Question

If $f\left(x\right)=\{\begin{array}{c}\frac{x^2}{a};0\le x<1\\ -1;1\le x<\sqrt{2}\\ \frac{2b^2-4b}{x^2};\sqrt{2}\le x<\infty \end{array}$ then find the value of $a$ and $b$ if $f\left(x\right)$ is continuous in $[0,\infty )$.Find $a+b$.

Answer 1

Given $f\left(x\right)$ is continuous

Apply continuity at $x=1$

We get $\frac{1}{a}=-1$, so the value of $a$ is $-1$.

Apply continuity at $x=\sqrt{2}$

We get $-1=b(b-2)$

So the value of $b$ is $1$.

$\therefore$ $a+b=-1+1=0$