HL Term 3 Revision – Worked Solutions
Q1 – Logs and Trig
Solve: log₂(x) + log₂(x − 2) = sin(π/2)
Since sin(π/2) = 1:
log₂[x(x − 2)] = 1 → x² − 2x = 2 → x² − 2x − 2 = 0
Use the quadratic formula:
x = [2 ± √(4 + 8)] / 2 = [2 ± √12] / 2 = 1 ± √3
Valid solution: x = 1 + √3 ≈ 2.73 (x must be > 2)
Q2 – Rational Function
f(x) = (x² + 3x + 2)/(x + 1) → factor numerator: (x + 1)(x + 2)
f(x) = (x + 2), except at x = −1 where it's undefined (hole).
Q3 – Composite Function
(f ∘ g)(x) = 4x² − 12x + 9 and g(x) = x − 1
Let u = x − 1 → x = u + 1
Then f(u) = 4(u + 1)² − 12(u + 1) + 9 = 4u² − 4u + 1
So f(x) = 4x² − 4x + 1
Q4 – Trig Equation
cos(x) = sin(2x), 0 ≤ x ≤ 2π
Use sin(2x) = 2sin(x)cos(x):
cos(x) = 2sin(x)cos(x) → 0 = cos(x)(2sin(x) − 1)
