A) (1/2, 2)
B) (4/3, 2)
C) (0, 2)
D) (0, 4/3)
Correct Answer: D
Solution :
[d] We have \[g'(x)=f'\left( \frac{x}{2} \right)-f'(2-x)\] Given \[f''(x)<0\forall \in (0,2)\] So, \[f'(x)\] is decreasing on (0, 2). Let \[\frac{x}{2}>2-x\] or \[f'\left( \frac{x}{2} \right)<f'(2-x)\]. Thus, \[\forall x>\frac{4}{3},g'(x)<0\]. Therefore, \[g(x)\] decreasing in \[\left( \frac{4}{3},2 \right)\] and increasing in \[\left( 0,\frac{4}{3} \right)\].You need to login to perform this action.
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