Question

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

Answer 1

Given:  f is continuous on  $$(0,\mathrm{\infty })$$

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

At x = 1, we have

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

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

Also, 

At x = $$\sqrt{2}$$,

we have

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

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

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

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

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

If a = 1, then

$$b^2-2b=1[\text{ From eq }.(1)]$$
$$\Rightarrow b^2-2b-1=0$$
$$\Rightarrow b=\frac{2\pm \sqrt{4+4}}{2}=\frac{2\pm 2\sqrt{2}}{2}=1\pm \sqrt{2}$$

If a = −1, then

$$b^2-2b=-1[\text{ From eq }.(1)]$$
$$\Rightarrow b^2-2b+1=0$$
$$\Rightarrow {(b-1)}^2=0$$
$$\Rightarrow b=1$$

Hence, the most suitable values of a and b are

a = −1, b = 1  or a = 1,

$$b=1\pm \sqrt{2}$$