Let f(x) = begin{cases} int_{0}^{x} |1-t| dt, | vedclass

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(A, D) Given $f(x) = \begin{cases} \int_{0}^{x} |1-t| dt, & x > 1 \ x - \frac{1}{2}, & x \leq 1 \end{cases}$.For $x > 1$, $\int_{0}^{x} |1-t| dt = \i

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$f(x)$ is not differentiable at $x=1$

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(A, D) Given $f(x) = \begin{cases} \int_{0}^{x} |1-t| dt, & x > 1 \ x - \frac{1}{2}, & x \leq 1 \end{cases}$.For $x > 1$, $\int_{0}^{x} |1-t| dt = \int_{0}^{1} (1-t) dt + \int_{1}^{x} (t-1) dt$.$= \left[ t - \frac{t^2}{2} ight]_{0}^{1} + \left[ \frac{t^2}{2} - t ight]_{1}^{x} = \left( 1 - \frac{1}{2} ight) + \left( \frac{x^2}{2} - x - (\frac{1}{2} - 1) ight) = \frac{1}{2} + \frac{x^2}{2} - x + \frac{1}{2} = \frac{x^2}{2} - x + 1$.Thus, $f(x) = \begin{cases} \frac{x^2}{2} - x + 1, & x > 1 \ x - \frac{1}{2}, & x \leq 1 \end{cases}$.Continuity at $x=1$:$LHL = \lim_{x 	o 1^-} (x - \frac{1}{2}) = \frac{1}{2}$.$RHL = \lim_{x 	o 1^+} (\frac{x^2}{2} - x + 1) = \frac{1}{2} - 1 + 1 = \frac{1}{2}$.$f(1) = 1 - \frac{1}{2} = \frac{1}{2}$.Since $LHL = RHL = f(1)$, $f(x)$ is continuous at $x=1$.Differentiability at $x=1$:$LHD = \lim_{h 	o 0} \frac{f(1-h) - f(1)}{-h} = \lim_{h 	o 0} \frac{(1-h - \frac{1}{2}) - \frac{1}{2}}{-h} = \lim_{h 	o 0} \frac{-h}{-h} = 1$.$RHD = \lim_{h 	o 0} \frac{f(1+h) - f(1)}{h} = \lim_{h 	o 0} \frac{(\frac{(1+h)^2}{2} - (1+h) + 1) - \frac{1}{2}}{h} = \lim_{h 	o 0} \frac{\frac{1+2h+h^2}{2} - h - \frac{1}{2}}{h} = \lim_{h 	o 0} \frac{\frac{h^2}{2}}{h} = 0$.Since $LHD 
eq RHD$, $f(x)$ is not differentiable at $x=1$.Therefore, both options $A$ and $D$ are correct.

Let $f(x) = \begin{cases} \int_{0}^{x} |1-t| dt, & x > 1 \\ x - \frac{1}{2}, & x \leq 1 \end{cases}$. Then:

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