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).

 

Answers (1)

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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