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 $g\left(x\right)$ increases in

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

Answer 1

Answer: (D)

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

$g\left(x\right)=2f\left(\frac{x}{2}\right)+f(2-x)f^{\prime \prime }\left(x\right)<0\forall x\in (0,2)g^{\prime }\left(x\right)=2\cdot \frac{1}{2}{f}^{\prime }\left(\frac{x}{2}\right)-f^{\prime }(2-x)g^{\prime }\left(x\right)=f^{\prime }\left(\frac{x}{2}\right)-f^{\prime }(2-x)$

For $g\left(x\right)$ to increase, $g^{\prime }\left(x\right)>0$ which implies

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

But since $f^{\prime \prime }\left(x\right)<0\forall xt(0,2)$

$f^{\prime }\left(x\right)$ id decreasing

$\therefore f^{\prime }\left(\frac{x}{2}\right)>f^{\prime }(2-x)$ implies $\frac{x}{2}<2-x\Rightarrow \frac{3x}{2}<2$

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

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