MCQ
If $f(x)=$  $7{e^{{{\sin }^2}x}} - {e^{{{\cos }^2}x}} + 2$, then,$\sqrt {7{f_{\min }} + {f_{\max }}}$ is equal to
  • A
    $0$
  • B
    $\sqrt {10}$
  • C
    $\sqrt {8}$
  • $8$

Answer

Correct option: D.
$8$
d
$f(t)=7 t-\frac{e}{t}+2$ for $t \in[1, e],$ where $t=e^{\sin ^{2} x}$

$\Rightarrow f^{\prime}(t)=7+\frac{e}{t^{2}}>0 \forall t \in R$

Hence $f$ is increasing function.

$\therefore f_{\min }=f(1)=7-e+2=9-e$

and $f_{\max }=f(e)=7 e-1+2=7 e+1$

Hence answer is 8 .

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