Match List-I with List-II List-I List-II (A) Maximum value of f(x) =…

The correct answer is Option (4) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

List-I	List-II
(A) Maximum value of $f(x) = \sin^2 x- \cos^2 x ∀x ∈ (π,2π)$ is	(II) 1
(B) Minimum value of $f(x) = \sin x.\cos x$	(III) $-\frac{1}{2}$
(C) Point of Minima of $f(x) = x^x (x > 0)$	(IV) $\frac{1}{e}$
(D) Maximum value of $f(x) = -x^{2026}$	(I) 0

$\text{(A) Maximum value of }f(x)=\sin^2 x-\cos^2 x\text{ for }x\in(\pi,2\pi)$

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

$x\in(\pi,2\pi)\Rightarrow 2x\in(2\pi,4\pi)$

$-\cos 2x$ takes its maximum value $1$.

(A) → (II)

$\text{(B) Minimum value of }f(x)=\sin x\cdot\cos x$

$\sin x\cos x=\frac{1}{2}\sin 2x$

Minimum value is $-\frac{1}{2}$.

(B) → (III)

$\text{(C) Point of minima of }f(x)=x^x,\;x>0$

$f(x)=x^x$

$\ln f=x\ln x$

$\frac{f'}{f}=\ln x+1=0$

$x=\frac{1}{e}$ gives minima.

(C) → (IV)

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

Since $2026$ is even, $x^{2026}\ge 0$.

$-x^{2026}$ is maximized at $x=0$ giving value $0$.

(D) → (I)