MCQ
Let $a,b \in R,\left( {a \ne 0} \right)$. if the function $f$ defined as$f\left( x \right)\left\{ \begin{array}{l} \frac{{2{x^2}}}{a}\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,0 \le x < 1\,\,\,\\ a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,1 \le x < \sqrt 2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ \frac{{2{b^2} - 4b}}{{{x^3}}}\,\,\,,\,\,\,\,\,\sqrt 2  \le x < \infty  \end{array} \right.,$  is continuous in the interval $\left[ {0,\infty } \right]$ , then an ordered pair $(a, b)$ is
  • A
    $\left( { - \sqrt 2 ,1 - \sqrt 3 } \right)$
  • B
    $\left( {\sqrt 2 , - 1 + \sqrt 3 } \right)$
  • C
    $\left( {\sqrt 2 ,1 - \sqrt 3 } \right)$
  • D
    $\left( { - \sqrt 2 ,1 + \sqrt 3 } \right)$

Answer

Continuity at $x=1$$\frac{2}{a} = a $
$\Rightarrow a = \pm \sqrt 2 $
Continuity at $x = \sqrt 2 \,a = \sqrt 2 $
$a = \frac{{2{b^2} - 4b}}{{2\sqrt 2 }}$
Put $a = \sqrt 2 $
$2 = {b^2} - 2b\,\,\,\,\,\,\,\,$
$ \Rightarrow {b^2} - 2b - 2 = 0$
$b = \frac{{2 \pm \sqrt {4 + 4.2} }}{2} = 1 \pm \sqrt 3 $
So, $\left( {a,b} \right) = \left( {\sqrt 2 ,1 - \sqrt 3 } \right)$

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