# The function

Question:

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

Solution:

(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 $b=1$