Q97 / 155 · WBJEE 2023 · Mathematics

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 \(2 e^{\frac{1}{\sqrt{2}}}\)
  • Correct answer
    C
    exists at infinitely many points
  • D
    exists at \(x=0\) only

Solution

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, \(f^{\prime \prime}(x) < 0\) at \(x=\pi / 4\) or \(x=n \pi+\pi / 4\) So, Global maximum exists at infinitely many points

Hence, the correct answer is option: 'C'.

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