Assertion (A): f(x) = |x| is differentiable a | vedclass

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(A) The function $f(x) = |x|$ is defined as:$f(x) = \begin{cases} x, & x \geq 0 \ -x, & x < 0 \end{cases}$At $x = 0$,the left-hand derivative is $\li

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$A$ is correct,$R$ is correct,$R$ is the correct explanation of $A$.

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(A) The function $f(x) = |x|$ is defined as:$f(x) = \begin{cases} x, & x \geq 0 \ -x, & x At $x = 0$,the left-hand derivative is $\lim_{h 	o 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^-} \frac{-h - 0}{h} = -1$.The right-hand derivative is $\lim_{h 	o 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h 	o 0^+} \frac{h - 0}{h} = 1$.Since the left-hand derivative $
eq$ right-hand derivative,$f(x)$ is not differentiable at $x = 0$.However,$f(x)$ is continuous everywhere,including $x = 0$.For any $x = a 
eq 0$,the function is locally linear (either $x$ or $-x$),so it is differentiable.Thus,Assertion $(A)$ is correct.Reason $(R)$ states a fundamental theorem in calculus: differentiability implies continuity,but continuity does not imply differentiability. This theorem explains why $f(x) = |x|$ is continuous at $x = 0$ but not differentiable there.Therefore,$R$ is the correct explanation of $A$.

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 the point. But the converse is not true.

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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 the point. But the converse is not true.