Question

The expression $x^2-4px+q^2>0$ for all real $x$ and also $r^2+p^2<pr$,then the range of $f\left(x\right)=\frac{x+r}{x^2+qx+p^2}$ is

• A. $[\frac{p}{2r},\frac{q}{2r}]$
• B. $(0,\infty )$
• C. $(-\infty ,0)$
• D. $(-\infty ,\infty )$

Answer 1

The correct option is D $(-\infty ,\infty )$

$x^2-4px+q^2>0,\forall x\in R$

$\Rightarrow 4p^2-q^2<0$ ...(i)

$r^2+p^2<qr$ ...(ii)

Let $y=\frac{x+r}{x^2+qx+p^2}$

$x^{2}y+x(qy-1)+p^{2}y-r=0$ ....(iii)

x is real,

$\Rightarrow (q^2-4p^2)y^2+y(-2q+4r)+1>0$

Eq. (i) $\Rightarrow$ Coefficient of $y^2$ is a positive

Discriminant $=(4r-2q{)}^2-4(q^2-4p^2)$

$=16(r^2+p^2-qr)<0$ [by Eq. (ii)]

Hence, Eq. (iii) is true for all real y or $y\in (-\infty ,\infty )$.