Q.If $f(x) = 2x$ and $g(x) = \dfrac{x^2}{2} + 1$, then which of the following can be a discontinuous function?
(A) $f(x) + g(x)$
(B) $f(x) - g(x)$
(C) $f(x) \cdot g(x)$
(D) $\dfrac{g(x)}{f(x)}$

Concept: Piecewise Function Evaluation — but here the real idea is continuity of combinations of continuous functions. Both $f(x)=2x$ and $g(x)=\frac{x^2}{2}+1$ are polynomials, hence continuous for all real $x$. …

We are given two functions:

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

Both are polynomials — $f$ is linear, $g$ is quadratic. Polynomials are continuous for all real numbers. So individually, each is continuous everywhere.

Now, when we combine continuous functions using addition, subtraction, or multiplication, the result is still continuous everywhere. That’s a fundamental theorem: the sum, difference, and product of continuous functions are continuous on their common domain. So options (A), (B), and (C) are all continuous for all $x \in \mathbb{R}$.

The only operation that can break continuity is division — because the quotient $\frac{g(x)}{f(x)}$ is defined only where $f(x) \neq 0$. At points where $f(x) = 0$, the function is not defined, and hence it is discontinuous there (unless the limit exists and we can remove the discontinuity, but here it's a simple zero in the denominator).

Let’s check:

1. Identify where $f(x) = 0$

   $f(x) = 2x = 0 \implies x = 0$.

2. Examine the quotient at $x = 0$

   $\frac{g(0)}{f(0)} = \frac{0^2/2 + 1}{0} = \frac{1}{0}$, which is undefined.

   The function $\frac{g(x)}{f(x)}$ has no value at $x=0$, so it cannot be continuous there.

3. Check if the discontinuity is removable

   For a removable discontinuity, the limit as $x \to 0$ must exist and be finite.

   $\lim_{x \to 0} \frac{x^2/2 + 1}{2x} = \lim_{x \to 0} \frac{1}{2x} + \frac{x}{4}$. …

Method: Continuity of Combinations (Sum, Difference, Product, Quotient)

This technique applies whenever you're asked which combination of two functions built from $f$ and $g$ can fail to be continuous.

Steps

Step 1: Confirm each building-block function is continuous on its own domain.

Most standard functions — polynomials, $\sin x$, $\cos x$, $e^x$ — are continuous everywhere they are defined. Establish this first, since the whole argument rests on it.

Step 2: Recall the algebra-of-continuity theorem.

If $f$ and $g$ are continuous at a point (or on an interval), then $f+g$, $f-g$, and $f\cdot g$ are automatically continuous there too — no separate check is ever needed for these three operations.

Step 3: Isolate the quotient as the only real suspect. …

Common Mistakes

Mistake 1: Assuming the sum, difference, or product options need to be checked too.

Why it's wrong: students sometimes try to test $f+g$, $f-g$, $f\cdot g$ individually for discontinuity, wasting time. Correct approach: recall that sums, differences, and products of continuous functions are always continuous — only division ever introduces a new possible discontinuity, so go straight to the quotient option.

Mistake 2: Thinking a quotient is discontinuous only when the limit fails to exist (an infinite blow-up), and missing that ANY point where the denominator is zero counts. …