Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | Maximum value of \(f(x) = \sin^2 x -\cos^2 x\ \forall x \in (\pi,2\pi)\) is | (I) | 0 |
| (B) | Minimum value of \(f(x) = \sin x.\cos x\) | (II) | 1 |
| (C) | Point of Minima of \(f(x) = x^x\ (x \gt 0)\) | (III) | \(\displaystyle-\frac{1}{2}\) |
| (D) | Maximum value of \(f(x) = -x^{2026}\) | (IV) | \(\displaystyle\frac{1}{e}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(A) Maximum value of \(f(x) = \sin^2 x -\cos^2 x\ \forall x \in (\pi,2\pi)\) is
This matches (II) 1. By the double-angle formula:
\(\sin^2 x - \cos^2 x = -\cos 2x\)
For \(x \in (\pi, 2\pi)\), \(2x \in (2\pi, 4\pi)\), and \(\cos 2x = -1\) at \(2x = 3\pi\), that is, \(x = \frac{3\pi}{2}\), which lies in the interval:
\(\displaystyle f\left(\frac{3\pi}{2}\right) = -\cos 3\pi = 1\)
(B) Minimum value of \(f(x) = \sin x.\cos x\)
This matches (III) \(-\frac{1}{2}\):
\(\displaystyle \sin x\cos x = \frac{1}{2}\sin 2x \ge -\frac{1}{2}\)
with equality when \(\sin 2x = -1\), for example at \(x = \frac{3\pi}{4}\).
(C) Point of Minima of \(f(x) = x^x\ (x \gt 0)\)
This matches (IV) \(\frac{1}{e}\). Writing \(f(x) = e^{x\log x}\):
\(f'(x) = x^x(\log x + 1)\)
Since \(x^x \gt 0\), \(f'(x) = 0\) when \(\log x = -1\), that is, \(x = \frac{1}{e}\). The sign of \(\log x + 1\) changes from negative to positive there, so by the first derivative test \(x = \frac{1}{e}\) is the point of minima.
(D) Maximum value of \(f(x) = -x^{2026}\)
This matches (I) 0. Since 2026 is even, \(x^{2026} \ge 0\), so
\(-x^{2026} \le 0\), with equality at \(x = 0\)
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).