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

Question

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

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Let $y=\frac{x^2+ax+b}{x^2+2x+3}$

$$x^2y+2xy+3y=x^2+ax+b$$ $$x^2(y-1)+x(2y-a)+3y-b=0$$ As $x$ is real,so the discriminant of the above quadratic equation has to be greater than or equal to zero.

$$(2y-a)^2-4(y-1)(3y-b)\geq0$$ $$-8y^2+y(-4a+4b+12)+a^2-4b\geq0$$ As above quadratic inequality is greater than or equal to zero,so its discriminant is less than or equal to zero.
$$(-4a+4b+12)^2+32(a^2-4b)\leq0$$ I am stuck here,i cant find $a^2+b^2$.

Answer 1

$$-5 \le \frac{x^2+ax+b}{x^2+2x+3} \le 4 $$ As the denominator is positive, this is equivalent to $$-5x^2-10x-15 \le x^2+ax+b \le 4x^2+8x + 12$$ which can be considered as two quadratic inequalities, $$6x^2+(a+10)x+(b+15) \ge 0, \quad 3x^2+(8-a)x+(12-b) \ge 0$$ Note that equality must happen as well for some $x$. For this to be true for all reals, the discriminants must be then zero, and $-15 \le b \le 12$ to ensure the quadratics are both above the x axis. So $$(a+10)^2 = 24(b+15), \quad (8-a)^2 = 12(12-b)$$ Solving, $a=14$ and $b = 9$ giving $a^2+b^2 = 277$.

Answer 2

Begin with $$g(y)=-8y^2+y(-4a+4b+12)+a^2-4b\geq 0,\quad\forall y\in[-5,4].$$ Then we wish to solve $g(-5)=g(4)=0$, that is, \begin{align} -200-5(-4a+4b+12)+a^2-4b=0,\\ -128+4(-4a+4b+12)+a^2-4b=0, \end{align} or \begin{align} a^2+20a-24b&=260,\\ a^2-16a+12b&=80. \end{align} Thus \begin{align} a-b=5\quad\mbox{and}\quad 0=a^2-4a-140=(a-14)(a+10) \end{align} and we obtain two sets of solutions $$a=14,\,b=9\quad\mbox{and}\quad a=-10,\,b=-15.$$ where the second solution does not fit because $a,$ $b$ are assumed to be positive integers. Hence $a^2+b^2=277.$