Let a,bR (a0).

Question

Let $a,b\in\mathbb{R}\ (a
e0)$. If the function $f$ defined as $f(x)=\begin{cases}\dfrac{2x^2}{a}, & 0\le x<1\a, & 1\le x<\sqrt2\\dfrac{2b^2-4b}{x^3}, & \sqrt2\le x<\infty\end{cases}$ is continuous in $[0,\infty)$, then an ordered pair $(a,b)$ is:

Accepted Answer

Continuity at $x=1$: $\dfrac2a=a\Rightarrow a=\pm\sqrt2$.

Continuity at $x=\sqrt2$: $a=\dfrac{2b^2-4b}{2\sqrt2}$. With $a=\sqrt2$: $2=b^2-2b\Rightarrow b^2-2b-2=0\Rightarrow b=1\pm\sqrt3$.

So $(a,b)=(\sqrt2,1-\sqrt3)$.

Solutions · 1