Question

If $g\left(x\right)=f\left(x\right)+f(1-x)$ and $f^{\prime \prime }\left(x\right)<0$ for $0\le x\le 1$, then

• A. $g\left(x\right)$ increases in $(-\infty ,\frac{1}{2})$
• B. $g\left(x\right)$ increases in $(0,\frac{1}{2})$
• C. $g\left(x\right)$ increases in $(\frac{1}{2},1)$
• D. $g\left(x\right)$ increases in $(\frac{1}{2},\infty )$

Answer 1

Answer: (B), (C)

The correct options are
B $g\left(x\right)$ increases in $(0,\frac{1}{2})$
C $g\left(x\right)$ increases in $(\frac{1}{2},1)$

We have, $f^{\prime \prime }\left(x\right)<0$, for $0\le x\le 1$

$\Rightarrow f^{\prime }\left(x\right)$ is decreasing function in $[0,1]$

Now, $g^{\prime }\left(x\right)=f^{\prime }\left(x\right)-f^{\prime }(1-x)$

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

$\Rightarrow f^{\prime }\left(x\right)>f^{\prime }(1-x)$

$\Rightarrow x<1-x$ $(\because f^{\prime }(x)$ is decreasing $)$

$\Rightarrow x<\frac{1}{2}$

$\therefore g\left(x\right)$ is an increasing function for $0<x<\frac{1}{2}$

Also, $g\left(x\right)$ is a decreasing if $g^{\prime }\left(x\right)<0$

$\Rightarrow f^{\prime }\left(x\right)<f^{\prime }(1-x)$

$x>1-x$ $(\because f^{\prime }(x)$ is decreasing $)$

$\Rightarrow x>\frac{1}{2}$

$\therefore g\left(x\right)$ is a decreasing function for $\frac{1}{2}<x<1$