JEE Advanced 2022 Paper 1Question 1
Consider the following lists: $$\begin{array}{ll}\textbf{List-I}&\textbf{List-II}\\ \text{(I) }\left\{x\in\left[-\frac{2\pi}{3},\frac{2\pi}{3}\right]:\cos x+\sin x=1\right\}&\text{(P) has two elements}\\ \text{(II) }\left\{x\in\left[-\frac{5\pi}{18},\frac{5\pi}{18}\right]:\sqrt{3}\tan 3x=1\right\}&\text{(Q) has three elements}\\ \text{(III) }\left\{x\in\left[-\frac{6\pi}{5},\frac{6\pi}{5}\right]:2\cos(2x)=\sqrt{3}\right\}&\text{(R) has four elements}\\ \text{(IV) }\left\{x\in\left[-\frac{7\pi}{4},\frac{7\pi}{4}\right]:\sin x-\cos x=1\right\}&\text{(S) has five elements}\\ &\text{(T) has six elements}\end{array}$$ The correct option is:
Trigonometric Ratio & Identities
Solution
Concept. Write each equation with one trigonometric function and count in its interval. Formulas used. (1) $\cos x+\sin x=\sqrt2\sin\left(x+\frac\pi4\right)$, $\sin x-\cos x=\sqrt2\sin\left(x-\frac\pi4\right)$(I) By (1), $\sin\left(x+\frac\pi4\right)=\frac1{\sqrt2}$: $x=0,\frac\pi2$ in $\left[-\frac{2\pi}3,\frac{2\pi}3\right]$. Two elements: (P). (II) $\tan3x=\frac1{\sqrt3}$: $x=\frac\pi{18}+\frac{k\pi}{3}$. In $\left[-\frac{5\pi}{18},\frac{5\pi}{18}\right]$: $\frac{\pi}{18},-\frac{5\pi}{18}$. Two elements: (P). (III) $\cos2x=\frac{\sqrt3}2$: $x=\pm\frac\pi{12}+k\pi$. In $\left[-\frac{6\pi}5,\frac{6\pi}5\right]$: $\pm\frac\pi{12},\pm\frac{11\pi}{12},\pm\frac{13\pi}{12}$. Six elements: (T). (IV) By (1), $\sin\left(x-\frac\pi4\right)=\frac{1}{\sqrt2}$: $x=\frac\pi2+2k\pi$ or $\pi+2k\pi$. In $\left[-\frac{7\pi}4,\frac{7\pi}4\right]$: $\frac\pi2,-\frac{3\pi}2,\pi,-\pi$. Four elements: (R). Answer: (B)