Question

If $x$ is real, then the maximum and minimum value of $\frac{x^2-6x+4}{x^2+2x+4}$ are given by

• A. $5,-\frac{1}{3}$
• B. $5,0$
• C. $5,\frac{1}{3}$
• D. $0,5$

Answer 1

Answer: (A)

$5,-\frac{1}{3}$

To find the maximum and minimum values of
$y=\frac{x^2-6x+4}{x^2+2x+4}$
On cross multiplication, we have
$y(x^2+2x+4)=x^2-6x+4$
$x^2(y-1)+x(2y+6)+4(y-1)=0$
Now, D = 0
$b^2-4ac=0$
$=(2y+6{)}^2-4(y-1)\times 2(y-1)$
$3y^2-14y-5=0$
On Solving this equation , we get
$y=5,y=-\frac{1}{3}$
Hence , Maximum value of the function is $y=5$
and Minimum value of the function $y=-\frac{1}{3}$