Mathematics 9709/12 May/June 2019: question paper and mark scheme

Mark scheme

1 2  5 2 2 B2,1 3 elements required. –1 for each error with or For  − 3x  term in x is 10 or 5C3 or 5C2 ×   × (−3x)³ or without x’s. Can be seen in an expansion. x   x 5 3 2  2  5.4.3  3 2  5 5.4  2     − x  or ( −3x )    x  3!  2  2!  3x 2  −1080 identified B1 Allow −1080x Allow if expansion stops at this term. Allow from expanding brackets. 3

Mark scheme

2 Midpoint of AB is (5, 1) B1  10 2  Can be seen in working, accept  ,  .  2 2 mAB = −½ oe B1 C to (5, 1) has gradient 2 *M1 Use of m1× m2 = −1. Forming equation of line (y = 2x − 9) DM1 Using their perpendicular gradient and their midpoint to form the equation. C (0, −9) or y = −9 A1 5

Mark scheme

3(i) dy dy dx M1 Multiply numerical gradient at x = 2 by ±0.05. = × = 7 × – 0.05 dt dx dt −0.35 (units/s) or Decreasing at a rate of (+) 0.35 A1 Ignore notation and omission of units 2 3(ii) x4 4 B1 Accept unsimplified ( y) = + (+c) oe 4 x Uses (2, 9) in an integral to find c. M1 The power of at least one term increase by 1. x4 4 A1 A0 if candidate continues to a final equation that is c = 3 or ( y = ) + + 3 oe a straight line. 4 x 3

Mark scheme

4(i) a 2 + 2 ab + b 2 , a 2 − 2 ab + b 2 B1 Correct expansions. sin²x + cos²x = 1 used → ( a + b ) + ( a − b ) = 1 M1 Appropriate use of sin²x + cos²x = 1 with ( a + b )2 2 2 and ( a − b ) 2 a² + b² = ½ A1 No evidence of ±2ab, scores 2/3 Alternative method for question 4(i) 2a = (s+c) & 2b = (s−c) or a = ½(s+c) & b = ½(s−c) B1 1 1 M1 Appropriate use of sin²x + cos²x = 1 a²+b² = ( s + c )2 + ( s − c )2 = ½(s²+c²) 4 4 a² + b² = ½ A1 Method using only (sinx − b)2 and (a − cosx)2 scores 0/3. 3 1 SC B1 for assuming θ is acute giving a = +b 5 or 2 5 − b 4(ii) sinx a+b M1 sinx tanx = → =2 Use of tanx = to form an equation in a and b cosx a−b cosx only a = 3b A1 2

Mark scheme

5 Perimeter of AOC = 2r + rθ B1 Angle COB = π – θ B1 Could be on the diagram. Condone 180 – θ. Perimeter of BOC = 2r + r(π – θ) B1 FT on angle COB if of form ( kπ − θ ) , k > 0. (2r +) πr – rθ = 2((2r) + rθ) M1 Sets up equation using r ( kπ − θ ) and ×2 on correct π −2 (2 + π – θ = 4 + 2θ → θ = ) side. Condone any omissions of OA, OB and/or 3 OC. θ = 0.38 A1 Equivalent answer in degrees scores A0. 5

Mark scheme

6(i) 3, –3 B1 Accept ± 3 −½ B1 2½ B1 3 Condone misuse of inequality signs. 6(ii) Only mark the curve from 0 → 2π. If the x axis is not labelled assume that 0 → 2π is the range shown. Labels on axes are not required. 2 complete oscillations of a cosine curve starting with a maximum at (0,a), a൐0 B1 Fully correct curve which must appear to level off at 0 and/or 2π. B1 Line starting on positive y axis and finishing below the x axis at 2π. Must be B1 straight. 3 6(iii) 4 B1 1

Mark scheme

7(i) x+2 B1 (f−1(x)) = oe 3 2x + 3 M1 Correct method to obtain x = , (or y = , if y= → (x – 1)y = 2x + 3 → x ( y − 2) = y + 3 interchanged) but condone + / − sign errors x −1 x+3  5  A1 Must be in terms of x (g−1(x) or y) = oe  eg + 1 x−2  x−2  x ≠ 2 only B1 FT for value of x from their denominator = 0 4 7(ii) 3 ( 2 x + 3) 7 B1 ( fg ( x ) = ) x −1 – 2 (= ) 3 18x + 27 = 13x – 13 or 3(4x + 11) = 7(x – 1) M1 7 (5x = – 40) Correct method from their fg= leading to a 3 linear equation and collect like terms. Condone omission of 2 ( x − 1) . Alternative method for question 7(ii) 7 13 B1 (f−1( )) = 3 9 2 x + 3 13 M1 13 = → 9(2x + 3)= 13(x – 1) (→ 5x = – 40) Correct method from g ( x ) = their leading to a x −1 9 9 linear equation and collect like terms. x = −8 A1 3

Mark scheme

8(i) 6×3+-2×k+-6×-3 = 0 M1 Use of scalar product = 0. JJJG JJJG JJJG JJJG JJJG JJJG (18 – 2k + 18 = 0) Could be AO . OB , AO . BO or OA . BO k = 18 A1 Alternative method for question 8(i) 76 + 18 + k2 = 18 + (k + 2)2 M1 Use of Pythagoras with appropriate lengths. k = 18 A1 2 8(ii) 36 + 4 + 36 = 9 +k² + 9 M1 Use of modulus leading to an equation and solve to k= or k2 = k = ±√58 or ±7.62 A1 Accept exact or decimal answers. Allow decimals to greater accuracy. 2 JJJG JJJG 8(iii)  −3   −2  M1 Complete method using AC = ± ⅔ AB JJJG   JJJG   JJJG JJJG JJJG JJJG AB =  6  → AC =  4  then OA + AC And then OA + their AC  3  2      4 A1 JJJG   OC =  2   −4    JJJG M1 Divides by modulus of their OC ÷ ( their 4 )2 + ( their 2 )2 + ( their − 4 )2  4 A1 1  1 =  2  or (4i + 2j – 4k) 6  6  −4  Alternative method for question 8(iii)  p  p − 6  3− p  M1 Correct method. Equates coefficients leading to JJJG   JJJG   JJJG   values for p, q, r Let OC =  q  → AC =  q + 2  &CB =  4 − q  r r+6        −3 − r  p – 6 = 2(3 – p); q+2 = 2(4 – q); r + 6 = 2 (–3 – r) A1 →p=4, q=2 & r= – 4 JJJG M1 Divides by modulus of their OC ÷ ( their 4 ) + ( their 2 ) + ( their − 4 ) 2 2 2  4 A1 1  1 =  2  or (4i + 2j – 4k) 6  6  −4  8(iii) Alternative method for question 8(iii) JJJG JJJG JJJG JJJG JJJG JJJG JJJG ( CB = OB − OC ∴ 2 OB − OC = OC – OA ) M1 Correct method. Gets to a numerical expression for JJJG JJJG JJJG k OC from OA & OB .  12  JJJG JJJG JJJG JJJG  → 2 OB + OA = 3 OC ∴ 3 OC =  6   −12     4 A1 JJJG   OC =  2   −4    JJJG M1 Divides by modulus of their OC ÷ ( their 4 ) + ( their 2 ) + ( their − 4 ) 2 2 2  4 A1 1  1 =  2  or (4i + 2j – 4k) 6  6  −4  4

Mark scheme

9 dy B1 For C1: = 2x – 4 → m = 2 dx y – ‘their 4’ = ‘their m’ (x – 3) or using y = mx + c M1 dy Use of : and (3, their 4) to find the tangent dx equation. y – 4 = 2( x – 3) or y = 2 x − 2 A1 If using = mx + c , getting c = −2 is enough. 2x – 2= 4 x + k (→ 4 x ² − 12 x + 4 − k = 0 ) *M1 Forms an equation in one variable using tangent & C2 Use of b² − 4ac = 0 on a 3 term quadratic set to 0. *DM1 Uses ‘discriminant = 0’ 144 = 16(4 – k) → k = − 5 A1 4 x² − 12 x + 4 − k = 0 → 4 x² − 12 x + 9 = 0 DM1 Uses k to form a 3 term quadratic in x 3  1 A1 Condone ‘correct’ extra solution. x= or , y = 1(or – 1). 2  2  Alternative method for question 9 dy B1 For C1: = 2x – 4 → m = 2 dx y – ‘their 4’ = ‘their m’ (x – 3) or using y = mx + c M1 dy Use of : and (3, their 4) to find the tangent dx equation. y – 4 = 2( x – 3) or y = 2 x − 2 A1 If using = mx + c , getting c = −2 is enough. dy − 1 *M1 dy − 1 For C2: = A(4 x + k ) 2 Finds for C2 in the form A(4 x + k ) 2 dx dx 9 − 1 1− k *DM1 dy At P: ‘their 2’ = A(4 x + k ) 2 " → ( x = or 4 x + k = 1 ) Equating ‘their 2’ to ‘their ’ and simplify to 4 dx form a linear equation linking 4x + k and a constant. ( 2 x − 2 ) = 4 x + k → ( 2 x − 2 )2 = 1 → ( 4 x 2 − 8 x + 3 = 0 ) DM1 Using their y = 2 x − 2 , y2 = 4x + k and their 2 4 x + k = 1 (but not =0) to form a 3 term quadratic in x. 3 1 A1 Needs correct values for x and k. x =  or  and from k = −5 ( or − 1) 2 2 from y2 = 4x + k, y = 1(or – 1). A1 Condone ‘correct’ extra solution. Alternative method for question 9 dy B1 For C1: = 2x – 4 → m = 2 dx y – ‘their 4’ = ‘their m’ (x – 3) or using y = mx + c M1 dy Use of : and (3, their 4) to find the tangent dx equation. y – 4 = 2( x – 3) or y = 2 x − 2 A1 If using = mx + c , getting c = −2 is enough. dy − 1 *M1 dy − 1 For C2: = A(4 x + k ) 2 Finds for C2 in the form A(4 x + k ) 2 dx dx − 1 1− k *DM1 dy At P: ‘their 2’ = A(4 x + k ) 2 " → ( x = or 4 x + k = 1 ) Equating ‘their 2’ to ‘their ’ and simplify to 4 dx form a linear equation linking 4x + k and a constant. From 4x + k = 1 and y2 = 4x + k → y2 = 1 DM1 Using their 4x + k = 1 (but not =0) and C2 to form y2 = a constant 9 3 1 A1 Needs correct values for y and x. y = 1(or – 1) and x =  or  2 2 From 4 x + k = 1 , k = –5 ( or – 1) A1 Condone ‘correct’ extra solution 8

Mark scheme

10(a)(i) S10 = S15 – S10 or S10 = S(11 to 15) M1 Either statement seen or implied. 5(2a + 9d) oe B1 5 A1 7.5(2a + 14d) – 5(2a + 9d) or [ (a + 10d) + (a+14d)] oe 2 a A1 Correct answer from convincing working d= AG 3 4 a Condone starting with d = and evaluating both 3 summations as 25a. 10(a)(ii) ( a + 9d ) = 36 + ( a + 3d ) M1 Correct use of a + ( n − 1) d twice and addition of ±36 a = 18 A1 2 Correct answer www scores 2/2 10(b) S∞ = 9 ×S4; a =9 ( a 1− r 4 ) or 9(a + ar+ ar +ar ) 2 3 B1 May have 12 in place of a. 1− r 1− r 9(1 – rn) = 1 where n = 3,4 or 5 M1 Correctly deals with a and correctly eliminates ‘1 – r’ 8 A1 r4 = oe 9 (5th term =) 10⅔ or 10.7 A1 4 Final answer of 10.6 suggests premature approximation – award 3/4 www.

Mark scheme

11(i) dy  1 −  1  9 −  3 B1B1B1 B1 B1 for each, without × 4. B1 for ×4 twice. =  ( 4 x + 1) 2  [× 4]  − ( 4 x + 1) 2  [× 4] dx  2   2    SC If no other marks awarded award B1 for both  2 18 8 x − 16  powers of (4x +1) correct.  4x + 1 − or 3  ( ) 3   4x + 1 ( 4 x + 1) 2   3   1  B1B1B1 B1 B1 for each, without ÷ 4. B1 for ÷4 twice.  ( 4 x + 1) 2   9 ( 4 x + 1) 2  + C not required. ∫ydx =   [÷ 4] +   [÷ 4] (+C) 3 1      2   2   ( 4x + 1) + 9  SC If no other marks awarded , B1 for both powers 3   6 2 ( )  4 x + 1 ( +C )  of (4x +1) correct.     6 11(ii) dy 2 18 M1 dy =0 → − 3 =0 Sets their to 0 (and attempts to solve dx 4x + 1 dx ( ) 4x + 1 2 4x + 1 = 9 or (4x + 1)2 = 81 A1 Must be from correct differential. x = 2, y = 6 or M is (2, 6) only. A1 Both values required. Must be from correct differential. 3 11(iii) Realises area is ∫y dx and attempt to use their 2 and sight of 0. *M1 Needs to use their integral and to see ‘their 2’ substituted. 1 DM1 Uses both 0 and ‘their 2’ and subtracts. Condone Uses limits 0 to 2 correctly → [4.5 + 13.5] – [ + 4.5] ( = 13⅓ ) wrong way round. 6 (Area =) 1⅓ or 1.33 A1 Must be from a correct differential and integral. 3 13⅓ or 1⅓ with little or no working scores M1DM0A0.