Question 14

Step-by-step evaluation of List-I:

(P) $\sin^6 x + \cos^4 x = 1$ in $[-\pi, \pi]$ Since $\sin^6 x \le \sin^2 x$ and $\cos^4 x \le \cos^2 x$, then $\sin^6 x + \cos^4 x \le \sin^2 x + \cos^2 x = 1$. Equality holds when $(\sin^2 x = 0 \text{ and } \cos^2 x = 1)$ or $(\sin^2 x = 1 \text{ and } \cos^2 x = 0)$. For $x \in [-\pi, \pi]$, $x \in \{0, \pi, -\pi, \pi/2, -\pi/2\}$. Total = 5 elements. So, P $\rightarrow$ 5.

(Q) $\sin^2 x + \cos^6 x = 1$ in $[-\pi/2, \pi/2]$ Similarly, equality holds when $(\sin^2 x = 0 \text{ and } \cos^2 x = 1)$ or $(\sin^2 x = 1 \text{ and } \cos^2 x = 0)$. For $x \in [-\pi/2, \pi/2]$, $x \in \{0, \pi/2, -\pi/2\}$. Total = 3 elements. So, Q $\rightarrow$ 3.

(R) $\cos^2(x/2) - \sin^2 x = 1/2$ in $[-\pi, \pi]$ $\frac{1 + \cos x}{2} - (1 - \cos^2 x) = 1/2 \implies 1 + \cos x - 2 + 2\cos^2 x = 1 \implies 2\cos^2 x + \cos x - 2 = 0$. $\cos x = \frac{-1 \pm \sqrt{17}}{4}$. Only $\cos x = \frac{-1 + \sqrt{17}}{4} \approx 0.78$ is possible. This gives 2 values in $[-\pi, \pi]$. So, R $\rightarrow$ 2.

(S) $6\sin^2(x/2) - \cos 3x = 3$ in $[-2\pi, 2\pi]$ $3(1 - \cos x) - \cos 3x = 3 \implies 3 - 3\cos x - (4\cos^3 x - 3\cos x) = 3 \implies -4\cos^3 x = 0 \implies \cos x = 0$. $x \in \{-\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}\}$. Total = 4 elements. So, S $\rightarrow$ 4.

Matching: P-5, Q-3, R-2, S-4. Correct Option: B