Question 8. 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 g is decreasing in ( 0, \alpha ) and increasing in ( \alpha, 3 ), then 8 \alpha is
(1) 24
(2) 0
(3) 18
(4) 20
Answer (3)
Explanation:
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) =\mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})If 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