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 $f(x) = x^x$ $(x > 0)$	(III) $-\frac{1}{2}$
(D) Maximum value of $f(x) = -x^{2026}$	(IV) $\frac{1}{e}$

Choose the correct answer from the options given below:

Let me solve each part systematically so you can understand exactly what's happening!

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(A) Maximum value of $f(x) = \sin^2 x - \cos^2 x$, $\forall x \in (\pi, 2\pi)$

Simplify using the trigonometric identity $\cos(2x) = \cos^2 x - \sin^2 x$

$\sin^2 x - \cos^2 x = -\cos(2x)$

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When $x \in (\pi, 2\pi)$, then $2x \in (2\pi, 4\pi)$

To maximize $-\cos(2x)$, we need to minimize $\cos(2x)$

The minimum value of $\cos(2x) = -1$

Therefore, maximum of $-\cos(2x) = -(-1) = 1$

(A) → (II)

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(B) Minimum value of $f(x) = \sin x \cos x$

Using the identity $\sin(2x) = 2\sin x \cos x$

$\sin x \cos x = \dfrac{1}{2}\sin(2x)$

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The range of $\sin(2x)$ is $[-1, 1]$

Minimum value of $\sin(2x) = -1$

Therefore, minimum of

$\dfrac{1}{2}\sin(2x) = \dfrac{1}{2} \times (-1) = -\dfrac{1}{2}$

(B) → (III)

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(C) Point of Minima of $f(x) = x^x$ $(x > 0)$

Let $y = x^x$

Taking natural log: $\ln(y) = x\ln(x)$

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Differentiate using implicit differentiation:

$\dfrac{1}{y} \times \dfrac{dy}{dx} = \ln(x) + x \times \dfrac{1}{x} = \ln(x) + 1$

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$\dfrac{dy}{dx} = x^x[\ln(x) + 1]$

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For critical points, set $\dfrac{dy}{dx} = 0$

$x^x[\ln(x) + 1] = 0$

Since $x^x > 0$ for all $x > 0$:

$\ln(x) + 1 = 0$

$\ln(x) = -1$

$x = e^{-1} = \dfrac{1}{e}$

(C) → (IV)

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(D) Maximum value of $f(x) = -x^{2026}$

$x^{2026}$ is always $\geq 0$ since $2026$ is even

When $x = 0$: $x^{2026} = 0$

When $x \neq 0$: $x^{2026} > 0$

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Since $x^{2026} \geq 0$, then $-x^{2026} \leq 0$

The minimum value of $x^{2026} = 0$ at $x = 0$

Therefore, maximum value of $-x^{2026} = 0$

(D) → (I)

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Final Matching:

(A) → (II)

(B) → (III)

(C) → (IV)

(D) → (I)

Look for the option that matches: A-II, B-III, C-IV, D-I