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If the Function F ( X ) = 2 X − Sin − 1 X 2 X + Tan − 1 X is Continuous at Each Point of Its Domain, Then the Value of F (0) is - Mathematics

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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 

विकल्प

  • 2

  • \[\frac{1}{3}\] 

  • \[- \frac{1}{3}\]  

  • \[\frac{2}{3}\]

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उत्तर

\[\frac{1}{3}\] 

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

If f(x) is continuous at x = 0, then

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

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अध्याय 9: Continuity - Exercise 9.4 [पृष्ठ ४५]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 9 Continuity
Exercise 9.4 | Q 26 | पृष्ठ ४५

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