JEE Main 2021 (Online) 3rd August Morning ShiftQuestion 1
The number of solutions of the equation $x^{2}=\cot x$ in $[0,2\pi]$ is $\_\_\_\_\_\_$ .
Trigonometric Ratio & Identities
Answer$2$
Solution
Concept. $x^2\ge0$, so roots can only be where $\cot x>0$: on $\left(0,\frac\pi2\right)$ and $\left(\pi,\frac{3\pi}{2}\right)$. Formulas used. (1) $\cot x$ decreases on each interval $(k\pi,(k+1)\pi)$; $\cot x<0$ on $\left(\frac\pi2,\pi\right)$ and $\left(\frac{3\pi}2,2\pi\right)$Step 1 - $\left(0,\frac\pi2\right)$. $\cot x$ falls from $+\infty$ to $0$ while $x^2$ rises from $0$: exactly one root. Step 2 - $\left(\pi,\frac{3\pi}2\right)$. $\cot x$ falls from $+\infty$ to $0$ while $x^2$ rises from about $9.9$: exactly one root. Step 3 - Elsewhere. On $\left(\frac\pi2,\pi\right)$ and $\left(\frac{3\pi}2,2\pi\right)$, $\cot x<0<x^2$; at $0,\pi,2\pi$, $\cot x$ is undefined. Step 4 - Total. $2$. Answer: 2