The left-hand derivative of f(x) = [x] sin(pi | vedclass

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(A) Given $f(x) = [x] \sin(\pi x)$.For $x = k$,where $k \in \mathbb{Z}$,the left-hand derivative $(LHD)$ is defined as:$LHD = \lim_{h 	o 0^+} \frac{f

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$(-1)^{k}(k-1) \pi$

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(A) Given $f(x) = [x] \sin(\pi x)$.For $x = k$,where $k \in \mathbb{Z}$,the left-hand derivative $(LHD)$ is defined as:$LHD = \lim_{h 	o 0^+} \frac{f(k-h) - f(k)}{-h}$Since $k$ is an integer,$[k-h] = k-1$ for small $h > 0$.Also,$f(k) = [k] \sin(k\pi) = k \cdot 0 = 0$.Substituting these into the limit:$LHD = \lim_{h 	o 0^+} \frac{(k-1) \sin(\pi(k-h)) - 0}{-h}$Using the identity $\sin(k\pi - \pi h) = \sin(k\pi)\cos(\pi h) - \cos(k\pi)\sin(\pi h)$:Since $\sin(k\pi) = 0$ and $\cos(k\pi) = (-1)^k$,we have $\sin(k\pi - \pi h) = -(-1)^k \sin(\pi h) = (-1)^{k+1} \sin(\pi h)$.$LHD = \lim_{h 	o 0^+} \frac{(k-1) (-1)^{k+1} \sin(\pi h)}{-h}$Using $\lim_{h 	o 0} \frac{\sin(\pi h)}{h} = \pi$:$LHD = (k-1) (-1)^{k+1} (-\pi) = (k-1) (-1)^{k+1+1} \pi = (-1)^k (k-1) \pi$.

The left-hand derivative of $f(x) = [x] \sin(\pi x)$ at $x = k$,where $k$ is an integer and $[\cdot]$ denotes the greatest integer function,is:

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