Question

Let $a,bϵR,(a\ne 0)$. If the function f defined as $\{\begin{array}{cc}\frac{2x^2}{a},& 0\le x<1\\ a,& 1\le x<\sqrt{2}\\ \frac{2b^2-4b}{x^3},& \sqrt{2}\le x<\infty \end{array}$ is continuous in the interval $[0,\infty )$, then an ordered pair $(a,b)$ is:

• A. $(-\sqrt{2},1-\sqrt{3})$
• B. $(\sqrt{2},-1+\sqrt{3})$
• C. $(\sqrt{2},1-\sqrt{3})$
• D. $(-\sqrt{2},1+\sqrt{3})$

Answer 1

Answer: (C)

$(\sqrt{2},1-\sqrt{3})$

Continuity at $x=1$

$\frac{2}{a}=a\Rightarrow a=\pm \sqrt{2}$

Continuity at $x=\sqrt{2}$

$a=\frac{2b^2-4b}{2\sqrt{2}}$

Case 1) Put $a=-\sqrt{2},$ we get $b$ as complex number.

Case 2) Put $a=\sqrt{2}$

$2=b^2-2b\Rightarrow b^2-2b-2=0$
$b=\frac{1\pm \sqrt{4+4.2}}{2}=1\pm \sqrt{3}$
$(\sqrt{2},1-\sqrt{3})$