Let $f:[0: 2] \rightarrow \mathrm{R}$ be a twice differentiable function such that $f^{\prime \prime}(x)>0$, for all $x \in(0,2)$. If $\phi(x)=f(x)+f(2-x)$, then $\phi$ is :
Correct Option: , 3
$\phi(x)=f(x)+f(2-x)$
Now, differentiate w.r.t. $x$,
$\phi^{\prime}(x)=f^{\prime}(x)-f^{\prime}(2-x)$
For $\phi(x)$ to be increasing $\phi^{\prime}(x)>0$
$\Rightarrow f^{\prime}(x)-f^{\prime}(2-x)>0$
$\Rightarrow f^{\prime}(x)>f^{\prime}(2-x)$
But $f^{\prime \prime}(x)>0 \Rightarrow f^{\prime}(x)$ is an increasing function
Then, $f^{\prime}(x)>f^{\prime}(2-x)>0$
$\Rightarrow x>2-x$
$\Rightarrow x>1$
Hence, $\phi(x)$ is increasing on $(1,2)$ and decreasing on $(0,1)$.
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