Match List I with List II

List I	List II
A. If $f(x) = 2x$ and $g(x) = \frac{x^2}{2} + 1$, then $\frac{g(x)}{f(x)}$ is	I. discontinuous at exactly three points.
B. The function $f(x) = \frac{4-x^2}{4x-x^3}$ is	II. continuous everywhere
C. The function $f(x) = \lvert x \rvert + \lvert x-1 \rvert$ is	III. discontinuous at $x = 0$.
D. The function $f(x) = \lvert \sin x \rvert$ is	IV. continuous at $x = 0$ and $x = 1$

Choose the correct answer from the options given below:

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$f(x) = 2x$ and $g(x) = \dfrac{x^2}{2} + 1$

$\dfrac{g(x)}{f(x)} = \dfrac{\dfrac{x^2}{2} + 1}{2x}$

$= \dfrac{x^2 + 2}{4x}$

$= \dfrac{x}{4} + \dfrac{1}{2x}$

The term $\dfrac{1}{2x}$ makes the function undefined at $x = 0$, so it is discontinuous at $x = 0$.

$A \to III$

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$f(x) = \dfrac{4 - x^2}{4x - x^3}$

Factor the denominator:

$4x - x^3 = x(4 - x^2) = x(2 - x)(2 + x)$

So,

$f(x) = \dfrac{4 - x^2}{x(4 - x^2)}$

Even though $(4 - x^2)$ cancels to give $\dfrac{1}{x}$, the original function is still undefined wherever the original denominator equals zero.

$x(2 - x)(2 + x) = 0$ gives $x = 0,\; x = 2,\; x = -2$

These are exactly three points of discontinuity.

$B \to I$

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$f(x) = |x| + |x - 1|$

$|x|$ is continuous everywhere and $|x - 1|$ is continuous everywhere. The sum of two continuous functions is always continuous.

A sharp corner (like in absolute value graphs) does not mean the function is discontinuous — it only means the function is not differentiable there.

So $f(x)$ is continuous everywhere, and in particular at $x = 0$ and $x = 1$.

$C \to IV$

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$f(x) = |\sin x|$

$\sin x$ is continuous everywhere. If any function $h(x)$ is continuous everywhere, then $|h(x)|$ is also continuous everywhere — the absolute value simply reflects the negative portions above the $x$-axis without creating any breaks.

$D \to II$

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Final matching:

$A \to III,\quad B \to I,\quad C \to IV,\quad D \to II$