A) \[1\]
B) \[0\]
C) \[\frac{1}{2}\]
D) \[-1\]
Correct Answer: C
Solution :
Given, \[f(x)=-\left\{ \begin{matrix} \frac{1-\cos x}{{{x}^{2}}} & , & x\ne 0 \\ k & , & x=0 \\ \end{matrix} \right.\] Since, x is continuous \[\therefore \] \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=f(0)\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos x}{{{x}^{2}}}=k\] \[\Rightarrow \] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{-(-\sin x)}{2x}=k\] [using L?Hospital?s rule] \[\Rightarrow \] \[\frac{1}{2}\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \,x}{x}=k\] \[\Rightarrow \] \[\frac{1}{2}.1=k\Rightarrow k=\frac{1}{2}\]
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