If f(x) = begin{cases} frac{1}{2}(b^2 - a^2), | vedclass

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(D) To check the differentiability of $f(x)$ at $x = a$,we calculate the left-hand derivative ($L$.$H$.$D$.) and right-hand derivative ($R$.$H$.$D$.).

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$f'(x)$ is not differentiable at $x = a$

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(D) To check the differentiability of $f(x)$ at $x = a$,we calculate the left-hand derivative ($L$.$H$.$D$.) and right-hand derivative ($R$.$H$.$D$.).$L$.$H$.$D$. at $x = a$ is $\lim_{h 	o 0} \frac{f(a-h) - f(a)}{-h} = \lim_{h 	o 0} \frac{\frac{1}{2}(b^2 - a^2) - \frac{1}{2}(b^2 - a^2)}{-h} = 0$.Now,$R$.$H$.$D$. at $x = a$ is $\lim_{h 	o 0} \frac{f(a+h) - f(a)}{h} = \lim_{h 	o 0} \frac{\frac{1}{2}b^2 - \frac{(a+h)^2}{6} - \frac{a^3}{3(a+h)} - \frac{1}{2}(b^2 - a^2)}{h}$.Simplifying the expression: $\lim_{h 	o 0} \frac{1}{h} \left[ \frac{1}{2}a^2 - \frac{(a+h)^3 + 2a^3}{6(a+h)} ight]$.As $h 	o 0$,the expression does not converge to a finite value,implying the derivative does not exist at $x = a$.Since $f(x)$ is not differentiable at $x = a$,$f'(x)$ is also not differentiable at $x = a$.

If $f(x) = \begin{cases} \frac{1}{2}(b^2 - a^2), & 0 \leq x \leq a \\ \frac{1}{2}b^2 - \frac{x^2}{6} - \frac{a^3}{3x}, & a < x \leq b \\ \frac{1}{3}\left(\frac{b^3 - a^3}{x}\right), & x > b \end{cases}$,then which of the following is true?

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