Q. The general solution of $e^{- 1/\sqrt{2}}\left( e^{\sin x} + e^{\cos x} \right) = 2$ is

A) $x = m\pi$
B) $x = \frac{(4m + 1)\pi}{4}$
C) $x = \frac{(4m + 1)\pi}{2}$
D) None of these
Explanation

(b)
$e^{\sin x} + e^{\cos x} = 2e^{1/2}$ …(i)
$\Rightarrow \ e^{\sin x} > 0$ and $e^{\cos x} > 0$
$\therefore e^{\sin x} + e^{\cos x} \geq 2\sqrt{e^{\sin{x + \cos x}}}$ $(\because AM \geq GM)$
$\Rightarrow \ e^{\sin x} + e^{\cos x} \geq 2e^{1/\sqrt{2}}\$…(ii)
Since, equality holds
$\Rightarrow \ e^{\sin x} = e^{\cos x}$
$\Rightarrow \sin{x = \cos x}$
$\Rightarrow \tan{x = 1\ \Rightarrow x = m\pi + \frac{\pi}{4}}$
$\Rightarrow x = (4m + 1)\ \frac{\pi}{4}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.