The function $f(x)=\left\{\begin{array}{cc}x^{2} / a & , \quad 0 \leq x<1 \\ a & , \quad 1 \leq x<\sqrt{2} \\ \frac{2 b^{2}-4 b}{r^{2}}, & \sqrt{2} \leq x<\infty\end{array}\right.$
is continuous for $0 \leq x<\infty$, then the most suitable values of $a$ and $b$ are
(a) $a=1, b=-1$
(b) $a=-1, b=1+\sqrt{2}$
(c) $a=-1, b=1$
(d) none of these
(c) $a=-1, b=1$
Given: $f(x)$ is continuous for $0 \leq x<\infty$.
This means that $f(x)$ is continuous for $x=1, \sqrt{2}$.
Now,
If $f(x)$ is continuous at $x=1$, then
$\lim _{x \rightarrow 1^{-}} f(x)=f(1)$
$\Rightarrow \lim _{h \rightarrow 0} f(1-h)=a$
$\Rightarrow \frac{(1-h)^{2}}{a}=a$
$\Rightarrow \frac{1}{a}=a$
$\Rightarrow a^{2}=1$
$\Rightarrow a=\pm 1$
If $f(x)$ is continuous at $x=\sqrt{2}$, then
$\lim _{x \rightarrow \sqrt{2}^{-}} f(x)=f(\sqrt{2})$
$\Rightarrow \lim _{h \rightarrow 0} f(\sqrt{2}-h)=\frac{2 b^{2}-4 b}{2}$
$\Rightarrow \lim _{h \rightarrow 0} a=b^{2}-2 b$
$\Rightarrow a=b^{2}-2 b$
$\Rightarrow b^{2}-2 b-a=0$
$\therefore$ For $a=1$, we have
$b^{2}-2 b-1=0$
$\Rightarrow b=\frac{2 \pm \sqrt{4-4(-1)}}{2}=1 \pm \sqrt{2}$
Also,
For $a=-1$, we have
$b^{2}-2 b+1=0$
$\Rightarrow(b-1)^{2}=0$
$\Rightarrow b=1$
Thus, $a=-1$ and $b=1$
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