The function $f(x)= \begin{cases}x^2 / a & , 0 \leq x<1 \\ a & , 1 \leq x<\sqrt{2} \\ \frac{2 b^2-4 b}{x^2}, & \sqrt{2} \leq x<\infty\end{cases}$ is continuous for $0 \leq x<\infty$, then the most suitable values of a and b are
Answer & explanation
Correct answer: option 3
It is given that f(x) is continuous for $0 \leq x<\infty$. So, it is continuous at x = 1 and $x=\sqrt{2}$.
∴ $\lim\limits_{x \rightarrow 1^{-}} f(x)=\lim\limits_{x \rightarrow 1^{+}} f(x)=f(1)$
and, $\lim\limits_{x \rightarrow(\sqrt{2})} f(x)=\lim\limits_{x \rightarrow(\sqrt{2})^{+}} f(x)=f(\sqrt{2})$
$\Rightarrow \lim\limits_{x \rightarrow 1} \frac{x^2}{a}=\lim\limits_{x \rightarrow 1} a=a$
and, $\lim\limits_{x \rightarrow(\sqrt{2})^{-}} a=\lim\limits_{x \rightarrow(\sqrt{2})^{+}} \frac{2 b^2-4 b}{x^2}=\frac{2 b^2-4 b}{2}$
$\Rightarrow \frac{1}{a}=a$ and $ a=\frac{2 b^2-4 b}{2}$
$\Rightarrow a= \pm 1$ and $b^2-2 b=a$
Now, a = 1 and $b^2-2 b=a$
⇒ a = 1, b = $1 \pm \sqrt{2}$
⇒ a = -1 and $b^2-2 b=a$
⇒ a = -1, b = 1