Question

The function $f\left(x\right)=\left\{\begin{array}{cc}{x}^2/a,& \mathrm{if}0\le x<1\\ a,& \mathrm{if}1\le x<\sqrt{2}\\ \frac{2b^2-4b}{x^2},& \mathrm{if}\sqrt{2}\le x<\infty \end{array}\right.$
is continuous on (0, ∞), then find the most suitable values of a and b.

Answer 1

Given: f is continuous on $\left(0,\infty \right)$

∴ f is continuous at x = 1 and $\sqrt{2}$


At x = 1, we have

$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1-h\right)=\underset{h\to 0}{\lim }\left[\frac{{\left(1-h\right)}^2}{a}\right]=\frac{1}{a}$

$\underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(1+h\right)=\underset{h\to 0}{\lim }\left(a\right)=a$

Also,

At x = $\sqrt{2}$, we have

$\underset{x\to {\sqrt{2}}^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(\sqrt{2}-h\right)=\underset{h\to 0}{\lim }\left(a\right)=a$

$\underset{x\to {\sqrt{2}}^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(\sqrt{2}+h\right)=\underset{h\to 0}{\lim }\left[\frac{2b^2-4b}{{\left(\sqrt{2}+h\right)}^2}\right]=\frac{2b^2-4b}{2}=b^2-2b$

f is continuous at x = 1 and $\sqrt{2}$

∴ $\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)\mathrm{and}\underset{x\to {\sqrt{2}}^{-}}{\lim }f\left(x\right)=\underset{x\to {\sqrt{2}}^{+}}{\lim }f\left(x\right)$

$$\begin{gathered} \Rightarrow \frac{1}{a}=a\mathrm{and}{b}^2-2b=a \\ \Rightarrow a^2=1\mathrm{and}{b}^2-2b=a \\ \Rightarrow a=\pm 1\mathrm{and}{b}^2-2b=a...\left(1\right) \end{gathered}$$

If a = 1, then

$$\begin{gathered} b^2-2b=1\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow b^2-2b-1=0 \\ \Rightarrow b=\frac{2\pm \sqrt{4+4}}{2}=\frac{2\pm 2\sqrt{2}}{2}=1\pm \sqrt{2} \end{gathered}$$

If a = −1, then

$$\begin{gathered} b^2-2b=-1\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right] \\ \Rightarrow b^2-2b+1=0 \\ \Rightarrow {\left(b-1\right)}^2=0 \\ \Rightarrow b=1 \end{gathered}$$

Hence, the most suitable values of a and b are

a = −1, b = 1 or a = 1, $b=1\pm \sqrt{2}$