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ZIMSEC A Level · 6042/2 · N2024

Pure Mathematics Paper 2 November 2024

Questions
61
Total marks
152
Syllabus code
6042/2

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Questions
61
Pass mark
37
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Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[2 marks]Partial fractions
Factorise 2x2−x−12x^2-x-1.

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Question 102

[2 marks]Partial fractions
Write 2x22x2−x−1\dfrac{2x^2}{2x^2-x-1} as 11 plus a proper fraction, before splitting it into partial fractions.

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Question 103

[2 marks]Partial fractions
Express 2x22x2−x−1\dfrac{2x^2}{2x^2-x-1} in partial fractions.

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Question 201

[3 marks]Integration by parts
Find ∫e2xcos⁡x dx\displaystyle\int e^{2x}\cos x\,dx, omitting the constant of integration.

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Question 202

[3 marks]Integration by parts
Evaluate ∫01e2xcos⁡x dx\displaystyle\int_0^1e^{2x}\cos x\,dx, giving the answer correct to 3 significant figures.

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Question 301

[3 marks]Binomial expansion for a rational index
Before expanding (3−x)−12(3-x)^{-\frac12} by the binomial series, it is written as a constant multiplied by (1−x3)−12\left(1-\dfrac{x}{3}\right)^{-\frac12}. State that constant.

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Question 302

[3 marks]Binomial expansion for a rational index
Expand (1−x3)−12\left(1-\dfrac{x}{3}\right)^{-\frac12} up to and including the term in x3x^3.

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Question 303

[2 marks]Binomial expansion for a rational index
State the range of values of xx for which the binomial expansion of (3−x)−12(3-x)^{-\frac12} is valid.

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Question 401

[3 marks]Cubic equations and inequalities
Solve the equation x3−4x2+x=−6x^3-4x^2+x=-6.

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Question 402

[2 marks]Cubic equations and inequalities
State the yy-intercept of the graph of y=x3−4x2+x+6y=x^3-4x^2+x+6.

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Question 403

[3 marks]Cubic equations and inequalities
Given that x3−4x2+x+6=(x+1)(x−2)(x−3)x^3-4x^2+x+6=(x+1)(x-2)(x-3), solve the inequality x3−4x2+x+6<0x^3-4x^2+x+6<0.

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Question 501

[3 marks]Inverse functions and their graphs
A function is defined by h:x⟶2x−3h:x\longrightarrow\dfrac{2}{x-3} for x>3x>3. Find h−1(x)h^{-1}(x).

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Question 502

[3 marks]Inverse functions and their graphs
State the domain of h−1h^{-1} where h:x⟶2x−3h:x\longrightarrow\dfrac{2}{x-3} for x>3x>3.

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Question 503

[2 marks]Inverse functions and their graphs
For h:x⟶2x−3h:x\longrightarrow\dfrac{2}{x-3} with x>3x>3, find the value of hh−1(2)hh^{-1}(2).

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Question 504

[2 marks]Inverse functions and their graphs
How are the graphs of y=h(x)y=h(x) and y=h−1(x)y=h^{-1}(x) related when they are drawn on the same axes?
  1. AEach is the reflection of the other in the line y=xy=x, so any point (a,b)(a,b) on one has a partner (b,a)(b,a) on the other.
  2. BEach is the reflection of the other in the yy-axis, since taking an inverse reverses the sign of every input.
  3. COne is the other rotated through a half turn about the origin, so the two curves lie in opposite quadrants.
  4. DOne is the other translated along the xx-axis, by a distance equal to the value taken by the horizontal asymptote of the first curve.

Question 505

[2 marks]Inverse functions and their graphs
State the horizontal asymptote of y=h−1(x)=3+2xy=h^{-1}(x)=3+\dfrac{2}{x}.

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Question 601

[2 marks]The R-alpha form, trigonometric equations and compound angles
Expressing 2cos⁡θ−sin⁡θ2\cos\theta-\sin\theta in the form Rcos⁡(θ+x)R\cos(\theta+x) with R>0R>0, find RR in exact form.

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Question 602

[2 marks]The R-alpha form, trigonometric equations and compound angles
Expressing 2cos⁡θ−sin⁡θ2\cos\theta-\sin\theta in the form Rcos⁡(θ+x)R\cos(\theta+x) with 0∘<x<90∘0^\circ<x<90^\circ, find xx to the nearest degree.

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Question 603

[2 marks]The R-alpha form, trigonometric equations and compound angles
In writing acos⁡θ−bsin⁡θa\cos\theta-b\sin\theta as Rcos⁡(θ+x)R\cos(\theta+x), how are RR and xx obtained?
  1. ARR is the sum a+ba+b and xx is the angle whose sine is ba+b\dfrac{b}{a+b}, taken in the first quadrant so that it comes out acute.
  2. BR=a2+b2R=\sqrt{a^2+b^2} and tan⁡x=ba\tan x=\dfrac{b}{a}, from matching the cos⁡θ\cos\theta and sin⁡θ\sin\theta terms after expanding.
  3. CRR is the larger of aa and bb, and xx is the angle whose tangent is the smaller divided by the larger.
  4. DR=a2+b2R=a^2+b^2 and tan⁡x=ab\tan x=\dfrac{a}{b}, from matching the two terms and then squaring both of the results.

Question 604

[3 marks]The R-alpha form, trigonometric equations and compound angles
Solve 10cos⁡θ−5sin⁡θ=310\cos\theta-5\sin\theta=3 for 0≤θ≤360∘0\le\theta\le360^\circ, giving the answers to the nearest degree.

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Question 605

[3 marks]The R-alpha form, trigonometric equations and compound angles
Find the maximum value of 12cos⁡θ−sin⁡θ+11\dfrac{1}{2\cos\theta-\sin\theta+11}, correct to 2 significant figures.

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Question 606

[2 marks]The R-alpha form, trigonometric equations and compound angles
Find the minimum value of 12cos⁡θ−sin⁡θ+11\dfrac{1}{2\cos\theta-\sin\theta+11}, correct to 2 significant figures.

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Question 607

[2 marks]The R-alpha form, trigonometric equations and compound angles
Given that sin⁡A=35\sin A=\dfrac35 for 0<A<90∘0<A<90^\circ, find tan⁡A\tan A.

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Question 701

[2 marks]Complex numbers: DeMoivre's theorem, roots and loci
Find the modulus of the complex number Z=1+iZ=1+i.

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Question 702

[2 marks]Complex numbers: DeMoivre's theorem, roots and loci
Find the argument of the complex number Z=1+iZ=1+i, in radians.

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Question 703

[3 marks]Complex numbers: DeMoivre's theorem, roots and loci
Each of the five fifth roots of 1+i1+i has the same modulus. Find it, correct to 3 significant figures.

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Question 704

[3 marks]Complex numbers: DeMoivre's theorem, roots and loci
The fifth roots of 1+i1+i have arguments π4+8k\dfrac{\pi}{4+8k} divided by nothing; more precisely each is π4+2kπ5\dfrac{\frac{\pi}{4}+2k\pi}{5} for k=0,1,2,3,4k=0,1,2,3,4. State the smallest positive one, in radians.

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Question 705

[3 marks]Complex numbers: DeMoivre's theorem, roots and loci
How many distinct fifth roots does a non-zero complex number have, and how are they arranged on an Argand diagram?
  1. AFive, but all lying on one straight line through the origin, each a fifth of the distance of the one before it.
  2. BTen, arranged in five conjugate pairs, since every root of a complex number is accompanied by its own conjugate.
  3. CFive, spaced 72∘72^\circ apart around a circle centred on the origin whose radius is the fifth root of the modulus.
  4. DOne, since a fifth root is an odd root and an odd root of any number is unique, exactly as it is for real numbers.

Question 706

[3 marks]Complex numbers: DeMoivre's theorem, roots and loci
On an Argand diagram the region 0<arg⁡(Z+1+i)<π30<\arg(Z+1+i)<\dfrac{\pi}{3} is a wedge. State the coordinates of its vertex.

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Question 801

[3 marks]Arithmetic progressions and proof by induction
In an arithmetic progression with first term aa and common difference dd, the eighth term is twice the third term. Express aa in terms of dd.

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Question 802

[3 marks]Arithmetic progressions and proof by induction
An arithmetic progression has a=3da=3d and the sum of its first eight terms is 3939. Find the first term.

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Question 803

[2 marks]Arithmetic progressions and proof by induction
An arithmetic progression has a=3da=3d and the sum of its first eight terms is 3939. Find the common difference.

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Question 804

[2 marks]Arithmetic progressions and proof by induction
An arithmetic progression has first term 94\dfrac94 and common difference 34\dfrac34. Find its ninth term.

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Question 805

[3 marks]Arithmetic progressions and proof by induction
For an arithmetic progression with a=94a=\dfrac94 and d=34d=\dfrac34, the sum of nn terms takes the form 38n(n+k)\dfrac38n(n+k). Find kk.

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Question 806

[3 marks]Arithmetic progressions and proof by induction
A proof by induction of a formula for ∑r=1nUr\displaystyle\sum_{r=1}^{n}U_r begins by checking the formula when n=1n=1. What must then be done to complete it?
  1. ASubstitute a large value of nn and confirm that both sides agree to within a small margin of rounding error.
  2. BAssume the formula holds for n=kn=k, then use that assumption to deduce that it also holds for n=k+1n=k+1.
  3. CCheck the formula for a second value such as n=2n=2, and for a third such as n=3n=3, until a clear pattern appears.
  4. DDifferentiate both sides with respect to nn and confirm that the two derivatives obtained are equal to each other.

Question 901

[3 marks]Vectors: unit vectors, angles and the equation of a plane
Given p=i+j+k\mathbf{p}=\mathbf{i}+\mathbf{j}+\mathbf{k}, q=i+2j+3k\mathbf{q}=\mathbf{i}+2\mathbf{j}+3\mathbf{k} and r=i−3j+2k\mathbf{r}=\mathbf{i}-3\mathbf{j}+2\mathbf{k}, find p+q+r\mathbf{p}+\mathbf{q}+\mathbf{r}.

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Question 902

[3 marks]Vectors: unit vectors, angles and the equation of a plane
Find the exact magnitude of the vector 3i+6k3\mathbf{i}+6\mathbf{k}.

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Question 903

[3 marks]Vectors: unit vectors, angles and the equation of a plane
Find a unit vector parallel to 3i+6k3\mathbf{i}+6\mathbf{k}.

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Question 904

[3 marks]Vectors: unit vectors, angles and the equation of a plane
Find the scalar product of 3i+6k3\mathbf{i}+6\mathbf{k} and i+j+k\mathbf{i}+\mathbf{j}+\mathbf{k}.

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Question 905

[2 marks]Vectors: unit vectors, angles and the equation of a plane
Points PP and QQ have position vectors i+j+k\mathbf{i}+\mathbf{j}+\mathbf{k} and i+2j+3k\mathbf{i}+2\mathbf{j}+3\mathbf{k}. Find PQ→\overrightarrow{PQ}.

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Question 906

[2 marks]Vectors: unit vectors, angles and the equation of a plane
Points P(1;1;1)P(1;1;1), Q(1;2;3)Q(1;2;3) and R(1;−3;2)R(1;-3;2) all lie on one plane. What is its cartesian equation?
  1. Ax+y+z=3x+y+z=3, because a plane through three points always has the sum of the coordinates of the first of them.
  2. Bx=1x=1, because every one of the three points has xx-coordinate 1, so the plane is the one at that fixed value.
  3. Cy=1y=1, because the middle coordinate is the one that fixes a plane when three points are given in this form.
  4. Dz=1z=1, because the third coordinate is the height, and a plane drawn through any three given points is always a level surface.

Question 1001

[3 marks]Proof by induction and group theory
To start an induction proof of ∑r=1n2r+1r2(r+1)2=1−1(n+1)2\displaystyle\sum_{r=1}^{n}\dfrac{2r+1}{r^2(r+1)^2}=1-\dfrac{1}{(n+1)^2}, evaluate the left-hand side when n=1n=1.

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Question 1002

[3 marks]Proof by induction and group theory
In the inductive step of that proof, the expression (k+2)2−(2k+3)(k+2)^2-(2k+3) arises. Simplify it into a squared bracket.

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Question 1003

[2 marks]Proof by induction and group theory
The set S={a;2;4;6}S=\{a;2;4;6\} forms a group under addition modulo 8. Write down the value of aa.

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Question 1004

[2 marks]Proof by induction and group theory
In the group {0;2;4;6}\{0;2;4;6\} under addition modulo 8, evaluate 6+46+4.

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Question 1005

[3 marks]Proof by induction and group theory
For f(x)=px+qf(x)=px+q and g(x)=mx+ng(x)=mx+n, find f(g(x))f(g(x)), simplified.

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Question 1006

[3 marks]Proof by induction and group theory
For the set of functions f(x)=px+qf(x)=px+q under composition, why is the condition p≠0p\ne0 imposed?
  1. ABecause composition of two such functions would stop being associative as soon as one of them had p=0p=0.
  2. BBecause qq would then be forced to be zero as well, and the set would collapse to a single element only.
  3. CBecause the inverse is f−1(x)=x−qpf^{-1}(x)=\dfrac{x-q}{p}, which does not exist when p=0p=0, so that axiom would fail.
  4. DBecause the identity function is e(x)=xe(x)=x, and allowing p=0p=0 would place a second identity element inside the very same set.

Question 1101

[2 marks]Completing the square, quadratic graphs and exponential equations
Express y=x2−2x−3y=x^2-2x-3 in the form (x−p)2−q(x-p)^2-q.

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Question 1102

[2 marks]Completing the square, quadratic graphs and exponential equations
Given x2−2x−3=(x−p)2−qx^2-2x-3=(x-p)^2-q, state the values of pp and qq.

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Question 1103

[3 marks]Completing the square, quadratic graphs and exponential equations
For y=x2−2x−3y=x^2-2x-3, state the coordinates of the turning point and of both xx-intercepts.

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Question 1104

[2 marks]Completing the square, quadratic graphs and exponential equations
State the range of values of xx for which y=x2−2x−3y=x^2-2x-3 is negative.

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Question 1105

[3 marks]Completing the square, quadratic graphs and exponential equations
Given y=3xy=3^x, write the equation 32x+13−x=23^{2x}+\dfrac{1}{3^{-x}}=2 as a quadratic equation in yy equal to zero.

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Question 1106

[2 marks]Completing the square, quadratic graphs and exponential equations
Solve the equation 32x+13−x=23^{2x}+\dfrac{1}{3^{-x}}=2.

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Question 1107

[2 marks]Completing the square, quadratic graphs and exponential equations
Solving 32x+13−x=23^{2x}+\dfrac{1}{3^{-x}}=2 through the substitution y=3xy=3^x gives y=−2y=-2 or y=1y=1. Why is y=−2y=-2 discarded?
  1. ABecause 3x3^x is positive for every real value of xx, so yy can never take a negative value at all.
  2. BBecause a quadratic equation with a negative constant term can only ever be satisfied by its positive root.
  3. CBecause y=−2y=-2 would make 13−x\dfrac{1}{3^{-x}} undefined, since a reciprocal cannot be taken of a negative number.
  4. DBecause the substitution y=3xy=3^x was only stated to hold for values of xx that are greater than or equal to zero.

Question 1201

[3 marks]Determinants, inverse matrices and matrix products
Expand the determinant of N=(111x+11xxx+11)N=\begin{pmatrix}1&1&1\\x+1&1&x\\x&x+1&1\end{pmatrix} as a simplified expression in xx.

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Question 1202

[2 marks]Determinants, inverse matrices and matrix products
Given that the determinant of a matrix simplifies to x2−x+1x^2-x+1 and equals 77, with x>0x>0, find xx.

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Question 1203

[3 marks]Determinants, inverse matrices and matrix products
For A=(1111−1221−1)A=\begin{pmatrix}1&1&1\\1&-1&2\\2&1&-1\end{pmatrix} and B=(022−20−11−1−1)B=\begin{pmatrix}0&2&2\\-2&0&-1\\1&-1&-1\end{pmatrix}, give the first row of ABAB.

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Question 1204

[2 marks]Determinants, inverse matrices and matrix products
Find the determinant of A=(1111−1221−1)A=\begin{pmatrix}1&1&1\\1&-1&2\\2&1&-1\end{pmatrix}.

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Question 1205

[2 marks]Determinants, inverse matrices and matrix products
Find the determinant of B=(022−20−11−1−1)B=\begin{pmatrix}0&2&2\\-2&0&-1\\1&-1&-1\end{pmatrix}.

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Question 1206

[2 marks]Determinants, inverse matrices and matrix products
Matrices AA and BB have determinants 77 and −2-2. State the determinant of ABAB.

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Question 1207

[2 marks]Determinants, inverse matrices and matrix products
The inverse of a 3×33\times3 matrix AA is 1det⁡A\dfrac{1}{\det A} multiplied by which matrix?
  1. AThe matrix AA itself with every entry replaced by its own reciprocal, taken one position at a time in order.
  2. BThe transpose of AA, formed by writing the rows of the original matrix down as the columns of the new one.
  3. CThe matrix of cofactors of AA, used exactly as it stands with no further rearrangement of any one of its entries at all.
  4. DThe adjoint of AA, which is the transpose of the matrix of cofactors, so rows and columns are exchanged first.

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