A) 0
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) \[-\frac{1}{2}\]
Correct Answer: A
Solution :
\[f(x)=\left\{ \begin{align} & \frac{1-\cos x}{x},x\ne 0 \\ & \,\,\,\,\,k\,\,\,\,\,\,\,\,\,\,\,,x=0 \\ \end{align} \right.\] continuous at \[x=0\] \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f(x)=f(0)\]Þ \[\underset{x\to 0}{\mathop{\lim }}\,\frac{2.{{\sin }^{2}}x/2}{x}=k\] Þ \[\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}x/2}{{{\left( \frac{x}{2} \right)}^{2}}}.\frac{x}{4}=k\Rightarrow k=0\].
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