Question

If  \[f\left( x \right) = \begin{cases}\frac{\log\left( 1 + ax \right) - \log\left( 1 - bx \right)}{x}, & x \neq 0 \\ k , & x = 0\end{cases}\] and f (x) is continuous at x = 0, then the value of k is

पर्याय

• a − b 

• a + b

• log a + log b 

• none of these

Answer 1

\[a + b\]

 Given:

\[f\left( x \right) = \binom{\frac{\log\left( 1 + ax \right) - \log\left( 1 - bx \right)}{x}, x \neq 0}{k, x = 0}\]

If f(x) is continuous at x = 0, then

\[\lim_{x \to 0} f\left( x \right) = f\left( 0 \right)\]

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

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

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

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

\[ \Rightarrow a \times 1 + b \times 1 = k \left[ \because \lim_{x \to 0} \frac{\log\left( 1 + x \right)}{x} = 1 \right]\]

\[ \Rightarrow k = a + b\]