A) 11
B) 12
C) 10
D) 14
Correct Answer: B
Solution :
Let \[f(x)=4{{e}^{2x}}+9{{e}^{-2x}}\] \[\therefore \] \[{f}'(x)=8{{e}^{2x}}-18{{e}^{-2x}}\] Put \[{f}'(x)=0\Rightarrow 8{{e}^{2x}}-18{{e}^{-2x}}=0\] \[{{e}^{2x}}=3/2\Rightarrow x=\log {{(3/2)}^{1/2}}\] Again \[{f}''(x)=16{{e}^{2x}}+36{{e}^{-2x}}>0\] Now \[f(\log {{(3/2)}^{1/2}})=4{{e}^{2.(\log {{(3/2)}^{1/2}})}}+9{{e}^{-2(\log {{(3/2)}^{1/2}})}}\] = \[4\times \frac{3}{2}+9\times \frac{2}{3}\] = \[6+6=12\] Hence minimum value = 12.You need to login to perform this action.
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