Question

Let $\phi \left(x\right)=f\left(x\right)+f(1-x)$ and $f"\left(x\right)<0$ in $[0,1],$ then

• A. $\phi$ is monotonic increasing in $[0,\frac{1}{2}]$ and monotonic decreasing in $[\frac{1}{2},1]$
• B. $\phi$ is monotonic increasing in $[\frac{1}{2},1]$ and monotonic decreasing in $[0,\frac{1}{2}]$
• C. $\phi$ is neither increasing nor decreasing in any sub interval of $[0,1]$
• D. $\phi$ is increasing in $[0,1]$

Answer 1

The correct option is A $\phi$ is monotonic increasing in $[0,\frac{1}{2}]$ and monotonic decreasing in $[\frac{1}{2},1]$

$\varphi \left(x\right)=f\left(x\right)+f(1-x){\varphi }^{\prime }\left(x\right)=f^{\prime }\left(x\right)-f^{\prime }(1-x)...\left(1\right)f"\left(x\right)<0\Rightarrow f^{\prime }\left(x\right)\text{is a decreasing function}\text{case 1:}0\le x\le \frac{1}{2}\Rightarrow x\le 1-x\Rightarrow f^{\prime }\left(x\right)\ge f^{\prime }(1-x)\Rightarrow f^{\prime }\left(x\right)-f^{\prime }(1-x)\ge 0\text{From eqn}\left(1\right)\Rightarrow {\varphi }^{\prime }\left(x\right)>0\Rightarrow \varphi \text{is increasing in}[0,\frac{1}{2}]\text{case 2:}\frac{1}{2}\le x\le 1\Rightarrow x\ge 1-x\Rightarrow f^{\prime }\left(x\right)\le f^{\prime }(1-x)\text{From eqn}\left(1\right)\Rightarrow f^{\prime }\left(x\right)-f^{\prime }(1-x)\le 0\Rightarrow \varphi \text{is decreasing in}[\frac{1}{2},1]$