JEE Main 2020 (Online) 4th September Evening ShiftQuestion 1
The minimum value of $2^{\sin x}+2^{\cos x}$ is :
Differentiation and its Application
Solution
Concept. AM-GM turns the sum into a power of $2$ with exponent $\frac{\sin x+\cos x}{2}$, whose least value is known. Formulas used. (1) $p+q\ge2\sqrt{pq}$(2) $\sin x+\cos x\ge-\sqrt2$Step 1 - AM-GM. By (1), $$2^{\sin x}+2^{\cos x}\ge2\cdot2^{\frac{\sin x+\cos x}{2}}.$$ Step 2 - Bound. By (2), this is at least $2\cdot2^{-\frac{1}{\sqrt2}}=2^{1-\frac{1}{\sqrt2}}$. Step 3 - Attained. Both equalities hold at $\sin x=\cos x=-\frac{1}{\sqrt2}$. Answer: (A)