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
(a) 2

(b) $\frac{1}{3}$

(c) $-\frac{1}{3}$

(d) $\frac{2}{3}$

Answer 1

(b) $\frac{1}{3}$

Given: $f\left(x\right)=\frac{2x-{\sin }^{-1}x}{2x+{\tan }^{-1}x}$

If f(x) is continuous at x = 0, then

​$$\begin{gathered} \underset{x\to 0}{\lim }f\left(x\right)=f\left(0\right) \\ \Rightarrow \underset{x\to 0}{\lim }\frac{2x-{\sin }^{-1}x}{2x+{\tan }^{-1}x}=f\left(0\right) \\ \Rightarrow \underset{x\to 0}{\lim }\frac{x\left(2-{\displaystyle \frac{{\sin }^{-1}x}{x}}\right)}{x\left(2+{\displaystyle \frac{{\tan }^{-1}x}{x}}\right)}=f\left(0\right) \\ \Rightarrow \underset{x\to 0}{\lim }\frac{\left(2-{\displaystyle \frac{{\sin }^{-1}x}{x}}\right)}{\left(2+{\displaystyle \frac{{\tan }^{-1}x}{x}}\right)}=f\left(0\right) \\ \Rightarrow \frac{2-\underset{x\to 0}{\lim }\left({\displaystyle \frac{{\sin }^{-1}x}{x}}\right)}{2+\underset{x\to 0}{\lim }\left({\displaystyle \frac{{\tan }^{-1}x}{x}}\right)}=f\left(0\right) \\ \Rightarrow \frac{2-1}{2+1}=f\left(0\right) \\ \Rightarrow f\left(0\right)=\frac{1}{3} \end{gathered}$$