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.
80 original questions across 14 selected topic areas. No account or payment required to practise.
10 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: Transformed roots and rational graphs · Matrices and invariant lines · Hyperbolic inverses and parametric curvature · Arc length, surface area and rectangle bounds
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Question 1The roots of x^3-6x^2+11x-6=0 are alpha,beta,gamma. Without solving for them, find their sum, pair-product sum and product; calculate alpha^2+beta^2+gamma^2; then obtain a polynomial with roots 1/alpha,1/beta,1/gamma.5 indicative marks
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The roots of x^3-6x^2+11x-6=0 are alpha,beta,gamma. Without solving for them, find their sum, pair-product sum and product; calculate alpha^2+beta^2+gamma^2; then obtain a polynomial with roots 1/alpha,1/beta,1/gamma.
Cambridge 2027: Direct conceptual match
Recomputed all three Vieta relations, the squared-root sum 14, and the reciprocal polynomial by substitution and by checking roots 1,1/2,1/3.
Squaring the sum introduces twice the pair-product sum, so subtract 22 to isolate the squared roots.
As an independent check only after the derivation, the original roots are 1,2,3; substituting their reciprocals into the transformed polynomial gives zero.
Self-marking points
Vieta gives alpha+beta+gamma=6, alpha*beta+beta*gamma+gamma*alpha=11, alpha*beta*gamma=6.
The sum of squares is 6^2-2*11=14.
The reciprocal substitution x=1/y is valid because the nonzero constant term rules out zero roots.
Multiplying by y^3 gives 1-6y+11y^2-6y^3=0.
An equivalent polynomial is 6y^3-11y^2+6y-1=0.
Examiner tip
A transformed-root equation can be multiplied by a nonzero constant without changing its roots.
Question 2For f(x)=(x^2+1)/(x-1), state the domain and find both asymptotes, axis intersections, turning points and range. Sketch f. Then describe the domain and zero behaviour of the reciprocal function 1/f(x), distinguishing it from an algebraically extended formula at x=1.6 indicative marks
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For f(x)=(x^2+1)/(x-1), state the domain and find both asymptotes, axis intersections, turning points and range. Sketch f. Then describe the domain and zero behaviour of the reciprocal function 1/f(x), distinguishing it from an algebraically extended formula at x=1.
Cambridge 2027: Direct conceptual match
Checked the quotient identity, derivative roots, turning values and discriminant range; independently evaluated the reciprocal domain and missing x=1 value.
The oblique asymptote follows because the remainder term tends to zero. The derivative locates the largest value on the left branch and smallest on the right.
Alternatively y(x-1)=x^2+1 gives a quadratic in x with discriminant y^2-4y-4; requiring it nonnegative reproduces the range. Domain bookkeeping prevents silently filling the reciprocal's hole.
Self-marking points
The domain excludes x=1; division gives f=x+1+2/(x-1), with vertical asymptote x=1 and oblique asymptote y=x+1.
There is no real x-axis crossing, and the y-axis intersection is (0,-1).
f'=1-2/(x-1)^2=0 gives x=1 +/-sqrt(2).
Turning points are (1-sqrt(2),2-2sqrt(2)), a local maximum, and (1+sqrt(2),2+2sqrt(2)), a local minimum.
The range is y<=2-2sqrt(2) or y>=2+2sqrt(2); a sketch must show the two branches, asymptotes and labelled points.
For x!=1, 1/f=(x-1)/(x^2+1), and it has no zero on that inherited domain. The simplified expression has value zero at x=1 only if extended; the reciprocal of the original function has a hole there.
Examiner tip
Algebraic cancellation or simplification does not automatically enlarge an originally specified domain.
Question 3Using standard sums of r and r^2, derive a formula for S_n=sum from r=1 to n of r(2r+1). Evaluate S_10 exactly and explain why the infinite series cannot converge.5 indicative marks
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Using standard sums of r and r^2, derive a formula for S_n=sum from r=1 to n of r(2r+1). Evaluate S_10 exactly and explain why the infinite series cannot converge.
Cambridge 2027: Direct conceptual match
Expanded the closed form and independently summed the first ten r(2r+1) terms to 825; checked cubic growth.
Factoring n(n+1)/6 leaves 2(2n+1)+3=4n+5; keep both contributions.
The summand itself grows, so it does not tend to zero. The cubic partial-sum expression independently establishes divergence.
Self-marking points
Split S_n=2*sum r^2+sum r.
Insert n(n+1)(2n+1)/6 and n(n+1)/2.
S_n=n(n+1)(4n+5)/6.
S_10=10*11*45/6=825.
The positive partial sums grow without bound, so there is no finite infinite sum.
Examiner tip
Do not confuse summing a quadratic term in r with squaring a sum.
Question 4Consider x+y+z=6, x+2y+3z=14, 2x+3y+a*z=b. Write a matrix equation. Determine when its coefficient matrix is singular; at that value of a classify solutions for b=20 and b!=20. Solve the case a=5,b=21.5 indicative marks
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Consider x+y+z=6, x+2y+3z=14, 2x+3y+a*z=b. Write a matrix equation. Determine when its coefficient matrix is singular; at that value of a classify solutions for b=20 and b!=20. Solve the case a=5,b=21.
Cambridge 2027: Direct conceptual match
Computed determinant a-4 by independent row reduction, checked the entire parameterised solution line at a=4,b=20 and substituted (-1,6,1) for a=5,b=21.
Subtract the first equation from the second to obtain y+2z=8. Subtract the sum of the first two equations from the third to obtain (a-4)z=b-20.
Three plane equations can share a line, one point or no common point. A zero determinant alone does not tell whether the singular system is consistent.
Self-marking points
The coefficient matrix is [[1,1,1],[1,2,3],[2,3,a]] and the column on the right is (6,14,b).
Subtracting rows shows determinant a-4, so singularity occurs at a=4.
At a=4 the third left side is the sum of the first two, so consistency requires b=20.
For a=4,b=20 there is a line of solutions (x,y,z)=(t-2,8-2t,t); if b!=20 no common point exists.
For a=5,b=21 row subtraction gives z=1, then y=6 and x=-1; nonsingularity makes this solution unique.
Examiner tip
Inspect the right-hand side relation as well as the determinant.
Question 5For A=[[2,1],[1,2]], find the characteristic polynomial, verify its matrix identity by multiplication, then use that identity to express A^-1 and A^3 as linear combinations of I and A.5 indicative marks
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For A=[[2,1],[1,2]], find the characteristic polynomial, verify its matrix identity by multiplication, then use that identity to express A^-1 and A^3 as linear combinations of I and A.
Cambridge 2027: Direct conceptual match
Independently multiplied A^2, A^3 and A*(4I-A)/3; all match the stated identity, cube and inverse.
A scalar polynomial can be evaluated at a matrix only with the constant term multiplied by I. Verify the identity before using it.
Multiplication with the proposed inverse gives I using A^2=4A-3I; this independently checks the inverse without an adjugate calculation.
Self-marking points
det(lambda*I-A)=lambda^2-4lambda+3.
Direct multiplication gives A^2=[[5,4],[4,5]]=4A-3I.
Thus A satisfies A^2-4A+3I=0, its own characteristic equation.
Since det A=3!=0, multiplying by A^-1 gives A^-1=(4I-A)/3.
A^3=4A^2-3A=13A-12I=[[14,13],[13,14]].
Examiner tip
Do not replace the constant 3I by a matrix whose every entry is 3.
Question 6Column vectors undergo A=[[2,0],[0,3]]. Find all invariant points and all invariant lines through the origin, explaining the distinction. If B=[[1,1],[0,1]], calculate (BA)^-1 in two ways to check the inverse-product order.5 indicative marks
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Column vectors undergo A=[[2,0],[0,3]]. Find all invariant points and all invariant lines through the origin, explaining the distinction. If B=[[1,1],[0,1]], calculate (BA)^-1 in two ways to check the inverse-product order.
Cambridge 2027: Direct conceptual match
Derived invariant slopes and vertical case, then independently multiplied BA and both candidate inverse orders.
A line being mapped to itself means its set of points is preserved, not that every point on it is fixed. Handle the vertical line separately from finite slopes.
Check both left and right products of BA with the proposed inverse. B^-1 A^-1 would have top-right entry -1/3 and is not the inverse of BA.
Self-marking points
Invariant points satisfy (2x,3y)=(x,y), so only (0,0) is fixed.
A line y=m*x maps to Y=(3m/2)X; it is invariant only for m=0, giving the x-axis.
The vertical y-axis is also invariant; its points generally move along it rather than remain fixed.
BA=[[2,3],[0,3]], whose inverse is [[1/2,-1/2],[0,1/3]].
A^-1 B^-1=[[1/2,0],[0,1/3]]*[[1,-1],[0,1]] gives the same matrix; the order reverses.
Examiner tip
Use column-vector composition order and reverse the order again when inverting a product.
Question 7Sketch y=cosh x on the real line with its minimum and symmetry. Restrict it to x>=0 and derive its inverse arcosh u in logarithmic form. State the inverse domain/range and differentiate it where the derivative is finite.5 indicative marks
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Sketch y=cosh x on the real line with its minimum and symmetry. Restrict it to x>=0 and derive its inverse arcosh u in logarithmic form. State the inverse domain/range and differentiate it where the derivative is finite.
Cambridge 2027: Direct conceptual match
Recomputed the positive inverse branch from its quadratic, checked cosh(arcosh u)=u for u>=1 and the derivative domain u>1.
The other quadratic root is the reciprocal and belongs to the nonpositive x branch. At u=1 the roots coincide at v=1.
The inverse graph reflects the restricted cosh graph in y=x. Implicit differentiation gives 1/sinh x, and sinh x>=0 on the selected branch.
Self-marking points
cosh x=(e^x+e^-x)/2 is even, has minimum (0,1) and tends to infinity in both directions.
The nonnegative branch is one-one with range [1,infinity).
u=cosh x gives v^2-2uv+1=0 with v=e^x>=1.
Choose v=u+sqrt(u^2-1), giving arcosh u=ln(u+sqrt(u^2-1)) for u>=1 and output x>=0.
For u>1, d(arcosh u)/du=1/sqrt(u^2-1); at u=1 the graph has a vertical tangent and no finite derivative.
Examiner tip
Do not state a finite inverse derivative at the branch endpoint u=1.
Question 8(a) For x=t^2+1,y=t^3 with t>0, find dy/dx and d^2y/dx^2 and evaluate both at t=2. (b) Derive the Maclaurin expansion of ln(1+x) through x^3 from derivatives at zero; give an open interval on which its infinite power series is valid.5 indicative marks
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(a) For x=t^2+1,y=t^3 with t>0, find dy/dx and d^2y/dx^2 and evaluate both at t=2. (b) Derive the Maclaurin expansion of ln(1+x) through x^3 from derivatives at zero; give an open interval on which its infinite power series is valid.
Cambridge 2027: Direct conceptual match
Independently differentiated the parametric curve twice and obtained coefficients 1,-1/2,1/3 from the first three logarithm derivatives.
Differentiating the first derivative with respect to t is not yet differentiation with respect to x; divide by dx/dt once more.
Insert the derivative values into f(0)+f'(0)x+f''(0)x^2/2!+f'''(0)x^3/3!. The polynomial is a truncation, not an equality to the logarithm for nonzero x.
Self-marking points
dy/dx=3t^2/(2t)=3t/2.
d^2y/dx^2=[d(3t/2)/dt]/(2t)=3/(4t).
At t=2 these derivatives are 3 and 3/8 respectively.
For f=ln(1+x), f(0)=0, f'(0)=1, f''(0)=-1 and f'''(0)=2.
The expansion is x-x^2/2+x^3/3+... and is valid on the open interval |x|<1.
Examiner tip
The requested open convergence interval does not assert that both endpoints diverge.
Question 9The curve x=t,y=t^2/2, 0<=t<=1, is rotated about the x-axis. Find its arc length and generated surface area exactly. You may use t=sinh u and identities cosh^2u-sinh^2u=1 and sinh(2u)=2sinh u cosh u.5 indicative marks
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The curve x=t,y=t^2/2, 0<=t<=1, is rotated about the x-axis. Find its arc length and generated surface area exactly. You may use t=sinh u and identities cosh^2u-sinh^2u=1 and sinh(2u)=2sinh u cosh u.
Cambridge 2027: Direct conceptual match
Independently integrated the length and surface integrands by numerical quadrature and by the hyperbolic primitives; no unrounded constant is substituted early.
For the length substitution sqrt(1+t^2)=cosh u and dt=cosh u du, giving the integral of cosh^2u.
For the surface use the local circumference 2pi*y, not pi*y^2: the latter belongs to a volume calculation. Then sinh^2u cosh^2u=(cosh(4u)-1)/8 supplies the stated primitive.
Self-marking points
Arc length is L=integral(0 to 1) sqrt(1+t^2) dt.
With U=arsinh1=ln(1+sqrt(2)), L=(sqrt(2)+U)/2.
Surface area is S=integral 2pi*y*sqrt((dx/dt)^2+(dy/dt)^2) dt=pi*integral(0 to 1) t^2 sqrt(1+t^2) dt.
The substitution gives integral sinh^2u cosh^2u du=(sinh(4u)/32-u/8).
At U, sinh U=1,cosh U=sqrt(2) and sinh(4U)=12sqrt(2), so S=pi*(3sqrt(2)-U)/8 square units.
Examiner tip
These surface formulae are for a Cartesian/parametric curve; no polar surface-of-revolution outcome is inferred.
Question 10Let H_n=sum from r=1 to n of 1/r. For integers n>=1, use decreasing-function rectangles for y=1/x to prove ln(n+1)<H_n<=1+ln n. Give the resulting bounds for H_10 and determine the limit as n tends to infinity of sum from r=1 to n of 1/(n+r), explaining the rectangle interpretation.5 indicative marks
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Let H_n=sum from r=1 to n of 1/r. For integers n>=1, use decreasing-function rectangles for y=1/x to prove ln(n+1)<H_n<=1+ln n. Give the resulting bounds for H_10 and determine the limit as n tends to infinity of sum from r=1 to n of 1/(n+r), explaining the rectangle interpretation.
Cambridge 2027: Direct conceptual match
Verified the upper-bound n=1 equality and the two interval comparisons; rewrote the scaled sum as a Riemann sum and independently evaluated its ln2 limit.
Each rectangle comparison has its own interval endpoints. The first is strictly below the rectangle because the curve is strictly decreasing, while the upper H_n bound permits equality at n=1.
The last sum has n rectangles of width 1/n and a continuous integrand, so taking their shrinking-width limit is justified; it is not the divergent harmonic sum itself.
Self-marking points
On [r,r+1], the decreasing curve 1/x lies below the left rectangle of area 1/r; summing gives ln(n+1)<H_n.
For r>=2, 1/r is no greater than the area under 1/x on [r-1,r]; adding the first term gives H_n<=1+ln n.
For n=10, ln11<H_10<=1+ln10, approximately 2.39790<H_10<=3.30259.
Rewrite sum 1/(n+r) as (1/n)*sum 1/(1+r/n), a right-endpoint rectangle sum on [0,1].
Its limit is integral(0 to 1) 1/(1+x) dx=ln2.
Examiner tip
Keep the rectangle width as well as its height when passing to a limit.
