1. The graph shows y = x(5 - x) on 0 ≤ x ≤ 5. What is the exact area enclosed by the curve and the x-axis?
The area is the integral from 0 to 5 of x(5 - x). This is [5x2/2 - x3/3] from 0 to 5 = 125/2 - 125/3 = 125/6.
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The area is the integral from 0 to 5 of x(5 - x). This is [5x2/2 - x3/3] from 0 to 5 = 125/2 - 125/3 = 125/6.
The input x - 3 must equal 2, so x = 5. The output becomes 5 - 2(-1) = 7, so the corresponding stationary point is (5, 7).
For a logistic model, growth is greatest at half the limiting value, so N = 50. Solve 100/(1 + 9e-0.3t) = 50 to get 9e-0.3t = 1, hence t = (10 ln 9)/3 hours.
The total unscaled area is the integral from 0 to 4 of 4x - x2, which is 32/3. The unscaled area from 2 to 4 is 16/3, so P(X > 2) = (16/3)/(32/3) = 1/2.
The probabilities place 52 one standard deviation below the mean and 68 one standard deviation above it. The mean is the midpoint (52 + 68)/2 = 60, and the standard deviation is 68 - 60 = 8.
Differentiate to get f'(x) = 3x2 - 8x + 1. At x = 3, f'(3) = 27 - 24 + 1 = 4.
HSC Year 12 Mathematics Advanced practice should make calculus, functions, modelling, probability, and statistics testable without losing the worked method. These examples preview that graph-backed style before the no-login Mathematics Advanced demo.
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