Given that (x * y = frac{x + y}{2}, x circ y = frac{x^{2}}{y}) and ((3 * b) circ 48 = frac{1}{3}), find b, where b > 0.
-
A.
8 -
B.
6 -
C.
5 -
D.
4
Correct Answer: Option C
Explanation
((x * y) = frac{x+y}{2})
((3 * b) = frac{3+b}{2})
(x circ y = frac{x^{2}}{y})
((frac{3+b}{2}) circ 48 = frac{(frac{3+b}{2})^{2}}{48} = frac{1}{3})
(frac{(3+b)^{2}}{48 times 4} = frac{1}{3})
((3 + b)^{2} = frac{48 times 4}{3} = 64)
(b^{2} + 6b + 9 = 64 implies b^{2} + 6b + 9 – 64 = 0)
(b^{2} + 6b – 55 = 0 implies b^{2} – 5b + 11b – 55 = 0)
(b(b – 5) + 11(b – 5) = 0 implies (b – 5) = text{0 or (} b + 11) = 0)
Since b > 0, b – 5 = 0
b = 5.