The following three sets A,B and C satisfy A ⊂ B and C ⊂ A:
A={x∣(x−3)(x^2−20x+75)>0},
B={x∣x>α},
C={x∣x>β}.
If the maximum value of α is a and the minimum value of β is b,
what is the value of a+b?
To find the value of a+b, we need to determine the maximum value of α and the minimum value of β from the given sets A, B, and C.
Let's start by analyzing set A = {x∣(x−3)(x^2−20x+75)>0}.
To determine the values of x that satisfy the inequality (x−3)(x^2−20x+75)>0, we can start by factoring the quadratic equation inside the parentheses:
x^2−20x+75 = (x−5)(x−15).
Now we have two critical values, x = 5 and x = 15, which divide the number line into three intervals. These intervals are (-∞, 5), (5, 15), and (15, ∞).
To find the intervals that make the inequality (x−3)(x^2−20x+75)>0 true, we can test values within each interval.
For the interval (-∞, 5), we choose x = 0. Evaluating the inequality, we have:
(0−3)(0^2−20(0)+75)= -3(0-0+75) = -3(75) = -225 < 0.
For the interval (5, 15), we choose x = 10. Evaluating the inequality, we have:
(10−3)(10^2−20(10)+75)= 7(100-200+75) = 7(-25) = -175 < 0.
For the interval (15, ∞), we choose x = 20. Evaluating the inequality, we have:
(20−3)(20^2−20(20)+75)= 17(400-400+75) = 17(75) = 1275 > 0.
From these evaluations, we can see that the solution is only satisfied in the interval (15, ∞). Therefore, set A becomes A = {x∣x > 15}.
Now let's move on to set B = {x∣x > α}. From set A ⊂ B, we know that all elements of set A are also elements of set B. Since we have already determined set A to be A = {x∣x > 15}, we can conclude that α = 15.
Finally, we have set C = {x∣x > β}. From set C ⊂ A, we know that all elements of set C are also elements of set A. Since we have determined set A to be A = {x∣x > 15}, we can conclude that β = 15 as well.
Therefore, the value of a+b is α + β = 15 + 15 = 30.