Question

If $f\left(x\right)=\{\begin{array}{ccc}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}& for& -1\le x<0\\ 2x^2+3x-2& for& 0\le x<1\end{array}$ is continuous at $x=0$, then k$=$?

• A. $-4$
• B. $-3$
• C. $-2$
• D. $-1$

Answer 1

The correct option is C $-2$

$\underset{x\to 0}{lim}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}$

$\underset{x\to 0}{lim}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\times \frac{(\sqrt{1+kx}+\sqrt{1-kx})}{(\sqrt{1+kx}+\sqrt{1-kx})}$

$\underset{x\to 0}{lim}\frac{1+kx-(1-kx)}{x(\sqrt{1+kx}+\sqrt{1-kx})}$

$\underset{x\to 0}{lim}\frac{2kx}{x(\sqrt{1+kx}+\sqrt{1-kx})}$

$=\frac{2x}{1+1}=\frac{2k}{2}=k$

$\underset{x\to 0}{lim}2x^2+3x-2=0-2=-2$

$k=-2$