Question

If $\textit{f}\left(\textit{x}\right)=\frac{\left(\textit{x}\right)^{2} + \textit{ax} + \textit{b}}{\left(\textit{x}\right)^{2} + 2 \textit{x} + 3}$ where, $a,b\in N$ , and range of the given function is $\left[\right.-5,4\left]\right.$ , then find the value of $\left(\right.a^{2}+b^{2}\left.\right)$ .

Answer 1

Given, $\textit{f}\text{(}\textit{x}\text{)=}\textit{y}=\frac{\left(\textit{x}\right)^{2} + \textit{ax} + \textit{b}}{\left(\textit{x}\right)^{2} + 2 \textit{x} + 3}$
Where, $\textit{D}\geq 0$
$\Rightarrow \left(2 \textit{y} - \textit{a}\right)^{2}-4\left(\textit{y} - 1\right)\left(3 \textit{y} - \textit{b}\right)\geq 0$
$\Rightarrow -8\left(\textit{y}\right)^{2}+\left(- 4 \textit{a} + 4 \textit{b} + 1 2\right)\textit{y}+\left(\textit{a}\right)^{2}-4\textit{b}\geq 0$
$\text{⇒}\left(\textit{y}\right)^{2}+\left(\frac{4 \textit{a} - 4 \textit{b} - 1 2}{8}\right)\textit{y}+\frac{4 \textit{b} - \left(\textit{a}\right)^{2}}{8}\leq 0....\left(\right.i\left.\right)$
We know that, $-5\leq \textit{y}\leq 4$
$\Rightarrow \left(\textit{y} + 5\right)\left(\textit{y} - 4\right)\leq 0$
$\text{⇒}\left(\textit{y}\right)^{2}+\textit{y}-20\leq 0....\left(\right.ii\left.\right)$
From $\left(\right.i\left.\right)$ and $\left(\right.ii\left.\right)$ , we can say that,
$\frac{4 \textit{a} - 4 \textit{b} - 1 2}{8}=1$
$\Rightarrow 4a-4b=20$
$\Rightarrow a-b=5$
and $\frac{4 \textit{b} - \textit{a}^{2}}{8}=-20$
$\Rightarrow 4\textit{b}-\left(\textit{b} + 5\right)^{2}=160$
$\text{⇒}\textit{b}^{2}+6\textit{b}-135=0$
$\Rightarrow \left(\textit{b} - 9\right)\left(\textit{b} + 1 5\right)=0$
If $b=9$ so $a=14$
Now value of $\textit{a}^{2}+\textit{b}^{2}=9^{2}+14^{2}=81+196=227$