Question

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

Answer 1

Answer: 227

Given, $\text{f}\text{(}\text{x}\text{)=}\text{y}=\frac{{\left(\text{x}\right)}^2+\text{ax}+\text{b}}{{\left(\text{x}\right)}^2+2\text{x}+3}$
Where, $\text{D}\ge 0$
$\Rightarrow {\left(2\text{y}-\text{a}\right)}^2-4\left(\text{y}-1\right)\left(3\text{y}-\text{b}\right)\ge 0$
$\Rightarrow -8{\left(\text{y}\right)}^2+\left(-4\text{a}+4\text{b}+12\right)\text{y}+{\left(\text{a}\right)}^2-4\text{b}\ge 0$
$\text{⇒}{\left(\text{y}\right)}^2+\left(\frac{4\text{a}-4\text{b}-12}{8}\right)\text{y}+\frac{4\text{b}-{\left(\text{a}\right)}^2}{8}\le \mathrm{0....}(i)$
We know that, $-5\le \text{y}\le 4$
$\Rightarrow \left(\text{y}+5\right)\left(\text{y}-4\right)\le 0$
$\text{⇒}{\left(\text{y}\right)}^2+\text{y}-20\le \mathrm{0....}(ii)$
From $(i)$ and $(ii)$ , we can say that,
$\frac{4\text{a}-4\text{b}-12}{8}=1$
$\Rightarrow 4a-4b=20$
$\Rightarrow a-b=5$
and $\frac{4\text{b}-{\text{a}}^2}{8}=-20$
$\Rightarrow 4\text{b}-{\left(\text{b}+5\right)}^2=160$
$\text{⇒}{\text{b}}^2+6\text{b}-135=0$
$\Rightarrow \left(\text{b}-9\right)\left(\text{b}+15\right)=0$
If $b=9$ so $a=14$
Now value of ${\text{a}}^2+{\text{b}}^2=9^2+14^2=81+196=227$