QUESTION 1: Calculus y = x³ - 6x² + 9x + 1 Part a) Stationary points and classification Differentiate: dy/dx = 3x² - 12x + 9 Set equal to zero: 3x² - 12x + 9 = 0 3(x² - 4x + 3) = 0 3(x - 1)(x - 3) = 0 x = 1 or x = 3 Find the y-coordinates: When x = 1: y = 1 - 6 + 9 + 1 = 5 When x = 3: y = 27 - 54 + 27 + 1 = 1 Second derivative: d²y/dx² = 6x - 12 At x = 1: 6(1) - 12 = -6, which is less than 0, so it is a local maximum at (1, 5) ✅ At x = 3: 6(3) - 12 = 6, which is greater than 0, so it is a local minimum at (3, 1) ✅ Part b) Point of inflection Set the second derivative equal to zero: 6x - 12 = 0, so x = 2 Check for a sign change: At x = 1: d²y/dx² = -6 (negative) At x = 3: d²y/dx² = 6 (positive) The sign changes, so it is a genuine point of inflection. When x = 2: y = 8 - 24 + 18 + 1 = 3 Point of inflection at (2, 3) ✅ Part c) Tangent at x = 0 When x = 0: y = 1, so the point is (0, 1) Gradient: dy/dx at x = 0 is 3(0) - 12(0) + 9 = 9 y - 1 = 9(x - 0) y = 9x + 1 ✅ QUESTION 2: Normal distribution Let X be the height of a Year 12 student in cm. X follows a Normal distribution with mean 168 and variance 81, so the standard deviation is 9. Part a) P(X > 175) Calculator: normalCDF(175, 9999, 168, 9) P(X > 175) = 0.2183 ✅ Sanity check: 175 is a little under one standard deviation above the mean, so a probability a bit above 0.16 makes sense. Part b) The height only 10% of students exceed P(X > h) = 0.10, so P(X < h) = 0.90 Calculator: invNorm(0.90, 168, 9) h = 179.5 cm ✅ Part c) Sample of 25, find P(sample mean > 170) By the Central Limit Theorem: Standard error = 9 divided by √25 = 9/5 = 1.8 The sample mean follows a Normal distribution with mean 168 and standard deviation 1.8. Calculator: normalCDF(170, 9999, 168, 1.8) P(sample mean > 170) = 0.1333 ✅ Common mistakes to avoid: Mistake 1: Q1 part b: never call f''(x) = 0 a point of inflection without checking the sign change. Mistake 2: Q1 part c: you need BOTH the point (0, 1) and the gradient 9 before writing the tangent.