Question

The minimum and maximum value of $y=\frac{x^2-x+4}{x^2+x+4}$ will be and respectively.

• A. $\frac{3}{5}$
• B. $\frac{5}{2}$
• C. $\frac{5}{3}$
• D. $\frac{2}{5}$

Answer 1

Answer: (A), (C)

$\frac{5}{3}$

$y=\frac{x^2-x+4}{x^2+x+4}\Rightarrow (y-1)x^2+(y+1)x+4y-4\Rightarrow (y+1{)}^2-16(y-1{)}^2\ge 0\Rightarrow (5-3y)(5y-3)\ge 0\Rightarrow (3y-5)(5y-3)\le 0\Rightarrow y\in \left(\frac{3}{5},\frac{5}{3}\right]$