# be a twice differentiable function such that

Question:

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 :

1. (1) increasing on $(0,1)$ and decreasing on $(1,2)$.

2. (2) decreasing on $(0,2)$

3. (3) decreasing on $(0,1)$ and increasing on $(1,2)$.

4. (4) increasing on $(0,2)$

Correct Option: , 3

Solution:

$\phi(x)=f(x)+f(2-x)$

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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