Q.State whether True or False: If $f \cdot g$ is continuous at $x = a$, then $f$ and $g$ are separately continuous at $x = a$.

The key idea is that continuity of a product does not imply continuity of the individual factors — the product can "cancel" a discontinuity.

Step 1: Recall that if $f$ and $g$ are both continuous at $a$, then $f \cdot g$ is continuous at $a$. The converse is false.

Step 2: A counterexample suffices. Let

Then $f$ is discontinuous at $0$ (jump from $1$ to $0$), and $g$ is also discontinuous at $0$. …

The heart of this question is about Continuity of Compositions — specifically, whether continuity of a product forces continuity of each factor. The answer is a clear no, and the reason is intuitive: multiplication can "cancel" discontinuities.

Think of it this way. If $f$ and $g$ both have a jump at $x = a$, but their jumps are such that the product $f(x)g(x)$ happens to land on the same value from both sides, the product can be smooth even though each piece is broken. The product function doesn't "remember" the individual discontinuities — it only cares about the combined output.

Step-by-step reasoning

1. Restate the claim precisely

The statement says:

If $f \cdot g$ is continuous at $x = a$, then $f$ and $g$ are each continuous at $x = a$.

We need to decide if this is always true. One counterexample is enough to show it's false.

2. Look for a simple counterexample

The simplest way: make one function discontinuous, and the other also discontinuous, but arrange their product to be constant (and thus continuous everywhere).

Let $a = 0$ for convenience. Define:

Both $f$ and $g$ have a jump at $x = 0$ — neither is continuous there.

3. Compute the product

For $x \ge 0$: $f(x)g(x) = (1)(-1) = -1$

For $x < 0$: $f(x)g(x) = (-1)(1) = -1$

So $f(x)g(x) = -1$ for all real $x$. That's a constant function, which is continuous at every point, including $x = 0$.

4. Conclude …

Method: Disproving a "Continuity Implies..." Claim Using a Counterexample

This method solves True/False questions of the form "if [some combination of two functions] is continuous, then each function must individually be continuous" — disproved with a single well-chosen counterexample.

Steps

Step 1: Recognise what disproves a universal claim

A claim of the form "if X then always Y" is false the moment you exhibit one case where X holds but Y fails. You do not need to rule out every possible function — one counterexample is sufficient.

Step 2: Design two individually-discontinuous functions whose combination is smooth

Choose $f$ and $g$ that each jump at the point in question, built so that their product (or sum, depending on the claim) cancels the jump. A standard choice is opposite-signed step functions, where $g=-f$, since then $f\cdot g$ is a constant.

Step 3: Verify each piece really is discontinuous at that point

Show the left-hand and right-hand values of $f$ (and of $g$) disagree at the point, or that the value there does not match the two-sided limit.

Step 4: Compute the combination explicitly …

Common Mistakes

Mistake 1: Assuming the converse of a true theorem is automatically also true

Students know "if $f,g$ are continuous at $a$, then $f\cdot g$ is continuous at $a$" is a real theorem, and wrongly assume the reverse direction must hold too. Why it's wrong: a true implication $P\Rightarrow Q$ says nothing about whether $Q\Rightarrow P$ — the product being continuous is a weaker condition than both factors being continuous, so it can happen for other reasons too (e.g. cancellation of jumps). Correct approach: test the converse directly with a counterexample rather than assuming symmetry.

Mistake 2: "Proving" it True via invalid division

A student may try $f=\frac{f\cdot g}{g}$ and argue this is continuous since it's a quotient of continuous functions. Why it's wrong: this requires $g$ to be continuous AND nonzero at $a$ — exactly the things not yet established, so the argument is circular and breaks down wherever $g(a)=0$. Correct approach: recognize that recovering $f$ from the product by dividing is not generally valid, and instead search for a genuine counterexample.

Mistake 3: Choosing a counterexample that doesn't actually disprove the claim …