Question

Let $f( x )= x ^2+6 x +1$ and $R$ denote the set of points $( x , y )$ in the coordinate plane such that $f( x )+f( y ) \leq 0$ and $f( x )-f( y ) \leq 0$. The area of $R$ is equal to

• A. $16 \pi$
• B. $12 \pi$
• C. $8 \pi$
• D. $4 \pi$

Answer 1

Answer: (C)

$ f(x)+f(y)=x^2+6 x+y^2+6 y+2=(x+3)^2+(y+3)^2-16 $
$\text { and } f(x)-f(y)=\left(x^2+6 x+1\right)-\left(y^2+6 y+1\right) $
$=x^2-y^2+6(x-y) $
$ f(x)-f(y)=(x-y)(x+y+6) $
$\text { now } (x+3)^2+(y+3)^2 \leq 16 $
$\text { and } (x-y)(x+y+6) \leq 0$
$x-y \geq 0 \text { and } x+y+6 \leq 0 \text { or }$
$x-y \leq 0 \text { and } x+y+6 \geq 0$
each of these inequality describes a half plane bounded by a line that passes through $(-3,-3)$ and has slope 1 or -1 .
Thus the set $R$ is half the area of the circle i.e. $\frac{\pi r ^2}{2}=8 \pi$