Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12
Question:
<p>Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$. If $f(x)$ is differentiable at $x=0$, then which of the following is true?</p>
<p>$f(x)$ is differentiable only at $x=0$</p>
<p>$f(x)$ is differentiable for all $x\in\mathbb{R}$</p>
<p>$f(x)$ is differentiable for all $x>0$</p>
<p>$f(x)$ is non-differentiable everywhere</p>
Step-by-Step Solution
Key Concept: General
<b>Additive Functions with Differentiability — JEE Advanced 2011</b><br>
From $f(x+y)=f(x)+f(y)$: set $x=y=0\Rightarrow f(0)=0$.<br>
$f'(x)=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\to0}\dfrac{f(x)+f(h)-f(x)}{h}=\lim_{h\to0}\dfrac{f(h)}{h}=\lim_{h\to0}\dfrac{f(h)-f(0)}{h}=f'(0)$.<br>
So $f'(x)=f'(0)=c$ (constant) for all $x$. Since $f$ is differentiable at 0 (given), it is differentiable everywhere, with $f(x)=cx$.<br>
<b>Answer: 1 (option 2 = differentiable for all $x\in\mathbb{R}$)</b><br>
(Note: answer key says "1" which corresponds to option B = differentiable for all $x$.)<br>
<b>Key concept:</b> For additive functions, differentiability at one point implies differentiability everywhere with constant derivative.<br>
<b>Trap:</b> Thinking the function might be the Cantor-type non-measurable function — but the differentiability condition at 0 forces linearity.
Correct Answer: 1