Given that y = 4 sin 2x cos 2x, find the value of dy/dx when x = π/6.

Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

1. [M1] Use of the product rule with correct structure e.g. dy/dx = 4(cos2x(2cos2x) + sin 2x(-2sin2x)) oe
2. [A1] dy/dx = 4(2cos²2x - 2sin²2x) oe, isw
3. [M1] dep on previous M1 FT their dy/dx providing it is of the form a cos² 2x-b sin² 2x or equivalent where a, b are positive integers
4. [A1] dy/dx|x=π/6 = -4
5. [B1] When x=π/6, y=√3
6. [M1] FT -1 / (their dy/dx|x=π/6) for gradient of normal
7. [M1] FT their -1/dy/dx and their y, providing their y ≠ 0 for equation of normal
8. [M1] FT their equation of normal (substituting y=0) providing gradient is -1 / (their dy/dx|x=π/6)
9. [A1] (-4√3,0) or exact equivalent; mark final answer (dep on all previous marks awarded and a fully correct solution to (a) or a correct derivative and gradient of tangent at x=π/6 stated in this part)