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.
83 original questions across 16 selected topic areas. No account or payment required to practise.
5 questions in this topic. Exact English revisions checked in an AI-assisted academic review; independent teacher approval is still pending.
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Original Cambridge 2027 concept practice for source-backed gaps. Indicative self-marking only, not an official paper or complete syllabus coverage.
Selected subtopics: Quadratic intersections and inverse graphs · Polynomial division and binomial coefficients · Logarithmic models and fixed-point iteration
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Question 1Let f(x)=2x^2-8x+5. Write f in completed-square form, locate its vertex and solve f(x)=0 exactly. Then find both intersections with y=2x+5 and relate the number of intersections to a discriminant.6 indicative marks
Show your reasoning6 indicative marksCalculator allowed
Let f(x)=2x^2-8x+5. Write f in completed-square form, locate its vertex and solve f(x)=0 exactly. Then find both intersections with y=2x+5 and relate the number of intersections to a discriminant.
Cambridge 2027: Direct conceptual match
Re-expanded the completed square, substituted both surd roots and independently solved the two graph intersections; discriminant is (-10)^2=100.
Expand 2(x-2)^2-3 to recover the original polynomial. The completed square makes the smallest value and the exact roots visible.
Equating the two y expressions is substitution in a linear/quadratic simultaneous pair. Use y=2x+5 after solving for x; a positive discriminant confirms two crossings, not a tangent.
Self-marking points
f(x)=2(x-2)^2-3.
The vertex is (2,-3), a minimum because the leading coefficient is positive.
f(x)=0 gives x=2 +/- sqrt(6)/2.
For intersections, 2x^2-8x+5=2x+5, so 2x(x-5)=0.
The points are (0,5) and (5,15).
The intersection quadratic has discriminant 100>0, confirming two distinct real intersections.
Examiner tip
A discriminant describes the equation formed by the intersection, not necessarily the original quadratic alone.
Question 2(a) Divide P(x)=2x^3+3x^2-5x+7 by x+2 and check the remainder by a theorem. (b) Find the coefficient of x^3 in (2+x)^5.5 indicative marks
Show your reasoning5 indicative marksCalculator allowed
(a) Divide P(x)=2x^3+3x^2-5x+7 by x+2 and check the remainder by a theorem. (b) Find the coefficient of x^3 in (2+x)^5.
Cambridge 2027: Direct conceptual match
Recombined divisor times quotient plus 13 and counted ten choices times four for the coefficient 40.
Long division removes successive leading terms 2x^3, -x^2 and -3x. Recombination gives the original cubic with remainder 13.
For the binomial term choose three of the five factors to contribute x; the other two contribute 2. This positive-integer expansion terminates and has no convergence restriction.
Self-marking points
Polynomial division gives quotient 2x^2-x-3.
The remainder is 13, so P(x)=(x+2)(2x^2-x-3)+13.
P(-2)=-16+12+10+7=13, independently confirming the remainder.
The x^3 term is C(5,3)*2^2*x^3.
Its coefficient is 40.
Examiner tip
Remainder evaluation uses the root -2, and the binomial exponent on 2 is 5-3.
Question 3Positive measurements follow y=A*x^n with A>0. Two exact model points are (x,y)=(2,12) and (4,48). Explain which logarithmic graph is straight, find its gradient and intercept, determine A and n, and predict y at x=3.5 indicative marks
Show your reasoning5 indicative marksCalculator allowed
Positive measurements follow y=A*x^n with A>0. Two exact model points are (x,y)=(2,12) and (4,48). Explain which logarithmic graph is straight, find its gradient and intercept, determine A and n, and predict y at x=3.
Cambridge 2027: Direct conceptual match
Verified the log-gradient using exact ratios, recovered A=3 and substituted all three x values into y=3x^2.
The logarithms are defined because x,y,A are positive. Dividing the original model values also gives 48/12=(4/2)^n, so 2^n=4 independently checks n.
An intercept of ln3 is not A=ln3: exponentiate it to obtain A=3. The prediction is exact for the stipulated model, not a claim about real data error.
Self-marking points
Taking logs gives ln y=ln A+n ln x.
Plot ln y vertically against ln x horizontally; the gradient is n.
n=(ln48-ln12)/(ln4-ln2)=ln4/ln2=2.
ln A=ln12-2ln2=ln3, so A=3.
At x=3 the model gives y=3*3^2=27.
Examiner tip
Name both transformed axes; a straight graph of ln y against x instead would represent an exponential model.
Question 4Let f(x)=x^3-x-1. Show that a real root lies in (1,2). Use x_(n+1)=(x_n+1)^(1/3), with x_0=1, for six iterations. Estimate the root to 2 decimal places and justify that rounding by evaluating f at the two boundaries of the rounding interval. Explain why numerical iterations alone do not prove an exact root.5 indicative marks
Show your reasoning5 indicative marksCalculator allowed
Let f(x)=x^3-x-1. Show that a real root lies in (1,2). Use x_(n+1)=(x_n+1)^(1/3), with x_0=1, for six iterations. Estimate the root to 2 decimal places and justify that rounding by evaluating f at the two boundaries of the rounding interval. Explain why numerical iterations alone do not prove an exact root.
Cambridge 2027: Direct conceptual match
Computed all six cube-root iterates independently; checked monotonicity and the signs at 1.315 and 1.325.
Apply the given rearrangement repeatedly without rounding the stored values. Each iterate is an approximation in a sequence, not a new algebraic equation satisfied exactly by the true root.
Use the endpoints 1.315 and 1.325 of the 1.32 rounding bin, not the rounded answer itself, for the sign test. Monotonicity removes any ambiguity about which root is bracketed.
Self-marking points
f(1)=-1 and f(2)=5; continuity gives a root between them.
At a fixed point, x=(x+1)^(1/3) is equivalent to x^3-x-1=0.
The six iterates are approximately 1.259921, 1.312294, 1.322354, 1.324269, 1.324633, 1.324702.
f(1.315)<0 and f(1.325)>0, so this root rounds to 1.32.
f'(x)=3x^2-1>0 on [1,2], so there is only one root in that interval; finite decimal iteration is an approximation, not an exact equality.
Examiner tip
Report the rounding supported by a sign bracket, not a string of apparently stable digits.
Question 5(a) For f(x)=x^2+1 with domain x>=0, find f^-1, its domain and the mirror line between their graphs. (b) Sketch y=2sin(2x) for 0<=x<=pi, labelling zeros, extrema, amplitude and period of the continuing function.5 indicative marks
Show your reasoning5 indicative marksCalculator allowed
(a) For f(x)=x^2+1 with domain x>=0, find f^-1, its domain and the mirror line between their graphs. (b) Sketch y=2sin(2x) for 0<=x<=pi, labelling zeros, extrema, amplitude and period of the continuing function.
Cambridge 2027: Direct conceptual match
Checked positive inverse branch, reflected example points and the five quarter-period sine values analytically.
Restricting the original domain selects the positive square-root branch. For instance, (2,5) and (5,2) are reflected points on the original and inverse curves.
The argument 2x runs from 0 to 2pi over the requested interval, giving one complete sine cycle. Its period is halved by the horizontal scale factor, not doubled.
Self-marking points
f is one-one on x>=0; solving y=x^2+1 gives f^-1(x)=sqrt(x-1).
The inverse domain is x>=1 and its graph is the reflection of f in y=x.
For the sine graph, zeros are x=0,pi/2,pi.
The maximum is (pi/4,2) and the minimum is (3pi/4,-2).
The amplitude is 2 and the full period is pi; the sketch smoothly joins the labelled points.
Examiner tip
Draw the smooth curve and y=x mirror line; a list of coordinates alone is not a completed sketch.
