Question

The value of $f\left(0\right)$ so that the function $f\left(x\right)=\frac{2x-{\sin }^{-1}x}{2x+{\tan }^{-1}x}$ is continuous at each point on its domain is-

• A. $2$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. $-\frac{1}{3}$

Answer 1

The correct option is B $\frac{1}{3}$

Given $f\left(x\right)=f\left(x\right)=\frac{2x-{\sin }^{-1}x}{2x+{\tan }^{-1}x}$
Since, $f\left(x\right)$ is continuous at each point its domain, so it is continuous at $x=0$ also.
$LHL=f\left(0\right)=RHL$
We have
${\sin }^{-1}x=x+\frac{1^2{x}^3}{3!}+\frac{1^2.3^2{x}^5}{5!}+\frac{1^2.3^2{5^{2}x}^7}{7!}+.....$
${\tan }^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}+.....$
$RHL=\underset{x\to }{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0+h)$
$=\underset{h\to 0}{lim}\frac{2h-{\sin }^{-1}h}{2h+{\tan }^{-1}h}$
$=\underset{h\to 0}{lim}\frac{2h-h-\frac{1^2{h}^3}{3!}-\frac{1^2.3^2{x}^5}{5!}-....}{2h+h+\frac{h^3}{3}+\frac{h^5}{5}+....}$
$=\underset{h\to 0}{lim}\left(\frac{h-\frac{1^2{h}^3}{3!}+\frac{1^2.3^2{x}^5}{5!}}{3h+\frac{h^3}{3}+\frac{h^5}{5}}\right)$
$\underset{h\to 0}{lim}\frac{h}{h}\left(\frac{1-hhigherterm}{3+hhigherterm}\right)$
$=\frac{1}{3}$
Hence, $f\left(0\right)=\frac{1}{3}$