Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{\sqrt{1+kx}-\sqrt{1-x}}{x}& for-l\le x<0\\ 2x^2+3x-2& for0\le x\le 1\end{array}$ is continuous at $x=0$ then $k$ is:

• A. $-4$
• B. $-3$
• C. $-5$
• D. $-1$

Answer 1

Answer: (C)

$-5$

Given, $f\left(x\right)$ is continuous at $x=0$

$\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$

Here $f\left(0\right)=-2$

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

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

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

$=\frac{(k+1)}{2}$

Since, $f\left(x\right)$ is continuous at $x=0$

So, $LHL=f\left(0\right)$

$\Rightarrow \frac{k+1}{2}=-2$

$\Rightarrow k=-5$

Hence option ${}^{\prime }{C}^{\prime }$ is the answer.