Question

If the function f(x) = ${\displaystyle \frac{2x-{\sin }^{-1}x}{2x+{\tan }^{-1}x}}$, (x ≠ 0) is continuous at each point of its domain, then the value of f(0) is ______.

पर्याय

• 2

• ${\displaystyle \frac{1}{3}}$

• ${\displaystyle \frac{2}{3}}$

• ${\displaystyle -\frac{1}{3}}$

Answer 1

If the function f(x) = ${\displaystyle \frac{2x-{\sin }^{-1}x}{2x+{\tan }^{-1}x}}$, (x ≠ 0) is continuous at each point of its domain, then the value of f(0) is ${\displaystyle \frac{1}{3}}$.

Explanation:

Since, f(x) is continuous at each point of its domain.

∴ it is continuous at x = 0.

∴ f(0) = ${\displaystyle \underset{x\to 0}{lim}\text{f}\left(x\right)}$

= ${\displaystyle \underset{x\to 0}{lim}\left(\frac{2x-{\sin }^{-1}x}{2x+{\tan }^{-1}x}\right)}$

Applying L'Hospital rule on R.H.S., we get

f(0) = ${\displaystyle \underset{x\to 0}{lim}\frac{(2-\frac{1}{\sqrt{1-x^2}})}{(2+\frac{1}{1+x^2})}}$

= ${\displaystyle \frac{2-1}{2+1}}$

= ${\displaystyle \frac{1}{3}}$