Question

If $\varphi \left(x\right)=f\left(x\right)+f(1-x)$ and $f^{\prime \prime }\left(x\right)<0$ in $(-1,1)$, then $\varphi \left(x\right)$ strictly increases in the interval

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

Answer 1

The correct option is A $(0,\frac{1}{2})$

We have,

$\Rightarrow \varphi \left(x\right)=f\left(x\right)+f(1-x)$

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

Which vanishes at the points given by

$x=1-x\Rightarrow x=\frac{1}{2}$

Now, we have,

${\varphi }^{\prime \prime }\left(x\right)=f^{\prime \prime }\left(x\right)+f^{\prime \prime }(1-x)$

$\therefore {\varphi }^{\prime \prime }\left(\frac{1}{2}\right)=f^{\prime \prime }\left(\frac{1}{2}\right)+f^{\prime \prime }\left(\frac{1}{2}\right)<0$

$[\because f^{\prime \prime }\left(x\right)<0$ in $(-1,1)]$

$\Rightarrow$ maxima at $x=\frac{1}{2}$

Hence, $\varphi \left(x\right)$ strictly increase in $(0,\frac{1}{2})$