Algebra — Paper 2 · A Level Mathematics 9709

1 Solve the inequality |x −3| > |x + 2|. [4]

4 marks

Mark scheme: 1 EITHER State or imply non-modular inequality (x – 3)2 > (x + 2)2, or corresponding equation M1 Expand and solve a linear inequality, or equivalent M1 1 Obtain critical value 2 A1 1 1 State correct answer x < (allow x Y ) 2 2 A1 OR State a correct linear equation for the critical value, e.g. 3 – x = x + 2, or corresponding correct inequality, e.g. –(x – 3) > (x + 2) M1 Solve the linear equation, or inequality M1 1 Obtain critical value 2 A1 1 State correct answer x < 2 A1 OR Make recognisable sketches of both y = x − 3 and y = x + 2 on a single diagram B1 Obtain a critical value from the intersection of the graphs M1 1 Obtain critical value 2 A1 1 State final answer x < 2 A1 [4] ( )

This question in 9709/21 May/June 2007

1 Solve the inequality |3x - 1| < 2. [3]

3 marks

Mark scheme: 1 EITHER State or imply non-modular inequality (3x – 1)2 < 22, or corresponding equation or pair of linear equations M1 1 Obtain critical values – and 1 A1 3 1 State correct answer – < x < 1 A1 3 OR State one critical value, e.g. x = 1, by solving a linear equation (or inequality) or from a graphical method or by inspection B1 State the other critical value correctly B1 1 State correct answer – < x < 1 B1 [3] 3

This question in 9709/21 May/June 2008

1 Solve the inequality [4] |2x + 3| < |x −3|.

4 marks

Mark scheme: 1 EITHER: Obtain a non-modular inequality from (2x + 3)2 < (x – 3)2, or corresponding quadratic equation, or pair of linear equations 2x + 3 = ±(x – 3) M1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 Obtain critical values x = –6 and x = 0 A1 State answer –6 I x I 0 A1 OR: obtain the critical value x = –6 from a graphical method or by inspection, or by solving a linear equation or inequality B1 Obtain the critical value x = 0 similarly B2 State answer –6 I x I 0 B1 [4] 2

This question in 9709/21 Oct/Nov 2009

1 Solve the inequality [4] |x + 3| > |2x|.

4 marks

Mark scheme: 1 EITHER: Obtain a non-modular inequality from (x + 3)2 > (2x)2, or corresponding equation, or pair of linear equations (x + 3) = ± 2x M1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 Obtain critical values x = –1 and x = 3 A1 State answer –1 I x I 3 A1 OR: Obtain critical value x = 3 from a graphical method, or by inspection, or by solving a linear inequality or linear equation B1 Obtain the critical value x = –1 similarly B2 State answer –1 I x I 3 B1 [4] 2

This question in 9709/22 Oct/Nov 2009

4 The polynomial x3 3x2 4x 2 is denoted by + + + f(x). (i) Find the quotient and remainder when is divided by x2 x [4] f(x) + −1. (ii) Use the factor theorem to show that is a factor of [2] (x + 1) f(x).

6 marks

Mark scheme: 4 (i) Commence division by x2 + x – 1 obtaining quotient of the form x + k M1 Obtain quotient x + 2 A1 Obtain remainder 3x + 4 A1 Identify the quotient and remainder correctly A1√ [4] (ii) Substitute x = –1 and evaluate expression M1 Obtain answer 0 A1 [2] 1 1

This question in 9709/21 May/June 2010

3 Solve the inequality [4] |2x −1| < |x + 4|. 1

4 marks

Mark scheme: 3 EITHER State or imply non-modular inequality (2x –1)2 < (x + 4)2, or corresponding equation or pair of linear equations M1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 Obtain critical values –1 and 5 A1 State correct answer –1 < x < 5 A1 [4] OR Obtain one critical value, e.g. x = 5, by solving a linear equation (or inequality) or from a graphical method or by inspection B1 Obtain the other critical value similarly B2 State correct answer –1 < x < 5 B1  1 

This question in 9709/22 May/June 2010

7 The polynomial 2x3 ax2 bx 6, where a and b are constants, is denoted by It is given that + + + p(x). when is divided by the remainder is 30, and that when is divided by the p(x) (x −3) p(x) (x + 1) remainder is 18. (i) Find the values of a and b. [5] (ii) When a and b have these values, verify that is a factor of and hence factorise (x −2) p(x) p(x) completely. [4]

9 marks

Mark scheme: 7 (i) Substitute x = 3 and equate to 30 M1 Substitute x = –1 and equate to 18 M1 Obtain a correct equation in any form A1 Solve a relevant pair of equations for a or for b M1 Obtain a = 1 and b = –13 A1 [5] (ii) Either show that f(2) = 0 or divide by (x – 2), obtaining a remainder of zero B1 Obtain quadratic factor 2x2 + 5x – 3 B1 Obtain linear factor 2x – 1 B1 Obtain linear factor x + 3 B1 [Condone omission of repetition that x – 2 is a factor.] [If linear factors 2x – 1, x + 3 obtained by remainder theorem or inspection, award B2 + B1.] [4]

This question in 9709/23 May/June 2010

1 Solve the inequality [3] |x + 1| > |x −4|.

3 marks

Mark scheme: 1 EITHER: State or imply non-modular inequality ( x + 1)2 > ( x − 4 )2 , or corresponding equation or pair of linear equations M1 3 Obtain critical value A1 2 3 State correct answer x > A1 2 OR: State a correct linear equation for the critical value, e.g. x + 1 = − x + 4 , or corresponding correct linear inequality, e.g. x + 1 > − ( x − 4 ) M1 3 Obtain critical value A1 2 3 State correct answer x > A1 [3] 2

This question in 9709/21 Oct/Nov 2010

1 Solve the inequality [3] |x + 1| > |x −4|.

3 marks

Mark scheme: 1 EITHER: State or imply non-modular inequality ( x + 1)2 > ( x − 4 )2 , or corresponding equation or pair of linear equations M1 3 Obtain critical value A1 2 3 State correct answer x > A1 2 OR: State a correct linear equation for the critical value, e.g. x + 1 = − x + 4 , or corresponding correct linear inequality, e.g. x + 1 > − ( x − 4 ) M1 3 Obtain critical value A1 2 3 State correct answer x > A1 [3] 2

This question in 9709/22 Oct/Nov 2010

1 Solve the equation [3] |3x + 4| = |2x + 5|.

3 marks

Mark scheme: 1 EITHER Attempt to square both sides obtaining three terms on each side M1 Attempt solution of three-term quadratic equation M1 9 Obtain 5 x + 4 x − 9 = 0 and hence − and 1 A1 5 OR Obtain value 1 from graphical method, inspection or linear equation B1 9 Obtain value − similarly B2 [3] 5 dx dy

This question in 9709/21 May/June 2011

2 Solve the inequality [4] |2x −3| ≤|3x|.

4 marks

Mark scheme: 2 EITHER State or imply non-modular inequality (2x – 3)2 Y (3x)2, or corresponding equation or pair of linear equations M1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 3 Obtain critical values –3 and A1 5 3 State correct answer x Y –3 or x [ A1 5 OR State one critical value, e.g. x = –3, by solving a linear equation (or inequality) or from a graphical method or by inspection B1 State the other critical value correctly B2 3 State correct answer x Y –3 or x [ B1 [4] 5 2

This question in 9709/23 Oct/Nov 2011

1 Solve the inequality [4] |x + 3| < |2x + 1|.

4 marks

Mark scheme: 1 Either: State or imply non-modular inequality (x + 3)2 < (2x + 1)2 or corresponding equation or pair of linear equations B1 Attempt solution of 3-term quadratic or of 2 linear equations M1 Obtain critical values − 43 and 2 A1 State answer x < − 43 , x > 2 A1 Or: Obtain critical value x = 2 from graphical method, inspection, equation B1 Obtain critical value x = − 43 similarly B2 State answer x < − 43 , x > 2 B1 [4] 2

This question in 9709/22 May/June 2012

1 Solve the inequality [3] |2x + 1| < |2x −5|.

3 marks

Mark scheme: 1 EITHER State or imply non-modular inequality (2 x + 1)2 < (2 x − 5 )2 , or M1 corresponding equation or pair of linear equations Obtain critical value 1 A1 State correct answer x < 1 A1 OR State the critical value x = 1, by solving a linear equation (or inequality) or from a graphical method or by inspection B2 State correct answer x < 1 B1 [3]

This question in 9709/22 Oct/Nov 2012

1 Solve the inequality [3] |x −2| ≥|x + 5|.

3 marks

Mark scheme: 1 EITHER State or imply non-modular inequality ( x − 2 )2 ≥ ( x + 5 )2 , or corresponding equation or pair of linear equations M1 3 Obtain critical value − A1 2 3 State correct answer x ≤ − A1 2 OR State a correct linear equation for the critical value, e.g. x – 2 = – x – 5, or corresponding correct linear inequality, e.g. x – 2 ≥ – x – 5 M1 3 Obtain critical value − A1 2 3 State correct answer x ≤ − A1 [3] 2

This question in 9709/23 Oct/Nov 2012

2 Solve the inequality x −8 > 2x −4 . [4]

4 marks

Mark scheme: 2 Either State or imply non-modular inequality (x – 8)2 > (2x – 4)2, or corresponding equation or pair of linear equations M1 Make reasonable solution attempt at a quadratic, or solve two linear equations M1 Obtain critical values 4 and – 4 A1 State correct answer – 4 < x < 4 A1 Or Obtain one critical value, e.g. x = 4, by solving a linear equation (or inequality) or from a graphical method or by inspection B1 Obtain the other critical value similarly B2 State correct answer – 4 < x < 4 B1 [4]

This question in 9709/22 May/June 2013

1 Solve the inequality 3x x 4 . [4] −2 ≥ +

4 marks

Mark scheme: 1 Either State or imply non-modular inequality (3 x − 2 )2 >( x + 4 )2 or corresponding equation or pair of linear equations B1 Attempt solution of 3-term quadratic equation or of 2 linear equations M1 Obtain critical values − 12 and 3 A1 State answer x < − 12 , x > 3 A1 [4] Or Obtain critical value x = 3 from graphical method, inspection, equation B1 Obtain critical value x = − 12 similarly B2 State answer x < − 12 , x > 3 B1 [4] 2

This question in 9709/21 May/June 2014

1 (i) Solve the equation x 2 x . [2] + = −13 (ii) Hence solve the equation 3y 2 3y , giving your answer correct to 3 significant figures. + = −13 [2]

4 marks

Mark scheme: 1 (i) Either Square both sides to obtain linear equation M1 Obtain x = 16530 or 336 or 112 A1 [2] Or Solve linear equation in which, initially, signs of x are different M1 Obtain x + 2 = − x + 13 or equivalent and hence 112 or equivalent A1 [2] (ii) Apply logarithms and use power law M1 Obtain y log 3 = log 112 and hence y = .155 A1 [2]

This question in 9709/22 May/June 2014

1 (i) Solve the equation x 2 x . [2] + = −13 (ii) Hence solve the equation 3y 2 3y , giving your answer correct to 3 significant figures. + = −13 [2]

4 marks

Mark scheme: 1 (i) Either Square both sides to obtain linear equation M1 Obtain x = 16530 or 336 or 112 A1 [2] Or Solve linear equation in which, initially, signs of x are different M1 Obtain x + 2 = − x + 13 or equivalent and hence 112 or equivalent A1 [2] (ii) Apply logarithms and use power law M1 Obtain y log 3 = log 112 and hence y = .155 A1 [2]

This question in 9709/23 May/June 2014

1 (i) Solve the equation 3x 4 3x . [3] + = −11 (ii) Hence, using logarithms, solve the equation 3 2y 4 3 2y , giving the answer correct × + = × −11 to 3 significant figures. [2]

5 marks

Mark scheme: 1 (i) State or imply equation (3 x + 4) 2 = (3 x − 11) 2 or 3 x + 4 = − (3 x − 11) B1 Attempt solution of ‘quadratic’ equation or linear equation M1 7 Obtain x = or equivalent (and no other solutions) A1 [3] 6 (ii) Use logarithms to solve equation of form 2y = their answer to (i) ( must be + ve) M1 Obtain 0.222 (and no other solutions) A1 [2]

This question in 9709/21 May/June 2015

1 (i) Solve the equation 3x 5. [3] −2 = (ii) Hence, using logarithms, solve the equation 3 5y 5, giving the answer correct to 3 significant figures. × −2 = [2]

5 marks

Mark scheme: 1 (i) Either Square both sides to obtain three-term quadratic equation M1 Solve three-term quadratic equation to obtain two values M1 Obtain –1 and 73 A1 Or Obtain 73 from graphical method, inspection or linear equation B1 Obtain –1 similarly B2 [3] (ii) Use logarithmic method to solve an equation of the form 5 y = k where k > 0 M1 Obtain 0.526 and no others A1 [2]

This question in 9709/22 Oct/Nov 2015

1 Solve the equation x a 2x , giving x in terms of the positive constant a. [3] + = −5a … … … … … … … … … … … …

3 marks

Mark scheme: 1 2 2 2 5 x a x a + = − or pair of linear equations B1 SR B1 for 6 x a Attempt solution of quadratic equation or of pair of linear equations M1 Allow M1 if 4 3 and 6 seen Obtain, as final answers, 6a and 4 3 a A1 Total: 3

This question in 9709/23 May/June 2017

1 (i) Solve the equation 9x 3x 2 . [3] −2 = + … … … … … … … … … … … … … … … (ii) Hence, using logarithms, solve the equation 2 , giving your answer correct to 3 significant figures. 3y+2 −2 = 3y+1 + [2] … … … … … … … …

5 marks

Mark scheme: 1(i) State or imply non-modular equation 2 2 (9 2) (3 2) x x − = + or pair of linear equations B1 Attempt solution of quadratic equation or of 2 linear equations M1 Obtain 0 and 2 3 A1 SC: B1 for one correct solution 3 1(ii) Apply logarithms and use power law for 3y k = where 0 k > M1 Must be using their answers to part (i) Obtain 0.369 − A1 2

This question in 9709/21 Oct/Nov 2018

6 8xn5 The sequence of values given by the iterative formula + with initial value 2 converges xn+1 = 8 x2 x1 = n + to !. (a) Use the iterative formula to find the value of correct to 4 significant figures. Give the result of ! each iteration to 6 significant figures. [3] … … … … … … … … … … … (b) State an equation satisfied by and hence determine the exact value of [2] ! !. … … … … … … … … … …

5 marks

Mark scheme: 5(a) Use iteration correctly at least once M1 Need to see 3 values including the starting values Obtain final answer 1.817 A1 Answer required to exactly 4 significant figures Show sufficient iterations to 6 significant figures to justify answer or show sign change in interval [1.8165, 1.8175] A1 3 5(b) State equation 2 6 8 8 x x x + = + or equivalent using α B1 Obtain 3 6 or exact equivalent B1 2

This question in 9709/23 Oct/Nov 2020

2 The solutions of the equation 5 x 5 are x a and x b, where a b. = −2x = = < Find the value of 3a 7b . [5] −1 + −1 … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 Solve 5 5 2 = − x x to obtain 5 7 = x B1 Allow AWRT 0.714 Attempt solution of linear equation where signs of 5x and 2x are the same M1 Obtain 5 3 = − x A1 Allow AWRT –1.67 Substitute their values correctly M1 Substitution must be seen unless implied by a correct answer. Their values must come from consideration of 5 5 2 = − x x Obtain 6 4 − + and hence 10 A1 Alternative method for Question 2 State or imply non-modulus equation 2 2 25 (5 2 ) = − x x B1 Attempt solution of 3-term quadratic equation M1 Obtain 5 3 − and 5 7 A1 Allow AWRT 0.714 and AWRT -1.67 Substitute their values correctly M1 Substitution must be seen unless implied by a correct answer. Their values must come from consideration of 5 5 2 = − x x Obtain 6 4 − + and hence 10 A1 5

This question in 9709/23 May/June 2021

1 Solve the equation 5x 4x 9 . [3] −2 = + … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 Solve 5 2 4 9 − = + x x to obtain 11 = x Attempt solution of linear equation where signs of 5x and 4x are different M1 Obtain final value 7 9 = − x A1 Alternative method for question 1 State or imply non-modulus equation 2 2 (5 2) (4 9) − = + x x B1 Attempt solution of 3-term quadratic equation M1 Obtain x = 7 9 − and x =11 A1 3

This question in 9709/22 Feb/March 2022

2 The solutions of the equation 4x −1 = x + 3 are x = p and x = q, where p < q. Find the exact values of p and q, and hence determine the exact value of p −2 − q −1 . [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 Solve 4 x −=1 x + 3 to obtain x = 43 B1 Attempt solution of linear equation where signs of 4x and x are different M1 Obtain final value x = − 52 A1 Substitute numerical values and apply modulus signs correctly to obtain M1 Allow their p and q, p < q. − 125 − 13 or equivalent, retaining exactness and with no subsequent squaring Obtain 1531 A1 or exact equivalent. Alternative method for Question 2 State or imply non-modulus equation (4 x − 1) 2 = ( x + 3) 2 B1 Attempt solution of 3-term quadratic equation M1 Obtain final values − 52 and 43 A1 2 Substitute numerical values and apply modulus signs correctly to obtain M1 Allow their p  q . − 125 − 13 or equivalent, retaining exactness and with no subsequent squaring Obtain 1531 A1 or exact equivalent. 5

This question in 9709/22 Oct/Nov 2022

4 (a) Sketch, on the same diagram, the graphs of y 3 and y 9 [2] = −x = −2x. (b) Solve the inequality 3 9 [3] −x > −2x. … … … … … (c) Use logarithms to solve the inequality 500. Give your answer in the form x a, where 23x−10 <figures. < [3] the value of a is given correct to 3 significant … … … … … (d) List the integers that satisfy both of the inequalities 3 9 and 500. [1] −x > −2x 23x−10 < … …

9 marks

Mark scheme: 4(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw (more or less) correct graph of y = 9 − 2 x with steeper negative B1 Dependent on first B mark, appropriately positioned gradient with respect to first graph. 2 4(b) Solve linear equation or inequality with signs of x and 2x different M1 Obtain critical value 4 A1 Conclude x  4 only A1 6 must be discounted. Alternative Method for Question 4(b) State or imply non-modulus equation (or inequality) (3 − x ) 2 = (9 − 2 x ) 2 B1 Attempt solution of three-term quadratic equation (or inequality) M1 Dependent on previous B1. Conclude x  4 only A1 6 must be discounted. 3 4(c) State or imply (3 x − 10)ln2  ln500 B1 Or equivalent perhaps involving different logarithm base. Obtain critical value 6.32 B1 Obtain x  6.32 B1 Or greater accuracy. 3 4(d) State 5 and 6 only B1 1

This question in 9709/22 Oct/Nov 2023

1 Solve the inequality 5x + 7 2 2x - 3 . [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1 Solve 5 7 2 3 x x to obtain 10 3  B1 Or inequality. Attempt solution of linear equation where 5x and 2x have different signs M1 Or inequality. Obtain 4 7  A1 State 10 4 3 7 , x x   A1 A0 if ‘… and …’ used. Alternative Method for Question 1 State or imply non-modulus equation 2 2 (5 7) (2 3)    x x (B1) Or inequality. Attempt solution of three-term quadratic equation (M1) Or inequality. Obtain 10 3  and 4 7  (A1) State 10 4 3 7 , x x   (A1) A0 if ‘… and …’ used. 4

This question in 9709/22 May/June 2024

6 (a) Find the quotient and remainder when 18x 3 - 6 x 2 - 30x + 4 is divided by ( 3x - 1 ) . [3] … … … … … … 5 3 2 18x - 6x - 30x + 4 (b) Hence find d x . Give your answer in the form a - ln b , where a and b are y 1 3x - 1 integers. [5] … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Carry out division at least as far as 6 x 2 + k1 M1 Obtain quotient 6 x 2 − 10 A1 Obtain remainder − 6 A1 3 6(b) 2 6 B1 FT Following their quotient and remainder. Identify integrand as 6 x − 10 − 3 x − 1 Integrate to obtain at least 2x 3 and term of the form k 2 ln(3 x − 1) *M1 Obtain 2 x 3 − 10 x − 2ln(3 x − 1) A1 FT Following their quotient and remainder. Apply limits and appropriate logarithm properties DM1 Obtain 208 − ln49 A1 5

This question in 9709/22 Feb/March 2025

2 (a) Sketch on the same diagram the graphs of y = 2x - 9 and y = 4x - 5 . [2] (b) Solve the inequality 2x - 9 1 4 x - 5 . [3] … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw correct graph of y = 4 x − 5 B1 Correctly placed with reference to modulus graph and with steeper gradient. 2 2(b) Solve linear equation or inequality with signs of 2x and 4x different M1 Obtain −2 x + 9 = 4 x − 5 and hence x = 73 A1 OE Conclude x  7 A1  7   7  3 OE, e.g. ,  , or ,  .      3   3  Alternative Method for Question 2(b) State or imply non-modulus equation (or inequality) (2 x − 9) 2 = (4 x − 5) 2 B1* Attempt solution of three-term quadratic equation (or inequality) DM1 12 x 2 − 4 x − 56 = 0 leading to 73 and −2. Obtain at least 7 and conclude x  7 A1 Must be from correct work. 3 3  7   7  OE, e.g. ,  or ,       3   3  3

This question in 9709/25 May/June 2025