Question

Let $f\left(x\right)=\{\begin{array}{c}\frac{x^2}{a};0\le x<1\\ a;1\le x<\sqrt{2}\\ \frac{2b^2-4b}{x^2};\sqrt{2}\le x<\infty \end{array}$
If $f\left(x\right)$ is continuous for $0\le x<\infty$, 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

Answer: (C)

$a=-1,b=1$

Given $f\left(x\right)=\{\begin{array}{c}\frac{x^2}{a}0\le x<1\\ a,1\le x<\sqrt{2}\\ \frac{2b^2-4b}{x^2},\sqrt{2}\le x<\infty \end{array}$
Since, $f\left(x\right)$ is continuous in $0\le x<\infty$. So $f\left(x\right)$ is continuous at $x=1$
$f\left(1\right)=LHL=RHL$
Here, $f\left(1\right)=a$
$LHL={lim}_{x\to 1^{-}}f\left(x\right)={lim}_{h\to 0}f(1-h)$
$={lim}_{h\to 0}\frac{(1-h{)}^2}{a}=\frac{1}{a}$
$\therefore$ $\frac{1}{a}=a\Rightarrow a^2=1$
$\Rightarrow$ $a=\pm 1$
Also, $f\left(x\right)$ is continuous at $x=\sqrt{2}$
$LHL={lim}_{x\to {\sqrt{2}}^{-}}f\left(x\right)$
$\Rightarrow LHL=a=\pm 1$
$f\left(\sqrt{2}\right)=\frac{2b^2-4b}{2}=b^2-2b$
For $a=-1$
$-1=b^2-2b$
$b^2-2b+1=0$
$b=1$
For $a=1$
$b^2-2b-1=0$
$\Rightarrow (b-1{)}^2=2$
$\Rightarrow b=1\pm \sqrt{2}$