Question

Let $f\left(x\right)=\{\begin{array}{cc}{x}^2-1,& 0<x<2\\ 2x+3,& 2\le x<3\end{array}$
If $\alpha =\underset{x\to 2^{-}}{lim}f\left(x\right)$ and $\beta =\underset{x\to 2^{+}}{lim}f\left(x\right),$ then the quadratic equation whose roots are $\alpha ,\beta$ is

• A. $x^2+10x+21=0$
• B. $x^2-10x+21=0$
• C. $x^2-7x+12=0$
• D. $x^2-12x+10=0$

Answer 1

The correct option is B $x^2-10x+21=0$

$\alpha =\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}(x^2-1)=3$

$\beta =\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}(2x+3)=7$
Required quadratic equation is
$x^2-(3+7)x+3\times 7=0$
$\Rightarrow x^2-10x+21=0$