Match List-I with List-IIList-IList-II(A) f(x)= […] (I)is continuous…

The correct answer is Option (1) → (A) - (IV), (B) - (III), (C) - (I), (D) - (II)

For a piecewise function, continuity at the joining point needs the left-hand limit, the right-hand limit and the value there to be equal.

(A) \( f(x)=x^{10}-1 \) for \( x\le 1 \), \( x^2 \) for \( x\gt 1 \)

This matches (IV) is discontinuous at \( x=1 \).

    \( \displaystyle \lim_{x\to 1^-}(x^{10}-1)=0=f(1) \)

    \( \displaystyle \lim_{x\to 1^+}x^2=1 \)

The two one-sided limits differ, so f is discontinuous at x = 1. (Near x = 0, f is the polynomial \( x^{10}-1 \), so f is also continuous at x = 0.)

(B) \( g(x)=[x] \)

This matches (III) is discontinuous at \( x=-2 \). The greatest integer function jumps at every integer:

    \( \displaystyle \lim_{x\to -2^-}[x]=-3,\quad g(-2)=-2 \)

(For the same reason g is also discontinuous at x = 1.)

(C) \( h(x)=\frac{\sin x}{x} \) for \( x\lt 0 \), \( x+1 \) for \( x\ge 0 \)

This matches (I) is continuous at all points of domain. Each piece is continuous on its own interval, and at the joining point x = 0:

    \( \displaystyle \lim_{x\to 0^-}\frac{\sin x}{x}=1 \)

    \( \displaystyle \lim_{x\to 0^+}(x+1)=1=h(0) \)

(D) \( p(x)=\frac{x}{|x|} \) for \( x\lt 0 \), \( -1 \) for \( x\ge 0 \)

This matches (II) is continuous at \( x=0 \). For x < 0, \( |x|=-x \), so

    \( \displaystyle \frac{x}{|x|}=\frac{x}{-x}=-1 \)

So \( p(x)=-1 \) for every x: a constant function, which is continuous at x = 0.

Each List-II entry is used once. (C) and (D) are continuous everywhere, so they can only take (I) and (II) between them; that leaves (IV) for (A) and (III) for (B). Option (1) is the only option with (A)-(IV) and (B)-(III), so the matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II).

Note: Some List-I items fit more than one List-II entry. (A) fits (II) and (IV), (B) fits (III) and (IV), and (C) and (D) each fit both (I) and (II). The question can still be solved because (C) and (D) must share (I) and (II), which leaves (IV) for (A) and (III) for (B), and Option (1) is the only option containing (A)-(IV) and (B)-(III).