Solve equations of the type — Paper 2 · IGCSE Mathematics - Additional 0606

6 (a) (i) Express - 8x 9 x -3 in the form axb, where a and b are constants to be found. [2] 3 9 6 -3 (ii) Hence solve the equation - 8x x =- 6250 . [2] ` `j j (b) It is given that y = log a ( ax) + 2 log a ( 4x - 3) - 1, where a is a positive integer. (i) Explain why x must be greater than 0.75. [1] - 24x + 9x) . [3] (ii) Show that y can be written as log a ( 16x 3 2 (iii) Find the value of x for which y = log a ( 9x) . [2]

10 marks

This question in 0606/22 Feb/March 2017

2 Without using a calculator, express KK OO in the form a + b 5 , where a and b are integers. [5] 3 - 5 L P

5 marks

Mark scheme: Question Answer Marks Partial Marks 1 5x + 3 = 3x – 1 oe or 5x + 3 = 1 – 3x oe M1 x = –2 and x = –0.25 only A2 nfww mark final answer A1 for x = –2 ignoring extras implies M1 if no extras seen If M0 then SC1 for any correct value with at most one extra value Alternative method ( 5 x + 3 ) 2 = (1 − 3 x ) 2 oe soi M1 16 x 2 + 36 x + 8 = 0 oe A1 x = –0.25, x = –2 only; mark final answer A1 2 Without using a calculator… Sufficient evidence must be seen to be convinced that a calculator has not been used. Withhold the mark for any step that is unsupported. deals with the negative index soi B1 2  3 − 5  e.g.      1 + 5  3 − 5 1 − 5 M1 1 + 5 3 + 5 rationalises × oe allow for × 1 + 5 1 − 5 3 − 5 3 + 5 3 − 4 5 + 5 A1 3 + 4 5 + 5 multiplies out correctly oe allow for 1 − 5 9 − 5 squares correct binomial A1 2 allow for 2 + 5 = 4 + 4 5 + 5 2 ( ) ( ) −+2 5 = 4 − 4 5 + 5 oe ( ) ( ) 9 − 4 5 cao A1 dep on all previous marks awarded 2 Alternative method 1: dealing with the negative index soi B1 correctly squaring with at least 3 terms in the B1 numerator and denominator 3 − 5 3 − 5 9 − 6 5 + 5 × = oe 1 + 5 1 + 5 1 + 2 5 + 5 14 − 6 5 6 − 2 5  M1 rationalising their ×  oe    6 + 2 5 6 − 2 5  multiplying out correctly; at least 3 terms in the A1 numerator but condone a single value for the 84 − 64 5 + 60 denominator oe 36 − 20 9 − 4 5 cao A1 Alternative method 2 dealing with the negative index soi B1 M1 9 − 6 5 + 5 = a + b 5 1 + 2 5 + 5 ( )( ) 14 = 6 a + 10b A1 oe −=6 2 a + 6b a = 9 cao A1 b = –4 cao A1 Alternative method 3 for dealing with the negative index soi B1 M1 [ 3 − 5 = c + d 5 1 + 5 leading to] ( )( ) c + 5d = 3 c + d = −1 c = –2 and d = 1 A1 2 A1 −+2 5 = 4 − 4 5 + 5 ( ) 9 − 4 5 cao A1

This question in 0606/22 May/June 2017

7 (a) Given that a 7 = b , where a and b are positive constants, find, (i) log a b , [1] (ii) log b a . [1] 1 (b) Solve the equation log 81 y =- . [2] 4 32 x 2 - 1 (c) Solve the equation 2 = 16 . [3] 4 x

7 marks

Mark scheme: 7(a)(i) 7 B1 7(a)(ii) 1 1 B1 FT their 7 must not be 1 if following or through 7 their 7 7(b) − 1 − 1 M1 Anti-logs y = 81 4 or y = 3−1 or y = 9 2 oe 1 A1 nfww; implies the M1; y = only or 0.333[3….] only y = …. must be seen at least once 3 1 4 1 If M0 then SC1 for e.g. 81−= as final 3 answer 7(c) 2 5( x 2 −1) 2 B1 converts the terms given left hand side to 2 5( x −1) 4 2 32 x × 32 −1 powers of 2 or 4; may have cross- 2 oe or 2 oe or 2 (2 2 ) x 4 x 4 x multiplied or log32 x 2 −−1 log4 x 2 = log16 oe or separates the power in the numerator correctly or applies a correct log law 3 x 2 5 2 M1 combines powers and takes logs or 2 −= 16 oe ⇒ 3 x − 5 = 4 oe 3 2 5 equates powers; x 2 2 3 2 5 or 4 −= 16 oe ⇒ x − = 2 oe 2 2 or brings down all powers for an equation 8 x 2 2 already in logs or = 16 oe ⇒ x log8 = log512 oe 32 2 2 condone omission of necessary brackets or ( x − 1)log32 − x log 4 = log16 oe for M1; condone one slip [ x = ] ± 3 isw cao A1 or ± 1.732050... rot to 3 or more figs. isw

This question in 0606/22 May/June 2017

2 Solve the equation 3x - 1 = 5 + x . [3]

3 marks

Mark scheme: 2 3 x −=1 5 + x x = 3 B1 3 x −=1 −−5 x oe M1 M1 not earned if incorrect equation(s) present x = –1 A1

This question in 0606/23 Oct/Nov 2017

3 Do not use a calculator in this question. (a) Simplify ( 2 + 2 5 )(4 2 - 3 5 ) , giving your answer in the form a + b c , where a, b and c are integers. [3] 4 - 3 6 (b) Simplify , giving your answer in the form p 3 + q 2 , where p and q are integers. [4] 3 + 2

7 marks

Mark scheme: 3(a) Expand 4 terms: M1 8 + 8 10 − 3 10 − 30 –22 A1 5 10 A1 3(b) M1 Multiply numerator and 4 − 3 6 3 − 2 ( ) ( ) × denominator by 3 − 2 ( ) 3 + 2 3 − 2 ( ) ( ) M1 Expand 4 3 − 3 18 − 4 2 + 3 12 3 − 2 10 3 −13 2 A2 A1for each term

This question in 0606/21 Oct/Nov 2018

3 (i) Sketch the graph of y = 5x - 3 on the axes below, showing the coordinates of the points where the graph meets the coordinate axes. y O x [3] (ii) Solve the equation 5x - 3 = 2 - x . [3] 2

6 marks

Mark scheme: 3(i) Correct shape 3 B1 correct shape must have cusp on x- 0.6 oe indicated on x-axis axis 3 indicated on y-axis B1 for each correct point There must be a sketch to award the marks for the intercepts and sketch should be continuous with one intersection only on each axis 3(ii) Solves 5 x − 3 = x − 2 oe M1 or (5 x − 3) 2 = (2 − x ) 2 1 A1 [ x = ] oe 4 5 B1 [ x = ] oe 6

This question in 0606/21 May/June 2019

1 (i) On the axes below, draw the graph of y = 2x - 3 . y 8 6 4 2 -2 0 2 4 x -2 -4 [2] (ii) Solve the equation 7 - 2x - 3 = 0 . [3]

5 marks

Mark scheme: Question Answer Marks Partial Marks 1(i) 6 y B2 B1 shape 4 B1 Correct intersection with axes. 2 x −2 2 4 −2 −4 1(ii) 7 = 2x – 3 → x = 5 B1 Uses 7 = 3 – 2x oe M1 x = –2 A1

This question in 0606/21 Oct/Nov 2019

5 (a) On the axes below, sketch the graph of y = 5x - 7 , showing the coordinates of the points where the graph meets the coordinate axes. [3] y O x (b) Solve 5 5x - 7 - 1 = 14 . [3]

6 marks

Mark scheme: 5(a) Correct V shape with vertex on positive x- B1 axis (0, 7) B1  7  B1  , 0   5  5(b) x = 2 B1 5 x − 7 = their ( − 3) oe, soi M1 or 25 x − 35 = their ( − 15) oe, soi 4 A1 x = oe 5 Alternative method 25 x 2 − 70 x + 40 = 0 oe (B1 factorising e.g. ( 5 x − 4 )( x − 2 ) M1 4 A1) x = 2, 5

This question in 0606/22 Feb/March 2020

10 y y = ( x - 3) 2 y = 4x 23 B ( b, 4) O A ( a, 0) x 2 The diagram shows part of the graphs of y = 4x 3 and y = ( x - 3 ) 2 . The graph of y = ( x - 3) 2 meets the x-axis at the point A(a, 0) and the two graphs intersect at the point B(b, 4). (a) Find the value of a and of b. [2]

7 marks

Mark scheme: 10(a) a = 3, b = 1 B2 B1 for each 10(b) their b 2 their a 2 M1 3 0 4 x d x + their b ( x − 3 ) d x their b 3 their a M2 M1 for each, soi 5    3 ( x − 3 )  × 4 x 3   soi   +  5  0  3  their b 12 12 M1 ( their b ) − ( 0 ) + 5 5 ( their a − 3 ) 3 ( their b − 3 ) 3 − 3 3 76 1 A1 or or 15 515 5.07 or 5.06 rot to four or more figs; cao

This question in 0606/21 May/June 2020

2 DO NOT USE A CALCULATOR IN THIS QUESTION. 10 + 2 5 The point 1 - 5, p lies on the curve y = 2 . Find the exact value of p, simplifying your ` j x answer. [5]

5 marks

Mark scheme: 2 2 B1 or rationalises Squares: (1 − 5 ) = 1 − 5 − 5 + 5 10 + 2 5 (1 + 5 ) 2 × (1 − 5 ) 2 (1 + 5 ) 2 Rationalises, e.g. B1 or squares 10 + 2 5 6 + 2 5 (1 + 5 ) 2 = 1 + 5 + 5 + 5 × 6 − 2 5 6 + 2 5 Multiplies out, e.g. M1 Multiplies out 60 + 20 5 + 12 5 + 4(5)  10 + 2 5 6 + 2 5   2 × 2 =  36 − 20  (1 − 5 ) (1 + 5 )  60 + 20 5 + 12 5 + 4(5) (1 − 5 ) 2 5 + 2 5 A2 A1 for k + 2 5 or 5 + k 5

This question in 0606/22 May/June 2020

1 DO NOT USE A CALCULATOR IN THIS QUESTION. 6 + x A curve has equation y = where x H 0 . Find the exact value of y when x = 6 . Give your 3 + x answer in the form a + b c , where a, b and c are integers. [3]

3 marks

Mark scheme: Question Answer Marks Partial Marks 1  6 3  6 M1  y  6  oe, soi 3  6 3  6 Correctly multiplies out correct expression: M1  6 6  3 6  6  y  18 oe 9  6  y  4  6 A1 not from wrong working

This question in 0606/22 May/June 2022

1 (a) Solve the equation = 1. [2] 7 (b) y 14 12 10 8 6 4 2 x – 10 – 8 – 6 – 4 – 2 0 2 4 6 8 10 – 2 – 4 – 6 The diagram shows the graph of y = 3 x + 9 . By drawing a suitable graph on the same diagram, solve the inequality 3x + 9 G x - 5 . [3]

5 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 4x – 5 = 7 and 4x – 5 = 7 oe, soi M1 1 A1 x = 3, x =  2 1(b) Correct graph 3 B1 for correct graph y 14 and 12 10 8 6 B2 dep for 7  x   1 ; 4 dependent on a correct graph for –7 ⩽ x ⩽ 2 –1 -10 -8 -6 -4 -2 0 2 4 6 8 10 x or -2 B1 STRICT FT for their critical values -4 from the two intersections of their straight-line section of graph providing it -6 has negative gradient AND 7  x   1

This question in 0606/23 May/June 2023

3 DO NOT USE A CALCULATOR IN THIS QUESTION. A cylinder has base radius ( 2 + 3) m and volume r ( 16 + 9 3) m3 . Find the exact value of its height, giving your answer in its simplest form. [4]

4 marks

Mark scheme: 3 16 + 9 3 B1 (2 + 3) 2 M1 16 + 9 3 7 − 4 3 c 16 + 9 3 ( )( ) ( ) FT where a, b and c are non- 7 + 4 3 7 − 4 3 a + b 3 ( )( ) zero constants 16 + 9 3 7 − 4 3 or  7 + 4 3 7 − 4 3 112 − 64 3 + 63 3 − 108 A1 −112 + 64 3 − 63 3 + 108 or −1 4 − 3 or − 3 + 4 cao, nfww A1 Alternative method 16 + 9 3 (B1) (2 + 3) 2 2 16 + 9 3 2 − 3 ) ( )( 16 + 9 3 ( 2 − 3 ) 2 (M1) 2 2 or 2  2 ( 2 + 3 ) ( 2 − 3 ) ( 2 + 3 ) ( 2 − 3 ) 112 − 64 3 + 63 3 − 108 (A1) or 64 − 32 3 − 32 3 + 48 + 36 3 − 54 − 54 + 27 3 4 − 3 or − 3 + 4 cao, nfww (A1)

This question in 0606/22 Oct/Nov 2023

1 (a) Solve the equation 2 8 - 4x + 5 = 25 . [3] 2 57 - 9x (b) Solve the inequality 16 x - 5x - 3 1 . [4] 6

7 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 8 – 4x = 10 oe soi M1 and 8 – 4x = −10 oe soi OR 16 x 2 − 64 x − 36  = 0 oe 1 9 A2 mark final answer x = − , x = 2 2 1 9 A1 for x = − or x = 2 2 1(b) −30 x 2 + 105 x − 75 *0  oe M1 condone one sign or arithmetic error where * is any inequality sign or = Critical values 2.5 and 1 2 M1 for factorises or solves a 3-term quadratic to find critical values x < 1 x > 2.5 A1 mark final answer

This question in 0606/22 Feb/March 2024

1 (a) On the axes, sketch the graph of y = 4x - 6 , showing the points where the graph meets the axes. [2] y O x (b) Solve the equation 4x - 6 = 2 x . [3]

5 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) Fully correct graph with intercepts marked B2 B1 for a graph of correct shape with vertex on x-axis y (0, 6) O (1.5, 0) x 1(b) 4x – 6 = 2x and 4x – 6 = –2x oe M1 x = 3 x = 1 A2 A1 for either correct Alternative method 12x2 – 48x + 36 = 0 oe (B1) Factorises or solves (M1) x = 1, x = 3 (A1)

This question in 0606/21 May/June 2024