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ZIMSEC A Level · 6042/1 · N2025

Pure Mathematics Paper 1 November 2025

Questions
50
Total marks
120
Syllabus code
6042/1

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Questions
50
Pass mark
30
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[1 marks]Variation
A ball's mass mm varies directly as the cube of its diameter dd (m=kd3m=kd^3). What power of dd does mm vary as?

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Question 102

[3 marks]Variation
A soccer ball's mass varies directly as the cube of its diameter. A ball of diameter 7 cm has mass 0.11 kg. Calculate the mass of a similar ball of diameter 9 cm, to 2 s.f.

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Question 201

[1 marks]Differential equations
Given dydθ+2sin⁡2θ=0\dfrac{dy}{d\theta}+2\sin2\theta=0, integrating gives y=cos⁡2θ+Cy=\cos2\theta+C. Using y=1y=1 when θ=π/6\theta=\pi/6, find CC.

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Question 202

[3 marks]Differential equations
Given dydθ+2sin⁡2θ=0\dfrac{dy}{d\theta}+2\sin2\theta=0, where y=1y=1 when θ=π/6\theta=\pi/6, find the equation connecting yy and θ\theta.

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Question 301

[2 marks]Complex numbers
For the equation z2+3z+4i+2=0z^2+3z+4i+2=0, the sum of the two complex roots equals −b/a-b/a. Find the sum of the roots.

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Question 302

[3 marks]Complex numbers
Solve z2+3z+4i+2=0z^2+3z+4i+2=0, giving one root in the form x+iyx+iy, correct to 2 d.p.

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Question 401

[2 marks]Small-angle approximations
For small θ\theta (radians, θ→0\theta\to0), find an approximate value, in terms of θ\theta, for sin⁡4θ+tan⁡2θ3+cos⁡2θ\dfrac{\sin4\theta+\tan2\theta}{3+\cos2\theta}, neglecting terms in θ2\theta^2 and higher.

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Question 402

[3 marks]Small-angle approximations
Given 1∘≈0.0181^\circ\approx0.018 radians and (0.018)2≈0.000324(0.018)^2\approx0.000324, find an approximate value for cos⁡2∘sin⁡1∘\dfrac{\cos2^\circ}{\sin1^\circ}, to 2 d.p.

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Question 501

[2 marks]Quadratic equations
The equation x2+2x+4=P(x2+4)x^2+2x+4=P(x^2+4) has equal roots. Rearranging gives (1−P)x2+2x+(4−4P)=0(1-P)x^2+2x+(4-4P)=0. Setting the discriminant to zero leads to (1−P)2=k(1-P)^2=k for some constant kk. Find kk.

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Question 502

[3 marks]Quadratic equations
Given that x2+2x+4=P(x2+4)x^2+2x+4=P(x^2+4) has equal roots, find the possible values of PP.

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Question 601

[3 marks]Matrix transformations
The matrix (4−1−32)\begin{pmatrix}4&-1\\-3&2\end{pmatrix} maps every point on y=3xy=3x onto itself. Verify this for the point (1,3)(1,3): find its image under the matrix.

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Question 602

[2 marks]Matrix transformations
For a general point (x,3x)(x,3x) on the line y=3xy=3x, applying the matrix (4−1−32)\begin{pmatrix}4&-1\\-3&2\end{pmatrix} gives back exactly (x,3x)(x,3x). What does this prove about the matrix and the line?
  1. AEvery point on the line y=3xy=3x is invariant (mapped onto itself) under this transformation.
  2. BThe line y=3xy=3x is the only line in the plane invariant under any matrix, so the result holds for every possible transformation matrix, not just this one.
  3. CThe matrix swaps the coordinates of every point it is applied to, for any line.
  4. DThe matrix has no effect on any point in the plane, not just points on y=3xy=3x.

Question 701

[3 marks]Exponential equations
Substituting y=3xy=3^x into 2(32x+1)−20(3x)=−62(3^{2x+1})-20(3^x)=-6 gives a quadratic in yy. Solve for yy.

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Question 702

[3 marks]Exponential equations
Solve the equation 2(32x+1)−20(3x)=−62(3^{2x+1})-20(3^x)=-6 for xx.

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Question 801

[2 marks]Proof by induction
To prove 12+16+⋯+1n(n−1)=1−1n\dfrac12+\dfrac16+\cdots+\dfrac1{n(n-1)}=1-\dfrac1n by induction for n>1n>1, check the base case n=2n=2: what is the value of the left-hand side?

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Question 802

[2 marks]Proof by induction
In the induction proof for 12+16+⋯+1n(n−1)=1−1n\dfrac12+\dfrac16+\cdots+\dfrac1{n(n-1)}=1-\dfrac1n, going from n=mn=m to n=m+1n=m+1 adds one extra term to the left-hand side. What is that extra term?

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Question 803

[2 marks]Proof by induction
Having verified the base case n=2n=2 and shown that truth at n=mn=m implies truth at n=m+1n=m+1, why does this prove the statement for every integer n>1n>1?
  1. ABecause the formula was already given in the question, so no proof of the inductive step is actually needed.
  2. BBy the principle of mathematical induction: truth at the base case plus the inductive step together guarantee truth for every subsequent integer.
  3. CBecause the series converges, and any convergent series formula is automatically true for all nn.
  4. DBecause checking just two cases, n=2n=2 and n=mn=m, is always enough to prove a statement true for all real numbers, without needing any further justification.

Question 901

[2 marks]Coordinate geometry
The line 2x−14y+6=02x-14y+6=0 passes through L(−3,0)L(-3,0) and M(k,1)M(k,1). Find the value of kk.

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Question 902

[2 marks]Coordinate geometry
Points L(−3,0)L(-3,0) and M(4,1)M(4,1). Find the length LMLM, in surd form.

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Question 903

[3 marks]Coordinate geometry
Find the equation of the line through M(4,1)M(4,1) parallel to the line 2x+y+5=02x+y+5=0.

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Question 1001

[2 marks]Completing the square, curve sketching
Express 10x−x2−2710x-x^2-27 in the form −(a−x)2+b-(a-x)^2+b, where aa and bb are integers. State aa and bb.

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Question 1002

[2 marks]Completing the square, curve sketching
Given 10x−x2−27=−(5−x)2−210x-x^2-27=-(5-x)^2-2, why is 10x−x2−2710x-x^2-27 always negative for every real xx?
  1. ABecause x2x^2 is always positive, so subtracting it from 10x−2710x-27 must always give a negative result.
  2. BBecause the coefficient of x2x^2 in the original expression is negative, which alone guarantees a negative value everywhere.
  3. CBecause 10x−2710x-27 is negative for all xx, regardless of the −x2-x^2 term.
  4. DBecause −(5−x)2≤0-(5-x)^2\le0 for every real xx, so the whole expression is at most −2-2, which is always negative.

Question 1003

[2 marks]Completing the square, curve sketching
State the coordinates of the maximum point of the curve y=10x−x2−27y=10x-x^2-27.

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Question 1004

[2 marks]Completing the square, curve sketching
State the coordinates of the point where the curve y=10x−x2−27y=10x-x^2-27 crosses the yy-axis.

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Question 1101

[2 marks]Rational functions, transformations
Given f(x)=2x+5x+1f(x)=\dfrac{2x+5}{x+1}, x≠−1x\ne-1, show that f(x)≡2+Ax+1f(x)\equiv2+\dfrac{A}{x+1} for some constant AA. Find AA.

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Question 1102

[3 marks]Rational functions, transformations
Which sequence of transformations maps the graph of y=1xy=\dfrac1x onto the graph of y=2x+5x+1=2+3x+1y=\dfrac{2x+5}{x+1}=2+\dfrac3{x+1}?
  1. ATranslate 2 units in the positive xx-direction, stretch vertically by scale factor 3, then translate 1 unit in the negative yy-direction.
  2. BStretch vertically by scale factor 2, translate 3 units in the positive xx-direction, then translate 1 unit in the negative yy-direction.
  3. CReflect in the xx-axis, translate 1 unit in the positive xx-direction, then stretch vertically by scale factor 2.
  4. DTranslate 1 unit in the negative xx-direction, stretch vertically by scale factor 3, then translate 2 units in the positive yy-direction.

Question 1103

[3 marks]Rational functions, transformations
For y=f(x)=2x+5x+1y=f(x)=\dfrac{2x+5}{x+1} with x≥−1x\ge-1, state the vertical and horizontal asymptotes.

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Question 1201

[3 marks]Graphical solution of equations
On the interval 0<x<π/20<x<\pi/2, y1=cot⁡xy_1=\cot x decreases from +∞+\infty to 00, while y2=2x2−1y_2=2x^2-1 increases from −1-1 to about 3.933.93. Why does this show the equation cot⁡x=2x2−1\cot x=2x^2-1 has exactly one root in this interval?
  1. ABoth curves are strictly increasing over the interval, so they can never actually cross at all.
  2. BOne curve is strictly decreasing and the other strictly increasing, their relative order reverses across the interval, so they cross exactly once.
  3. CSince y1y_1 starts above y2y_2 and stays above it throughout the interval, the curves never meet.
  4. DThe two curves have the same value at both endpoints of the interval, which is what proves they cross exactly once regardless of how they behave in between.

Question 1301

[1 marks]Group theory
In (G,∗)=({0,1,2,3},+ mod 4)(G,*)=(\{0,1,2,3\},+\bmod4), state the identity element.

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Question 1302

[1 marks]Group theory
In (G,∗)=({0,1,2,3},+ mod 4)(G,*)=(\{0,1,2,3\},+\bmod4), state the inverse of 1.

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Question 1303

[1 marks]Group theory
In (G,∗)=({0,1,2,3},+ mod 4)(G,*)=(\{0,1,2,3\},+\bmod4), state the inverse of 3.

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Question 1304

[2 marks]Group theory
(G,∗)=({0,1,2,3},+ mod 4)(G,*)=(\{0,1,2,3\},+\bmod4) has closure, an identity element (0), and every element has an inverse within GG. What additional axiom must hold for (G,∗)(G,*) to be a group?
  1. ACommutativity, which is not actually required for a group and does not need to be checked here or in any other group.
  2. BThat GG has infinitely many elements, since finite sets can never form groups.
  3. CAssociativity: (a∗b)∗c=a∗(b∗c)(a*b)*c=a*(b*c) for all a,b,ca,b,c in GG, which is inherited here from ordinary integer addition.
  4. DThat every element equals its own inverse, which is required for any set to form a group.

Question 1305

[1 marks]Group theory
Is (G,∗)=({0,1,2,3},+ mod 4)(G,*)=(\{0,1,2,3\},+\bmod4) a group under addition modulo 4?

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Question 1401

[3 marks]Inverse and composite functions
For f:x↦e3xf:x\mapsto e^{3x}, find f−1(x)f^{-1}(x) and state its domain.

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Question 1402

[3 marks]Inverse and composite functions
Given f:x↦e3xf:x\mapsto e^{3x} and g:x↦ln⁡(x+2)g:x\mapsto\ln(x+2), x>−2x>-2, find fg(x)fg(x) in terms of xx.

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Question 1403

[2 marks]Inverse and composite functions
The graphs of y=f(x)=e3xy=f(x)=e^{3x} and y=f−1(x)=13ln⁡xy=f^{-1}(x)=\tfrac13\ln x are sketched on the same axes. What is the geometric relationship between the two curves?
  1. AThey are identical curves, since ff and f−1f^{-1} always coincide exactly.
  2. BThey are translations of each other by 1 unit along the xx-axis.
  3. CThey are reflections of each other in the line y=xy=x.
  4. DThey are reflections of each other in the yy-axis.

Question 1501

[2 marks]Newton-Raphson, root location
Rewriting e2x+4x−5=0e^{2x}+4x-5=0 as e2x=5−4xe^{2x}=5-4x, why does this equation have exactly one root?
  1. Ay1=e2xy_1=e^{2x} is always greater than y2=5−4xy_2=5-4x for every value of xx, so no root can actually exist.
  2. By1=e2xy_1=e^{2x} is strictly increasing while y2=5−4xy_2=5-4x is strictly decreasing, so an increasing and a decreasing curve can meet at most once, and here they do meet once.
  3. CBoth y1=e2xy_1=e^{2x} and y2=5−4xy_2=5-4x are strictly increasing functions of xx, so by the general rule for monotonic curves they are guaranteed to intersect at exactly one point somewhere on the real line.
  4. DThe equation is a straight line, so it can only ever have one root by definition.

Question 1502

[1 marks]Newton-Raphson, root location
For h(x)=e2x+4x−5h(x)=e^{2x}+4x-5, what is the sign of h(0)h(0)?

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Question 1503

[1 marks]Newton-Raphson, root location
For h(x)=e2x+4x−5h(x)=e^{2x}+4x-5, what is the sign of h(1)h(1)?

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Question 1504

[2 marks]Newton-Raphson, root location
Using Newton-Raphson with x0=0.5x_0=0.5 on h(x)=e2x+4x−5h(x)=e^{2x}+4x-5, one iteration gives x1=x0−h(x0)/h′(x0)x_1=x_0-h(x_0)/h'(x_0). Find x1x_1, to 4 d.p.

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Question 1505

[3 marks]Newton-Raphson, root location
Use the Newton-Raphson method, starting from x0=0.5x_0=0.5, to find the root of e2x+4x−5=0e^{2x}+4x-5=0, correct to 2 decimal places.

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Question 1601

[2 marks]Stationary points, curve sketching
Find dydx\dfrac{dy}{dx} for the curve y=2x3+3x2−12x+1y=2x^3+3x^2-12x+1.

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Question 1602

[3 marks]Stationary points, curve sketching
Find the stationary points of y=2x3+3x2−12x+1y=2x^3+3x^2-12x+1, using dydx=6x2+6x−12=6(x+2)(x−1)\dfrac{dy}{dx}=6x^2+6x-12=6(x+2)(x-1).

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Question 1603

[2 marks]Stationary points, curve sketching
At the stationary point (−2,21)(-2,21) of y=2x3+3x2−12x+1y=2x^3+3x^2-12x+1, the second derivative d2ydx2=12x+6\dfrac{d^2y}{dx^2}=12x+6 equals −18-18. What kind of stationary point is this?
  1. AA local maximum, since the second derivative is negative there.
  2. BA point of inflection, since the second derivative is negative there.
  3. CA local maximum, but only because x=−2x=-2 is negative, not because of the second derivative.
  4. DA local minimum, since the second derivative is negative there.

Question 1604

[2 marks]Stationary points, curve sketching
For the curve y=2x3+3x2−12x+1y=2x^3+3x^2-12x+1, state the value of d2ydx2\dfrac{d^2y}{dx^2} at the stationary point x=1x=1, and whether it is a maximum or minimum.

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Question 1605

[1 marks]Stationary points, curve sketching
State the coordinates of the point where the curve y=2x3+3x2−12x+1y=2x^3+3x^2-12x+1 crosses the yy-axis.

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Question 1701

[2 marks]Matrix inverse, simultaneous equations
Find the determinant of P=(1122−111−23)P=\begin{pmatrix}1&1&2\\2&-1&1\\1&-2&3\end{pmatrix}.

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Question 1702

[1 marks]Matrix inverse, simultaneous equations
Given det⁡P=−12\det P=-12 for P=(1122−111−23)P=\begin{pmatrix}1&1&2\\2&-1&1\\1&-2&3\end{pmatrix}, and that the top-left cofactor is −1-1, state the value of the top-left entry of P−1P^{-1}.

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Question 1703

[3 marks]Matrix inverse, simultaneous equations
Solve the simultaneous equations x+y+2z=0x+y+2z=0, 2x−y+z=−32x-y+z=-3, x−2y+3z=5x-2y+3z=5 for xx, yy and zz.

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Question 1704

[2 marks]Matrix inverse, simultaneous equations
For the solution x=−3x=-3, y=−1y=-1, z=2z=2 of the simultaneous equations, find x+y+zx+y+z.

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The answers, and why they are the answers

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