Question

Match each entry in List-I to the correct entry in List-II and choose the correct option.

• A. \(P \to (2),\ Q \to (5),\ R \to (3),\ S \to (4)\)
• B. \(P \to (5),\ Q \to (3),\ R \to (2),\ S \to (4)\)
• C. \(P \to (5),\ Q \to (4),\ R \to (1),\ S \to (3)\)
• D. \(P \to (4),\ Q \to (3),\ R \to (2),\ S \to (5)\)

Hint:

Useful identities: \[ \cos3x=4\cos^3x-3\cos x \] and \[ \cos^2\frac{x}{2}=\frac{1+\cos x}{2} \]

Answer 1

The Correct Option is D

Step 1: Solve part (P).
Given: \[ \sin^6x+\cos^4x=1 \] Let: \[ t=\sin^2x \] Then: \[ \cos^2x=1-t \] Hence: \[ t^3+(1-t)^2=1 \] \[ t^3+t^2-2t=0 \] \[ t(t^2+t-2)=0 \] \[ t(t-1)(t+2)=0 \] Thus: \[ t=0\quad \text{or}\quad t=1 \] So: \[ \sin x=0 \] or \[ \sin^2x=1 \] In: \[ [-\pi,\pi] \] solutions are: \[ x=-\pi,\ 0,\ \pi,\ -\frac{\pi}{2},\ \frac{\pi}{2} \] Total: \[ 5 \] Therefore: \[ (P)\to(5) \]

Step 2: Solve part (Q).
Given: \[ \sin^2x+\cos^6x=1 \] Using: \[ \sin^2x=1-\cos^2x \] \[ 1-\cos^2x+\cos^6x=1 \] \[ \cos^6x-\cos^2x=0 \] \[ \cos^2x(\cos^4x-1)=0 \] Thus: \[ \cos x=0 \] or \[ \cos^2x=1 \] In: \[ \left[-\frac{\pi}{2},\frac{\pi}{2}\right] \] solutions are: \[ -\frac{\pi}{2},\ 0,\ \frac{\pi}{2} \] Total: \[ 3 \] Therefore: \[ (Q)\to(3) \]

Step 3: Solve part (R).
Using: \[ \cos^2\frac{x}{2}=\frac{1+\cos x}{2} \] Equation becomes: \[ \frac{1+\cos x}{2}-\sin^2x=\frac12 \] \[ \cos x-2\sin^2x=0 \] Using: \[ \sin^2x=1-\cos^2x \] \[ \cos x-2(1-\cos^2x)=0 \] \[ 2\cos^2x+\cos x-2=0 \] Let: \[ u=\cos x \] \[ 2u^2+u-2=0 \] \[ u=\frac{-1\pm\sqrt{17}}4 \] Only: \[ u=\frac{-1+\sqrt{17}}4 \] lies in: \[ [-1,1] \] Hence: \[ \cos x=c \] has exactly: \[ 2 \] solutions in: \[ [-\pi,\pi] \] Therefore: \[ (R)\to(2) \]

Step 4: Solve part (S).
Using: \[ \sin^2\frac{x}{2}=\frac{1-\cos x}{2} \] Equation: \[ 6\cdot\frac{1-\cos x}{2}-\cos3x=3 \] \[ 3-3\cos x-\cos3x=3 \] \[ 3\cos x+\cos3x=0 \] Using: \[ \cos3x=4\cos^3x-3\cos x \] \[ 3\cos x+4\cos^3x-3\cos x=0 \] \[ 4\cos^3x=0 \] \[ \cos x=0 \] In: \[ [-2\pi,2\pi] \] solutions are: \[ -\frac{3\pi}{2},\ -\frac{\pi}{2},\ \frac{\pi}{2},\ \frac{3\pi}{2} \] Total: \[ 4 \] Therefore: \[ (S)\to(4) \]

Step 5: Identify the correct option.
Hence: \[ \boxed{\mathrm{(D)}} \]