Question

If $ f(x)=\left\{ \begin{matrix} \frac{2x-1}{\sqrt{1+x}-1}, & -1\le x<\infty ,x\ne 0 \\ k, & x=0 \\ \end{matrix} \right. $ is continuous everywhere, then k is equal to :

• A. $ \frac{1}{2}\log 2 $
• B. $ \log 4 $
• C. $ \log 8 $
• D. $ \log 2 $
• E. none of these

Answer 1

Answer: (B)

$ \because $ $ f(x)=\left\{ \begin{matrix} \frac{{{2}^{x}}-1}{\sqrt{1+x-1}}, & x\ne 0 \\ k, & x=0 \\ \end{matrix} \right. $ $ \therefore \,\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{2}^{x}}-1}{\sqrt{1+x}-1}\,=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{2}^{x}}\,{{\log }_{e}}2}{\frac{1}{2\sqrt{1+x}}} $ (By LHospitals rule) $ =2{{\log }_{e}}2={{\log }_{e}}4 $ $ \because $ $ f(x) $ is continuous at $ k=0 $ $ \therefore $ $ \underset{x\to 0}{\mathop{\lim }}\,f(x)=f(0) $ $ \Rightarrow $ $ {{\log }_{e}}4=k $