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The left hand derivative of $f(x)=[x]\sin(\pi x)$ at $x=k$, where $k$ is an integer is

$\begin{array}{1 1}(a)\;(-1)^k(k-1)\pi&(b)\;(-1)^{k-1}(k-1)\pi\\(c)\;(-1)^kk\pi&(d)\;(-1)^{k-1}k\pi\end{array}$

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At LHD=$\lim\limits_{h\to 0}\large\frac{f(k)-f(k-h)}{h}$ [ k-integer]
$\Rightarrow \lim\limits_{h\to 0}\large\frac{[k]\sin k\pi-[k-h]\sin(k-h)\pi}{h}$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{-(k-1)\sin(k-h)\pi}{h}$
$\sin k\pi=0$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{-(k-1)-(1)^{k-1}\sin h\pi}{h\pi}$$\times \pi$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{-(k-1)(-1)^{k-1}\sin h\pi}{h\pi}$$\times \pi$
$\Rightarrow \pi(k-1)(-1)^k$
Hence (a) is the correct answer.
answered Dec 31, 2013 by sreemathi.v

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