A) \[-1<x<1\]
B) \[x<-1\]
C) \[x>-1\]
D) \[x>0\]
Correct Answer: D
Solution :
[d]: \[f(x)={{e}^{ax}}+{{e}^{-ax}}\] \[f'(x)=a{{e}^{ax}}-a{{e}^{-ax}}=a({{e}^{ax}}-{{e}^{-ax}})\] If f(x) is monotonically increasing, then \[f'(x)>0\] \[\Rightarrow \] \[a({{e}^{ax}}-{{e}^{-ax}})>0=({{e}^{ax}}-{{e}^{-ax}})>0(\because a>0)\] \[\Rightarrow \] \[{{e}^{ax}}>{{e}^{-ax}}={{e}^{2ax}}>1={{e}^{0}}\Rightarrow 2ax>0\] \[\Rightarrow \] \[x>0\] \[(\because a>0)\] \[\therefore \] Function f (x) is increasing when x > 0.You need to login to perform this action.
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