Question

$E=\frac{x^2+y^2}{x^2+xy+4y^2}$ have the maximum and minimum as $M$ and $m$ respectively. Their average is denoted as $A$. Then $2007A$ is-

• A. $1400$
• B. $1338$
• C. $5014$
• D. $24$

Answer 1

The correct option is B $1338$

$E=\frac{x^2+y^2}{x^2+xy+y^2}$ will have minimum

at $y=0$

$E=\frac{x^2}{x^2}$

E will have maximum value

at $x=y$

$E=\frac{2x^2}{6x^2}=\frac{1}{3}$

E will have minimum value

$A=\frac{M+m}{2}$ (Given )

$A=\frac{1+1/3}{2}$

$A=\frac{4}{6}\Rightarrow \frac{2}{3}$

Now, $2007\times \frac{2}{3}\Rightarrow 2\times 669\Rightarrow 1338$

Therefore, $2007A$ equals $1338$ Answer