Question

If $f:[-2,2]\to R$ is defined by
$f\left(x\right)=\{\begin{array}{cccc}\frac{\sqrt{1+cx}-\sqrt{1-cx}}{x}& ,& for& -2\le x<0\\ \frac{x+3}{x+1}& ,& for& 0\le x\le 2\end{array}$ is continuous on $[-2,2]$ then $c=$

• A. $3$
• B. $\frac{3}{2}$
• C. $\frac{3}{\sqrt{2}}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

The correct option is A $3$

Since the function is continuous on $[-2,2]$, then it is also continuous at $x=0$.

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

or, $\underset{x\to 0}{lim}\frac{\sqrt{1+cx}-\sqrt{1-cx}}{x}=3$ [Since $f\left(0\right)=3$]

or, $\underset{x\to 0}{lim}\frac{2c}{\sqrt{1+cx}+\sqrt{1-cx}}=3$ ........ rationalisation

or, $\frac{2c}{2}=3$

or, $c=3$.