WBJEE · Maths · Application of Derivatives
Given \(f(x)=e^{\sin x}+e^{\cos x}\). The global maximum value of \(f(x)\)
- A does not exist
- B exists at a point in \(\left(0, \frac{\pi}{2}\right)\) and its value is \(2 e^{\frac{1}{\sqrt{2}}}\)
- C exists at infinitely many points
- D exists at \(x=0\) only
Answer & Solution
Correct Answer
(C) exists at infinitely many points
Step-by-step Solution
Detailed explanation
Hint : \(f(x)=e^{\sin x}+e^{\cos x}\) \(f^{\prime}(x)=0\) \(\begin{aligned} & \Rightarrow e^{\sin x} \cdot \cos x+e^{\cos x}(-\sin x)=0 \\ & \Rightarrow e^{\sin x-\cos x}=\tan x \end{aligned}\) \(\mathrm{x}=\pi / 4\) or in general \(\mathrm{x}=\mathrm{n} \pi+\frac{\pi}{4}\) Now,…
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