Question

Let $g\left(x\right)=2f\left(\frac{x}{2}\right)+f(2-x)$ and $f^{\prime \prime }\left(x\right)<0$ for every $x$ belongs to $(0,2)$.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 that $f^{\prime \prime }\left(x\right)<0\forall x\in (0,2)$

$\Rightarrow f^{\prime }\left(x\right)$ is a decreasing function.

We have $g\left(x\right)=2f\left(x/2\right)+f(2-x)$

Differentiating both sides wrt x, we get

$g^{\prime }\left(x\right)=2.f^{\prime }\left(x/2\right)\frac{d}{dx}\left(\frac{x}{2}\right)+f^{\prime }(2-x)\frac{d}{dx}(2-x)$

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

To get an interval of increase put:

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

$\Rightarrow f^{\prime }\left(\frac{x}{2}\right)-f^{\prime }(2-x)>0$

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

$\Rightarrow \frac{x}{2}<2-x$ as $f^{\prime }\left(x\right)$ is decreasing function, hence inequality reverses

$\Rightarrow x<4-2x$

$\Rightarrow x+2z<4$

$\Rightarrow 3x<4$

$\Rightarrow x<4/3$

Also $x\in (0,2)$

Hence $g\left(x\right)$ increases in $(0,4/3)$.