The diagram shows the curve y = f(x) defined for x > 0. The curve has a minimum point at A and crosses the x-axis at B and C. It is given that dy/dx = 2x - 2/x³ and that the curve passes through the point (4, 189/16). [Figure 10.1]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2017). The mark or mark type is shown in brackets.

1. [M1] Set = 0. 2x - 2/x³ = 0
2. [A1] Allow 'spotted' x = 1 x⁴ = 1 ⇒ x = 1 at A cao
3. [B1] f(x) = x² + 1/x² (+c) cao
4. [M1] Sub (4, 189/16). c must be present. Dep. on integration 189/16 = 16 + 1/16 + c
5. [A1] c = -17/4
6. [M1] Multiply by 4x² (or similar) to transform into 3-term quartic. x² + 1/x² - 17/4 = 0 ⇒ 4x⁴ - 17x² + 4 (= 0)
7. [M1] Treat as quadratic in x² and attempt solution or factorisation. (4x² - 1)(x² - 4) (=0)
8. [A1A1] Not necessary to distinguish. Ignore negative values. No working scores 0/4 x = 1/2, 2
9. [B2,1,0] Mark final integral ∫(x² + x⁻² − 17/4)dx = x³/3 - 1/x - 17x/4
10. [M1] Apply their limits from (iii) (Seen). Dep. on integration of at least 1 term of y (8/3 - 1/2 - 17/2) - (1/24 - 2 - 17/8)
11. [A1] Mark final answer. [y² scores 0/4] Area = 9/4