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 $\mathrm{x} \in(0,3)$. If $\mathrm{g}$ 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 $f^{\prime \prime}(x)>0 \forall x \in(0,3) \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})$ 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 $\mathrm{g}$ is decreasing in $(0, \alpha)$
$\mathrm{g}^{\prime}(\mathrm{x})<0$
$\mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)+\mathrm{f}^{\prime}(3-\mathrm{x})<0$
$\mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)<\mathrm{f}^{\prime}(3-\mathrm{x})$
$\Rightarrow \frac{\mathrm{x}}{3}<3-\mathrm{x}$
$\Rightarrow \mathrm{x}<\frac{9}{4}$
Therefore $\alpha=\frac{9}{4}$
Then $8 \alpha=8 \times \frac{9}{4}=18$
