Question

If $ f(x)=\frac{\log (1+ax)-\log (1-bx)}{x} $ for $ x\ne 0 $ and $ f(0)=k $ and $ f(x) $ is continuous at $ x=0 $ then k is equal to:

• A. $ a+b $
• B. $ a-b $
• C. $ a $
• D. $ b $
• E. $ b-a $

Answer 1

Answer: (A)

$ f(x)=\frac{\log (1+ax)-\log (1-bx)}{x} $ $ f(x) $ is continuous at $ x=k $ and $ f(0)=k $ . $ \underset{x\to 0}{\mathop{\lim }}\,f(x)=\underset{x\to 0}{\mathop{\lim }}\,\frac{\log (1+ax)-\log (1-bx)}{x} $ $ \left( \frac{0}{0}form \right) $ $ =\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1}{1+ax}.a+\frac{b}{1-bx} \right) $ $ =\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{a}{1+ax}+\frac{b}{1-bx} \right)=a+b $ $ \therefore $ $ a+b=f(0)=k $