If f(x) = 7e^{sin^2 x} - e^{cos^2 x} + 2,then | vedclass

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(D) Let $t = e^{\sin^2 x}$. Since $0 \le \sin^2 x \le 1$,we have $t \in [e^0, e^1] = [1, e]$.Using $\cos^2 x = 1 - \sin^2 x$,we have $e^{\cos^2 x} = e

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$8$

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(D) Let $t = e^{\sin^2 x}$. Since $0 \le \sin^2 x \le 1$,we have $t \in [e^0, e^1] = [1, e]$.Using $\cos^2 x = 1 - \sin^2 x$,we have $e^{\cos^2 x} = e^{1 - \sin^2 x} = \frac{e}{e^{\sin^2 x}} = \frac{e}{t}$.Thus,$f(t) = 7t - \frac{e}{t} + 2$ for $t \in [1, e]$.Calculating the derivative: $f'(t) = 7 + \frac{e}{t^2}$.Since $f'(t) > 0$ for all $t \in [1, e]$,the function $f(t)$ is strictly increasing.Therefore,$f_{\min} = f(1) = 7(1) - \frac{e}{1} + 2 = 9 - e$.And $f_{\max} = f(e) = 7(e) - \frac{e}{e} + 2 = 7e - 1 + 2 = 7e + 1$.Now,calculate $\sqrt{7f_{\min} + f_{\max}} = \sqrt{7(9 - e) + (7e + 1)} = \sqrt{63 - 7e + 7e + 1} = \sqrt{64} = 8$.

If $f(x) = 7e^{\sin^2 x} - e^{\cos^2 x} + 2$,then $\sqrt{7f_{\min} + f_{\max}}$ is equal to

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