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प्रश्न
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
\[\frac{1}{3}\]
\[- \frac{1}{3}\]
\[\frac{2}{3}\]
उत्तर
\[\frac{1}{3}\]
If f(x) is continuous at x = 0, then
\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0} \frac{2x - \sin^{- 1} x}{2x + \tan^{- 1} x} = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0} \frac{x\left( 2 - \frac{\sin^{- 1} x}{x} \right)}{x\left( 2 + \frac{\tan^{- 1} x}{x} \right)} = f\left( 0 \right)\]
\[ \Rightarrow \lim_{x \to 0} \frac{\left( 2 - \frac{\sin^{- 1} x}{x} \right)}{\left( 2 + \frac{\tan^{- 1} x}{x} \right)} = f\left( 0 \right)\]
\[ \Rightarrow \frac{2 - \lim_{x \to 0} \left( \frac{\sin^{- 1} x}{x} \right)}{2 + \lim_{x \to 0} \left( \frac{\tan^{- 1} x}{x} \right)} = f\left( 0 \right)\]
\[ \Rightarrow \frac{2 - 1}{2 + 1} = f\left( 0 \right)\]
\[ \Rightarrow f\left( 0 \right) = \frac{1}{3}\]
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