Match List-I with List-II. List-I(Function)List-II(Interval in which…

The correct answer is Option (3) → (A)-(IV), (B)-(I), (C)-(III), (D)-(II)

A function is increasing where its derivative is positive.

(A) \(\frac{x}{\log_e x}\)

This matches (IV) \((e,\infty)\).

The function is defined for x > 0, x ≠ 1. By the quotient rule:

    \( \displaystyle f'(x) = \frac{\log_e x\cdot 1 - x\cdot\frac{1}{x}}{(\log_e x)^2} = \frac{\log_e x - 1}{(\log_e x)^2} \)

The denominator is positive, so f′(x) > 0 when

    \(\log_e x \gt 1,\quad x \gt e\)

(B) \(\frac{x}{2}+\frac{2}{x},\ x \ne 0\)

This matches (I) \((-\infty,-2)\cup(2,\infty)\).

    \( \displaystyle f'(x) = \frac{1}{2} - \frac{2}{x^2} = \frac{x^2 - 4}{2x^2} \)

f′(x) > 0 when \(x^2 \gt 4\), that is, x < −2 or x > 2. On (−∞, −2) the values of f stay below f(−2) = −2 and on (2, ∞) they stay above f(2) = 2, so f is increasing on the union as well as on each part.

(C) \(x^x,\ x \gt 0\)

This matches (III) \(\left(\frac{1}{e},\infty\right)\).

Taking logarithms, \(\log_e y = x\log_e x\), and differentiating (logarithmic differentiation):

    \( \displaystyle \frac{1}{y}\frac{dy}{dx} = \log_e x + 1 \)

    \( f'(x) = x^x(1 + \log_e x) \)

Since \(x^x \gt 0\), f′(x) > 0 when

    \( \displaystyle \log_e x \gt -1,\quad x \gt \frac{1}{e} \)

(D) \(\sin x-\cos x\)

This matches (II) \(\left(-\frac{\pi}{4},\frac{\pi}{4}\right)\).

    \( \displaystyle f'(x) = \cos x + \sin x = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right) \)

For \(-\frac{\pi}{4} \lt x \lt \frac{\pi}{4}\), \(0 \lt x + \frac{\pi}{4} \lt \frac{\pi}{2}\), so f′(x) > 0.

So the matching is (A)-(IV), (B)-(I), (C)-(III), (D)-(II).

Note: Some List-I items fit more than one List-II entry. Since e > 2, the function in (B) is also increasing on (e, ∞), so (B) fits (I) and also (IV); since e > 1/e, the function in (C) is also increasing on (e, ∞), so (C) fits (III) and also (IV). The question can still be solved because (A) has only one possible match, (IV), and Option (3) is the only option containing (A)-(IV).