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 converse is not true. .

• A. Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A).
• B. Both Assertion (A) and Reason (R) are true, but Reason (R) is not correct explanation for Assertion (A).
• C. Assertion (A) is true, but Reason (R) is false.
• D. Assertion (A) is false, but Reason (R) is true.

Hint:

Differentiability at a point \(\implies\) Continuity at that point.
Continuity at a point \(\not\implies\) Differentiability at that point (e.g., sharp corners, cusps).
\(f(x)=|x|\) has a sharp corner at \(x=0\), making it continuous but not differentiable there. Away from \(x=0\), it's a simple linear function (\(x\) or \(-x\)) and is differentiable.

Answer 1

The Correct Option is A

Assertion (A): \(f(x) = |x|\). \begin{itemize} \item For \(x=a \neq 0\): If \(a>0\), then in a neighborhood of \(a\), \(f(x)=x\), so \(f'(x)=1\). Differentiable. If \(a<0\), then in a neighborhood of \(a\), \(f(x)=-x\), so \(f'(x)=-1\). Differentiable. So, \(f(x)=|x|\) is differentiable at \(x=a \neq 0\). \item At \(x=0\): \(f(x)=|x|\) is continuous at \(x=0\) since \(\lim_{x\to 0} |x| = 0 = f(0)\). For differentiability at \(x=0\): LHD = \( \lim_{h \to 0^-} \frac{|0+h|-|0|}{h} = \lim_{h \to 0^-} \frac{|h|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1 \). RHD = \( \lim_{h \to 0^+} \frac{|0+h|-|0|}{h} = \lim_{h \to 0^+} \frac{|h|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 \). Since LHD \(\neq\) RHD ( \(-1 \neq 1 \)), \(f(x)=|x|\) is not differentiable at \(x=0\). \end{itemize} So, Assertion (A) is TRUE: \(f(x)=|x|\) 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 converse is not true." This is a standard theorem in calculus. Differentiability implies continuity. However, continuity does not imply differentiability (e.g., \(f(x)=|x|\) at \(x=0\) is continuous but not differentiable). So, Reason (R) is TRUE. Explanation: Does Reason (R) explain Assertion (A)? Assertion (A) states two things: differentiability at \(a \neq 0\), and continuity but not differentiability at \(x=0\). Reason (R) states that differentiability implies continuity, and the converse is not true. The "converse is not true" part of Reason (R) is directly illustrated by the behavior of \(f(x)=|x|\) at \(x=0\) as stated in Assertion (A) (continuous but not differentiable). The first part of Assertion (A) (differentiable at \(a \neq 0\)) implies it's also continuous at \(a \neq 0\) (by the first part of Reason R). Reason (R) provides the general mathematical principles that underpin the behavior described in Assertion (A). The fact that \(|x|\) is continuous but not differentiable at \(x=0\) is a classic example of "converse is not true". The differentiability away from zero and its implied continuity also aligns. So, Reason (R) is a correct explanation for the properties stated in Assertion (A). This corresponds to option (a). \[ \boxed{\text{A is correct, R is correct, R is correct explanation of A}} \]