Question

Let $g\left(x\right)=2f\left(\frac{x}{2}\right)+f(2-x)$ and $f^{\prime \prime }\left(x\right)<0,\forall x\in (0,2).$ Then which of the following is (are) TRUE?

• A. $g\left(x\right)$ is decreasing in $(0,\frac{4}{3})$
• B. $g\left(x\right)$ is decreasing in $(\frac{4}{3},2)$
• C. $g\left(\frac{1}{10}\right)<g\left(\frac{11}{10}\right)$
• D. $g\left(x\right)$ has a local minimum at $x=\frac{4}{3}$

Answer 1

Answer: (B), (C)

$g\left(\frac{1}{10}\right)<g\left(\frac{11}{10}\right)$

$g\left(x\right)=2f\left(\frac{x}{2}\right)+f(2-x)$
$\Rightarrow g^{\prime }\left(x\right)=f^{\prime }\left(\frac{x}{2}\right)-f(2-x)$
Given, $f^{\prime \prime }\left(x\right)<0,\forall x\in (0,2)$
$\Rightarrow f^{\prime }\left(x\right)$ is decreasing in $(0,2).$

Case $1:$ $\frac{x}{2}<2-x$ i.e., $x<\frac{4}{3}$
$f^{\prime }\left(\frac{x}{2}\right)>f^{\prime }(2-x)$
$\Rightarrow g^{\prime }\left(x\right)>0$
$\therefore g\left(x\right)$ is increasing in $(0,\frac{4}{3})$

Case $2:$ $\frac{x}{2}>2-x$ i.e., $x>\frac{4}{3}$
$f^{\prime }\left(\frac{x}{2}\right)<f^{\prime }(2-x)$
$\Rightarrow g^{\prime }\left(x\right)<0$
$\therefore g\left(x\right)$ is decreasing in $(\frac{4}{3},2)$

Since $g\left(x\right)$ is increasing in $(0,\frac{4}{3})$ and $\frac{1}{10}<\frac{11}{10}$
$\therefore g\left(\frac{1}{10}\right)<g\left(\frac{11}{10}\right)$

$g^{\prime }\left(x\right)>0$ for $x\in (0,\frac{4}{3})$ and $g^{\prime }\left(x\right)<0$ for $x\in (\frac{4}{3},2).$
So, $g\left(x\right)$ has a local maximum at $x=\frac{4}{3}$