GCE Mathematics A H230/02: Pure Mathematics and Mechanics Advanced Subsidiary GCE Mark Scheme for Autumn 2021. 100% Approved Pass Rate.

Document Content and Description Below

Question Answer Marks AO Guidance 1 2 2 2 4 0 ( )3 + − = k M1 1.1 Sets f 2 ( ) equal to 0 Or any other complete method k = −6 A1 1.1 [2] 2 y x ≤ − + 2 4 B1 1.1 SC All 3 but with (correct... ) strict inequalities B2 2 y x ≥ − 4 B1 1.1 x ≥ 0 B1 1.1 Only 2 but with (correct) strict inequalities B1 [3] 3 (a) V at b = + 2375 6 , 2825 12 = + = + a b a b M1* 3.3 Using a linear model to set up two equations with the values given Or: correct attempt to find gradient M1dep* 3.1a Solving their two equations to find a linear model, possibly BC Using their gradient to find a linear model V t = + 75 1925 A1 1.1 cao [3] 3 (b) When t = 0, V = 1925 B1ft 3.4 Uses model and states initial investment is their ‘b’ Only ft a linear model from part (a) Compare their “1925” with 1900 and make a sensible comment about whether the straight-line model in part (a) is supported or not B1ft 3.5a [2]H230/02 Mark Scheme October 2021 7 Question Answer Marks AO Guidance 4 Considers f (2 ) f (2) h h + − B1 2.1 Or considers f ( ) f ( ) x h x h + − with x = 2 substituted at some point ( ) ( )2 2 f 2 2 2 3 2 8 8 3 h h h h + = + − = + + − M1 1.1 Considers f 2 ( + h) and attempts to expand Or considers f (x h + ) and attempts to expand f 2 f (2) 2 8 5 5 2 8 ( + − = + + − = + h h h h h ) ( 2 2 ) A1 1.1 Correct simplified expression for f 2 f (2) ( + − h) Correct simplified expression for f f ( ) ( x h x + − ) f (2 ) f (2) h 2 8 h h + − = + A1 1.1 Correct simplified expression for f (2 ) f (2) h h + − Correct simplified expression for f ( ) f ( ) x h x h + − ( ) ( ) 0 f 2 f (2) f 2 lim 8 h h h → + − ′ = = A1 2.2a cao – must be explicit that the limit (and not simply h = 0) is considered [5]H230/02 Mark Scheme October 2021 8 Question Answer Marks AO Guidance 5 (a) 2cos tan 3 1 cos x x x 2 = + ( ) ( ) 2 2 sin 2cos 3 1 cos cos x x x x = + M1 3.1a Replaces tan2 x with 2 2 sin cos x x ( ) 2 2 1 cos 2cos 3 1 cos cos x x x x   −     = +   M1 3.1a Replaces sin2 x with 1 cos − 2 x 2 1 cos 3cos 1 cos ( − = + 2 x x x ) ( ) 2 2 2 2cos 3cos 3cos − = + x x x 2 5cos 3cos 2 0 x x + − = A1 2.1 AG – correct working throughout Must show enough working to justify the given answer [3] 5 (b) DR (5cos3 2 cos3 1 0 θ θ − + = )( ) M1 1.1a Attempt to solve 3-term quadratic cos3 1 θ = − and 2 cos3θ = 5 A1 2.1 May be implied θ = − 13 arccos 1 ( ) ,θ = 1 2 3 5 arccos( ) , 60 M1 A1 1.1 1.1 Correct order of operation to find one value of θ (or all values of 3θ correct) (3 66.42..., ) 180, 293.57... θ = 22.1 A1 1.1 Correct value (to at least 1 dp) 97.9 A1 1.1 Correct value (to at least 1 dp) Any additional values in the range loses final A mark if earned [6] [Show More]

Last updated: 1 year ago

Preview 1 out of 16 pages