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Let F ( X ) = Log ( 1 + X a ) − Log ( 1 − X B ) X X ≠ 0. Find the Value of F at X = 0 So that F Becomes Continuous at X = 0.

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प्रश्न

Let  \[f\left( x \right) = \frac{\log\left( 1 + \frac{x}{a} \right) - \log\left( 1 - \frac{x}{b} \right)}{x}\] x ≠ 0. Find the value of f at x = 0 so that f becomes continuous at x = 0.

 

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उत्तर

Given: 

\[f\left( x \right) = \frac{\log\left( 1 + \frac{x}{a} \right) - \log\left( 1 - \frac{x}{b} \right)}{x}, x \neq 0\]

If f(x) is continuous at x = 0, then

\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]
\[\Rightarrow \lim_{x \to 0} \left( \frac{\log\left( 1 + \frac{x}{a} \right) - \log\left( 1 - \frac{x}{b} \right)}{x} \right) = f\left( 0 \right)\]
\[\Rightarrow \lim_{x \to 0} \left( \frac{\log\left( 1 + \frac{x}{a} \right)}{\frac{ax}{a}} - \frac{\log\left( 1 - \frac{x}{b} \right)}{\frac{bx}{b}} \right) = f\left( 0 \right)\]
\[\Rightarrow \frac{1}{a} \lim_{x \to 0} \left( \frac{\log\left( 1 + \frac{x}{a} \right)}{\frac{x}{a}} \right) - \left( - \frac{1}{b} \right) \lim_{x \to 0} \left( \frac{\log\left( 1 - \frac{x}{b} \right)}{\frac{- x}{b}} \right) = f\left( 0 \right)\]

\[\Rightarrow \frac{1}{a} \times 1 - \left( - \frac{1}{b} \right) \times 1 = f\left( 0 \right) \left[ \text{ Using } : \lim_{x \to 0} \frac{\log\left( 1 + x \right)}{x} = 1 \right]\]
\[ \Rightarrow \frac{1}{a} + \frac{1}{b} = f\left( 0 \right)\]
\[ \Rightarrow \frac{a + b}{ab} = f\left( 0 \right)\]

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अध्याय 8: Continuity - Exercise 9.1 [पृष्ठ १९]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 8 Continuity
Exercise 9.1 | Q 30 | पृष्ठ १९

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