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$\sqrt {8}$
- ✓$8$
$\Rightarrow f^{\prime}(t)=7+\frac{e}{t^{2}}>0 \forall t \in R$
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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$\mathrm{Q}=\mathrm{A}^{\mathrm{T}} \mathrm{BA}$, then the inverse of the matrix $\mathrm{A} \mathrm{Q}^{2021} \mathrm{~A}^{\mathrm{T}}$ is equal to :