prime}(3-x)$
$\Rightarrow \frac{x}{3}<3-x$
$\Rightarrow x<\frac{9}{4}$
Therefore $\alpha=\frac{9}{4}$
Then $8 \alpha=8 \times \frac{9}{4}=18$
