Assertion (A): If y = f(x) = (|x| - |x - 1|)^ | vedclass

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(D) For $x > 1$,$|x| = x$ and $|x - 1| = x - 1$.Then $f(x) = (x - (x - 1))^2 = (1)^2 = 1$.For $0 < x < 1$,$|x| = x$ and $|x - 1| = -(x - 1) = 1 - x$.T

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$A$ is false,$R$ is true.

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(D) For $x > 1$,$|x| = x$ and $|x - 1| = x - 1$.Then $f(x) = (x - (x - 1))^2 = (1)^2 = 1$.For $0 Then $f(x) = (x - (1 - x))^2 = (2x - 1)^2 = 4x^2 - 4x + 1$.The left-hand derivative at $x = 1$ is $LHD = \lim_{h ightarrow 0^-} \frac{f(1+h) - f(1)}{h} = \lim_{h ightarrow 0^-} \frac{(2(1+h) - 1)^2 - 1}{h} = \lim_{h ightarrow 0^-} \frac{(2h + 1)^2 - 1}{h} = \lim_{h ightarrow 0^-} \frac{4h^2 + 4h}{h} = 4$.The right-hand derivative at $x = 1$ is $RHD = \lim_{h ightarrow 0^+} \frac{f(1+h) - f(1)}{h} = \lim_{h ightarrow 0^+} \frac{1 - 1}{h} = 0$.Since $LHD 
eq RHD$,the derivative at $x = 1$ does not exist.Thus,Assertion $(A)$ is false.The Reason $(R)$ is the standard definition of the derivative,which is true.Therefore,$(A)$ is false and $(R)$ is true.

Assertion $(A)$: If $y = f(x) = (|x| - |x - 1|)^2$,then $\left(\frac{dy}{dx}\right)_{x=1} = 1$.
Reason $(R)$: If $\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$ exists,then it is called the derivative of $f(x)$ at $x = a$.
Then:

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Assertion $(A)$: If $y = f(x) = (|x| - |x - 1|)^2$,then $\left(\frac{dy}{dx}\right)_{x=1} = 1$.Reason $(R)$: If $\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$ exists,then it is called the derivative of $f(x)$ at $x = a$.Then: