Question

Let $a,b\in R,(a\ne 0)$. If the function f defined as
$f(x)=\{\begin{array}{ll}\frac{2x^2}{a}& ,0\le x<1\\ a& ,1\le x<\sqrt{2}\\ \frac{2b^2-4b}{x^3}& ,\sqrt{2}\le x<\infty \end{array}$
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: (A)

$f(x)=\{\begin{array}{ll}\frac{2x^2}{a}& 0\le x<1\\ a& 1\le x<\sqrt{2}\\ \frac{2b^2-4b}{x^3}& \sqrt{2}\le x<\infty \end{array}$
$\Rightarrow$ continuous at $x=1$ and $x=\sqrt{2}$
$\underset{x\to T}{\lim }f(x)=\underset{x\to 1^{+}}{\lim }f(x)=f(1)$
$\Rightarrow \frac{2}{a}=a\Rightarrow a^2=2,-----(1)$
and $\underset{x\to \sqrt{2}}{\lim }f(x)=\underset{x\to {\sqrt{2}}^{+}}{\lim }f(x)=f(\sqrt{2})$
$\Rightarrow a=\frac{2b^2-4b}{2\sqrt{2}}$
$\Rightarrow b^2-2b=\sqrt{2}a$
If $a=\sqrt{2}$ then $b^2-2b-2=0$
$\Rightarrow b=1\pm \sqrt{3}$
If $a=-\sqrt{2}$ then $b^2-2b+2=0$
$\Rightarrow b$ is imaginary which is not possible
$\Rightarrow (a,b)=(\sqrt{2},1+\sqrt{3})$ or $(\sqrt{2},1-\sqrt{3})$