GCE Mathematics A H230/01: Pure Mathematics and Statistics Advanced Subsidiary GCE Mark Scheme for Autumn 2021. Graded A+

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Question Answer Marks Guidance 1 (a) 12 × 3 × 4 × sin 30 M1 = 3 A1 [2] (b) AC2 = 32 + 42 – 2×3×4×cos 30o (= 4.22) M1 or AC = 3 4 2 3 4 cos30 2 2 + − × × × o Correct expression AC2 or ... AC AC = 2.05 (3 sf) A1 [2] (c) sin sin30 3 their2.05 C = oe eg sin sin30 3C = 4.22 M1 Correct sin rule using their (b) ACB = sin–1 ( 3sin30 their2.05 ) (= sin−10.73…) M1 Attempt inverse sine of their(b) 3sin30 . May be implied by answer 32 = ‘2.05’2 + 42 − 2ב2.05’×4×cosACB M1 Correct cos rule using their (b) ACB = cos−1 2 2 2 '2.05' 4 3 2 '2.05' 4 + − × × M1 Attempt inverse cos of 2 2 2 '2.05' 4 3 2 '2.05' 4 + − × × . May be implied by answer ACB = 46.9o or 47.0° or 47.1o (3 sf) A1ft Allow 47° Condone premature rounding. FT their (b) [3] 2 dndt M1 Attempt differentiate = –5000t–2 A1 At t = 5, d ndt = 5000 25 − ISW M1 Substitute t = 5 into their d ndt which must include t−2 (Rate of change of n is ) –200 A1 [4] 3 (a) x < 1 B1 Any notation Allow 0.4 < x < 1 2 < x < 3 B1 Any notation. Lose only one B1 if < seen instead of < {x: x < 1}  {x: 2 < x < 3} B1f ft their (i) dep at least two ranges; allow < [3] SC: (MR ‘>’) 1 < x < 2; x > 3 B0B1 {x: 1 < x < 2}  {x: x > 3} B1 3 (b) Attempt reflect in y-axis M1 Approx correct shape, orientation & location Through (-3, 0.5), (-2, 0.5), (-1, 0.5), (-0.5, -1.5) A1 Allow ± 2mm SC. All four points plotted ± 2mm: B1 [2]H230/01 Mark Scheme October 2021 7 3 (c) Curve intersects x-axis only once B1 oe [1] 4 (a) 15 12       − or 15i − 12j B1 B1 B1 for each element. Allow i, j notation without “squiggles” [1] 4 (b) 6r – s = 0 M1 s = 6r 02 + (9r)2 = 9 M1 Attempt |a|2 = 9, or |a| = 3; allow in terms of both r and s 81r2 = 9 A1 r = ± 1 3 A1 Allow just r = 13 r = 1 3 and s = 2 or r = – 13 and s = –2 A1 Correctly paired [5] 5 (a) Maximum speed of the car B1 or, eg, doesn't drive faster than 80, or speed limit or model will show consumption eventually becoming negative Condone eg “Maximum number of miles car can drive” or model may not apply for above 80 mph [1] 5 (b) d 5 125 d 3 v ( ) 12 v v − 2 = 0 ( ⇒ − = 12 5 125 6v 0 ) M1 Attempt differentiate C & equate to 0 v = 50 A1 2 2 d 3 12 2 5 125 d ( ) v v v − = 6 − 125 when v = 50 or any correct method showing that SP is a maximum M1 Must be correct Maximum speed is 50 mph A1 Units essential. Dep only on 1st M1H230/01 Mark Scheme October 2021 8 5 (b) ctd Alternative method 1 v = b2a − (= 12 5 2 ( 3) 125 × − − ) Attempt complete square M1 v = 50 A1 Coefficient of v2 negative, hence stationary point is a maximum M1 Maximum speed is 50 mph A1 Units essential Alternative method 2 12 3 2 5 125 v v − = 0 (v = 0 or 100) & M1 Working must be seen Correct sketch graph & v = 50 B1 NB. This mark can be gained without working to justify the graph. v = 50 seen on graph as giving maximum M1 Maximum speed is 50 mph A1 Units essential [4] 5 (c) v = 0 does not give C = 0 oe B1 They will not consume fuel at 0 mph oe [1] 5 (d) eg k( 12 5 125 v v − 3 2 ) B1 or "Increase both constants by the same factor" B1B1 with any k > 1 B1 or with numerical value of k (> 1) B1B1 SC: “Increase both constants” B1 [Show More]

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