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 ) ≤ 0 and f( x )-f( y ) ≤ 0. The area of R is equal to

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 ) ≤ 0 and f( x )-f( y ) ≤ 0. The area of R is equal to (A) 16 π (B) 12 π (C) 8 π (D) 4 π

Tardigrade Admin · Accepted Answer

f(x)+f(y)=x2+6 x+y2+6 y+2=(x+3)2+(y+3)2-16 text and f(x)-f(y)=(x2+6 x+1)-(y2+6 y+1) =x2-y2+6(x-y) f(x)-f(y)=(x-y)(x+y+6) text now (x+3)2+(y+3)2 ≤ 16 text and (x-y)(x+y+6) ≤ 0 x-y ≥ 0 text and x+y+6 ≤ 0 text or x-y ≤ 0 text and x+y+6 ≥ 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. (π r 2/2)=8 π

Q. 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

$16 π$

$12 π$

$8 π$

$4 π$

Q. Let f(x)=x2+6x+1 and R denote the set of points (x,y) in the coordinate plane such that f(x)+f(y)≤0 and f(x)−f(y)≤0. The area of R is equal to

16π

12π

8π

4π

Q. 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

$16 π$

$12 π$

$8 π$

$4 π$