Hint: Let’s find out limits at all the critical points and then we’ll compare to find out the desired values.
Complete step by step answer:
\[
{\text{We have been given f}}(x){\text{ }} \\
f\left( x \right) = \left\{ \begin{gathered}
\dfrac{{{{\text{x}}^2}}}{a}{\text{ ,}}if{\text{ }}0 \leqslant x < 1 \\
a{\text{ ,}}if{\text{ }}1 \leqslant x < \sqrt 2 \\
\dfrac{{2{b^2} - 4b}}{{{{\text{x}}^2}}}{\text{ ,}}if{\text{ }}\sqrt 2 \leqslant x < \infty \\
\end{gathered} \right\} \\
\]
So,
At x = 1, we can say that $\mathop {\lim }\limits_{{x^ - } \to 1} f\left( x \right) = f\left( 1 \right) = \mathop {\lim }\limits_{{x^ + } \to 1} f\left( x \right)$
Taking left hand limit,
$\mathop {\lim }\limits_{{x^ - } \to 1} f\left( x \right) = \mathop {\lim }\limits_{{x^ - } \to 1} \dfrac{{{{\text{x}}^2}}}{a}$
$\Rightarrow \mathop {\lim }\limits_{{x^ - } \to 1} f\left( x \right) = \dfrac{1}{a}$
Taking right hand limit,
$\mathop {\lim }\limits_{{x^ + } \to 1} f\left( x \right) = \mathop {\lim }\limits_{{x^ + } \to 1} a$
$\Rightarrow \mathop {\lim }\limits_{{x^ + } \to 1} f\left( x \right) = a$
And $f\left( 1 \right) = a$
Comparing them you get,
$\Rightarrow a = \dfrac{1}{a}$
$\Rightarrow {a^2} = 1$
$\Rightarrow a = \pm 1$
Now at x = \sqrt 2 we can say that $\mathop {\lim }\limits_{{x^ - } \to \sqrt 2 } f\left( x \right) = f\left( {\sqrt 2 } \right) = \mathop {\lim }\limits_{{x^ + } \to \sqrt 2 } f\left( x \right)$
Taking left hand limit,
$\mathop {\lim }\limits_{{x^ - } \to \sqrt 2 } f\left( x \right) = \mathop {\lim }\limits_{{x^ - } \to \sqrt 2 } a$
$\Rightarrow \mathop {\lim }\limits_{{x^ - } \to \sqrt 2 } f\left( x \right) = a$
Taking right hand limit,
$\mathop {\lim }\limits_{{x^ + } \to \sqrt 2 } f\left( x \right) = \mathop {\lim }\limits_{{x^ + } \to \sqrt 2 } \dfrac{{2{b^2} - 4b}}{{{{\text{x}}^2}}}$
Now at
$f\left( {\sqrt 2 } \right) = {b^2} - 2b$
Comparing them you get,
For a = 1
${b^2} - 2b = 1$
$\Rightarrow {b^2} - 2b - 1 = 0$
Using $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$\Rightarrow b = \dfrac{{2 \pm \sqrt 8 }}{2} = 1 \pm \sqrt 2$
For a = - 1
${b^2} - 2b = - 1$
$\Rightarrow {b^2} - 2b + 1 = 0$
$\Rightarrow$ b = 1,1
So, we have $a = \pm 1,b = 1 \pm \sqrt 2 ,1$
NOTE: Be careful while calculating the value of b. Since we are using a quadratic formula, there will be 3 values for different values of a.