The left-hand derivative of f(x) = {x} sin(pi | vedclass

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(B) The left-hand derivative $(LHD)$ at $x = k$ is defined as $\lim_{h 	o 0^+} \frac{f(k-h) - f(k)}{-h}$.Given $f(x) = \{x\} \sin(\pi x)$,we have $f(

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$(-1)^k 	imes \pi$

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(B) The left-hand derivative $(LHD)$ at $x = k$ is defined as $\lim_{h 	o 0^+} \frac{f(k-h) - f(k)}{-h}$.Given $f(x) = \{x\} \sin(\pi x)$,we have $f(k) = \{k\} \sin(\pi k) = 0 	imes 0 = 0$.For a small $h > 0$,$k-h$ lies in the interval $(k-1, k)$,so $\{k-h\} = k-h - (k-1) = 1-h$.Thus,$f(k-h) = (1-h) \sin(\pi(k-h)) = (1-h) \sin(k\pi - \pi h) = (1-h) (\sin(k\pi)\cos(\pi h) - \cos(k\pi)\sin(\pi h))$.Since $\sin(k\pi) = 0$ and $\cos(k\pi) = (-1)^k$,we get $f(k-h) = (1-h) (0 - (-1)^k \sin(\pi h)) = -(1-h) (-1)^k \sin(\pi h) = (1-h) (-1)^{k+1} \sin(\pi h)$.Now,$LHD = \lim_{h 	o 0^+} \frac{(1-h) (-1)^{k+1} \sin(\pi h) - 0}{-h} = \lim_{h 	o 0^+} \frac{(1-h) (-1)^{k+1} \sin(\pi h)}{-h}$.Using $\lim_{h 	o 0} \frac{\sin(\pi h)}{\pi h} = 1$,we get $LHD = (-1)^{k+1} 	imes (-1) 	imes \pi = (-1)^{k+2} \pi = (-1)^k \pi$.

The left-hand derivative of $f(x) = \{x\} \sin(\pi x)$ at $x = k$ ($k$ is an integer) is (where $\{ \}$ denotes the fractional part function).

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