Question
The function $\text{f(x)}=\begin{cases}\frac{\text{x}^2}{\text{a}},&0\leq\text{x}<1\\\text{a},&1\leq\text{x}<\sqrt{2}\\\frac{2\text{b}^2-4\text{b}}{\text{x}^2},&\sqrt{2}\leq\text{x}<\infty\end{cases}$ is continuous for $0\leq\text{x}<\infty,$ then the most suitable values of a and b are:
- $\text{a}=1,\text{ b}=-1$
- $\text{a}=-1,\text{ b}=1+\sqrt{2}$
- $\text{a}=-1,\text{ b}=1$
- $\text{None os these}.$