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: Check Option (A).
Given: \[ g(x)=xf(x) \] Take: \[ f(x)=\frac1x,\qquad x\neq0 \] Then: \[ g(x)=1,\qquad x\neq0 \] If: \[ g(0)=0 \] then: \[ \lim_{x\to0}g(x)=1\neq g(0) \] Thus \(g\) need not be continuous at \(x=0\). Therefore: \[ \Rightarrow \mathrm{Option\ (A)\ is\ Incorrect} \]

Step 2: Check Option (B).
If \(f\) is continuous at \(x=0\), then: \[ \lim_{x\to0}f(x)=f(0) \] Now: \[ g'(0)=\lim_{x\to0}\frac{g(x)-g(0)}{x} \] Since: \[ g(x)=xf(x) \] and: \[ g(0)=0 \] we get: \[ g'(0)=\lim_{x\to0}f(x) \] Since the limit exists, \[ g'(0) \] exists. Thus: \[ \Rightarrow \mathrm{Option\ (B)\ is\ Correct} \]

Step 3: Check Option (C).
Suppose: \[ g(x)=x \] Then: \[ f(x)=1,\qquad x\neq0 \] Take: \[ f(0)=5 \] Then: \[ g(x)=xf(x)=x,\qquad \forall x \] So \(g\) is differentiable at \(0\), but \(f\) is not continuous at \(0\). Therefore: \[ \Rightarrow \mathrm{Option\ (C)\ is\ Incorrect} \]

Step 4: Check Option (D).
If \(g\) is differentiable at \(0\), then: \[ g'(0)=\lim_{x\to0}\frac{g(x)-g(0)}{x} \] Using: \[ g(x)=xf(x),\qquad g(0)=0 \] \[ g'(0)=\lim_{x\to0}f(x) \] Since differentiability implies existence of derivative, \[ \lim_{x\to0}f(x) \] exists. Thus: \[ \Rightarrow \mathrm{Option\ (D)\ is\ Correct} \]

Step 5: Identify the correct options.
Hence: \[ \boxed{\mathrm{(B)\ and\ (D)}} \]