A) \[36\]
B) \[49\]
C) \[64\]
D) \[81\]
Correct Answer: C
Solution :
[c] \[f(x)=7{{e}^{{{\sin }^{2}}x}}-{{e}^{{{\cos }^{2}}x}}+2\] Clearly, period of \[f(x)\]is \[\pi .\] Now, \[f'(x)=(7{{e}^{{{\sin }^{2}}x}}+{{e}^{{{\cos }^{2}}x}})\sin 2x\] \[f'(x)=0\] for \[2x=n\pi \] or \[x=n\pi /2,\] \[n\in Z.\] Now, \[f(0)=9-e\] \[f\left( \frac{\pi }{2} \right)=7e+11\] and \[f(\pi )=9-e\] So, \[7{{f}_{\min .}}+{{f}_{\max .}}=7(9-e)+7e+1=64\]You need to login to perform this action.
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