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Cambridge 2027 mathematics: core methods and applications

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AI-assisted English content review; independent teacher approval pending.

Practice scope and review

Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.

115 original questions across 17 selected topic areas. No account or payment required to practise.

Cambridge International AS & A Level Mathematics (9709), examination year 2027: AI-assisted conceptual alignment to official sources, not teacher or exam-board certification. Syllabus mapping and remaining coverage.

24 questions in this topic

Exact English revisions checked in an AI-assisted academic review; independent teacher approval is still pending.

All topics

Original written questions with indicative worked solutions. Conceptual mappings require teacher review; this is not a complete course or an official paper.

Subtopics in this topic (24)
  • Straight lines, perpendiculars and distances
  • Radian definition and degree conversion
  • Principal inverse trigonometric values
  • Derivative as a limiting chord gradient
  • Rational-power integration and initial conditions
  • Exponential and logarithm inverse graphs (Papers 2/3)
  • Logarithmic equations and inequalities (Papers 2/3)
  • Squared reciprocal trigonometric identities (Papers 2/3)
  • Compound-angle expansions (Papers 2/3)
  • Double-angle identities in integration (Papers 2/3)
  • Two- and three-dimensional vector notation
  • Formulating a differential equation from rates
  • Complex components, conjugation and Argand representation
  • Cartesian complex arithmetic
  • Conjugate roots of a real polynomial
  • Polar arithmetic and geometric transformations
  • Cartesian square roots of complex numbers
  • Complex loci and boundary conventions
  • Smooth contact and Newton's third law
  • Position slopes, velocity slopes and signed areas
  • Constant acceleration vertically and on an incline
  • Data-display choice and limitations
  • Equiprobable sample spaces and counting selections
  • Normal approximation in a discrete hypothesis test

24 of 24 questions shown

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Question 1A survey uses points A(1,2) and B(5,4). (a) Find AB in point-gradient, slope-intercept and general forms. (b) Find its midpoint M and length. (c) Find the perpendicular bisector and its intersection C with y=2x-1. Show that C is equidistant from A and B.4 indicative marks
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A survey uses points A(1,2) and B(5,4). (a) Find AB in point-gradient, slope-intercept and general forms. (b) Find its midpoint M and length. (c) Find the perpendicular bisector and its intersection C with y=2x-1. Show that C is equidistant from A and B.

Cambridge 2027: Direct conceptual match

Straight lines, perpendiculars and distances: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. m=(4-2)/(5-1)=1/2. Thus y-2=(x-1)/2, y=x/2+3/2 and x-2y+3=0.
  2. M=((1+5)/2,(2+4)/2)=(3,3); AB=sqrt(4^2+2^2)=sqrt(20)=2sqrt(5).
  3. Perpendicular slopes have product -1, so the bisector has gradient -2 and equation y-3=-2(x-3), or y=-2x+9.
  4. At the intersection, 2x-1=-2x+9 gives x=5/2,y=4. CA^2=(3/2)^2+2^2=25/4; CB^2=(-5/2)^2+0^2=25/4. Both distances are 5/2.

Self-marking points

View marking points
  1. Obtain gradient 1/2 and all three equivalent line forms.
  2. Find M=(3,3) and AB=2sqrt(5).
  3. Use perpendicular gradient -2 to obtain y=-2x+9.
  4. Solve the intersection C=(5/2,4) and verify CA=CB=5/2.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 2A circular sector has radius 12 cm and arc length 10pi cm. (a) Use the definition of radian measure to find its angle in radians and degrees. (b) A second sector subtends 225 degrees. Convert this to radians and find its arc length for radius 8 cm.3 indicative marks
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A circular sector has radius 12 cm and arc length 10pi cm. (a) Use the definition of radian measure to find its angle in radians and degrees. (b) A second sector subtends 225 degrees. Convert this to radians and find its arc length for radius 8 cm.

Cambridge 2027: Direct conceptual match

Radian definition and degree conversion: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. One radian is the angle subtending an arc equal in length to the radius. Therefore theta=10pi/12=5pi/6.
  2. Since a full turn is 2pi radians=360 degrees, pi radians=180 degrees and (5pi/6)(180/pi)=150 degrees.
  3. 225 degrees gives 225pi/180=5pi/4 radians. The second arc is r theta=8(5pi/4)=10pi cm. Checking the fraction of a full circumference gives (225/360)(2pi*8)=10pi cm independently.

Self-marking points

View marking points
  1. Use theta=s/r, explaining one radian.
  2. Obtain 5pi/6 radians=150 degrees.
  3. Convert 225 degrees to 5pi/4 radians and obtain arc 10pi cm.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 3Work in radians. (a) Evaluate sin^-1(sin(5pi/6)), cos^-1(cos(4pi/3)) and tan^-1(tan(3pi/4)), stating the principal range used each time. Here sin^-1 denotes inverse sine, not a reciprocal. (b) Solve sin x=sin(5pi/6) for 0<=x<=2pi and explain why inverse sine alone does not give both answers.3 indicative marks
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Work in radians. (a) Evaluate sin^-1(sin(5pi/6)), cos^-1(cos(4pi/3)) and tan^-1(tan(3pi/4)), stating the principal range used each time. Here sin^-1 denotes inverse sine, not a reciprocal. (b) Solve sin x=sin(5pi/6) for 0<=x<=2pi and explain why inverse sine alone does not give both answers.

Cambridge 2027: Direct conceptual match

Principal inverse trigonometric values: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. sin(5pi/6)=1/2. The inverse sine returns the unique angle in [-pi/2,pi/2], hence pi/6, not 5pi/6.
  2. cos(4pi/3)=-1/2; the angle in [0,pi] is 2pi/3. tan(3pi/4)=-1; the angle in (-pi/2,pi/2) is -pi/4.
  3. Sine is positive in quadrants I and II. Thus x=pi/6 or pi-pi/6=5pi/6. A principal inverse selects one angle from its restricted range, whereas the original sine function is not one-to-one on [0,2pi].

Self-marking points

View marking points
  1. Use sine principal range [-pi/2,pi/2] to obtain pi/6.
  2. Use cosine range [0,pi] and tangent range (-pi/2,pi/2) to obtain 2pi/3 and -pi/4.
  3. Find x=pi/6,5pi/6 and distinguish a principal value from the full equation solution.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 4For f(x)=x2x^{2}+3x, P has x=2 and Q has x=2+h, h nonzero. (a) Derive the chord gradient PQ. (b) Calculate it for h=0.1,-0.1,0.01,-0.01 and explain its limit. (c) Give f'(2), f'(x), f''(x) and the tangent equation at P.3 indicative marks
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For f(x)=x2x^{2}+3x, P has x=2 and Q has x=2+h, h nonzero. (a) Derive the chord gradient PQ. (b) Calculate it for h=0.1,-0.1,0.01,-0.01 and explain its limit. (c) Give f'(2), f'(x), f''(x) and the tangent equation at P.

Cambridge 2027: Direct conceptual match

Derivative as a limiting chord gradient: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. f(2)=10 and f(2+h)=10+7h+h2h^{2}. The chord gradient is (f(2+h)-f(2))/h=7+h for h!=0.
  2. The four gradients are 7.1,6.9,7.01,6.99. From either side they approach 7 as h approaches zero; the chord tends to the tangent.
  3. Thus f'(2)=7. At a general x, the quotient is 2x+3+h, so f'(x)=2x+3 and f''(x)=2. At P(2,10), y-10=7(x-2), giving y=7x-4.

Self-marking points

View marking points
  1. Expand the difference quotient to 7+h.
  2. Obtain 7.1,6.9,7.01,6.99 and limiting gradient 7.
  3. Use derivative notation correctly and find tangent y=7x-4.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 5A curve on x>0 has dy/dx=3sqrt(x)-2/x2x^{2}+4(2x+1)^3 and passes through (1,5). (a) Find y, showing the antiderivative of every term and the integration constant. (b) Find y(4) and verify your result for y by differentiation.4 indicative marks
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A curve on x>0 has dy/dx=3sqrt(x)-2/x2x^{2}+4(2x+1)^3 and passes through (1,5). (a) Find y, showing the antiderivative of every term and the integration constant. (b) Find y(4) and verify your result for y by differentiation.

Cambridge 2027: Direct conceptual match

Rational-power integration and initial conditions: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. Write the derivative as 3x^(1/2)-2x^-2+4(2x+1)^3. Reverse the power rule: integrals of the first two terms are 2x^(3/2) and 2x^-1.
  2. The derivative of (2x+1)^4 is 8(2x+1)^3. Thus y=2x^(3/2)+2/x+(2x+1)^4/2+C.
  3. At x=1, 5=2+2+81/2+C, so C=-79/2.
  4. At x=4, y=16+1/2+6561/2-79/2=6515/2. Differentiating gives 3sqrt(x)-2/x2x^{2}+4(2x+1)^3, and substitution at x=1 gives 5.

Self-marking points

View marking points
  1. Integrate rational powers to 2x^(3/2)+2/x.
  2. Integrate the linear composite to (2x+1)^4/2.
  3. Use (1,5) to obtain C=-79/2.
  4. Obtain y(4)=6515/2 and verify all derivative terms.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 6Let f(x)=e^(2x). (a) Find f−1f^{-1}(x), including its domain and range. (b) Sketch f and f−1f^{-1} on labelled axes, marking intercepts, asymptotes and the images of (0,1) and (ln(2)/2,2) under reflection in y=x. (c) On separate axes sketch y=e^(-2x), giving its monotonicity and end behaviour.4 indicative marks
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Let f(x)=e^(2x). (a) Find f−1f^{-1}(x), including its domain and range. (b) Sketch f and f−1f^{-1} on labelled axes, marking intercepts, asymptotes and the images of (0,1) and (ln(2)/2,2) under reflection in y=x. (c) On separate axes sketch y=e^(-2x), giving its monotonicity and end behaviour.

Cambridge 2027: Direct conceptual match

Exponential and logarithm inverse graphs (Papers 2/3): the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review. Shared Paper 2/3 techniques are evidenced separately from qualification eligibility: Paper 2 is AS pure-only (1+2), not A-level carry-forward; Paper 3 is compulsory A-level content.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. From y=e^(2x), ln y=2x, so f−1f^{-1}(x)=(ln x)/2, domain (0,infinity), range R. The original has domain R and range (0,infinity).
  2. Draw f increasing through (0,1) and (ln(2)/2,2), approaching y=0 as x tends to -infinity. It has no x-intercept. Reflect across y=x to draw the increasing inverse through (1,0) and (2,ln(2)/2); its vertical asymptote is x=0 and it has no y-intercept.
  3. For e^(-2x), use a decreasing positive curve through (0,1). It tends to 0 as x tends to infinity and grows without bound as x tends to -infinity. Its horizontal asymptote is y=0.

Self-marking points

View marking points
  1. Obtain inverse (ln x)/2 on x>0 with range R.
  2. Specify increasing reflected graphs, intercepts and asymptotes.
  3. Reflect both given points correctly.
  4. Describe decreasing e^(-2x), intercept 1 and both end limits.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 7A model gives C(t)=80e^(-0.3t), t>=0. (a) Solve C(t)=18 exactly using logarithms, then give t to 3 significant figures. (b) Solve C(t)<18 and find the first integer t satisfying it. (c) Solve 3^(2x-1)=7, leaving an exact logarithmic answer.3 indicative marks
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A model gives C(t)=80e^(-0.3t), t>=0. (a) Solve C(t)=18 exactly using logarithms, then give t to 3 significant figures. (b) Solve C(t)<18 and find the first integer t satisfying it. (c) Solve 3^(2x-1)=7, leaving an exact logarithmic answer.

Cambridge 2027: Direct conceptual match

Logarithmic equations and inequalities (Papers 2/3): the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review. Shared Paper 2/3 techniques are evidenced separately from qualification eligibility: Paper 2 is AS pure-only (1+2), not A-level carry-forward; Paper 3 is compulsory A-level content.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. 80e^(-0.3t)=18 gives e^(-0.3t)=9/40. Taking ln gives -0.3t=ln(9/40), so t=(10/3)ln(40/9)=4.97 to 3 significant figures.
  2. For C(t)<18, -0.3t<ln(9/40). Division by the negative number reverses the sign: t>(10/3)ln(40/9). This threshold lies between 4 and 5, so the first integer is 5.
  3. Taking ln of 3^(2x-1)=7 gives (2x-1)ln3=ln7. Hence x=(1+ln7/ln3)/2.

Self-marking points

View marking points
  1. Derive t=(10/3)ln(40/9), about 4.97.
  2. Reverse the inequality when dividing by -0.3, and obtain first integer 5.
  3. Obtain x=(1+ln(7)/ln(3))/2.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 8(a) Derive sec^2 x=1+tan^2 x and cosec^2 x=1+cot^2 x from sin^2 x+cos^2 x=1, stating restrictions. (b) Solve 2sec^2 x-3tan x-3=0 for 0<=x<pi. (c) Simplify (cosec^2 x-1)/cot x wherever the original expression is defined.4 indicative marks
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(a) Derive sec^2 x=1+tan^2 x and cosec^2 x=1+cot^2 x from sin^2 x+cos^2 x=1, stating restrictions. (b) Solve 2sec^2 x-3tan x-3=0 for 0<=x<pi. (c) Simplify (cosec^2 x-1)/cot x wherever the original expression is defined.

Cambridge 2027: Direct conceptual match

Squared reciprocal trigonometric identities (Papers 2/3): the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review. Shared Paper 2/3 techniques are evidenced separately from qualification eligibility: Paper 2 is AS pure-only (1+2), not A-level carry-forward; Paper 3 is compulsory A-level content.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. Dividing by cos^2 x gives tan^2 x+1=sec^2 x when cos x!=0. Dividing by sin^2 x gives 1+cot^2 x=cosec^2 x when sin x!=0.
  2. Put u=tan x: 2(1+u2u^{2})-3u-3=0, or 2u^2-3u-1=0. Thus u=(3+sqrt17)/4 or (3-sqrt17)/4.
  3. The positive root gives x=tan^-1((3+sqrt17)/4) in quadrant I. The negative root gives x=pi+tan^-1((3-sqrt17)/4) in quadrant II. These are all solutions on [0,pi), with x=pi/2 excluded.
  4. (cosec^2 x-1)/cot x=cot^2 x/cot x=cot x. The original requires sin x!=0 and cot x!=0, equivalently sin x cos x!=0; simplification must retain these exclusions.

Self-marking points

View marking points
  1. Divide by cos^2 x and sin^2 x with nonzero denominators.
  2. Reduce to 2tan^2 x-3tan x-1=0.
  3. Give both interval solutions via (- or +) 17\sqrt{17}.
  4. Reduce the final expression to cot x with sin x cos x nonzero.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 9Angles A and B are acute with sin A=3/5 and cos B=12/13. (a) Use compound-angle formulae to find sin(A+B), cos(A+B) and tan(A-B) exactly. (b) Decide whether A+B exceeds pi/2, justifying without inverse-trigonometric approximations.4 indicative marks
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Angles A and B are acute with sin A=3/5 and cos B=12/13. (a) Use compound-angle formulae to find sin(A+B), cos(A+B) and tan(A-B) exactly. (b) Decide whether A+B exceeds pi/2, justifying without inverse-trigonometric approximations.

Cambridge 2027: Direct conceptual match

Compound-angle expansions (Papers 2/3): the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review. Shared Paper 2/3 techniques are evidenced separately from qualification eligibility: Paper 2 is AS pure-only (1+2), not A-level carry-forward; Paper 3 is compulsory A-level content.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. Acuteness fixes the positive square roots: cos A=sqrt(1-9/25)=4/5 and sin B=sqrt(1-144/169)=5/13.
  2. sin(A+B)=sin A cos B+cos A sin B=36/65+20/65=56/65. cos(A+B)=cos A cos B-sin A sin B=48/65-15/65=33/65.
  3. tan A=3/4 and tan B=5/12. tan(A-B)=(3/4-5/12)/(1+(3/4)(5/12))=(1/3)/(21/16)=16/63.
  4. Because 0<A+B<pi and cos(A+B)>0, A+B is in quadrant I. It does not exceed pi/2.

Self-marking points

View marking points
  1. Find cos A=4/5 and sin B=5/13.
  2. Expand to sin(A+B)=56/65 and cos(A+B)=33/65.
  3. Obtain tan(A-B)=16/63.
  4. Use positive cosine and 0<A+B<pi to show A+B<pi/2.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 10(a) Use a double-angle identity to integrate cos^2(2x). (b) Evaluate integral from 0 to pi/4 of [3cos^2(2x)-sin^2 x] dx exactly, showing the identity used for each square.4 indicative marks
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(a) Use a double-angle identity to integrate cos^2(2x). (b) Evaluate integral from 0 to pi/4 of [3cos^2(2x)-sin^2 x] dx exactly, showing the identity used for each square.

Cambridge 2027: Direct conceptual match

Double-angle identities in integration (Papers 2/3): the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review. Shared Paper 2/3 techniques are evidenced separately from qualification eligibility: Paper 2 is AS pure-only (1+2), not A-level carry-forward; Paper 3 is compulsory A-level content.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. cos4x=2cos^2(2x)-1 gives cos^2(2x)=(1+cos4x)/2. Integrating gives x/2+sin4x/8+C.
  2. Also sin^2 x=(1-cos2x)/2. Therefore 3cos^2(2x)-sin^2 x=1+(3/2)cos4x+(1/2)cos2x.
  3. An antiderivative is x+(3/8)sin4x+(1/4)sin2x. At pi/4 it is pi/4+0+1/4; at 0 it is 0, so the integral is pi/4+1/4.

Self-marking points

View marking points
  1. Use cos^2(2x)=(1+cos4x)/2.
  2. Obtain x/2+sin4x/8+C.
  3. Rewrite the combined integrand as 1+(3/2)cos4x+(1/2)cos2x.
  4. Evaluate to pi/4+1/4.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 11O is the origin, A has position vector 2i-j+3k and B has column position vector (5,3,-1)^T. (a) Write a=OA and b=OB in column and i,j,k notation and write AB in both. (b) In a separate xy-plane, U=(2,-1), V=(-1,3). Write UV as a column vector and in i,j notation. Explain how AB differs from OB.3 indicative marks
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O is the origin, A has position vector 2i-j+3k and B has column position vector (5,3,-1)^T. (a) Write a=OA and b=OB in column and i,j,k notation and write AB in both. (b) In a separate xy-plane, U=(2,-1), V=(-1,3). Write UV as a column vector and in i,j notation. Explain how AB differs from OB.

Cambridge 2027: Direct conceptual match

Two- and three-dimensional vector notation: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. a=(2,-1,3)^T=2i-j+3k; b=(5,3,-1)^T=5i+3j-k. The superscript T means the displayed row is to be read as a column.
  2. AB=b-a=(5-2,3-(-1),-1-3)^T=(3,4,-4)^T=3i+4j-4k.
  3. UV=(-1-2,3-(-1))^T=(-3,4)^T=-3i+4j. AB describes movement from A to B; OB is B's position relative to O. Their tails are different, and a+AB=b.

Self-marking points

View marking points
  1. Translate a and b between the two 3D notations.
  2. Obtain AB=(3,4,-4)^T=3i+4j-4k.
  3. Obtain UV=(-3,4)^T=-3i+4j and distinguish displacement from position.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 12A liquid's temperature T degrees C falls at a rate proportional to its excess above a constant room temperature of 20 degrees C. Initially T=80; initially it is cooling at 6 degrees C per minute. (a) Form a differential equation and determine the proportionality constant with units. (b) Solve with the initial condition. (c) Find when T=35 and explain the long-term model prediction.3 indicative marks
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A liquid's temperature T degrees C falls at a rate proportional to its excess above a constant room temperature of 20 degrees C. Initially T=80; initially it is cooling at 6 degrees C per minute. (a) Form a differential equation and determine the proportionality constant with units. (b) Solve with the initial condition. (c) Find when T=35 and explain the long-term model prediction.

Cambridge 2027: Direct conceptual match

Formulating a differential equation from rates: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. The excess is T-20; cooling means dT/dt is negative, so dT/dt=-k(T-20). Initially -6=-k(80-20), giving k=1/10 min^-1.
  2. Separate: dT/(T-20)=-dt/10. Integration gives ln(T-20)=-t/10+C on this branch. Since T(0)-20=60, T=20+60e^(-t/10).
  3. 35=20+60e^(-t/10) gives e^(-t/10)=1/4 and t=10ln4 minutes. The exponential stays positive and tends to zero; the model approaches 20 degrees C without crossing it, assuming room temperature stays constant.

Self-marking points

View marking points
  1. Form dT/dt=-k(T-20), k>0, and find k=0.1 per minute.
  2. Separate variables and find T=20+60e^(-t/10).
  3. Obtain t=10ln4 minutes and limit 20 from above.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 13Let z=-3+3sqrt(3)i. (a) State Re z, Im z and z*, then find |z| and principal arg z in (-pi,pi]. (b) Draw an Argand diagram for z and z*, with equal scales on the real and imaginary axes. Label exact coordinates, modulus and both arguments. (c) If (a-1)+(2b+1)i=z for real a,b, find a,b.4 indicative marks
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Let z=-3+3sqrt(3)i. (a) State Re z, Im z and z*, then find |z| and principal arg z in (-pi,pi]. (b) Draw an Argand diagram for z and z*, with equal scales on the real and imaginary axes. Label exact coordinates, modulus and both arguments. (c) If (a-1)+(2b+1)i=z for real a,b, find a,b.

Cambridge 2027: Direct conceptual match

Complex components, conjugation and Argand representation: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. Re z=-3, Im z=3sqrt3 (a real number), z*=-3-3sqrt3 i. |z|=sqrt(9+27)=6.
  2. z is in quadrant II with reference angle pi/3, so arg z=2pi/3. The conjugate is in quadrant III with principal argument -2pi/3.
  3. Draw a horizontal Re axis and vertical Im axis with the same unit scale. Mark Z=(-3,3sqrt3) and Z*=(-3,-3sqrt3), draw their radius segments of length 6 from O, and label anticlockwise angle 2pi/3 and clockwise angle -2pi/3 from the positive real axis. These points are mirror images across the real axis.
  4. Equality of complex numbers requires a-1=-3 and 2b+1=3sqrt3. Thus a=-2 and b=(3sqrt3-1)/2.

Self-marking points

View marking points
  1. State components and conjugate correctly.
  2. Find modulus 6 and argument 2pi/3.
  3. Specify points reflected in real axis with arguments +/-2pi/3.
  4. Equate real and imaginary parts to find a=-2 and b=(3sqrt3-1)/2.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 14Let z=2+3i and w=1-2i. (a) Calculate z+w, z-w and zw in Cartesian form. (b) Calculate z/w by showing the conjugate method. (c) Verify that w(z/w)=z.3 indicative marks
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Let z=2+3i and w=1-2i. (a) Calculate z+w, z-w and zw in Cartesian form. (b) Calculate z/w by showing the conjugate method. (c) Verify that w(z/w)=z.

Cambridge 2027: Direct conceptual match

Cartesian complex arithmetic: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

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Worked guidance

  1. z+w=(2+1)+(3-2)i=3+i; z-w=(2-1)+(3+2)i=1+5i. zw=2-4i+3i-6i^2=8-i.
  2. z/w=((2+3i)(1+2i))/((1-2i)(1+2i))=(2+7i+6i^2)/(1+4)=(-4+7i)/5.
  3. (1-2i)(-4+7i)/5=(-4+15i-14i^2)/5=(10+15i)/5=2+3i, confirming the division.

Self-marking points

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  1. Find 3+i, 1+5i and 8-i using i2i^{2}=-1.
  2. Multiply numerator and denominator by 1+2i to obtain (-4+7i)/5.
  3. Multiply back to obtain 2+3i.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 15The real-coefficient polynomial p(z)=z3z^{3}-5z^2+11z-15 has root 1+2i. (a) Explain why another root is known without division. (b) Construct the corresponding real quadratic factor and factor p fully. (c) List all roots and verify their sum and product against the coefficients.4 indicative marks
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The real-coefficient polynomial p(z)=z3z^{3}-5z^2+11z-15 has root 1+2i. (a) Explain why another root is known without division. (b) Construct the corresponding real quadratic factor and factor p fully. (c) List all roots and verify their sum and product against the coefficients.

Cambridge 2027: Direct conceptual match

Conjugate roots of a real polynomial: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

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Worked guidance

  1. Conjugating p(1+2i)=0 leaves its real coefficients unchanged, hence p(1-2i)=0.
  2. The pair gives (z-(1+2i))(z-(1-2i))=(z-1)^2-(2i)^2=z2z^{2}-2z+5.
  3. Polynomial division by z2z^{2}-2z+5 gives quotient z-3 and remainder zero: (z-3)(z2z^{2}-2z+5)=z3z^{3}-5z^2+11z-15. Thus the roots are 3,1+2i,1-2i.
  4. The sum is 3+(1+2i)+(1-2i)=5, matching minus the z2z^{2} coefficient. The product is 3*((1+2i)(1-2i))=3*5=15, matching minus the constant term for a monic cubic.

Self-marking points

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  1. Use real coefficients to infer root 1-2i.
  2. Construct (z-1)^2+4=z2z^{2}-2z+5.
  3. Factor p=(z-3)(z2z^{2}-2z+5) and list roots.
  4. Check sum 5 and product 15.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 16Let z=2e^(i*pi/6) and w=3e^(i*pi/3). (a) Calculate zw and z/w in exponential and Cartesian forms, showing modulus and argument rules. (b) Describe geometrically the maps u -> wu and u -> u/w. Draw the image of z under each on equal-scale Argand axes, labelling coordinates.4 indicative marks
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Let z=2e^(i*pi/6) and w=3e^(i*pi/3). (a) Calculate zw and z/w in exponential and Cartesian forms, showing modulus and argument rules. (b) Describe geometrically the maps u -> wu and u -> u/w. Draw the image of z under each on equal-scale Argand axes, labelling coordinates.

Cambridge 2027: Direct conceptual match

Polar arithmetic and geometric transformations: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

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Worked guidance

  1. zw=(2*3)e^(i(pi/6+pi/3))=6e^(i*pi/2)=6i. Multiplication multiplies distances from O and adds angles.
  2. z/w=(2/3)e^(i(pi/6-pi/3))=(2/3)e^(-i*pi/6)=sqrt3/3-i/3.
  3. Multiplication by w enlarges from O by factor 3 and rotates anticlockwise pi/3. Division by w enlarges by factor 1/3 and rotates clockwise pi/3.
  4. On horizontal Re and vertical Im axes with equal unit scales, start at z=(sqrt3,1). Mark the multiplication image (0,6) and division image (sqrt3/3,-1/3), with radius lengths 6 and 2/3 and arguments pi/2 and -pi/6 respectively.

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  1. Multiply moduli and add arguments to obtain zw=6i.
  2. Divide moduli and subtract arguments to obtain z/w=sqrt3/3-i/3.
  3. Describe enlargement factors 3 and 1/3 and rotations +/-pi/3 about the origin.
  4. Label image points (0,6) and (sqrt3/3,-1/3).

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 17(a) Find both square roots of -7+24i in exact Cartesian form by writing the root as a+bi for real a,b. (b) Check each by squaring and explain why taking a positive square root separately of real and imaginary parts would be invalid.4 indicative marks
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(a) Find both square roots of -7+24i in exact Cartesian form by writing the root as a+bi for real a,b. (b) Check each by squaring and explain why taking a positive square root separately of real and imaginary parts would be invalid.

Cambridge 2027: Direct conceptual match

Cartesian square roots of complex numbers: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

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Worked guidance

  1. (a+bi)^2=a2a^{2}-b2b^{2}+2abi, so a2a^{2}-b2b^{2}=-7 and ab=12. Also |a+bi|^2=|-7+24i|=sqrt(49+576)=25, giving a2a^{2}+b2b^{2}=25.
  2. Adding equations gives 2a^2=18, hence a=+/-3. Then b=12/a, giving b=4 or -4 respectively. The two roots are 3+4i and -3-4i.
  3. (3+4i)^2=9-16+24i=-7+24i; (-3-4i)^2 is identical. Separate square roots of the components ignore the -b^2 and 2ab terms, so they do not solve the complex square equation.

Self-marking points

View marking points
  1. Equate a2a^{2}-b2b^{2}=-7 and 2ab=24.
  2. Use a2a^{2}+b2b^{2}=25 to obtain a2a^{2}=9,b2b^{2}=16.
  3. Apply ab=12 to obtain 3+4i and -3-4i.
  4. Square both roots and explain the cross term.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 18Write z=x+iy. (a) Draw |z-(1+2i)|=3. (b) Derive and draw |z-1|<=|z+3|. (c) On a separate diagram draw arg(z-(1+i))=pi/4, using the principal argument. Specify all boundaries and excluded points precisely.3 indicative marks
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Write z=x+iy. (a) Draw |z-(1+2i)|=3. (b) Derive and draw |z-1|<=|z+3|. (c) On a separate diagram draw arg(z-(1+i))=pi/4, using the principal argument. Specify all boundaries and excluded points precisely.

Cambridge 2027: Direct conceptual match

Complex loci and boundary conventions: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

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Worked guidance

  1. |z-(1+2i)|=sqrt((x-1)^2+(y-2)^2), so draw the circle (x-1)^2+(y-2)^2=9, centre (1,2), radius 3, on equal-scale real and imaginary axes. This is only the circumference, not a disk.
  2. Both distances are nonnegative, so squaring preserves <=: (x-1)^2+y2y^{2}<=(x+3)^2+y2y^{2}. Cancelling gives -8x<=8, hence x>=-1. Draw the solid vertical line x=-1 and shade its right-hand side.
  3. arg(z-(1+i))=pi/4 means the displacement from (1,1) has equal positive components. Thus y-1=x-1>0: draw the ray y=x for x>1 with an open endpoint at (1,1). The zero displacement has no argument and is excluded; the opposite ray has argument -3pi/4, not pi/4.

Self-marking points

View marking points
  1. Give circle centre (1,2), radius 3, full boundary included.
  2. Square distances and derive half-plane x>=-1 including x=-1.
  3. Give ray y=x, x>1, with endpoint (1,1) excluded.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 19Blocks A (2 kg) and B (3 kg) touch on a smooth horizontal table. A horizontal force 20 N pushes A towards B; contact is maintained. (a) Draw a separate force diagram for each block, labelling all action-reaction pairs across different bodies. (b) Find acceleration and contact force. (c) Explain why contact forces cancel for the combined system but not for A alone, and one limitation of the smooth-table model. Take g=10 m s^-2.4 indicative marks
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Blocks A (2 kg) and B (3 kg) touch on a smooth horizontal table. A horizontal force 20 N pushes A towards B; contact is maintained. (a) Draw a separate force diagram for each block, labelling all action-reaction pairs across different bodies. (b) Find acceleration and contact force. (c) Explain why contact forces cancel for the combined system but not for A alone, and one limitation of the smooth-table model. Take g=10 m s^-2.

Cambridge 2027: Direct conceptual match

Smooth contact and Newton's third law: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

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Worked guidance

  1. Draw A with weight 20 N down, table normal 20 N up, 20 N push right and B's contact force R left. Draw B with weight 30 N down, table normal 30 N up and A's contact force R right. Smooth table contact has no tangential friction.
  2. For A+B, 20=(2+3)a, so a=4 m s−2s^{-2} right. For B alone, R=3a=12 N. For A, 20-R=2a verifies R=12 N.
  3. A-on-B and B-on-A are an equal, opposite third-law pair on different bodies. Each table normal has its partner acting on the table; each weight has its gravitational partner acting on Earth. Normal and weight on one block are not a third-law pair.
  4. The block contact forces are internal and cancel only in the combined-system sum. A alone still experiences B's 12 N force. Smoothness neglects friction at the table; a rough real table would exert tangential forces and could change the acceleration.

Self-marking points

View marking points
  1. Draw weights, upward normals, applied force and equal/opposite horizontal block forces.
  2. Find a=4 m s−2s^{-2} and contact force 12 N.
  3. Identify third-law partners on different bodies, not normal and weight.
  4. Explain cancellation and neglected tangential friction.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 20A particle has displacement s(t)=t2t^{2}-4t metres for 0<=t<=3 seconds. (a) Find velocity and acceleration, explaining the gradient meaning of each. (b) Sketch its velocity-time graph with intercepts. Use signed areas under it, not just s(3), to find displacement over the interval. (c) Find distance travelled and explain its difference from displacement.4 indicative marks
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A particle has displacement s(t)=t2t^{2}-4t metres for 0<=t<=3 seconds. (a) Find velocity and acceleration, explaining the gradient meaning of each. (b) Sketch its velocity-time graph with intercepts. Use signed areas under it, not just s(3), to find displacement over the interval. (c) Find distance travelled and explain its difference from displacement.

Cambridge 2027: Direct conceptual match

Position slopes, velocity slopes and signed areas: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

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Worked guidance

  1. Velocity is the slope of the displacement-time curve: v=ds/dt=2t-4 m s−1s^{-1}. Acceleration is the slope of the velocity-time graph: a=dv/dt=2 m s−2s^{-2}.
  2. Draw a straight velocity-time line through (0,-4), crossing the time axis at (2,0), and ending at (3,2). The negative velocity before t=2 means motion in the negative direction.
  3. Area from 0 to 2 is -1/2*2*4=-4 m. Area from 2 to 3 is +1/2*1*2=+1 m. Their signed sum is displacement -3 m, agreeing with s(3)-s(0)=9-12=-3.
  4. Distance uses absolute areas: 4+1=5 m. The particle reverses at t=2; distance counts both legs positively, while displacement records their signed net effect.

Self-marking points

View marking points
  1. Interpret ds/dt=2t-4 as position-time slope and dv/dt=2 as velocity-time slope.
  2. Sketch line through (0,-4),(2,0),(3,2).
  3. Use signed triangular areas -4 and +1 to obtain -3 m.
  4. Add magnitudes for distance 5 m.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 21Take g=10 m s−2s^{-2} and neglect air resistance. (a) A ball is projected vertically upwards at 18 m s−1s^{-1} from a platform 5 m above ground. Find its maximum height above ground and time to reach the ground. (b) Separately, a particle is projected up a smooth plane inclined at 30 degrees to the horizontal with speed 6 m s−1s^{-1}. Resolve Newton's second law along the plane, then find time and distance to its first rest.4 indicative marks
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Take g=10 m s−2s^{-2} and neglect air resistance. (a) A ball is projected vertically upwards at 18 m s−1s^{-1} from a platform 5 m above ground. Find its maximum height above ground and time to reach the ground. (b) Separately, a particle is projected up a smooth plane inclined at 30 degrees to the horizontal with speed 6 m s−1s^{-1}. Resolve Newton's second law along the plane, then find time and distance to its first rest.

Cambridge 2027: Direct conceptual match

Constant acceleration vertically and on an incline: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. With upward positive, v2v^{2}=u2u^{2}+2as at the highest point gives 0=18^2-20s. The rise is 16.2 m, so maximum height is 5+16.2=21.2 m.
  2. Ground corresponds to displacement -5 from launch: -5=18t-5t^2, or 5t^2-18t-5=0. The positive root is t=(18+sqrt424)/10 seconds, approximately 3.86 s. The other root is negative and not physical after launch.
  3. For the incline take uphill positive. Smooth contact removes friction, and the normal has no along-plane component. Thus ma=-mg sin30 and a=-5 m s−2s^{-2}.
  4. 0=6-5t gives t=1.2 s. Also 0=626^{2}-10s gives s=3.6 m uphill. Both results concern the instant before the subsequent reversal.

Self-marking points

View marking points
  1. Use upward-positive a=-10 to obtain rise 16.2 m and height 21.2 m.
  2. Select positive root (18+sqrt424)/10 seconds.
  3. Resolve a=-g sin30=-5 m s−2s^{-2} along the plane.
  4. Find stopping time 1.2 s and distance 3.6 m.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 22Waiting times in minutes are grouped as [0,2):8 people, [2,5):12, [5,10):10. (a) Choose and specify an appropriate graph, calculating every bar height. Explain why raw frequencies as heights mislead. (b) Estimate the median using a cumulative-frequency graph and identify the assumption. (c) Explain what a stem-and-leaf display would preserve but why these grouped data cannot supply it exactly.4 indicative marks
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Waiting times in minutes are grouped as [0,2):8 people, [2,5):12, [5,10):10. (a) Choose and specify an appropriate graph, calculating every bar height. Explain why raw frequencies as heights mislead. (b) Estimate the median using a cumulative-frequency graph and identify the assumption. (c) Explain what a stem-and-leaf display would preserve but why these grouped data cannot supply it exactly.

Cambridge 2027: Direct conceptual match

Data-display choice and limitations: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. Waiting time is continuous and class widths are unequal. Draw a histogram over the three stated intervals, heights frequency/width: 8/2=4,12/3=4,10/5=2 people per minute. Bar areas recover 8,12,10.
  2. Using heights 8,12,10 would give areas 16,36,50, falsely representing relative frequencies. Label the horizontal axis minutes and the vertical axis frequency density.
  3. Cumulative-frequency coordinates are (0,0),(2,8),(5,20),(10,30). A straight segment interpolation at cumulative frequency 15 gives 2+((15-8)/12)*3=3.75 minutes. This assumes uniform spread in [2,5); it is an estimate, not the exact raw-data median.
  4. A stem-and-leaf display could preserve each original value and reveal gaps or outliers. Group counts do not determine individual times, so an exact stem-and-leaf cannot be reconstructed. The histogram also hides within-class detail.

Self-marking points

View marking points
  1. Choose a histogram with densities 4,4,2 people per minute.
  2. Explain area, not height, represents frequency for unequal widths.
  3. Use cumulative points (0,0),(2,8),(5,20),(10,30) to estimate median 3.75.
  4. Discuss uniform-within-class interpolation and lost individual values.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 23A bag contains four distinct counters labelled A,B,C,D. Two counters are selected without replacement, each remaining counter equally likely at each draw. (a) List the ordered sample space and justify equal probabilities. (b) Find the probability that the pair contains exactly one of A,B by counting. (c) Rework using unordered selections and explain why mixing the two denominators is invalid.4 indicative marks
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A bag contains four distinct counters labelled A,B,C,D. Two counters are selected without replacement, each remaining counter equally likely at each draw. (a) List the ordered sample space and justify equal probabilities. (b) Find the probability that the pair contains exactly one of A,B by counting. (c) Rework using unordered selections and explain why mixing the two denominators is invalid.

Cambridge 2027: Direct conceptual match

Equiprobable sample spaces and counting selections: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. The ordered outcomes are AB,AC,AD,BA,BC,BD,CA,CB,CD,DA,DB,DC. Each has probability (1/4)(1/3)=1/12.
  2. Exactly one of A,B means one member from {A,B} and one from {C,D}. The favourable ordered outcomes are AC,AD,BC,BD,CA,DA,CB,DB: 8 of 12, so probability 2/3.
  3. Unordered pairs are {A,B},{A,C},{A,D},{B,C},{B,D},{C,D}. Each combines two ordered outcomes and has probability 2/12=1/6. Four are favourable, giving 4/6=2/3.
  4. Using 4/12 or 8/6 mixes sample spaces: the numerator and denominator must count objects with the same ordering convention.

Self-marking points

View marking points
  1. List all 12 ordered outcomes, each probability 1/12.
  2. Identify 8 favourable ordered outcomes and probability 2/3.
  3. Count 6 equally likely unordered pairs with 4 favourable.
  4. Explain the ordered/unordered denominator consistency.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

Question 24A machine is claimed to produce faulty items independently with probability p=0.20. In a random sample of 200, 51 are faulty. (a) State hypotheses for testing an increase in p at 5%. (b) Use a normal approximation with continuity correction to test the claim, explaining suitability. Use Phi(1.856)=0.9683 and the upper 5% critical value 1.645. (c) Find the smallest integer fault count in the approximate rejection region and interpret the decision without claiming proof.4 indicative marks
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A machine is claimed to produce faulty items independently with probability p=0.20. In a random sample of 200, 51 are faulty. (a) State hypotheses for testing an increase in p at 5%. (b) Use a normal approximation with continuity correction to test the claim, explaining suitability. Use Phi(1.856)=0.9683 and the upper 5% critical value 1.645. (c) Find the smallest integer fault count in the approximate rejection region and interpret the decision without claiming proof.

Cambridge 2027: Direct conceptual match

Normal approximation in a discrete hypothesis test: the prompt explicitly requires the mapped operations and the worked solution demonstrates them. Conceptual placement only, pending teacher review.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. H0:p=0.20; H1:p>0.20. Under H0, X~B(200,0.20). Independent Bernoulli trials and a random sample are assumed.
  2. np=40 and n(1-p)=160 are both large. Approximate X by Y~N(40,32), where 32 is variance, not standard deviation.
  3. P(X>=51) is approximated by P(Y>50.5). z=(50.5-40)/sqrt32=1.856 to 3 decimals, giving upper-tail probability 1-0.9683=0.0317. This is below 0.05, so reject H0: evidence supports an increased fault probability, not certainty.
  4. For rejection at X>=k, require (k-0.5-40)/sqrt32>=1.645. Thus k>=40.5+1.645sqrt32, approximately 49.806. Smallest integer k=50. For k=49 the corrected z is only 1.503; for k=50 it is 1.679. This is an approximate, not exact binomial, critical region.

Self-marking points

View marking points
  1. State H0:p=0.20 and H1:p>0.20; under H0 X~B(200,0.2).
  2. Use mean 40, variance 32 and check np=40,n(1-p)=160.
  3. Correct X>=51 to Y>50.5, obtain z=1.856 and p-value about 0.0317.
  4. Reject at 5% and find approximate critical count 50 using the half-unit boundary.

Examiner tip

Show the reasoning and retain exact values until the final numerical answer. Marking points are indicative, not an official Cambridge mark scheme.

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