Mark Scheme Pure Mathematics Year 1 (AS) Unit Test 3 .

Mark schemeMarksAOsPearsonProgression Stepand ProgressdescriptorCorrectly shows that either 1 f(3) 0, f( 2) 0 or f 0 2 M13.1a4thDraws the conclusion that (x – 3), (x 2) or (2x 1) musttherefore be a factor.M12.2aEither makes an attempt at long division by setting up thelong division, or makes an attempt to find the remainingfactors by matching coefficients. For example, statingM11.1bA12.2aA11.1bA11.1bQ1Pure Mathematics Year 1 (AS) Unit Test 3: Further AlgebraSchemeDividepolynomials bylinear expressionswith noremainder x 3 ax2 bx c 2x3 x2 13x 6or x 2 rx2 px q 2 x3 x2 13x 6or 2 x 1 ux2 vx w 2 x3 x2 13x 6For the long division, correctly finds the the first twocoefficients.For the matching coefficients method, correctly deduces thata 2 and c 2 or correctly deduces that r 2 and q 3 orcorrectly deduces that u 1 and w –6For the long division, correctly completes all steps in thedivision.For the matching coefficients method, correctly deduces thatb 5 or correctly deduces that p 5 or correctly deducesthat v –1States a fully correct, fully factorised final answer:(x – 3)(2x 1)(x 2)(6 marks) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.1

Mark schemePure Mathematics Year 1 (AS) Unit Test 3: Further AlgebraNotesOther algebraic methods can be used to factorise h(x). For example, if (x – 3) is known to be a factor then2 x3 x 2 13x 6 2 x 2 ( x 3) 5 x( x 3) 2( x 3) by balancing (M1) (2 x 2 5 x 2)( x 3) by factorising (M1) (2 x 1)( x 2)( x 3) by factorising (A1) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.2

Mark schemePure Mathematics Year 1 (AS) Unit Test 3: Further AlgebraQSchemeMarksAOsPearsonProgression Stepand Progressdescriptor2aStates or implies the expansion of a binomial expression tothe 8th power, up to and including the x3 term.M11.1a5thUnderstand anduse the generalbinomialexpansion forpositive integer n(a b) C0 a C1a b C2 a b C3a b .8888786 285 3or(a b)8 a8 8a 7b 28a 6b 2 56a 5b3 .Correctly substitutes 1 and 3x into the formula:M11.1bM1dep1.1bA11.1b(1 3x)8 18 8 17 3x 28 16 3x 56 15 3 x .23Makes an attempt to simplify the expression (2 correctcoefficients (other than 1) or both 9x2 and 27x3).(1 3x)8 18 24 x 28 9 x 2 56 27 x3 .States a fully correct answer:(1 3x)8 1 24 x 252 x 2 1512 x3 .(4)2bStates x 0.01 or implies this by attempting the g thebinomialexpansion forpositive integer nA11.1b1 24 0.01 252 0.01 1512 0.01 .23Attempts to simplify this expression (2 calculated termscorrect):1 0.24 0.0252 0.0015121.266712 1.2667 (5 s.f.)(3)(7 marks)Notes Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.3

Mark schemePure Mathematics Year 1 (AS) Unit Test 3: Further AlgebraQSchemeMarksAOsPearsonProgression Stepand Progressdescriptor3aStates or implies the expansion of a binomial expression tothe 9th power, up to and including the x3 term.M11.1a5thUse the binomialexpansion to findarbitrary termsfor positiveinteger n(a b) C0 a C1a b C2 a b C3a b .9999897 296 3or (a b)9 a9 9a8b 36a 7b 2 84a 6b3 .Correctly substitutes 2 and px into the formula.M11.1bM1dep1.1bA11.1b(2 px)9 29 9 28 px 36 27 px 84 26 px .23Makes an attempt to simplify the expression (at least onepower of 2 calculated and one bracket expanded correctly).(2 px)9 512 9 256 px 36 128 p 2 x 2 84 64 p3 x3 .States a fully correct answer:(2 px)9 512 2304 px 4608 p 2 x 2 5376 p3 x3 .(4)3biStates that 5376 p3 84M1ft2.2aCorrectly solves for p:A1ft1.1bCorrectly find the coefficient of the x term: 1 2304 576 4 B1ft1.1bCorrectly find the coefficient of the x2 term:2 1 4608 288 4 B1ftp3 3bii11 p 6441.1b5thUnderstand anduse the generalbinomialexpansion forpositive integer n5thUnderstand anduse the generalbinomialexpansion forpositive integer n(4)(8 marks) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.4

Mark schemePure Mathematics Year 1 (AS) Unit Test 3: Further AlgebraNotesft marks – pursues a correct method and obtains a correct answer or answers from their 5376 from part a. Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.5

Mark schemePure Mathematics Year 1 (AS) Unit Test 3: Further AlgebraQScheme4aAttempt is made at expanding p q . Accept seeing thecoefficients 1, 5, 10, 10, 5, 1 or seeing5MarksAOsPearsonProgression Stepand ProgressdescriptorM11.1a5thUnderstand anduse the generalbinomialexpansion forpositive integer n p q 5 5C0 p5 5C1 p 4 q 5C2 p3q 2 5C3 p 2 q3 5C4 pq 4 5C5 q5o.e.Fully correct answer is stated:A11.1b p q 5 p5 5 p 4 q 10 p3q 2 10 p 2q3 5 pq 4 q5(2)4bStates that p, or the probability of rolling a 4, isB114States that q, or the probability of not rolling a 4, is34States or implies that the sum of the first 3 terms (or 1 thesum of the last 3 terms) is the required probability.3.35thB13.3Use the binomialexpansion to findarbitrary termsfor positiveinteger nM12.2aM11.1bA11.1bFor example,p5 5 p 4 q 10 p 3q 2 or 1 (10 p 2 q3 5 pq 4 q5 )543 1 1 3 1 3 5 10 4 4 4 4 4 or211590 1024 1024 102445 1 2 3 3 1 3 3 or 1 10 5 4 4 4 4 4 270 405 243 or 1 1024 1024 1024 Either53o.e. or awrt 0.104512(5) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.6

Mark schemePure Mathematics Year 1 (AS) Unit Test 3: Further Algebra(7 marks)Notes Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.7

Mark schemeMarksAOsPearsonProgression Stepand ProgressdescriptorMakes an attempt to interpret the meaning of f(5) 0.For example, writing 125 25 5p q 0M12.2a5th5p q 150A11.1bMakes an attempt to interpret the meaning of f( 3) 8.For example writing 27 9 – 3p q 8M12.2aSolve non-linearsimultaneousequations incontext 3p q 26A11.1bMakes an attempt to solve the simultaneous equations.M1ft1.1aSolves the simultaneous equations to find that p 22A1ft1.1bSubstitutes their value for p to find that q 40A1ft1.1bQ5aPure Mathematics Year 1 (AS) Unit Test 3: Further AlgebraScheme(7)5bDraws the conclusion that (x – 5) must be a factor.Either makes an attempt at long division by setting up thelong division, or makes an attempt to find the remainingfactors by matching coefficients. For example, stating:M12.2aM1ft1.1bA12.2aA11.1bA11.1b x 5 ax2 bx c x3 x2 22x 405thDividepolynomials bylinear expressionswith a remainder(ft their 22 or 40)For the long division, correctly finds the the first twocoefficients.For the matching coefficients method, correctly deduces thata 1 and c 8For the long division, correctly completes all steps in thedivision.For the matching coefficients method, correctly deduces thatb 6States a fully correct, fully factorised final answer:(x – 5)(x 4)(x 2)(5) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.8

Mark schemePure Mathematics Year 1 (AS) Unit Test 3: Further Algebra(12 marks)NotesAward ft through marks for correct attempt/answers to solving their simultaneous equations.In part b other algebraic methods can be used to factorise:x – 5 is a factor (M1)x3 x 2 22 x 40 x 2 ( x 5) 6 x( x 5) 8( x 5) by balancing (M1) ( x 2 6 x 8)( x 5) by factorising (M1) ( x 4)( x 2)( x 5) by factorising (A1 A1) (i.e. A1 for each factor other than (x – 5)) Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.9

Mark schemeSchemeMarksAOsPearsonProgression Stepand Progressdescriptor13x 16 either on its own or as2part of an inequality/equation with 0 on the other side.M13.1a6thMakes an attempt to complete the square.For example, stating:M11.1bCompletealgebraic proofsin unfamiliarcontexts usingdirect orexhaustivemethodsA11.1bA12.1Q6Pure Mathematics Year 1 (AS) Unit Test 3: Further AlgebraConsiders the expression x 2 213 169 256 (ignore any (in)equation) x 4 1616 States a fully correct answer:213 87 (ignore any (in)equation) x 4 16 Interprets this solution as proving the inequality for all values213 of x. Could, for example, state that x 0 as a number4 squared is always positive or zero, therefore213 87 0 . Must be logically connected with the x 4 16 statement to be proved; this could be in the form of an1additional statement. So x 2 6 x 18 2 x (for all x) or by2a string of connectives which must be equivalent to “if andonly if”s.(4)(4 marks)NotesAny correct and complete method (e.g. finding the discriminant and single value, finding the minimum point bydifferentiation or completing the square and showing that it is both positive and a minimum, sketching the graph13supported with appropriate methodology etc) is acceptable for demonstrating that x 2 x 16 0 for all x.2 Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.10

Mark schemePure Mathematics Year 1 (AS) Unit Test 3: Further AlgebraQ7aSchemeMakes an attempt to expand the binomial expression 1 x 3MarksAOsPearsonProgression Stepand ProgressdescriptorM11.1a6th(must be terms in x0, x1, x2, x3 and at least 2 correct).1 3x 2 x3 1 3x 3x 2 x3A11.1b0 3xA11.1bx 0* as required.A1*2.2aSolve problemsusing thebinomialexpansion (forpositive integer n)in unfamiliarcontexts(including the linkto binomialprobabilities)(4)7bPicks a number less than or equal to zero, e.g. x 1, andattempts a substitution into both sides. For example,M11.1a1 3 1 1 1 3 1 3 1 1 2323Correctly deduces for their choice of x that the inequaltity doesnot hold. For example, 3 0A12.2a5thUse the binomialexpansion to findarbitrary terms forpositive integer n(2)(6 marks)Notes Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material is not copyright free.11

Mark scheme Pure Mathematics Year 1 (AS) Unit Test 3: Further Algebra Pearson Education Ltd 2017. Copying permitted for purchasing institution only. This material .