Functions Question 565
Question: Function $ f(x)=\frac{1-\cos 4x}{8x^{2}}, $ where $ x\ne 0 $ and $ f(x)=k $ where $ x=0 $ is a continous function at $ x=0 $ then the value of k will be
[AMU 2005]
Options:
A) $ k=0 $
B) $ k=1 $
C) $ k=-1 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)= \begin{cases} & \frac{1-\cos 4x}{8x^{2}},x\ne 0 \\ & k,,,x=0 \\ \end{cases} . $ If $ f(x) $ is continous function at point $ x=0 $ then $ \underset{x\to 0+}{\mathop{\lim }},[f(x)=\underset{x\to 0-}{\mathop{\lim }},[f(x)] $ $ \underset{x\to 0}{\mathop{\lim }},[f(x)]=\underset{h\to 0-}{\mathop{\lim }},[f(0+h)] $ $ =\underset{h\to 0}{\mathop{\lim }},[f(h)]=\underset{h\to 0}{\mathop{\lim }},\frac{1-\cos 4h}{8h^{2}}=\underset{h\to 0}{\mathop{\lim }},\frac{2{{\sin }^{2}}2h}{8h^{2}}=\underset{h\to 0}{\mathop{\lim }},\frac{{{\sin }^{2}}2h}{4h^{2}} $ $ =\underset{h\to 0}{\mathop{\lim }},{{( \frac{\sin 2h}{2h} )}^{2}}={{(1)}^{2}}=1 $ $ \underset{x\to {0^{-}}}{\mathop{\lim }},f(x)\underset{,h\to 0}{\mathop{=\lim }},[f(0-h) $ $ =\underset{h\to 0}{\mathop{\lim }},[f(-h)]=\underset{h\to 0}{\mathop{\lim }},\frac{1-\cos 4(-h)}{8{{(-h)}^{2}}} $ $ =\underset{h\to 0}{\mathop{\lim }},\frac{1-\cos 4h}{8h^{2}} $ $ =1 $ $ f(0)=1\Rightarrow k=1 $ .
BETA
