Question

If the function  \[f\left( x \right) = \frac{2x - \sin^{- 1} x}{2x + \tan^{- 1} x}\] is continuous at each point of its domain, then the value of f (0) is 

पर्याय

• 2

• \[\frac{1}{3}\] 

• \[- \frac{1}{3}\]  

• \[\frac{2}{3}\]

Answer 1

\[\frac{1}{3}\] 

Given: 

\[f\left( x \right) = \frac{2x - \sin^{- 1} x}{2x + \tan^{- 1} x}\]

If f(x) is continuous at x = 0, then

\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0} \frac{2x - \sin^{- 1} x}{2x + \tan^{- 1} x} = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0} \frac{x\left( 2 - \frac{\sin^{- 1} x}{x} \right)}{x\left( 2 + \frac{\tan^{- 1} x}{x} \right)} = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0} \frac{\left( 2 - \frac{\sin^{- 1} x}{x} \right)}{\left( 2 + \frac{\tan^{- 1} x}{x} \right)} = f\left( 0 \right)\]
\[ \Rightarrow \frac{2 - \lim_{x \to 0} \left( \frac{\sin^{- 1} x}{x} \right)}{2 + \lim_{x \to 0} \left( \frac{\tan^{- 1} x}{x} \right)} = f\left( 0 \right)\]
\[ \Rightarrow \frac{2 - 1}{2 + 1} = f\left( 0 \right)\]
\[ \Rightarrow f\left( 0 \right) = \frac{1}{3}\]