A) \[{{(-1)}^{k}}\,\,(k-1)\,\pi \]
B) \[{{(-1)}^{k-1}}(k-1)\,\pi \]
C) \[{{(-1)}^{k}}k\pi \]
D) \[{{(-1)}^{k-1}}k\,\pi \]
Correct Answer: A
Solution :
\[{f}'(k-0)=\underset{h\to 0}{\mathop{\text{lim}}}\,\frac{[k-h]\sin \pi (k-h)-[k]\sin \pi k}{-h}\] \[=\underset{h\to 0}{\mathop{\text{lim}}}\,\frac{{{(-1)}^{k-1}}(k-1)\sin \pi h-k\times 0}{-h}\] \[=\underset{h\to 0}{\mathop{\text{lim}}}\,\frac{{{(-1)}^{k-1}}(k-1)\sin \pi h}{-h}\]\[={{(-1)}^{k}}.(k-1)\pi \].
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