Use substitution to form and solve a quadratic — Paper 2 · IGCSE Mathematics - Additional 0606

11 y 4y = 5x + 20 Q y = 5+ 10x P O R x The diagram shows part of the curve y = 5 + 10 x and the line 4y = 5x + 20 . The line and curve intersect at the points P(0, 5) and Q. The line QR is parallel to the y-axis. (i) Find the coordinates of Q. [4] (ii) Find the area of the shaded region. You must show all your working. [6]

10 marks

Mark scheme: 11(i) 5 x + 20 M1 or better; equates and solves as far as 5 + 10 x = → 20 + 4 10 x = 5 x + 20 clearing the fraction 4  x  4 10 M1 Simplifies as far as x =L = x = oe    x  5 x = 6.4 cao A1 squares and simplifies to 6.4 [ y = ]13 B1 11(ii) (area of trapezium = ) their 57.6 B1 FT x = their 6.4, y = their 13 using any valid method 6.4 M1 ∫0 ( 5 + 10 x ) d x 1 3 M1 1 3 2 10 x ) 2 dx = k (10 x ) 10 x 2 d x = k 10 ( x ) ∫ ( 2 or or ∫  3  A1  1 3  2 10 x ) 10 ) ( x ) 2   2 ( 2   2 ( 5 x + or 5 x +  3 × 10   3       3  M1 limits used correctly or correct FT and 10 × 6.4 )  2 ( 2  subtraction of trapezium; their 5(6.4) + − their 57.6 oe  3 × 10  992   their − their 57.6 15 128 A1 allow 8.5333333 … rot to 4 or more sf or 8.53 oe 15

This question in 0606/23 May/June 2017

2 Solve the equation 0 .5 -0 .5 = x . [5] x + 5 x

5 marks

Mark scheme: 2 2 x1.5 + 6 x −0.5 = x x 0.5 + 5 x −0.5 M1 Attempt to multiply by x 0.5 + 5 x− 0.5 ( ) or x 0.5 or divide by x 0.5 2 x1.5 + 6 x − 0.5 = x1.5 + 5 x 0.5 or A1 Simplified numerical powers x1.5 − 5 x 0.5 + 6 x− 0.5 = 0 6 2 2 x + 2 x + 6 x or = x or = x x + 5 5 1 + x x 2 − 5 x + 6 = 0 M1 M1dep obtain a three term quadratic. Allow errors in signs and coefficients but not powers ( x − 3 )( x − 2 ) = 0 M1 Solve a three term quadratic x = 3 or 2 only A1

This question in 0606/22 Oct/Nov 2017

3 (i) Write 8 + 7x - x 2 in the form a - (x - b) 2 , where a and b are constants. [3] (ii) Hence state the maximum value of 8 + 7 x - x 2 and the value of x at which it occurs. [2] (iii) Using your answer to part (i), or otherwise, solve the equation 8 + 7z 2 - z 4 = 0 . [3]

8 marks

Mark scheme: 3(i) 2 3 7 81  7  B1 b = −  x −  2 4  2   7  2 M1 ± 8 ±   seen  2  or expand given form and equate for 8 or 7 A1 fully correct 3(ii) 81 2 maximum their B1 4 7 when x = their B1 2 from their correct form 3(iii)  2 7  2 81 M1 replace x by z 2 in their (i) and  z −  = oe  2  4 equate to zero. 7 9 M1 z2 = ± 2 2 z = ± 8 A1

This question in 0606/23 Oct/Nov 2018

10 y B y = 1 + x + 5 x y - 3x = 3 A O D C x The diagram shows the curve y = 1 + x + 5 x and the straight line y - 3x = 3 . The curve and line intersect at the points A and B. The lines BC and AD are perpendicular to the x-axis. (i) Using the substitution u2 = x, or otherwise, find the coordinates of A and of B. You must show all your working. [6]

12 marks

Mark scheme: 10(i) Eliminates x or y e.g. M1 3 x + 3 = x + 5 x + 1 or 3 + 3u 2 = u 2 + 5u + 1 Rearranges to a 3-term quadratic e.g. A1 0 = 2 x − 5 x + 2 or 0 = 2u 2 − 5u + 2 Factorises or solves 0 = 2 x − 5 x + 2 oe M1 or 0 = 2u 2 − 5u + 2 oe x = 0.5 , x = 2 A1 or u = 0.5 , u = 2 A(0.25, 3.75) B (4, 15) oe A2 A1 for each or for x = 0.25 and x = 4 10(ii) Method 1: Finding the area of the trapezium and subtracting Valid method to find the area of the trapezium soi M1 1125 5 A1 or 35 or 35.2 or 35.15625 rot to 4 or 32 32 more figs, soi Attempts to integrate M1 their 4 ∫their 0.25( x + 5 x + 1)dx [ −their 35.2] their 4 3 A1    x 2 5 x 2  + + x [ − their 35.2] oe   2 3  2    their 0.25 F(their 4) – F(their 0.25) [ −their 35.2] M1 45 13 A1 or or 2.8125 isw 16 216 or 2.81, or 2.812 Method 2: Finding the difference of two integrals Attempts to integrate M2 M1 for an attempt to form the their 4 difference with at most one error and ∫their 0.25( x + 5 x + 1 − (3 + 3 x ))dx attempts to integrate their 4 or ∫their 0.25( −2 x + 5 x − 2)dx oe their 4 A1 FT 3       −2 x 2 5 x 2   dep on at least M1 already awarded; their + − 2 x oe  must be at least 3 terms and, if FT, must     2 3    2   be of equivalent difficulty   their 0.25 F(their 4) – F(their 0.25) M1 45 13 A2 or or 2.81, 2.812 or 2.8125 16 216

This question in 0606/22 Feb/March 2019

4 (i) Given that y = 2x 2 - 4x - 7 , write y in the form a (x - b) 2 + c , where a, b and c are constants. [3] (ii) Hence write down the minimum value of y and the value of x at which it occurs. [2] (iii) Using your answer to part (i), solve the equation 2p - 4 p - 7 = 0 , giving your answer correct to 2 decimal places. [3]

8 marks

Mark scheme: 4(i) y = 2(x – 1)2 – 9 B3 a = 2, b = 1, c = –9 in correct form. B1 for each 4(ii) minimum their –9 B1 FT from their correct form, with a > 0 when x = their 1 B1 FT from their correct form, with a > 0 4(iii) x = p or p = x 2 soi B1 9 M1 − c ( x − 1) = ( x − b ) = 2 a 9 − c or p − 1 = oe p − b = ( ) ( ) 2 a using their values of a, b, c from (i) p = 9.74 A1 completion not involving use of quadratic formula

This question in 0606/23 Oct/Nov 2019

3 Find the exact solution of 3 2 x - 3 x + 1 - 4 = 0. [4]

4 marks

Mark scheme: 3 Rewrites in quadratic form soi M1 e.g. y = 3x then y 2 − 3 y − 4 = 0 or (3 x ) 2 − 3(3 x ) − 4 = 0 Factorises or solves their 3-term quadratic M1 e.g. (y + 1)(y – 4) [= 0] or (3x + 1)(3x – 4) [= 0] 3x = 4 A1 ignore 3x = −1 ln 4 A1 x = log 3 4 or oe, only ln3

This question in 0606/22 Feb/March 2020

7 (a) Solve the simultaneous equations 10 x + 2 y = 5, 10 3 x + 4 y = 50 , giving x and y in exact simplified form. [4] 3 - x 3 - 10 = 0. [3] (b) Solve 2x 2 1

7 marks

Mark scheme: 7(a) x + 2y = lg5 or B1 3x + 4y = lg50 Solves their linear simultaneous M1 equations x = lg2 or equivalent simplified form A1 1 5 A1 If A0 A0 then SC1 for a correct pair of y = lg or equivalent simplified unsimplified values or a correct pair of 2 2 decimal values correct to at least 3sf form 7(b)  1  1  M1  x 3 + 2  2 x 3 − 5  oe       1 5 M1 x 3 = −,2 2 125 A1 x = –8, 8

This question in 0606/21 May/June 2020

9 The following functions are defined for x 2 1. x + 3 2 f( )x = g( )x = 1 + x x - 1 (a) Find f g ( )x . [2] (b) Find g -1 ( )x . [2] (c) DO NOT USE A CALCULATOR IN THIS PART OF THE QUESTION. Solve the equation f( x) = g( x) . [5]

9 marks

Mark scheme: 9(a) x 2 + 4 2 B1 for an attempt at the correct oe, final answer [ fg( x ) = ] order of composition with at x 2 most one error 9(b) Complete, correct method to find the inverse M1  −1 A1 g ( x) =  x − 1 final answer 9(c) x 3 − x 2 − 4 = 0 M1 condone one sign or arithmetic error Shows x – 2 is a factor or shows that x = 2 is a M1 solution Uses x – 2 is a factor to find x 2 + x + 2 B2 B1 for a quadratic factor with 2 terms correct Indicates that x 2 + x + 2 has no real roots and A1 dep on all previous marks states x = 2 as the only solution awarded

This question in 0606/21 Oct/Nov 2021

1 (a) Solve the equation 5 w - 1 = 12, giving your answer correct to 2 decimal places. [2] (b) Solve the equation x 2 1 3 - 5x 3 + 6 = 0. [3]

5 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) w 1 log 5 12 M1 log12 or w 1 log5 w  2.54 cao A1 1(b) Rewrites in quadratic form e.g.: M1 1 y  x 3 y 2  5 y  6  0 2 1 1 3   5 x 3  6  0 or  x and factorises or solves e.g. : M1 Factorising their 3 term (y – 2)(y – 3) = 0 quadratic x = 8 and A1 x = 27

This question in 0606/21 May/June 2022

4 Solve the following equations, giving your answers to 3 significant figures. (a) 2 3 x + 1 = 5 x - 2 [3] 2y + 1 6 (b) e = 1 + 2 y + 1 [4] e

7 marks

Mark scheme: 4(a) (3x + 1)log2 = (x – 2)log5 oe B1 (3log2 – log5)x = −log2 – 2log5 M1 FT if of equivalent difficulty x = −8.32 A1 4(b) Writes as a quadratic in e2y + 1 M1 condone one error or states u = e2y + 1 and writes as a quadratic in u oe, soi 2 y +1 ( e ) 2 − e 2 y +1 − 6  = 0 oe A1 or u 2 − u − 6 [ = 0] oe (e 2 y +1 + 2)(e 2 y +1 − 3)  = 0  leading to A1 e 2 y +1 = 3 or (u + 2)(u – 3) [=0] leading to e 2 y +1 = 3 y = 0.0493 and no other solutions A1

This question in 0606/21 Oct/Nov 2022

6 (a) Write 3x 2 + 15 x - 20 in the form a ( x + b) 2 + c where a, b and c are rational numbers. [4] (b) State the minimum value of 3x 2 + 15x - 20 and the value of x at which it occurs. [2] 2 1 (c) Use your answer to part (a) to solve the equation 3y 3 + 15y 3 - 20 = 0 , giving your answers correct to three significant figures. [3]

9 marks

Mark scheme: 6(a) 2 B4 2  5  155  5  3  x +  − B2 for 3  x +  or 3 ( x + 2.5 )2  2  4  2   5  2 or B1 for  x +  or ( x + 2.5 )2  2  155 B2 for c = − or −38.75 4 25 or B1 for − −3 20 oe 4 6(b) 155 5 B2 FT their c from part a and −their b from Min value − when x is − 4 2 (a) B1 for either without contradiction 6(c) 1 2 M1 FT an expression of correct form from (a)   155 3 5 3  y +  = soi  2  4 1 A1 3 5 155 Rearranges as far as: y = −  2 12 soi y = 1.31 or −226 A1

This question in 0606/21 Oct/Nov 2022

5 DO NOT USE A CALCULATOR IN THIS QUESTION. (a) Show that x - 1 is a factor of the expression x 3 - 2 x 2 - 19x + 20 . [1] (b) Hence write x 3 - 2x 2 - 19x + 20 as a product of its linear factors. [3] (c) Hence find the exact solutions of the equation e 3 y - 2e 2 y - 19e y + 20 = 0 . [2]

6 marks

Mark scheme: 5(a) 13 – 2(12) – 19 + 20 = 0 1 5(b) (x – 1)(x2 – x – 20) M2 M1 for two terms correct in the quadratic factor (x – 1)(x + 4)(x – 5) A1 5(c) e y = 1, e y = 5 M1 y = 0, y = ln5 mark final answer A1 1.61 or decimal equivalent for ln5 seen is A0 as calculator use not permitted

This question in 0606/22 Feb/March 2023

1 (a) Write 19 - 12x - 3x 2 in the form a ( x + b) 2 + c where a, b and c are integers. [4] (b) Hence find the maximum value of 19 - 12x - 3 x 2 and the value of x at which this maximum occurs. [2] (c) Use your answer to part (a) to solve the equation 19 - 12 u - 3 u = 0 . [3]

9 marks

Mark scheme: Question Answer Marks Partial marks 1(a) −3(x + 2)2 + 31 B4 B2 for −3(x + 2)2 or B1 for (x + 2)2 or a = –3 and b = 2 B2 for c = 31 or B1 for –4  –3 + 19 soi 1(b) Maximum value 31 when x = –2 B2 Strict FT their c from part (a) and −their b from part (a) B1 for either without contradiction 1(c) 2 M1 FT an expression of correct form from part (a) −3 u + 2 = −31 oe ( ) 31 A1 Rearranges as far as u = −2 3 1.48 cao or 1.475[13…] rot to 3 or more dp A1 43 − 4 93 or isw 3

This question in 0606/21 Oct/Nov 2023