Question

Let \(a,b\in\mathbf{R}\), \((a\neq0)\). If the function \(f\) defined as
\(f(x)=\begin{cases}\frac{2x^2}{a}, & 0\leq x<1\\ a, & 1\leq x<\sqrt{2}\\ \frac{2b^2-4b}{x^3}, & \sqrt{2}\leq x<\infty\end{cases}\)
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. \((-\sqrt{2},\,1-\sqrt{3})\)

Hint:

Impose continuity at the break points \(x=1\) and \(x=\sqrt{2}\).

Answer 1

The Correct Option is A

At \(x=1\): \(\lim_{x\to1^-}\frac{2x^2}{a}=\frac{2}{a}\) must equal \(f(1)=a\), so \(a^2=2\).
At \(x=\sqrt{2}\): \(\lim_{x\to\sqrt2^-}f(x)=a\) must equal \(f(\sqrt{2})=\frac{2b^2-4b}{2\sqrt{2}}\), so \(2\sqrt{2}\,a=2b^2-4b\), i.e. \(b^2-2b=\sqrt{2}\,a\).
If \(a=-\sqrt{2}\): \(b^2-2b+2=0\) has no real root. So \(a=\sqrt{2}\) and \(b^2-2b-2=0\Rightarrow b=1\pm\sqrt{3}\).
Among the options, \((\sqrt{2},1-\sqrt{3})\) fits.
Answer: (1) \((\sqrt{2},\,1-\sqrt{3})\)