Question

Let $f\left(x\right)=\varphi (2-x)+\varphi \left(x\right)$ and $p^{\prime \prime }\left(x\right)<0$ for $x\in [0,2],$ then

• A. $f\left(x\right)$ is m.i. in $f[0,1]$
• B. $f\left(x\right)$ is m.d. in $f[0,1]$
• C. $f\left(x\right)$ is m.i. in $f[1,2]$
• D. $f\left(x\right)$ is m.d. in $f[1,2]$

Answer 1

Answer: (A), (D)

$f\left(x\right)$ is m.i. in $f[0,1]$
$f\left(x\right)$ is m.d. in $f[1,2]$

It is given that ${\varphi }^{\prime }\left(x\right)<0$ for all $xϵ[0,2]$.
Hence $\varphi \left(x\right)$ is a decreasing function in [0,2].
$f^{\prime }\left(x\right)={\varphi }^{\prime }\left(x\right)-{\varphi }^{\prime }(2-x)$
For monotonically increasing function
$f^{\prime }\left(x\right)>0$
Or
${\varphi }^{\prime }\left(x\right)-{\varphi }^{\prime }(2-x)>0$
$\varphi \left(x\right)>{\varphi }^{\prime }(2-x)$
Or
$x<2-x$ ...since $\varphi \left(x\right)$ is a deceasing function in [0,2].
$2x<2$
$x<1$
Hence
$f\left(x\right)$ is increasing in $[0,1]$.
Conversely $f\left(x\right)$ will be decreasing in $[1,2]$.