A) 0
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) \[-\frac{1}{2}\]
Correct Answer: A
Solution :
Given that \[f(x)=\left\{ \begin{matrix} \frac{1-\cos x}{x}, & x\ne 0 \\ k, & x=0 \\ \end{matrix} \right.\] \[\underset{x\to 0}{\mathop{\lim }}\,\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{1-\cos x}{x}\] \[=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{2{{\sin }^{2}}x/2}{4{{(x/2)}^{2}}}\,.\,x=0\] and \[f(0)=k\] \[\therefore \] function is continuous at \[x=0\] \[\therefore \] \[\underset{x\to 0}{\mathop{\lim }}\,\,\,f\left( x \right)=f\left( 0 \right)\] \[0=k\] or \[k=0\]You need to login to perform this action.
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