The function $f(x)=\left\{\begin{array}{cc}x^{2} / a, & \text { if } 0 \leq x<1 \\ a, & \text { if } 1 \leq x<\sqrt{2} \\ \frac{2 b^{2}-4 b}{x^{2}}, & \text { if } \sqrt{2} \leq x<\infty\end{array}\right.$
is continuous on $(0, \infty)$, then find the most suitable values of $a$ and $b$.
Given: $f$ is continuous on $(0, \infty)$
$\therefore f$ is continuous at $x=1$ and $\sqrt{2}$
At $x=1$, we have
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0}\left[\frac{(1-h)^{2}}{a}\right]=\frac{1}{a}$
$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{h \rightarrow 0} f(1+h)=\lim _{h \rightarrow 0}(a)=a$
Also,
At $x=\sqrt{2}$, we have
$\lim _{x \rightarrow \sqrt{2}^{-}} f(x)=\lim _{h \rightarrow 0} f(\sqrt{2}-h)=\lim _{h \rightarrow 0}(a)=a$
$\lim _{x \rightarrow \sqrt{2}^{+}} f(x)=\lim _{h \rightarrow 0} f(\sqrt{2}+h)=\lim _{h \rightarrow 0}\left[\frac{2 b^{2}-4 b}{(\sqrt{2}+h)^{2}}\right]=\frac{2 b^{2}-4 b}{2}=b^{2}-2 b$
$f$ is continuous at $x=1$ and $\sqrt{2}$
$\therefore \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \quad$ and $\quad \lim _{x \rightarrow \sqrt{2}^{-}} f(x)=\lim _{x \rightarrow \sqrt{2}^{+}} f(x)$
$\Rightarrow \frac{1}{a}=a$ and $b^{2}-2 b=a$
$\Rightarrow a^{2}=1$ and $b^{2}-2 b=a$
$\Rightarrow a=\pm 1$ and $b^{2}-2 b=a$ .....(1)
If $a=1$, then
$b^{2}-2 b=1 \quad[$ From eq. (1) $]$
$\Rightarrow b^{2}-2 b-1=0$
$\Rightarrow b=\frac{2 \pm \sqrt{4+4}}{2}=\frac{2 \pm 2 \sqrt{2}}{2}=1 \pm \sqrt{2}$
If $a=-1$, then
$b^{2}-2 b=-1 \quad$ [From eq. (1)]
$\Rightarrow b^{2}-2 b+1=0$
$\Rightarrow(b-1)^{2}=0$
$\Rightarrow b=1$
Hence, the most suitable values of a and b are
$a=-1, b=1$ or $a=1, b=1 \pm \sqrt{2}$