Question

$
   {\text{The function }}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\}{\text{ is continuous on}}\left[ {0,\infty } \right). \\
  {\text{Find the most suitable values of }}a{\text{ and }}b. \\
$

Answer 1

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.