Question

The function $f\left(x\right)=\{\begin{array}{c}{x}^2/a,0\le x<1\\ a,1\le x<\sqrt{2}\\ \frac{2b^2-4b}{x^2},\sqrt{2}\le x<\infty \end{array}$
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 the above

Answer 1

The correct option is C $a=-1,b=1$

$f\left(x\right)$ is continuous for $0\le x<\infty$
$\underset{x\to 1^{-}}{lim}\frac{x^2}{a}=\underset{x\to 1^{+}}{lim}a=\underset{x\to {\sqrt{2}}^{-}}{lim}\frac{2b^2-4b}{x^2}$
$\Rightarrow \frac{1}{a}=a=b^2-2b$
So,
$\frac{1}{a}=a\Rightarrow a=\pm 1$
$b^2-2b=a$
For $a=1$,
$b^2-2b=1\Rightarrow b=1\pm \sqrt{2}$
For $a=-1$,
$b^2-2b=-1\Rightarrow b=1$