Match List-I with List-II

Choose the correct answer from the options given below:

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

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

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

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

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

$-\cos 2x$ 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)