Answer only one of the following two alternatives. EITHER The curve C has parametric equations x = t², y = (2 - t)³⁄², for 0 ≤ t ≤ 2. Find (i) d²y/dx² in terms of t, (ii) the mean value of y with respect to x over the interval 0 ≤ x ≤ 4, (iii) the y-coordinate of the centroid of the region enclosed by C, the x-axis and the y-axis. OR The curve C has equation y = (ax² + bx + c) / (x + d), where a, b, c and d are constants. The curve cuts the y-axis at (0, −2) and has asymptotes x = 2 and y = x + 1. (i) Write down the value of d. (ii) Determine the values of a, b and c. (iii) Show that, at all points on C, either y ≤ 3 - 2√6 or y ≥ 3 + 2√6.

Mark scheme

Marking points from the official mark scheme (May/June 2014). The mark or mark type is shown in brackets.

1. [M1A1] (EITHER)(i) dx/dt = 2t, dy/dt = -3/2 (2-t)^(1/2) ⇒ dy/dx = -3(2-t)^(1/2) / (8t)
2. [M1A1] d/dt(dy/dx) = d/dt [ -3/8 (2-t)^(1/2) t⁻¹ ] = 3(4-3t) / (16t²(2-t)^(1/2))
3. [A1] d²y/dx² = (d/dt(dy/dx)) / (dx/dt) = 3(4-3t) / (16t³(2-t)²) (any correct simplified form)
4. [M1A1] (EITHER)(ii) ∫ydx = ∫(from 0 to 2) (2-t)^(3/2) (2t) dt (LNR) for x=0 to x=4 means t=0 to t=2
5. [M1A1] = 2∫(from 0 to 2) t(2-t)^(3/2) dt. Using integration by parts (or substitution u=2-t) = 2 [0 + [-4/15(2-t)^(5/2)]] (from 0 to 2)
6. [M1A1] = 32√2/15 (LR). Mean Value = (32√2/15) / (4-0) = 8√2/15 or 0.754.
7. [M1A1] (EITHER)(iii) ∫y²dx = ∫(from 0 to 2) (2-t)³ (2t) dt
8. [A1] = 16/5. ybar = (1/(2 * Area)) ∫y²dx = (1/(2 * (32√2/15))) * (16/5)
9. [A1] = 5√2/16 or 0.442
10. [B1] (OR)(i) d = -2
11. [M1A1] (OR)(ii) x=0, y=-2 ⇒ -2 = c/d => c = -2d. From x=2 asymptote, d=-2. So c=4.
12. [M1A1] Oblique asymptote y = x+1. `(ax² + bx + c) / (x + d) = ax + (b-ad) + (c-d(b-ad))/(x+d)`. Comparing `ax+(b-ad)` with `x+1`
13. [A1A1] a = 1, b - ad = 1 => b - (1)(-2) = 1 => b = -1. So a=1, b=-1, c=4.
14. [M1] (OR)(iii) y = (x² - x + 4) / (x - 2) ⇒ y(x-2) = x² - x + 4 ⇒ x² - (y+1)x + (2y+4) = 0
15. [M1] For real x, discriminant ≥ 0: (y+1)² - 4(2y+4) ≥ 0
16. [A1] ⇒ y² - 6y - 15 ≥ 0
17. [M1A1] (y-3)² - 24 ≥ 0 (or for solving y² – 6y – 15 = 0, roots are 3 ± 2√6)
18. [M1] y ≤ 3 - 2√6 or y ≥ 3 + 2√6 (AG)
19. [A1] (Note: if done by differentiation, max and min must be demonstrated for full marks.)