Question

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

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

Since, $f$ is continuous for $0\le x<\infty ,f$ is continuous at $x=1$.

$\underset{x\to 1^{-}}{lim}\frac{x^2}{a}=a$

$\therefore a=\pm 1$

Since, $f$ is continuous at $x=\sqrt{2}$.

$\underset{x\to \sqrt{2}}{lim}\frac{2b^2-4b}{x^2}=a$

$\therefore \frac{2b^2-4b}{2}=a$

$\Rightarrow b^2-2b=a$

When $a=1,b^2-2b=1$

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

When $a=-1,b^2-2b=-1$

$\Rightarrow (b-1{)}^2=0\Rightarrow b=1$

Hence, $a=-1$ and $b=1$ are most suitable values.