Find the Range of the following relation
y=a^2x-ax^2, x≥1/2a, a<0
To find the range of the given relation, we need to determine the set of all possible y-values for a given range of x-values. In this case, we have the equation:
y = a^2x - ax^2
Given the conditions x ≥ 1/2a and a < 0, we can proceed as follows:
1. Solve the equation for x in terms of y:
Rearranging the equation, we get:
y = ax(a - x)
Dividing both sides by a and rearranging, we have:
y/a = x(a - x)
y/a = ax - x^2
x^2 - ax + (y/a) = 0
2. Determine the discriminant:
The discriminant (D) is given by:
D = b^2 - 4ac
In this case, a = 1, b = -a, and c = y/a. Substituting these values into the discriminant formula, we have:
D = (-a)^2 - 4(1)(y/a)
D = a^2 + 4y/a
3. Analyzing the discriminant:
Since a < 0, we know that a^2 is positive. Additionally, we have the condition a < 0, which implies that 4y/a is positive as well. Therefore, the discriminant D will always be positive.
4. Determine the range of y-values:
Since the discriminant D is positive, the quadratic equation must have two real solutions for x. This implies that for a given y-value, there are two possible solutions for x.
The range of y-values will be the set of all possible values of y, given the conditions. Since we have not specified any restrictions on y, the range is the set of all real numbers, R.
Hence, the range of the given relation is the set of all real numbers (R).