1. Tardigrade
  2. Question
  3. Mathematics
  4. Let a, b ∈ R, (a ≠ 0). If the function f defined as f(x) = begincases (2x2/a) , 0 ≤ x < 1 a &, 1≤ x < √2 (2b2 - 4b/x3) , √2 ≤ x < ∞ endcases is continuous in the interval [0, ∞), then an ordered pair (a, b) is :

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Q. Let $a, b \in \ R, (a \neq 0)$. If the function f defined as
$ f(x) = \begin{cases} \frac{2x^2}{a} & , \quad 0 \leq x < 1\\ a &, \quad 1\leq x < \sqrt{2} \\ \frac{2b^2 - 4b}{x^3} & , \quad \sqrt{2} \leq x < \infty \end{cases} $
is continuous in the interval $[0, \infty)$, then an ordered pair $(a, b)$ is :

JEE MainJEE Main 2016Continuity and Differentiability

Solution:

$f(x)=\begin{cases}\frac{2 x^{2}}{a} & 0 \leq x<1 \\ a & 1 \leq x<\sqrt{2} \\ \frac{2 b^{2}-4 b}{x^{3}} & \sqrt{2} \leq x<\infty\end{cases}$
$\Rightarrow $ continuous at $x=1$ and $x=\sqrt{2}$
$\displaystyle\lim_{x \rightarrow T} f(x)=\displaystyle\lim_{x \rightarrow 1^{+}} f(x)=f(1)$
$\Rightarrow \frac{2}{a}=a \Rightarrow a^{2}=2\,\,\,\,\,\,,-----(1)$
and $\displaystyle\lim_{x \rightarrow \sqrt{2}} f(x)=\displaystyle\lim_{x \rightarrow \sqrt{2}^{+}} f(x)=f(\sqrt{2})$
$\Rightarrow a=\frac{2 b^{2}-4 b}{2 \sqrt{2}}$
$\Rightarrow b^{2}-2 b=\sqrt{2} a$
If $a=\sqrt{2}$ then $b^{2}-2 b-2=0$
$\Rightarrow b=1 \pm \sqrt{3}$
If $a=-\sqrt{2}$ then $b^{2}-2 b+2=0$
$ \Rightarrow b$ is imaginary which is not possible
$\Rightarrow (a, b)=(\sqrt{2}, 1+\sqrt{3})$ or $(\sqrt{2}, 1-\sqrt{3})$