> GCE Further Mathematics B (MEI) Y436/01: Further pure with technology Advanced GCE Mark Scheme for Autumn 2021

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Annotations and abbreviations Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark aw... arded 0, 1 B0, B1 Independent mark awarded 0, 1 E Explanation mark 1 SC Special case ^ Omission sign MR Misread BP Blank page Highlighting Other abbreviations in mark scheme Meaning E1 Mark for explaining a result or establishing a given result dep* Mark dependent on a previous mark, indicated by *. The * may be omitted if only previous M mark. cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working AG Answer given awrt Anything which rounds to BC By Calculator DR This indicates that the instruction In this question you must show detailed reasoning appears in the question.Y436/01 Mark Scheme October 2021 3 Question Answer Marks AOs Guidance 1 (a) x y 2 2 + = 4 B1 [1] 1.2 (b) Let d be the distance concerned. Then d is as in the diagram below. By reference to a diagram using the cosine rule 2 2 2 2 2 2 2 2cos( ) 8(1 cos( )) d c b c b = + − × × − = − − . Therefore d = − − 2 2 1 cos( ) c b . M1 A1 [2] 2.1 1.1 Diagram not necessary. Alternatives here include 2 2 2 (2cos( ) 2cos( )) (2sin( ) 2sin( )) d c b c b = − + −Y436/01 Mark Scheme October 2021 4 Question Answer Marks AOs Guidance (c) For the circles to touch, need d = 2 in the above. Then 2 2 2 1 cos( ) 1 2 1 cos( ) 1 1 cos( ) 2 1 1 cos( ) 2 1 cos( ) 2 , since 0 3 c b c b c b c b c b c b π b c π = − − ⇒ = − − ⇒ = − − ⇒ = − − ⇒ − = ⇒ − = ≤ < < M1 A1 [2] 2.1 1.1 Or just state that it’s an equilateral triangle in this case. (d) B1 B1 [2] 1.1 1.1 Equations are not required in this part.Y436/01 Mark Scheme October 2021 5 Question Answer Marks AOs Guidance (e) 2 2 2 2 1 9 x y x y + = + = B1 B1 [2] 1.2 1.2 2 (a) Gradient of the line through (0, a) and (1, a2) is 2 (1 ) 1 0 a a a a − = − − . The line crosses the y – axis at (0, a) so the equation of the line is y a a x a = − + ( 1) M1 A1 [2] 1.1a 1.1b (b) The two straight lines are y b b x b = − + ( 1) and y c c x c = − + ( 1) . These are parallel if 2 2 ( 1) ( 1) 0 0 ( )( ) ( ) 0 ( )( 1) 0 1 (since ) 1 b b c c c b b c c b c b c b c b c b c b c b c b − = − ⇒ = − + − ⇒ = − + − − ⇒ = − + − ⇒ = + − ≠ ⇒ + = M1 M1 A1 [3] 3.1a 2.4 2.1 Note that equation can be solved using CAS which is an acceptable method.Y436/01 Mark Scheme October 2021 6 Question Answer Marks AOs Guidance (c) B1 B1 [2] 1.1b 1.1b One mark for each branch of the curve, check for shape, asymptotes, max/min and quadrants and and points on the axes. Note that question requires ‘sketch’. [Show More]

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