JEE Main 2022 (Online) 29th July Evening ShiftQuestion 1
The number of elements in the set $S=\left\{x\in\mathbb{R}:2\cos\left(\frac{x^2+x}{6}\right)=4^x+4^{-x}\right\}$ is :
Trigonometric Ratio & Identities
Solution
Concept. The right side is at least $2$ and the left side at most $2$; both must equal $2$. Formulas used. (1) AM-GM: $4^x+4^{-x}\ge2$, equality only at $x=0$Step 1 - Bounds. $2\cos\left(\frac{x^2+x}{6}\right)\le2$ and, by (1), $4^x+4^{-x}\ge2$. Step 2 - Equality. The right side is $2$ only at $x=0$, and there the left side is $2\cos0=2$. So $x=0$ is the only solution. Step 3 - Count. $1$. Answer: (A)