Mathematics · May / June 2026 · Paper 42 · Question 18

1 2 × (2 )( )sin30 x x 1 2 leading to 2 x [1]

Must show sin 30. M1 Accept 1 2 × 2 × 2 x sin30 Allow if in stages, e.g. h = x sin 30 then area = 1 ×x × 2 x . with no errors 2 2 Ignore units throughout Q18.

1 2 x = 1 2 2 (2x – 6)(x – 4) Correct expansion of (2x – 6)(x – 4): 2x2 – 8x – 6x + 24 or better leading to x2 – 14x + 24 = 0 [10]

M1 Accept brackets expanded for M1 Accept for M1 1 = 1 2 × 2 (2x – 6)(x – (2 )( )sin30 x x 4)[sin 90]. M1 may be done in stages. B1 Accept correct expansion of (x – 3)(x – 4) oe: B1 M1 can be awarded in any order. x2 – 4x – 3x + 12 or better A1 dep on M1 B1 with no errors or omissions isw attempts to solve/factorise.

(x – 2)(x – 12) [= 0] 72 cao [15]

M2 M1 for (x + a)(x + b) Working may be seen in part (b), allow credit for where a + b = –14 or ab = 24 this if it is used in part (c). or for partial factors If factorisation shown isw other methods. x(x – 2) – 12(x – 2) or x(x – 12) – 2(x – 12)

B2 B1 for x = 12 For B1, accept 12 and 2 but no other values. If x-values are not shown, B1 may be implied by answer 72 and 2. Correct partial factors leading to answer 72 implies M2 B2. Correct partial factors leading to x = 12 implies M2 B1. isw answer 10 < t ⩽ 20 after correct answer seen.