# The function $f(x)=\left\{\begin{array}{1 1}\large\frac{x^2}{a}&0\leq x <1\\a&1\leq x < \sqrt 2\\\large\frac{2b^2-4b}{x^2}&\sqrt 2\leq x < \infty\end{array}\right.$ is continuous for $0\leq x < \infty$ then the most suitable values of a and b are
$\begin{array}{1 1}(a)\;a=1,b=-1&(b)\;a=-1,b=1+\sqrt 2\\(c)\;a=-1,b=1&(d)\;\text{None of these}\end{array}$