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  • Let <img alt="f' \left ( x \right )>0\; \forall \; x \; \epsilon\; R" src="https://entrancecorner.oncodecogs.com/gif.latex?f%27%20%5Cleft%20%28%20x%20%5Cright%20%29%3E0%5C%3B%20%5Cforall%20%5C%3

Let f' \left ( x \right )>0\; \forall \; x \; \epsilon\; R and g\left ( x \right )=f\left ( 2-x \right )+f\left ( 4+x \right ). Then  g\left ( x \right ) is increasing in

Option: 1

(1)\; \left ( -\infty ,-1 \right )


Option: 2

(2)\; \left ( -\infty ,0 \right )


Option: 3

(3)\; \left ( -1,\infty \right )

 


Option: 4

(4)\; \left ( -4,2 \right )


Answers (1)

best_answer

Given f''\left ( x \right )>0

\Rightarrow f'\left ( x \right ) is an increasing function

Now g'\left ( x \right )=f'\left ( 4+x \right )-f'\left ( 2-x \right )

For increasing function

\\g'\left ( x \right )>0\\f'\left ( 4+x \right )>f'\left ( 2-x \right )\\4+x>2-x (f'\left (x \right ) is an increasing)

x>-1

Posted by

Nehul

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