A-Level Equations and inequalities: worked solution

Question

Curve C is y=(1+3x−18x²)/(9x+3). Without a calculator, find the set of values y can take.

Method: treat y as a fixed number and ask for which y the equation y = (1+3x-18x²)/(9x+3) has a real solution x. If no real x gives that y, the curve never reaches it. Step 1: Multiply both sides by (9x+3). This is allowed because x ≠ -1/3 on the curve. y(9x + 3) = 1 + 3x - 18x² 9xy + 3y = 1 + 3x - 18x² Step 2: Collect everything on one side, arranged as a quadratic in x. 18x² + 9xy - 3x + 3y - 1 = 0 18x² + (9y - 3)x + (3y - 1) = 0 Step 3: The coefficient of x² is 18, which is never zero, so this is genuinely a quadratic in x. A quadratic has a real root exactly when its discriminant b² - 4ac ≥ 0, with a = 18, b = 9y - 3, c = 3y - 1. Step 4: Form and expand the discriminant. b² = (9y - 3)² = 81y² - 54y + 9 4ac = 4(18)(3y - 1) = 72(3y - 1) = 216y - 72 b² - 4ac = 81y² - 54y + 9 - 216y + 72 = 81y² - 270y + 81 Step 5: Impose the condition and simplify by dividing by 27 (81/27 = 3, 270/27 = 10, 81/27 = 3). 81y² - 270y + 81 ≥ 0 3y² - 10y + 3 ≥ 0 Step 6: Factorise 3y² - 10y + 3. Split the middle term using two numbers with product 3 × 3 = 9 and sum -10, namely -1 and -9. 3y² - y - 9y + 3 = y(3y - 1) - 3(3y - 1) = (3y - 1)(y - 3) So the condition is (3y - 1)(y - 3) ≥ 0. Step 7: Solve the quadratic inequality. The roots are y = 1/3 and y = 3, and the parabola opens upwards, so the expression is ≥ 0 outside the roots: y ≤ 1/3 or y ≥ 3. Step 8: Check the endpoints really are attained (they must be, since ≥ allows equality). y = 1/3 makes the quadratic 18x² + 0x + 0 = 0, so x = 0, and (0, 1/3) is on C. y = 3 makes it 18x² + 24x + 8 = 2(3x + 2)² = 0, so x = -2/3, and (-2/3, 3) is on C. Set of values: {y ∈ ℝ : y ≤ 1/3 or y ≥ 3}.