Question

Assertion (A): \[ f(x)=|x| \] is differentiable at \(x=a\neq 0\) and continuous but not differentiable at \(x=0\). Reason (R): If a function is differentiable at a point, then it is continuous at that point. But the converse is not true.

• A. A is correct, R is correct, R is correct explanation of A
• B. A is correct, R is correct, but R is not correct explanation of A
• C. A is correct, R is false
• D. A is false, R is correct

Hint:

Always remember: \[ \text{Differentiable} \Rightarrow \text{Continuous} \] but \[ \text{Continuous} \nRightarrow \text{Differentiable}. \] The function \[ f(x)=|x| \] is the most common example of a function that is continuous but not differentiable at a point.

Answer 1

The Correct Option is A

Step 1: Examine the assertion \(A\).
The absolute value function is defined as \[ |x|= \begin{cases} x, & x\ge 0,\\ -x, & x\lt 0. \end{cases} \] For every point \[ a\neq 0, \] the function is either \(x\) or \(-x\) in a neighbourhood of \(a\).
Since both \(x\) and \(-x\) are differentiable functions, \(f(x)=|x|\) is differentiable at every point \[ a\neq 0. \] Now consider \[ x=0. \] The left-hand derivative is \[ \lim_{h\to 0^-}\frac{|h|-|0|}{h} = \lim_{h\to 0^-}\frac{-h}{h} = -1. \] The right-hand derivative is \[ \lim_{h\to 0^+}\frac{|h|-|0|}{h} = \lim_{h\to 0^+}\frac{h}{h} = 1. \] Since \[ LHD\neq RHD, \] the derivative at \(x=0\) does not exist.
However, \[ \lim_{x\to 0}|x|=0=f(0), \] so the function is continuous at \(x=0\).
Therefore, Assertion \(A\) is correct.

Step 2: Examine the reason \(R\).
A fundamental theorem of calculus states that \[ \text{Differentiability } \Rightarrow \text{ Continuity}. \] Thus, if a function is differentiable at a point, it must be continuous there.
But the converse is not always true: \[ \text{Continuity } \nRightarrow \text{ Differentiability}. \] A standard example is \[ f(x)=|x|, \] which is continuous at \(x=0\) but not differentiable there.
Hence, Reason \(R\) is also correct.

Step 3: Determine whether \(R\) explains \(A\).
The assertion states that \(f(x)=|x|\) is continuous at \(x=0\) but not differentiable there.
The reason explains exactly why this is possible: differentiability implies continuity, but continuity alone does not guarantee differentiability.
Since \(|x|\) is a classic example illustrating this fact, the reason correctly explains the assertion.

Step 4: Final conclusion.
Therefore, \[ \boxed{\text{A is correct, R is correct, and R is the correct explanation of A.}} \] Hence, the correct option is \[ \boxed{(1)} \]