Functions Question 535
Question: In the function $ f(x)=\frac{2x-{{\sin }^{-1}}x}{2x+{{\tan }^{-1}}x},\ (x\ne 0) $ is continuous at each point of its domain, then the value of $ f(0) $ is
[RPET 2000]
Options:
A) 2
B) $ 1/3 $
C) $ 2/3 $
D) $ -1/3 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)=\underset{x\to 0}{\mathop{\lim }}( \frac{2x-{{\sin }^{-1}}x}{2x+{{\tan }^{-1}}x} )=f(0) $ , $ ( \frac{0}{0} ) $ Applying L-Hospital?s rule, $ f(0)=\underset{x\to 0}{\mathop{\lim }},\frac{( 2-\frac{1}{\sqrt{1-x^{2}}} )}{( 2+\frac{1}{1+x^{2}} )} $ $ =\frac{2-1}{2+1}=\frac{1}{3} $ Trick : f (0) = $ \underset{x\to 0}{\mathop{\lim }},\frac{2-\frac{{{\sin }^{-1}}x}{x}}{2+\frac{{{\tan }^{-1}}x}{x}}=\frac{2-1}{2+1}=\frac{1}{3} $ .
BETA
