Discussion
| List-I | List-II |
|---|---|
| (P) The number of elements in the set \(\{x \in [-\pi, \pi] : \sin^{6}x + \cos^{4}x = 1\}\) | (1) is 1 |
| (Q) The number of elements in the set \(\{x \in [-\frac{\pi}{2}, \frac{\pi}{2}] : \sin^{2}x + \cos^{6}x = 1\}\) | (2) is 2 |
| (R) The number of elements in the set \(\{x \in [-\pi, \pi] : \cos^{2}(\frac{x}{2}) - \sin^{2}x = \frac{1}{2}\}\) | (3) is 3 |
| (S) The number of elements in the set \(\{x \in [-2\pi, 2\pi] : 6\sin^{2}(\frac{x}{2}) - \cos 3x = 3\}\) | (4) is 4 |
| (5) is 5 |
Answer: B
The correct match is (P) → (5), (Q) → (3), (R) → (2), (S) → (4).
(P) Put \(s = \sin^{2}x\): \(s^{3} + (1 - s)^{2} = 1\) reduces to \(s(s + 2)(s - 1) = 0\), so \(s = 0\) or \(s = 1\). In \([-\pi, \pi]\), \(\sin x = 0\) at \(-\pi, 0, \pi\) and \(\sin^{2}x = 1\) at \(\pm\frac{\pi}{2}\): 5 elements.
(Q) Put \(c = \cos^{2}x\): \(1 - c + c^{3} = 1\) gives \(c(c - 1)(c + 1) = 0\), so \(c = 0\) or \(1\). In \([-\frac{\pi}{2}, \frac{\pi}{2}]\) that is \(x = \pm\frac{\pi}{2}\) and \(x = 0\): 3 elements.
(R) Using \(\cos^{2}\frac{x}{2} = \frac{1 + \cos x}{2}\) and \(\sin^{2}x = 1 - \cos^{2}x\):
\[ \begin{aligned} 2\cos^{2}x + \cos x - 2 &= 0 \\ \cos x &= \frac{-1 \pm \sqrt{17}}{4} \end{aligned} \]
Only \(\frac{-1 + \sqrt{17}}{4} \approx 0.78\) lies in \([-1, 1]\), and it is taken once on each side of 0 in \([-\pi, \pi]\): 2 elements.
(S) \(6\sin^{2}\frac{x}{2} = 3 - 3\cos x\), so the equation becomes \(3\cos x + \cos 3x = 0\). With \(\cos 3x = 4\cos^{3}x - 3\cos x\) this is \(4\cos^{3}x = 0\), so \(\cos x = 0\): \(x = \pm\frac{\pi}{2}, \pm\frac{3\pi}{2}\), 4 elements.
Only the second option has 5, 3, 2, 4 in order.
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