A) \[x>0\]
B) \[x<0\]
C) \[x>1\]
D) \[x<1\]
Correct Answer: B
Solution :
Let \[f(x)={{e}^{ax}}+{{e}^{-ax}},a<0\] \[\because \] \[f(x)\]is monotonic decreasing function. \[\therefore \] \[f'(x)=a{{e}^{ax}}+{{e}^{-ax}},a<0\] \[\Rightarrow \] \[{{e}^{ax}}>{{e}^{-ax}}\] \[(\because a<0)\] \[\Rightarrow \] \[{{e}^{2ax}}>1\] \[\Rightarrow \] \[2ax>0\] \[\Rightarrow \] \[2x<0\] \[(\because a<0)\] \[\Rightarrow \] \[x<0\]You need to login to perform this action.
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