Question

Let \(\mathbb R\) denote the set of all real numbers. Let \[ f:\mathbb R\to\mathbb R \] be an arbitrary function and let \[ g:\mathbb R\to\mathbb R \] be the function defined by \[ g(x)=xf(x),\qquad \forall x\in\mathbb R \] Then which of the following statements is (are) TRUE?

• A. The function \(g\) is always continuous at \(x=0\)
• B. If \(f\) is continuous at \(x=0\), then \(g\) is differentiable at \(x=0\)
• C. If \(g\) is differentiable at \(x=0\), then \(f\) is continuous at \(x=0\)
• D. If \(g\) is differentiable at \(x=0\), then \[\lim_{x\to0}f(x)\] exists

Hint:

If: \[ g(x)=xf(x) \] then: \[ g'(0)=\lim_{x\to0}f(x) \] provided the limit exists.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question tests definitions of continuity and differentiability at a point. For \( g(x) = xf(x) \), \( g(0) = 0 \cdot f(0) = 0 \).
Continuity at \( 0 \) means \( \lim_{x \to 0} g(x) = g(0) = 0 \).
Differentiability at \( 0 \) means \( \lim_{h \to 0} \frac{g(h) - g(0)}{h} \) exists.
Step 3: Detailed Explanation:
1. Statement (A):
Is \( \lim_{x \to 0} xf(x) \) always \( 0 \)? No.
Consider \( f(x) = 1/x \) for \( x \neq 0 \) and \( f(0) = 0 \). Then \( g(x) = 1 \) for \( x \neq 0 \).
\( \lim_{x \to 0} g(x) = 1 \neq g(0) \). So (A) is False.
2. Statement (B):
If \( f \) is continuous at \( 0 \), then \( \lim_{h \to 0} f(h) = f(0) \).
Checking \( g'(0) = \lim_{h \to 0} \frac{hf(h) - 0}{h} = \lim_{h \to 0} f(h) \).
Since \( f \) is continuous, the limit exists and equals \( f(0) \). So \( g \) is differentiable. Statement (B) is True.
3. Statement (D):
If \( g \) is differentiable at \( 0 \), then by definition, \( \lim_{h \to 0} \frac{g(h) - g(0)}{h} = \lim_{h \to 0} \frac{hf(h)}{h} = \lim_{h \to 0} f(h) \) must exist.
Therefore, if \( g \) is differentiable, \( \lim_{x \to 0} f(x) \) exists. Statement (D) is True.
4. Statement (C):
If \( g \) is differentiable, we know \( \lim_{x \to 0} f(x) = L \) exists. For \( f \) to be continuous, we need \( L = f(0) \).
But \( f \) is an "arbitrary" function. If \( f(x) = 1 \) for \( x \neq 0 \) and \( f(0) = 2 \), then \( g(x) = x \) for \( x \neq 0 \), which is differentiable at \( 0 \) (slope 1), yet \( f \) is not continuous at \( 0 \). Statement (C) is False.
Step 4: Final Answer:
The true statements are (B) and (D).