Question

If x is real, find the maximum and minimum values of $\frac{x^2+14x+9}{x^2+2x+3}$.
(b) Prove that for any real values of x the expression $\frac{x+a}{x^2+bx+c^2}$ will have any value if $b^2>4c^2$ and $a^2+c^2<ab$.

Answer 1

(a) Let $\frac{x^2+14x+9}{x^2+2x+3}=y$

$\therefore$ $x^2(1-y)+2x(7-y)+3(3-y)=0$........(1)

For real values of x, $B^2-4AC$ of (1) should be $\ge 0$

$\therefore$ $4{(7-y)}^2-12(1-y)(3-y)\ge 0$

or $(40-14y+y^2)-3(3-4y+y^2)\ge 0$

or $-2y^2-2y+40\ge 0$

or $-2(y^2+y-20)\ge 0$

or $(y+5)(y-4)$ is $\le 0$

or $[y-(-5\left)\right](y-4)$ is $-ive$.

By $\S 5$ y should lie between $-5$ and $4$.

Therefore the maximum value is $4$ and minimum is $-5$.

(b) $y=\frac{x+a}{x^2+bx+c^2}$

$\therefore$ $y{x}^2+x(by-1)+(c^{2}y-a)=0$

Since x is real $\therefore \Delta \ge 0$

$\therefore$ ${(by-1)}^2-4y(c^{2}y-a)\ge 0$

or $y^2(b^2-4c^2)+2y(2a-b)+1=+ive$.........(1)

The sign of a quadratic expression is same as of its first terms i.e. of $b^2-4c^2$ provided its discriminant $D$ is $-ive$

or $4{(2a-b)}^2-4(b^2-4c^2)=-ive$

or $16(a^2+c^2-ab)=-ive$ or $a^2+c^2<ab$

Hence the sign of expression (1) is same as of $b^2-4c^2$ i.e. $+ive$. Hence y will have any value.