Question

The function $f\left(x\right)=\left\{\begin{array}{cc}{x}^2/a,& 0\le x<1\\ a,& 1\le x<\sqrt{2}\\ \frac{2b^2-4b}{x^2},& \sqrt{2}\le x<\infty \end{array}\right.$
is continuous for 0 ≤ x < ∞, then the most suitable values of a and b are
(a) a = 1, b = −1
(b) a = −1, b = 1 + $\sqrt{2}$
(c) a = −1, b = 1
(d) none of these

Answer 1

(c) a = -1, b = 1

Given: $f\left(x\right)$ is continuous for 0 ≤ x < ∞.

This means that $f\left(x\right)$ is continuous for $x=1,\sqrt{2}$.

Now,

If $f\left(x\right)$ is continuous at x = 1, then
$$\begin{gathered} \underset{x\to 1^{-}}{\lim }f\left(x\right)=f\left(1\right) \\ \Rightarrow \underset{h\to 0}{\lim }f\left(1-h\right)=a \\ \Rightarrow \frac{{\left(1-h\right)}^2}{a}=a \\ \Rightarrow \frac{1}{a}=a \\ \Rightarrow a^2=1 \\ \Rightarrow a=\pm 1 \end{gathered}$$

If $f\left(x\right)$ is continuous at x = $\sqrt{2}$, then​

$$\begin{gathered} \underset{x\to {\sqrt{2}}^{-}}{\lim }f\left(x\right)=f\left(\sqrt{2}\right) \\ \Rightarrow \underset{h\to 0}{\lim }f\left(\sqrt{2}-h\right)=\frac{2b^2-4b}{2} \\ \Rightarrow \underset{h\to 0}{\lim }a=b^2-2b \\ \Rightarrow a=b^2-2b \\ \Rightarrow b^2-2b-a=0 \end{gathered}$$

∴ For a = 1, we have

$$\begin{gathered} b^2-2b-1=0 \\ \Rightarrow b=\frac{2\pm \sqrt{4-4\left(-1\right)}}{2}=1\pm \sqrt{2} \end{gathered}$$

Also,
For a = −1, we have

$$\begin{gathered} b^2-2b+1=0 \\ \Rightarrow {\left(b-1\right)}^2=0 \\ \Rightarrow b=1 \end{gathered}$$

Thus, $a=-1\mathrm{and}b=1$