Question

Consider a rational function $f(x)=\frac{x^2-6x+4}{x^2+2x+4}$ and a quadratic function $g(x)=(1+m)x^2-2(1+3m)x-2(1+m)$ where $m$ is a parameter.
If the range of the function $f(x)$ lies between the roots of $g(x)=0$ then the number of integral values of $m$ equals

• A. 7
• B. 8
• C. 9
• D. 10

Answer 1

Answer: (C)

$f(x)=\frac{x^2-6x+4}{x^2+2x+4}=1-\frac{8x}{x^2+2x+4}$
$f^{\prime }(x)=-8\left[\frac{\left(x^2+2x+4\right)-x(2x+2)}{{\left(x^2+2x+4\right)}^2}\right]=-8\left[\frac{-x^2+4}{{\left(x^2+2x+4\right)}^2}\right]=\frac{8\left(x^2-4\right)}{{\left(x^2+2x+4\right)}^2}$
$f^{\prime }(x)=0\Rightarrow x=2\text{ or }-2$
$f^{(2)}=\frac{4-12+4}{4+4+4}=\frac{-4}{12}=\frac{-1}{3}$
$f(-2)=\frac{4+12+4}{4-4+4}=5$
Hence range of $f(x)$ is $\left[-\frac{1}{3},5\right]$
the graph of $y=f(x)$ is as shown

$\text{ Let }y=\frac{8x}{x^2+2x+4}$
$x^{2}y+2(y-4)x+4y=0$
$x\in R\Rightarrow D\ge 0$
$4(y-4{)}^2-16y^2\ge 0$
$(y-4{)}^2-(2y{)}^2\ge 0$
$(3y-4)(y+4)\le 0$

$f(x)=\frac{x^2-6x+4}{x^2+2x+4}=1-\frac{8x}{x^2+2x+4}$
$f^{\prime }(x)=-8\left[\frac{\left(x^2+2x+4\right)-x(2x+2)}{{\left(x^2+2x+4\right)}^2}\right]=-8\left[\frac{-x^2+4}{{\left(x^2+2x+4\right)}^2}\right]=\frac{8\left(x^2-4\right)}{{\left(x^2+2x+4\right)}^2}$
$f^{\prime }(x)=0\Rightarrow x=2\text{ or }-2$
$f^{(2)}=\frac{4-12+4}{4+4+4}=\frac{-4}{12}=\frac{-1}{3}$
$f(-2)=\frac{4+12+4}{4-4+4}=5$
Hence range of $f(x)$ is $\left[-\frac{1}{3},5\right]$
the graph of $y=f(x)$ is as shown

$\text{ Let }y=\frac{8x}{x^2+2x+4}$
$x^{2}y+2(y-4)x+4y=0$
$x\in R\Rightarrow D\ge 0$
$4(y-4{)}^2-16y^2\ge 0$
$(y-4{)}^2-(2y{)}^2\ge 0$
$(3y-4)(y+4)\le 0$


now $g(x)=0$
$x^2-\frac{2(1+3m)}{1+m}x-\frac{2(1+m)}{1+m}=0$
or$g(x)=x^2-\frac{2(1+3m)}{1+m}x-2=0$
as range of $f(x)$ is $\left[-\frac{1}{3},5\right]$
so one root of $g(x)$ is less than $\left(-\frac{1}{3}\right)$ and other is greater than 5
$g\left(-\frac{1}{3}\right)<0\text{ and }g(5)<0$
$g\left(-\frac{1}{3}\right)=\frac{1}{9}+\frac{2}{3}\frac{(1+3m)}{1+m}-2<0=1+\frac{6(1+3m)}{1+m}-18<0=\frac{6(1+3m)}{1+m}-17<0$
$=\frac{6+18m-17-17m}{1+m}<0=\frac{m-11}{1+m}<0$
$g(5)<0$
$25-\frac{10(1+3m)}{1+m}-2<0;23-\frac{10(1+3m)}{1+m}<0;\frac{23+23m-10-30m}{1+m}<0$
$\frac{13-7m}{1+m}<0;\frac{7m-13}{m+1}>0$