Question

Let $g\left(x\right)=2f\left(\frac{x}{2}\right)+f(x-2)$ and $f^{\prime \prime }\left(x\right)<0\forall x\in (0,2)$, 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 $(0,\frac{4}{3})$

$g\left(x\right)=2f\left(\frac{x}{2}\right)+f(x-2)$ & $f^{\prime \prime }\left(x\right)<0\forall xϵ(0,2)$

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

$g^{\prime }\left(x\right)>0$

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

$f^{\prime \prime }\left(x\right)<0$, so, $f^{\prime }\left(x\right)$ is a decreasing function

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

$2-x>\frac{x}{2}$

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

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

$g\left(x\right)$ increases in $(0,\frac{4}{3})$