Mathematics > MARK SCHEMES > OCR GCE Further Mathematics A Y540/01: Pure Core 1 Advanced GCE Mark Scheme for Autumn 2021 (All)

OCR GCE Further Mathematics A Y540/01: Pure Core 1 Advanced GCE Mark Scheme for Autumn 2021

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Mark scheme October 2021 3 Question Answer Marks AO Guidance 1 (a) (i) Circle Centre 1−2i, Radius 3 B1 B1 1.1 2.2a Be generous over circles drawn freehand If the axes are scaled then a mar ... k at (1, -2) will do. For radius, an indication that the radius is 3 will do (e.g. passing through (4, -2) etc if marked will do.) [2] (ii) Straight vertical line B1 1.1 12 x = B1 2.2a Can be seen by x = ½ being labelled on the axis and vertical line through it [2] (b) Inside circle B1 1.1 And to the left of 1 2 x = B1 2.2a Or their line if it is vertical. [2]Y540/01 Mark scheme October 2021 4 Question Answer Marks AO Guidance 2 (a) (i) f (0) π 4 = B1 1.1 Not for 450 [1] (ii) ( )2 1 1 f '( ) f '(0) 1 1 2 x x = ⇒ = + + M1 A1 2.1 1.1 Diffn – Must be seen 2 1 f '( ) 1 x x = + is M0 [2] (iii) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 1 1 f '( ) 1 1 2 2 1 f ''( ) 1 2 2 2 2 2 2 2 2 2 1 f ''(0) 4 2 x x x x x x x x x x x = = + + + + ⇒ = × − × + + + − + = + +   − ⇒ = = −     M1 A1 A1 2.1 2.1 2.1 Diffn their f’(x) oe, e.g. ( ) ( ) ( ) ( )2 2 2 1 f 1 1 x x x + ′′ = − + + f’’(0) must be seen. The substitution must be seen (implied by 2 4 − ) AG [3] (b) 2 f ( ) f (0) f '(0) f ''(0) x2 x x = + + M1 1.1 Using the formula and substituting their value for f;(0) 2 2 1 1 4 2 2 2 4 2 4 x x x x π π = + − × = + − A1 2.2a ft their values from (a) [2]Y540/01 Mark scheme October 2021 5 Question Answer Marks AO Guidance 3 (a) e.g. α β γ 2 2 2 + + = −5 means that at least one root is complex But complex roots come in complex pairs so there are 2 complex roots. Given that there are 3 roots and 2 are complex one is real. B1 B1 B1 2.4 2.4 2.4 [3] (b) α β γ + + = 3 B1 1.1 ( ) ( ) ( ) 2 2 2 2 2 9 5 2 But 7 k k α β γ α β γ αβ βγ γα αβ βγ γα αβ βγ γα + + = + + + + + = − + + + = + + ⇒ = M1 A1 3.1a 1.1 Attempt to obtain identity and substitute Condone missing 2 and sign errors [3] (c) 1 1 1 3 2 3 7 5 0 u u u             − + − =       M1 1.1 For the substitution “=0” not necessary here but needed for A1 ⇒ − + − + = 5 7 3 1 0 u u u 3 2 oe A1 1.1 Allow in terms of z Allow ft from their k in (b) [Show More]

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