A-LevelMathematicsIntegration9709/12 · May/June 2019 · Q1112 Marks

The diagram shows part of the curve y = √(4x + 1) + 9/√(4x + 1) and the minimum point M. [Figure 11.1]

(i)Find expressions for dy/dx and ∫y dx.[6]
(ii)Find the coordinates of M.[3]
(iii)The shaded region is bounded by the curve, the y-axis and the line through M parallel to the x-axis. Find, showing all necessary working, the area of the shaded region.[3]

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Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

[B1B1B1] dy/dx = (1/2)(4x+1)^(-1/2)[×4] - 9(1/2)(4x+1)^(-3/2)[×4]
[SC If no other marks awarded award B1 for both powers of (4x +1) correct.] or 2/√(4x+1) - 18/(4x+1)√(4x+1) or (8x-16)/((4x+1)√(4x+1))
[B1B1B1] ∫y dx = (4x+1)^(3/2) / (3/2) [÷4] + 9(4x+1)^(1/2) / (1/2) [÷4] (+C)
[SC If no other marks awarded, B1 for both powers of (4x +1) correct.] or (√(4x+1))^3 / 6 + 9√(4x+1) / 2 (+C)
[M1] Sets their dy/dx to 0 (and attempts to solve)
[A1] 4x + 1 = 9 or (4x + 1)² = 81
[A1] x = 2, y = 6 or M is (2, 6) only.
[*M1] Realises area is ∫y dx and attempt to use their 2 and sight of 0.
[DM1] Uses limits 0 to 2 correctly → [4.5 + 13.5] - [1/6 + 4.5] (= 13 1/3)
[A1] (Area =) 1 1/3 or 1.33

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