Question

If $f(x)=\{\begin{array}{cc}\frac{2x-1}{\sqrt{1+x}-1},& -1\le x<\infty ,x\ne 0\\ k,& x=0\end{array}$ 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)

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