The function $f(x)= \begin{cases}\frac{x^2}{a} & , 0 \leq x<1 \\ a & , 1 \leq x<\sqrt{2} \\ \frac{2 b^2-4 b}{x^2} & , \sqrt{2} \leq x<\infty\end{cases}$ is continuous for $0 \leq x<\infty$. Then which of the following statements is correct?
(a) The number of all possible ordered pairs (a, b) is 3
(b) The number of all possible ordered pairs (a, b) is 4
(c) The product of all possible values of b is -1
(d) The product of all possible values of b is 1
Answer & explanation
Correct answer: option 1
It is given that f(x) is continuous on $[0, \infty)$. So, it is continuous at $x=1$ and $x=\sqrt{2}$.
∴ $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$
and, $\lim _{x \rightarrow \sqrt{2}^{-}} f(x)=\lim _{x \rightarrow \sqrt{2}^{+}} f(x)=f(\sqrt{2})$
$\Rightarrow \lim _{x \rightarrow 1^{-}} \frac{x^2}{a}=\lim _{x \rightarrow 1} a$ and $\lim _{x \rightarrow \sqrt{2}-} a=\lim _{x \rightarrow \sqrt{2}^{+}} \frac{2 b^2-4 b}{x^2}$
$\Rightarrow \frac{1}{a}=a$ and $a=\frac{2 b^2-4 b}{2}$
$\Rightarrow a^2=1$ and $b^2-2 b=a$
$\Rightarrow a= \pm 1$ and $b^2-2 b=a$
$\Rightarrow a=1, b^2-2 b-1=0$ or, $a=-1, b^2-2 b+1=0$
$\Rightarrow a=1, b=1 \pm \sqrt{2}$ or, $a=-1, b=1$
$\Rightarrow (a, b)=(1,1+\sqrt{2}),(1,1-\sqrt{2}),(-1,1)$