Answer only one of the following two alternatives. **EITHER** Let `ω = cos(π/5) + i sin(π/5)`. Show that `ω⁵ + 1 = 0` and deduce that `ω⁴ – ω³ + ω² – ω = -1`. [2] Show further that `ω - ω⁴ = 2 cos(π/5)` and `ω³ - ω² = 2 cos(2π/5)`. [4] Hence find the values of `cos(π/5) + cos(2π/5)` and `cos(π/5) cos(2π/5)`. [4] Find a quadratic equation having roots `cos(π/5)` and `cos(2π/5)` and deduce the exact value of `cos(π/5)`. [4] **OR** Given that `x² d²y/dx² + 4x(1+x) dy/dx + 2(1 + 4x + 2x²)y = 8x²` and that `x²y = z`, show that `d²z/dx² + 4 dz/dx + 4z = 8x²`. [4] Find the general solution for y in terms of x. [8] Describe the behaviour of y as x → ∞. [2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2011). The mark or mark type is shown in brackets.

1. [B1] Verifies that ω is a root. (cos(π/5) + i sin(π/5))⁵ + 1 = cos π + i sin π + 1 = 0
2. [B1] Factorises. (ω⁵ + 1) = (ω+1)(ω⁴ – ω³ + ω² – ω + 1) = 0 ⇒ ω ≠ -1 ⇒ ω⁴ – ω³ + ω² – ω + 1 = 0 ⇒ ω⁴ – ω³ + ω² – ω = -1
3. [M1] Finds ω⁴ and subtracts. ω = cos(π/5) + i sin(π/5) ω⁴ = cos(4π/5) + i sin(4π/5) = -cos(π/5) + i sin(π/5) ⇒ ω - ω⁴ = 2 cos(π/5)
4. [A1] Finds ω⁴ and subtracts. ω⁴ = cos(4π/5) + i sin(4π/5) ⇒ ω - ω⁴ = 2 cos(π/5)
5. [M1] Finds ω³ and ω² and subtracts. ω³ = cos(3π/5) + i sin(3π/5) ω² = cos(2π/5) + i sin(2π/5) ⇒ ω³ – ω² = 2 cos(2π/5)
6. [A1] Finds ω³ and ω² and subtracts. ω³ – ω² = 2 cos(2π/5)
7. [M1A1] -2cos(π/5) - 2cos(2π/5) = -1 ⇒ cos(π/5) + cos(2π/5) = 1/2
8. [M1A1] cos(π/5)cos(2π/5) = 1/4 (ω-ω^4)(ω³-ω²) = 1/4 (2cos(π/5))(2cos(2π/5))
9. [M1] Finds required quadratic equation. x² - ½x - ¼ = 0 or 4x²-2x-1=0
10. [M1A1] Solves for x. ⇒ x = (2±2√5)/8 = (1±√5)/4
11. [A1] States required value. ⇒ cos(π/5) = (1+√5)/4 (since 0 < cos(π/5) < 1)
12. [M1] Differentiates z = x²y ⇒ dz/dx = x² dy/dx + 2xy
13. [A1] Differentiates twice. ⇒ d²z/dx² = x² d²y/dx² + 4x dy/dx + 2y
14. [M1] Rearranges LHS of DE. x² d²y/dx² + 4x(1+x) dy/dx + (2+8x+4x²)y = (x² d²y/dx² + 4x dy/dx + 2y) + 4x² dy/dx + 8xy + 4x²y
15. [A1] Rearranges LHS of DE. x² d²y/dx² + 4x(1+x) dy/dx + (2+8x+4x²)y = d²z/dx² + 4(dz/dx - 2xy) + 4z = d²z/dx² + 4 dz/dx + 4z = 8x² (AG)
16. [M1] Finds and solves AQE. m² + 4m + 4 = 0 ⇒ (m+2)² = 0 ⇒ m = -2
17. [A1] States CF CF: z = Ae⁻²ˣ + Bxe⁻²ˣ
18. [M1] States form of PI. PI: z = ax² + bx + c
19. [A1] Differentiates twice, ⇒ z' = 2ax+b ⇒ z'' = 2a
20. [M1] substitutes 2a + 8ax + 4b + 4ax² + 4bx + 4c = 8x²
21. [A1] and equates coefficients. 2a + 4b + 4c = 0 ; 8a + 4b = 0 ; 4a = 8
22. [A1] Solves. a=2 b=-4 c=3
23. [M1] States GS for z-x. z = Ae⁻²ˣ + Bxe⁻²ˣ + 2x² - 4x + 3
24. [A1] States GS for y-x. y = (Ae⁻²ˣ + Bxe⁻²ˣ)/x² + 2 - 4/x + 3/x²
25. [M1] Considers the effect of x → ∞. As x → ∞, e⁻²ˣ → 0, 1/x → 0 and 1/x² → 0
26. [A1] Considers the effect of x → ∞. ∴ y → 2