Functions Question 474
Question: If $ f(x)= \begin{cases} & \frac{1-\cos x}{x}, x\ne 0 \\ & k, x=0 \\ \end{cases} $ is continuous at $ x=0 $ then $ k= $
[Karnataka CET 2004]
Options:
A) 0
B) $ \frac{1}{2} $
C) $ \frac{1}{4} $
D) $ -\frac{1}{2} $
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Answer:
Correct Answer: A
Solution:
$ f(x)= \begin{cases} & \frac{1-\cos x}{x},x\ne 0 \\ & k ,x=0 \\ \end{cases} . $ 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}{{{( \frac{x}{2} )}^{2}}}.\frac{x}{4}=k\Rightarrow k=0 $ .
BETA
