The maximum value of f(x)=e^{ x}+e^{ x} ; x R is

Correct Answer is : 2e122e^{{1}{{2}}}2e21

Correct Answer is : 2e122e^{{1}{{2}}}2e21

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WBJEE 2022 Math Solved Paper

Question : 59 of 75

Marks: +1, -0

f(x)=e^{\sin x}+e^{\cos x} ; x \in R

Solution:

Hint

: f(x)=e^{\sin x}+e^{\cos x}

f^{'}(x)=e^{\sin x} \cdot \cos x - e^{\cos x} \cdot \sin x

f^{''}(x) \; \; = e^{\sin x} \cos^{2} x + e^{\cos x} \cdot \sin^{2} x - \sin x \cdot e^{\sin x} - \cos x \cdot e^{\cos x}

\; \; = e^{\sin x}(1-\sin x-\sin^{2} x)+e^{\cos x}(1-\cos^{2} x-\cos x)

f^{'}(\pi/4) \; \; =0 \; \text{ and } \; f^{''}(\pi/4) < 0

f^{'}(x) \; \; =0 \; \text{ at } \; x=\pi/4+2n\pi \; \text{ or } \; 5\pi/4+2n\pi (n \in Z)

f^{'}(\pi/4+2n\pi)=0

and

f^{''}(\pi/4+2n\pi) < 0

\therefore f_{\max}=f(\pi/4+2n\pi), n \in Z = 2e^{\frac{1}{\sqrt{2}}}

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