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ZIMSEC A Level · 6042/2 · N2025

Pure Mathematics Paper 2 November 2025

Questions
47
Total marks
152
Syllabus code
6042/2

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Questions
47
Pass mark
29
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[3 marks]Modulus functions and inequalities
Solve the inequality ∣x2−2∣<−x|x^2-2|<-x.

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Question 102

[2 marks]Modulus functions and inequalities
In solving ∣x2−2∣<−x|x^2-2|<-x by cases, why is only x=−2x=-2 (not x=1x=1) accepted from the case x2−2≥0x^2-2\ge0, and only x=−1x=-1 (not x=2x=2) accepted from the case x2−2<0x^2-2<0?
  1. ABecause the original inequality is only ever true for negative values of xx, by definition.
  2. BBecause x=−2x=-2 and x=−1x=-1 are always the two smallest roots, regardless of which case produced them.
  3. CBecause each case's algebraic solution must also satisfy that case's own defining condition on xx, and only one root from each case does so.
  4. DBecause x=1x=1 and x=2x=2 are not real numbers, so they cannot be solutions to any inequality.

Question 201

[1 marks]Graph transformations
The graph y=f(x)y=f(x) has yy-intercept (0,16)(0,16). State the yy-intercept of y=f(x)−6y=f(x)-6.

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Question 202

[2 marks]Graph transformations
The graph y=f(x)y=f(x) has xx-intercepts (−2,0)(-2,0), (2,0)(2,0) and (4,0)(4,0). State the xx-intercepts of y=f(x+1)y=f(x+1).

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Question 203

[1 marks]Graph transformations
The graph y=f(x)y=f(x) has yy-intercept (0,16)(0,16). State the yy-intercept of y=−12f(x)y=-\tfrac12f(x).

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Question 204

[2 marks]Graph transformations
The graph y=f(x)y=f(x) has xx-intercepts (−2,0)(-2,0), (2,0)(2,0) and (4,0)(4,0). State the xx-intercepts of y=f(2x)y=f(2x).

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Question 301

[3 marks]Differentiation from first principles, tangents
Given f(x)=2x−3x2f(x)=\dfrac{2x-3}{x^2}, differentiate from first principles to find f′(x)f'(x).

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Question 302

[3 marks]Differentiation from first principles, tangents
For f(x)=2x−3x2f(x)=\dfrac{2x-3}{x^2}, with f′(x)=−2x+6x3f'(x)=\dfrac{-2x+6}{x^3}, find the equation of the tangent to the curve at x=2x=2.

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Question 401

[2 marks]Complex numbers: division, modulus/argument, loci
The complex number u=5+4i3−2iu=\dfrac{5+4i}{3-2i}. Express uu in the form x+iyx+iy.

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Question 402

[1 marks]Complex numbers: division, modulus/argument, loci
For u=713+2213iu=\dfrac7{13}+\dfrac{22}{13}i, find the modulus ∣u∣|u|.

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Question 403

[1 marks]Complex numbers: division, modulus/argument, loci
For u=713+2213iu=\dfrac7{13}+\dfrac{22}{13}i, find the argument of uu in degrees.

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Question 404

[2 marks]Complex numbers: division, modulus/argument, loci
The complex number u≈0.54+1.69iu\approx0.54+1.69i. What does the locus ∣z−u∣=2|z-u|=2 represent on an Argand diagram?
  1. AA circle of radius 2, centred at the point representing uu.
  2. BA circle of radius 2, centred at the origin, regardless of the value of uu.
  3. CA single point, located a distance 2 from uu along the positive real axis.
  4. DA straight line perpendicular to the line joining the origin to uu, at distance 2 from uu.

Question 501

[1 marks]Functions: domain, simultaneous equations, inverse
The function f:x↦mx+nf:x\mapsto\dfrac{m}{x}+n. State the domain of f(x)f(x).

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Question 502

[2 marks]Functions: domain, simultaneous equations, inverse
The function f:x↦mx+nf:x\mapsto\dfrac{m}{x}+n satisfies f(−1)=112f(-1)=1\tfrac12 and f(2)=9f(2)=9. Find the values of mm and nn.

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Question 503

[2 marks]Functions: domain, simultaneous equations, inverse
Given f(x)=5x+132f(x)=\dfrac5x+\dfrac{13}2, find an expression for f−1(x)f^{-1}(x).

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Question 504

[2 marks]Functions: domain, simultaneous equations, inverse
Given f−1(x)=102x−13f^{-1}(x)=\dfrac{10}{2x-13}, evaluate f−1(4)f^{-1}(4).

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Question 601

[3 marks]Rational and modulus inequalities, simultaneous equations
Solve the inequality 2x−3<3x−4\dfrac2{x-3}<\dfrac3{x-4}.

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Question 602

[2 marks]Rational and modulus inequalities, simultaneous equations
The point (1,5)(1,5) satisfies the simultaneous equations n2x+my=14n^2x+my=14 and ny−mx=8ny-mx=8. Find one valid pair of values for nn and mm.

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Question 603

[3 marks]Rational and modulus inequalities, simultaneous equations
Solve the inequality ∣2x−3∣<5−∣x∣|2x-3|<5-|x|.

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Question 701

[3 marks]Differential equations: separable, exponential decay
Find the equation of the curve which satisfies dydx=−xy\dfrac{dy}{dx}=-\dfrac{x}{y} and passes through (−4,3)(-4,3).

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Question 702

[1 marks]Differential equations: separable, exponential decay
The curve x2+y2=25x^2+y^2=25 satisfies dydx=−xy\dfrac{dy}{dx}=-\dfrac{x}{y} and passes through (−4,3)(-4,3). Describe this solution curve.

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Question 703

[2 marks]Differential equations: separable, exponential decay
The rate of destruction of a drug by the kidney is proportional to the amount xx present at time tt. Form the differential equation relating xx and tt.

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Question 704

[3 marks]Differential equations: separable, exponential decay
A drug decays according to x=Ae−ktx=Ae^{-kt}. At t=0t=0, x=200x=200 mg. At t=1t=1 hour, x=180x=180 mg. Find the quantity of drug remaining after 8 hours.

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Question 801

[3 marks]Series summation, Maclaurin and binomial expansions
The nnth term of a series is 4n−4+6n4^{n-4}+6n. Find the sum SNS_N of the first NN terms.

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Question 802

[2 marks]Series summation, Maclaurin and binomial expansions
Find Maclaurin's series for f(x)=(1−x)−1f(x)=(1-x)^{-1} up to and including the term in x4x^4.

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Question 803

[3 marks]Series summation, Maclaurin and binomial expansions
Find the expansion of (9−x)3/2(1+3x)3/2(9-x)^{3/2}(1+3x)^{3/2} up to and including the term in x2x^2.

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Question 804

[2 marks]Series summation, Maclaurin and binomial expansions
The expansion (9−x)3/2(1+3x)3/2(9-x)^{3/2}(1+3x)^{3/2} requires ∣x/9∣<1|x/9|<1 and ∣3x∣<1|3x|<1. Find the set of values of xx for which the expansion is valid.

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Question 901

[3 marks]Trigonometric identities: harmonic form, double angle
Express 2cos⁡θ−3sin⁡θ2\cos\theta-\sqrt3\sin\theta in the form Rsin⁡(θ−α)R\sin(\theta-\alpha), with R>0R>0 and 0∘<α<360∘0^\circ<\alpha<360^\circ.

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Question 902

[3 marks]Trigonometric identities: harmonic form, double angle
Solve 2cos⁡θ−3sin⁡θ=12\cos\theta-\sqrt3\sin\theta=1 for 0∘≤θ≤360∘0^\circ\le\theta\le360^\circ, to the nearest degree.

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Question 903

[1 marks]Trigonometric identities: harmonic form, double angle
State the maximum value of 2cos⁡θ−3sin⁡θ2\cos\theta-\sqrt3\sin\theta, for 0∘≤θ≤360∘0^\circ\le\theta\le360^\circ.

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Question 904

[1 marks]Trigonometric identities: harmonic form, double angle
State the minimum value of 2cos⁡θ−3sin⁡θ2\cos\theta-\sqrt3\sin\theta, for 0∘≤θ≤360∘0^\circ\le\theta\le360^\circ.

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Question 905

[1 marks]Trigonometric identities: harmonic form, double angle
Express 2cos⁡2θ−12\cos^2\theta-1 in terms of cos⁡2θ\cos2\theta.

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Question 1001

[3 marks]Numerical integration, integration by parts, Newton-Raphson
Use the trapezium rule with 5 ordinates (h=0.25h=0.25) to find an approximate value of ∫01x2ex dx\displaystyle\int_0^1x^2e^x\,dx, to 2 d.p.

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Question 1002

[3 marks]Numerical integration, integration by parts, Newton-Raphson
Evaluate the exact value of ∫01x2ex dx\displaystyle\int_0^1x^2e^x\,dx using integration by parts.

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Question 1003

[2 marks]Numerical integration, integration by parts, Newton-Raphson
The trapezium rule gives 0.760.76 for ∫01x2ex dx\int_0^1x^2e^x\,dx, whose exact value is e−2≈0.718e-2\approx0.718. Is this an over-estimation or under-estimation, and why?
  1. AAn over-estimation, since x2exx^2e^x is convex (curves upward) on [0,1][0,1], so the chord segments used by the trapezium rule lie above the curve.
  2. BAn over-estimation, but only because too few ordinates (5) were used; more ordinates would make it an under-estimation instead.
  3. CAn under-estimation, since x2exx^2e^x is concave (curves downward) on [0,1][0,1], so the chord segments lie below the curve.
  4. DNeither: the trapezium rule always gives the exact value for any smooth, continuous function.

Question 1004

[3 marks]Numerical integration, integration by parts, Newton-Raphson
Use the Newton-Raphson method, starting from x0=1x_0=1, to find a root of x3+3x−7=0x^3+3x-7=0, correct to 2 d.p.

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Question 1101

[3 marks]Small errors, partial fractions and integration
An error of 3% is made in measuring the radius of a sphere (volume =43πr3=\tfrac43\pi r^3). Find the percentage error in the volume.

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Question 1102

[3 marks]Small errors, partial fractions and integration
Express 2x+1(x−2)(x2+1)\dfrac{2x+1}{(x-2)(x^2+1)} in partial fractions.

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Question 1103

[3 marks]Small errors, partial fractions and integration
Using 2x+1(x−2)(x2+1)=1x−2−xx2+1\dfrac{2x+1}{(x-2)(x^2+1)}=\dfrac1{x-2}-\dfrac{x}{x^2+1}, evaluate ∫012x+1(x−2)(x2+1) dx\displaystyle\int_0^1\dfrac{2x+1}{(x-2)(x^2+1)}\,dx.

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Question 1201

[1 marks]Group theory (Cayley table), matrix transformations
In the group R=({O,N,I,T},r)R=(\{O,N,I,T\},r), the row and column for OO reproduce the header row/column exactly. State the identity element of this group.

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Question 1202

[1 marks]Group theory (Cayley table), matrix transformations
In the group R=({O,N,I,T},r)R=(\{O,N,I,T\},r), with identity OO, state the inverse of the element NN.

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Question 1203

[2 marks]Group theory (Cayley table), matrix transformations
The Cayley table for R=({O,N,I,T},r)R=(\{O,N,I,T\},r) shows closure, an identity element OO, and every element has an inverse. What additional property must be checked to confirm RR is a group?
  1. AThat every element is its own inverse, which is required for any group.
  2. BAssociativity: that (a∗b)∗c=a∗(b∗c)(a*b)*c=a*(b*c) for all elements a,b,ca,b,c in the set.
  3. CCommutativity: that a∗b=b∗aa*b=b*a for all elements, since this alone is sufficient to define a group.
  4. DThat the set contains an even number of elements, which is required for any group.

Question 1204

[3 marks]Group theory (Cayley table), matrix transformations
Which matrix MM describes an enlargement by factor 2 followed by a shear, factor 3 in the xx-axis direction with the yy-axis invariant?
  1. AM=(6202)M=\begin{pmatrix}6&2\\0&2\end{pmatrix}
  2. BM=(2302)M=\begin{pmatrix}2&3\\0&2\end{pmatrix}
  3. CM=(2062)M=\begin{pmatrix}2&0\\6&2\end{pmatrix}
  4. DM=(2602)M=\begin{pmatrix}2&6\\0&2\end{pmatrix}

Question 1205

[1 marks]Group theory (Cayley table), matrix transformations
Using M=(2602)M=\begin{pmatrix}2&6\\0&2\end{pmatrix}, find the image A′A' of the point A(3,2)A(3,2).

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Question 1206

[1 marks]Group theory (Cayley table), matrix transformations
Using M=(2602)M=\begin{pmatrix}2&6\\0&2\end{pmatrix}, find the image B′B' of the point B(1,1)B(1,1).

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Question 1207

[1 marks]Group theory (Cayley table), matrix transformations
Using M=(2602)M=\begin{pmatrix}2&6\\0&2\end{pmatrix}, find the image C′C' of the point C(5,1)C(5,1).

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Question 1208

[2 marks]Group theory (Cayley table), matrix transformations
Triangle ABCABC (area 2 square units) is transformed by M=(2602)M=\begin{pmatrix}2&6\\0&2\end{pmatrix} to give triangle A′B′C′A'B'C'. Find the area of A′B′C′A'B'C'.

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