A-LEVEL MATHEMATICS 7357/1 Paper 1 Mark scheme

Q Marking instructions AO Mark Typical solution 1 Circles the correct response 1.1b B1 Total 1 Q Marking instructions AO Mark Typical solution 2 Circles the correct response 1.1b B1 Total 1 Q Marking instructions AO Mark Typical solution 3 Circles the correct response 1.1b B1 6.4 cm2 Total 1 ( )104log a− d d kxy ke x = 6 MARK SCHEME – A-LEVEL MATHEMATICS – 7357/1 – JUNE 2019 Q Marking instructions AO Mark Typical solution 4 Uses negative reciprocal to obtain equation with correct gradient 3.1a M1 45 1 5 4 17 3 4 1 5 3 11 5 4 11 4 11 55 x y k x y y k yx yx − + = = ⇒ + = ⇒= =− × + × = −= = + Obtains correct x coordinate of midpoint Or obtains correct equations of lines through A and B perpendicular to AB 5 4 31.5 5 4 9.5 y x y x − = − =− OE 1.1b B1 Substitutes their mid-point value of x to obtain value of y coordinate of midpoint (not in terms of a or b) Or Finds a value for their 2 ab + Or Finds k by adding correct equations of lines through A and B perpendicular to AB Or equating intercepts. 1.1a M1 Obtains correct equation ACF Eg 4 , 2.2 5 y x c c = + = ISW once correct answer seen. 1.1b A1 Total 4 7 MARK SCHEME – A-LEVEL MATHEMATICS – 7357/1 – JUNE 2019 Q Marking instructions AO Mark Typical solution 5(a) Uses =260 for arithmetic sequence with n=16 to form a correct equation PI by ( ) 8 2 15 260 ad += 1.1a M1 Completes rigorous argument with correct algebraic manipulation to show required result Must see at least one line of simplification after ( ) 8 2 15 260 ad += before given answer. 2.1 R1 5(b) Forms a second equation in a and d using and solves simultaneously to find a or d 3.1a M1 ( ) ( )41 . 0.5 410 2 ad ad a d S += += = =− = × − × = Obtains correct a and d 1.1b A1 Uses their a and d to obtain their value of =41 820ad + Follow through provided one of their a or d is correct. 1.1b A1F 5(c) Explains that values of are positive n < 41 Or Explains that values of are negative for Or Uses quadratic manipulation or differentiation of formula for Sn to obtain n = 40.5 CSO 2.4 M1 The terms before the 41st term are all positive. The terms after the 41st term are all negative so the sum of the first 41 terms must be a maximum value. Completes a valid argument explaining all terms positive before 41 and negative after 41 Or Completes argument linking 40.5 with the sum to 40 terms and the sum to 41 terms. CSO 2.1 R1 Total 7 nS ( ) ( ) ( ) ( ) 2 2 15 65 4 30 65 ad ad ad ad + − = += += += 60 315S = 41S nU nU 41n > 8 MARK SCHEME – A-LEVEL MATHEMATICS – 7357/1 – JUNE 2019 Q Marking Instructions AO Marks Typical Solution 6(a) Deduces the range of Accept ( ) 1 f 2 x ≥ , 1 [0.5, ) 2 y or ≥∞ OE AO2.2a B1 1 2 {y: } y≥ 6(b)(i) Rearranges formula, isolating squared term with at least one correct step seen. 1.1a M1 x ≥ 1 2 Obtains inverse function in any correct form. 1.1b A1 Obtains correct inverse function using 1f ( ) ... x− = and states correct domain 2.5 A1 6 (b)(ii) States correct range Accept 1f ( ) 0 x− ≥ OE 1.1b B1 {y: 0} y≥ 6(c) Recalls correct transformation 1.2 B1 Reflection in y = x 6(d) Forms equation using two of the three expressions x =