A) 1
B) -2
C) 2
D) ½
Correct Answer: D
Solution :
\[f(x)=\left\{ \begin{matrix} \frac{\sin 5x}{{{x}^{2}}+2x}\,\,\,, & x\ne 0 \\ k+\frac{1}{2}\,\,\,\,\,\,\,\,, & x=0 \\ \end{matrix} \right.\] \[L.H.L.\,f({{0}^{-}})=\underset{h\to 0}{\mathop{\lim }}\,\,f(0-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{\sin \,\,5\,\,(0-h)}{(0-{{h}^{2}})+2\,(0-h)}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{\sin \,(-5h)}{{{h}^{2}}-2h}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{\frac{\sin \,5h}{5h}}{\frac{1}{5}(h-2)}=-\frac{1}{\frac{1}{5}(-2)}\] \[=\frac{5}{2}\] Since it is continuous at \[x=0\] \[\therefore \,\,L.H.L=f(0)\] \[\Rightarrow \] \[\frac{5}{2}=k+\frac{1}{2}\]
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