JEE Advanced 2022 Paper 1 Q15 Mathematics Trigonometry Trigonometric Identities & Equations Medium

JEE Advanced 2022 Paper 1 · Q15 · Trigonometric Identities & Equations

Consider the following lists:

List-I: (I) $\{x\in[-2\pi/3,2\pi/3]:\cos x+\sin x=1\}$ (II) $\{x\in[-5\pi/18,5\pi/18]:\sqrt{3}\tan 3x=1\}$ (III) $\{x\in[-6\pi/5,6\pi/5]:2\cos(2x)=\sqrt{3}\}$ (IV) $\{x\in[-7\pi/4,7\pi/4]:\sin x-\cos x=1\}$

List-II: (P) has two elements (Q) has three elements (R) has four elements (S) has five elements (T) has six elements

The correct option is:

  1. A. (I)$\to$(P); (II)$\to$(S); (III)$\to$(P); (IV)$\to$(S)
  2. B. (I)$\to$(P); (II)$\to$(P); (III)$\to$(T); (IV)$\to$(R)
  3. C. (I)$\to$(Q); (II)$\to$(P); (III)$\to$(T); (IV)$\to$(S)
  4. D. (I)$\to$(Q); (II)$\to$(S); (III)$\to$(P); (IV)$\to$(R)
Reveal answer + step-by-step solution

Correct answer:B

Solution

(I) $\cos x+\sin x=1\Rightarrow\sqrt{2}\sin(x+\pi/4)=1\Rightarrow\sin(x+\pi/4)=1/\sqrt{2}\Rightarrow x+\pi/4\in\{\pi/4,3\pi/4\}+2k\pi$, so $x\in\{0,\pi/2\}+2k\pi$. In $[-2\pi/3,2\pi/3]$: $x=0,\pi/2$ — 2 elements (P). (II) $\tan 3x=1/\sqrt{3}\Rightarrow 3x=\pi/6+k\pi\Rightarrow x=\pi/18+k\pi/3$. In $[-5\pi/18,5\pi/18]$: $k=0\Rightarrow\pi/18\approx 0.175$ in range; $k=-1\Rightarrow\pi/18-\pi/3=-5\pi/18$ in range. Two elements (P). (III) $\cos 2x=\sqrt{3}/2\Rightarrow 2x=\pm\pi/6+2k\pi\Rightarrow x=\pm\pi/12+k\pi$. In $[-6\pi/5,6\pi/5]\approx[-3.77,3.77]$: $x=\pi/12, -\pi/12, \pi/12+\pi=13\pi/12, -\pi/12+\pi=11\pi/12, \pi/12-\pi=-11\pi/12, -\pi/12-\pi=-13\pi/12$. All 6 in range (T). (IV) $\sin x-\cos x=1\Rightarrow\sqrt{2}\sin(x-\pi/4)=1\Rightarrow x-\pi/4\in\{\pi/4,3\pi/4\}+2k\pi\Rightarrow x\in\{\pi/2,\pi\}+2k\pi$. In $[-7\pi/4,7\pi/4]$: $x=\pi/2,\pi,\pi/2-2\pi=-3\pi/2,\pi-2\pi=-\pi$. 4 elements (R). Match: (I)$\to$P, (II)$\to$P, (III)$\to$T, (IV)$\to$R — option (B).

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