# Let $f:[0,2]\rightarrow \mathbb{R}$ be a twice differentiable function such that $f"(x)>0,$ for all $x\epsilon \left ( 0,2 \right ).$ If $\phi (x)=f(x)+f(2-x),$ then $\phi$ is :   Option 1) decreasing on $(0,1)$ and increasing on $(1,2)$ Option 2) increasing on $(0,2)$ Option 3) increasing on $(0,1)$ and decreasing on $(1,2).$   Option 4) decreasing on $(0,2).$

$f :[0,2]\rightarrow \mathbb{R}$ twice differentiable

Such that. $f"(x)>0\; \; \vee \; \; x\epsilon (0,2)$

If $\phi (x)=f(x)+f(2-x)\; then\; \phi$

for $\phi (x)$ to be increasing $\phi^{1} (x)>0$

$\phi^{1} (x)=f^{1}(x)-f^{1}(2-x)$

Since $f"(x)>0\Rightarrow f^{1}(x)$ in increasing

So for $x\epsilon (0,1)$

$f^{1}(x)-f^{1}(2-x)<0$

hence $\phi ^{1}(x)<0$

and for $x\epsilon (1,2)$

$f^{1}(x)-f^{1}(2-x)>0$

$\Rightarrow \phi ^{1}(x)>0$

hence $\phi (a)$ increasing

Option 1)

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

Option 2)

increasing on $(0,2)$

Option 3)

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

Option 4)

decreasing on $(0,2).$

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