Match the LIST-I with LIST-IILIST-ILIST-IIA. f(x)=log_e(x^2+1) I.not…

The correct answer is Option (2) → A-III, B-IV, C-II, D-I

(A) \( f(x)=\log_e(x^2+1) \)

This matches (III) Continuous everywhere.

Since \( x^2+1\ge 1\gt 0 \) for every real x, the logarithm is defined and continuous on all of R.

(B) \( f(x)=|x| \)

This matches (IV) Continuous everywhere but not differentiable at x = 0.

At x = 0 the left derivative is −1 and the right derivative is 1:

    \( \displaystyle \lim_{h\to 0^-}\frac{|h|-0}{h}=-1 \)

    \( \displaystyle \lim_{h\to 0^+}\frac{|h|-0}{h}=1 \)

(C) \( f(x)=x^2 \)

This matches (II) is continuous as well as differentiable at x = 0.

A polynomial is continuous and differentiable everywhere; f′(0) = 2(0) = 0.

(D) \( f(x)=[x] \)

This matches (I) not continuous at all integral values.

At an integer n, the left-hand limit is n − 1 while f(n) = n.

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

Note: Some List-I items fit more than one List-II entry. \( \log_e(x^2+1) \) in (A) and \( x^2 \) in (C) are both continuous everywhere and also continuous and differentiable at x = 0, so each fits (II) and (III); \( |x| \) in (B) is also continuous everywhere, so it fits (III) and (IV). The question can still be solved because (D) has only one possible match, (I), which leaves Options (1) and (2); Option (1) matches (C) \( x^2 \) with (IV), which is wrong because \( x^2 \) is differentiable at x = 0. So Option (2) is the only option in which every item has a valid match.