Question

Let $g\left(x\right)=2f\left(\frac{x}{2}\right)+f(2-x)$ and $f^{\prime \prime }x)<0\forall xϵ(0,2),$ then $g\left(x\right)$ increases in

• A. $(\frac{1}{2},2)$
• B. $(\frac{4}{3},2)$
• C. $(0,2)$
• D. $(0,\frac{4}{3})$

Answer 1

The correct option is D

$(0,\frac{4}{3})$

Given : $g\left(x\right)=2f\left(\frac{x}{2}\right)+f(2-x)$

Differentiating :

$g^{\prime }\left(x\right)=f^{\prime }\left(\frac{x}{2}\right)-f^{\prime }(2-x)$

$\because f^{\prime \prime }\left(x\right)<0\Rightarrow f^{\prime }\left(x\right)$ is a decreasing function in $(0,2)$

For $g\left(x\right)$ to be increasing $g^{\prime }\left(x\right)>0$

$\Rightarrow f^{\prime }\left(\frac{x}{2}\right)>f^{\prime }(2-x)$

$\Rightarrow \frac{x}{2}<2-x$

$\Rightarrow \frac{3x}{2}<2$

$\Rightarrow x<\frac{4}{3}$

$\Rightarrow xϵ(0,\frac{4}{3})$