Question

Let $f\left(x\right)=\varphi (2-x)+\varphi \left(x\right)$ and $\varphi "\left(x\right)<0$ for $xϵ[0,2]$, then

• A. $f\left(x\right)$ is monotonically increasing in $(0,1)$
• B. $f\left(x\right)$ is monotonically decreasing in $(0,1)$
• C. $f\left(x\right)$ is monotonically increasing in $(1,2)$
• D. $f\left(x\right)$ is monotonically decreasing in $(0,2)$

Answer 1

Answer: (A)

$f\left(x\right)$ is monotonically increasing in $(0,1)$

Given $f\left(x\right)=\varphi (2-x)+\varphi \left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)=-{\varphi }^{\prime }(2-x)+{\varphi }^{\prime }\left(x\right)$
Since, $\varphi "\left(x\right)<0$
$\Rightarrow {\varphi }^{\prime }\left(x\right)$ is monotonicaly decreasing.
For $0<x<1$,
$\Rightarrow x<2-x$
$\Rightarrow {\varphi }^{\prime }\left(x\right)>{\varphi }^{\prime }(2-x)$
$\Rightarrow {\varphi }^{\prime }\left(x\right)-{\varphi }^{\prime }(2-x)>0$
$\therefore f^{\prime }\left(x\right)>0$
So $f\left(x\right)$ is monotonocally increasing in (0,1).
For $1<x<2$, $2-x<x;$
${\varphi }^{\prime }(2-x)>{\varphi }^{\prime }\left(x\right)$
$\Rightarrow 0>{\varphi }^{\prime }\left(x\right)-{\varphi }^{\prime }(2-x)$
$\Rightarrow f^{\prime }\left(x\right)<0$
So $f\left(x\right)$ is monotonically decreasing in (1,2).