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ZIMSEC A Level · 6042/1 · J2023

Pure Mathematics Paper 1 June 2023

Questions
58
Total marks
120
Syllabus code
6042/1

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Questions
58
Pass mark
35
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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]Sequences
A sequence UnU_n is defined by Un=(−1)n+1+2U_n=(-1)^{n+1}+2. Find the first 3 terms of the sequence.

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Question 102

[1 marks]Sequences
A sequence UnU_n is defined by Un=(−1)n+1+2U_n=(-1)^{n+1}+2, whose first terms are 3, 1, 3, 1,…3,\ 1,\ 3,\ 1,\dots. State the behaviour of the sequence.
  1. AIt oscillates between 3 and 1, with period 2.
  2. BIt increases without limit as nn increases, growing steadily larger and larger.
  3. CIt decreases steadily towards 0 as nn increases, never quite reaching zero.
  4. DIt converges to the limit 2 as nn becomes large, because the terms settle down.

Question 201

[1 marks]Circular measure
A and B are two points on the circumference of a circle centre O and radius rr. The minor arc AB subtends an angle of 2θ2\theta radians at O. Write down the area of the minor sector AOB in terms of rr and θ\theta.

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Question 202

[2 marks]Circular measure
A and B are two points on the circumference of a circle centre O and radius rr, and the minor arc AB subtends an angle of 2θ2\theta radians at O. Which expression gives the area of the minor segment cut off by the chord AB?
  1. A12r2θ−r2sin⁡2θ\tfrac12r^2\theta-r^2\sin2\theta
  2. Br2θ−r2sin⁡θr^2\theta-r^2\sin\theta
  3. Cr2θ−12r2sin⁡2θr^2\theta-\tfrac12r^2\sin2\theta
  4. Dr2θ+12r2sin⁡2θr^2\theta+\tfrac12r^2\sin2\theta

Question 301

[2 marks]Variation
A variable pp is inversely proportional to the square of 2q+12q+1, so that p=k(2q+1)2p=\dfrac{k}{(2q+1)^2}. Given that p=34p=\dfrac34 when q=2q=2, find the value of kk.

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Question 302

[2 marks]Variation
A variable pp is inversely proportional to the square of 2q+12q+1, and p=34p=\dfrac34 when q=2q=2. Find the positive value of qq for which p=13p=\dfrac13.

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Question 401

[3 marks]Exponential inequalities
Solve the inequality (1.05)n−4<60(1.05)^{n-4}<60, giving the bound on nn correct to 3 significant figures.

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Question 402

[1 marks]Exponential inequalities
Given that (1.05)n−4<60(1.05)^{n-4}<60, state the largest integral value of nn.

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Question 501

[1 marks]Simultaneous equations
Substituting y=3−2xy=3-2x from 2x+y=32x+y=3 into xy=1xy=1 produces a quadratic in xx. Write that quadratic in the form 2x2+bx+c=02x^2+bx+c=0.

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Question 502

[1 marks]Simultaneous equations
The simultaneous equations xy=1xy=1 and 2x+y=32x+y=3 reduce to 2x2−3x+1=02x^2-3x+1=0. Find the two values of xx.

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Question 503

[2 marks]Simultaneous equations
Solve the simultaneous equations xy=1xy=1 and 2x+y=32x+y=3, giving both pairs (x,y)(x,y).

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Question 601

[2 marks]Differentiation
Differentiate e2xsin⁡3xe^{2x}\sin 3x with respect to xx.

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Question 602

[2 marks]Differentiation
Differentiate ln⁡(1+x2)\ln(1+x^2) with respect to xx.

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Question 701

[2 marks]Parametric equations
A curve C has parametric equations x=2t+12tx=2t+\dfrac{1}{2t} and y=2t−12ty=2t-\dfrac{1}{2t}, where tt is a parameter. Find x2x^2 in terms of tt.

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Question 702

[2 marks]Parametric equations
A curve C has parametric equations x=2t+12tx=2t+\dfrac{1}{2t} and y=2t−12ty=2t-\dfrac{1}{2t}, where tt is a parameter. Find y2y^2 in terms of tt.

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Question 703

[2 marks]Parametric equations
A curve C has parametric equations x=2t+12tx=2t+\dfrac{1}{2t} and y=2t−12ty=2t-\dfrac{1}{2t}. By evaluating x2−y2x^2-y^2, find the cartesian equation of C.

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Question 801

[2 marks]Indices
Express 1(x)43\dfrac{1}{(\sqrt{x})^{\frac43}} in the form xnx^n.

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Question 802

[3 marks]Indices
Solve 21,527×16=2x\dfrac{2^{1,5}}{2^7\times16}=2^x.

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Question 901

[2 marks]Polynomials
Divide a3−b3a^3-b^3 by a−ba-b.

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Question 902

[2 marks]Polynomials
Given that x4+x2+x+1≡(x2+A)(x2−1)+Bx+Cx^4+x^2+x+1\equiv(x^2+A)(x^2-1)+Bx+C, determine the numerical value of A.

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Question 903

[2 marks]Polynomials
Given that x4+x2+x+1≡(x2+A)(x2−1)+Bx+Cx^4+x^2+x+1\equiv(x^2+A)(x^2-1)+Bx+C and that A=2A=2, determine the numerical values of B and C.

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Question 1001

[2 marks]Inverse functions
The function hh is defined by h(x)=3x+7h(x)=3x+7 for x∈Rx\in\mathbb{R}. Write down h−1(x)h^{-1}(x) in terms of xx.

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Question 1002

[2 marks]Inverse functions
The function gg is defined by g(x)=62x−4g(x)=\dfrac{6}{2x-4}, x≠2x\neq2, for x∈Rx\in\mathbb{R}. Write down g−1(x)g^{-1}(x) in terms of xx.

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Question 1003

[1 marks]Inverse functions
For g(x)=62x−4g(x)=\dfrac{6}{2x-4}, x≠2x\neq2, the inverse is g−1(x)=3x+2g^{-1}(x)=\dfrac{3}{x}+2. State the value of xx for which g−1(x)g^{-1}(x) is not defined.

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Question 1004

[2 marks]Inverse functions
The graphs of h(x)=3x+7h(x)=3x+7 and h−1(x)=x−73h^{-1}(x)=\dfrac{x-7}{3} are drawn on the same axes. What is the relationship between the two graphs?
  1. AEach is the reflection of the other in the yy-axis.
  2. BEach is the reflection of the other in the line y=xy=x.
  3. CEach is the reflection of the other in the xx-axis, so one of them is upside down.
  4. DThey are parallel straight lines of equal gradient that never meet at any point.

Question 1101

[2 marks]Proof by induction
In proving by induction that ∑r=1napr−1=a(pn−1)p−1\displaystyle\sum_{r=1}^{n}ap^{r-1}=\frac{a(p^n-1)}{p-1}, the base case is n=1n=1. State the common value that both sides take when n=1n=1.

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Question 1102

[3 marks]Proof by induction
Assume that ∑r=1kapr−1=a(pk−1)p−1\displaystyle\sum_{r=1}^{k}ap^{r-1}=\frac{a(p^k-1)}{p-1}. Adding the next term apkap^{k}, write a(pk−1)p−1+apk\dfrac{a(p^k-1)}{p-1}+ap^{k} as a single fraction in its simplest form.

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Question 1201

[2 marks]Matrices
Given the matrix A=(1252−534p7)A=\begin{pmatrix}1&2&5\\2&-5&3\\4&p&7\end{pmatrix}, find det⁡A\det A in terms of pp.

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Question 1202

[2 marks]Matrices
Given the matrix A=(1252−534p7)A=\begin{pmatrix}1&2&5\\2&-5&3\\4&p&7\end{pmatrix}, find the value of pp for which det⁡A=54\det A=54.

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Question 1203

[2 marks]Matrices
Find the cofactor of the element in the first row and first column of (1252−534−17)\begin{pmatrix}1&2&5\\2&-5&3\\4&-1&7\end{pmatrix}.

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Question 1204

[2 marks]Matrices
The matrix A=(1252−534−17)A=\begin{pmatrix}1&2&5\\2&-5&3\\4&-1&7\end{pmatrix} has determinant 54, and the cofactor of its element in the first row and first column is −32-32. Find the element in the first row and first column of A−1A^{-1}, as a fraction in its lowest terms.

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Question 1301

[2 marks]Taylor series
Let f(x)=1xf(x)=\dfrac1x. Find f′′′(a)f'''(a), the third derivative evaluated at x=ax=a.

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Question 1302

[3 marks]Taylor series
The function f(x)=1xf(x)=\dfrac1x is expanded by Taylor series in ascending powers of (x−a)(x-a). State the term in (x−a)3(x-a)^3.

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Question 1303

[1 marks]Taylor series
The function f(x)=1xf(x)=\dfrac1x is expanded by Taylor series in ascending powers of (x−a)(x-a). State the coefficient of (x−a)2(x-a)^2.

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Question 1304

[2 marks]Taylor series
Using the expansion 1x=1a−(x−a)a2+(x−a)2a3−(x−a)3a4\dfrac1x=\dfrac1a-\dfrac{(x-a)}{a^2}+\dfrac{(x-a)^2}{a^3}-\dfrac{(x-a)^3}{a^4} with a=1a=1, evaluate 11,01\dfrac{1}{1,01} correct to 3 significant figures.

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Question 1401

[1 marks]Trigonometry
Express 4cos⁡θ−3sin⁡θ4\cos\theta-3\sin\theta in the form Rsin⁡(θ−α)R\sin(\theta-\alpha), where R>0R>0. Find the value of RR.

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Question 1402

[2 marks]Trigonometry
Express 4cos⁡θ−3sin⁡θ4\cos\theta-3\sin\theta in the form Rsin⁡(θ−α)R\sin(\theta-\alpha), where R>0R>0 and 0∘<α<360∘0^\circ<\alpha<360^\circ. Find α\alpha to the nearest degree.

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Question 1403

[3 marks]Trigonometry
Given that 4cos⁡θ−3sin⁡θ=5sin⁡(θ−233,13∘)4\cos\theta-3\sin\theta=5\sin(\theta-233,13^\circ), solve 4cos⁡θ−3sin⁡θ=34\cos\theta-3\sin\theta=3 for 0∘≤θ≤360∘0^\circ\leq\theta\leq360^\circ, giving your answers correct to 1 decimal place.

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Question 1404

[2 marks]Trigonometry
Given that 4cos⁡θ−3sin⁡θ=5sin⁡(θ−233∘)4\cos\theta-3\sin\theta=5\sin(\theta-233^\circ), find the least value of 14cos⁡θ−3sin⁡θ+9\dfrac{1}{4\cos\theta-3\sin\theta+9}.

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Question 1501

[2 marks]Complex numbers
Given that z1=3+4iz_1=3+4i and z2=1+iz_2=1+i, find z1−z2z_1-z_2.

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Question 1502

[2 marks]Complex numbers
Given that z1=3+4iz_1=3+4i and z2=1+iz_2=1+i, find the argument of z1−z2z_1-z_2 in degrees, correct to 1 decimal place.

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Question 1503

[3 marks]Complex numbers
Solve the equation z2−4z+53=0z^2-4z+53=0, expressing the roots in the form a+bia+bi, where a,b∈Ra,b\in\mathbb{R}.

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Question 1504

[1 marks]Complex numbers
State the sum of the roots of z2−4z+53=0z^2-4z+53=0.

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Question 1505

[1 marks]Complex numbers
State the product of the roots of z2−4z+53=0z^2-4z+53=0.

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Question 1601

[3 marks]Vectors
The points A, B and C have position vectors 2i−k2\mathbf{i}-\mathbf{k}, 3i+2j−k3\mathbf{i}+2\mathbf{j}-\mathbf{k} and 5i−3j+4k5\mathbf{i}-3\mathbf{j}+4\mathbf{k} respectively. Calculate the angle ABC correct to the nearest 0,1∘0,1^\circ.

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Question 1602

[3 marks]Vectors
The point A has position vector 2i−k2\mathbf{i}-\mathbf{k} and the point C has position vector 5i−3j+4k5\mathbf{i}-3\mathbf{j}+4\mathbf{k}. N lies on AC so that AN→:NC→=1:2\overrightarrow{AN}:\overrightarrow{NC}=1:2. Find the position vector of N.

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Question 1603

[3 marks]Vectors
The point N has position vector 3i−j+23k3\mathbf{i}-\mathbf{j}+\tfrac23\mathbf{k} and the point C has position vector 5i−3j+4k5\mathbf{i}-3\mathbf{j}+4\mathbf{k}. Find the unit vector in the direction of NC→\overrightarrow{NC}.

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Question 1604

[1 marks]Vectors
The point N has position vector 3i−j+23k3\mathbf{i}-\mathbf{j}+\tfrac23\mathbf{k} and the point C has position vector 5i−3j+4k5\mathbf{i}-3\mathbf{j}+4\mathbf{k}. Find ∣NC→∣\left|\overrightarrow{NC}\right| correct to 2 decimal places.

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Question 1701

[1 marks]Coordinate geometry
State the gradient of the straight line 2y−x−5=02y-x-5=0.

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Question 1702

[3 marks]Coordinate geometry
The straight line ℓ\ell has equation 2y−x−5=02y-x-5=0. Another straight line mm passes through the point P(−2,6)P(-2,6) and is perpendicular to ℓ\ell. Find the equation of mm, giving the answer in the form ax+by+c=0ax+by+c=0.

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Question 1703

[3 marks]Coordinate geometry
Find the co-ordinates of the point of intersection of the lines 2y−x−5=02y-x-5=0 and 2x+y−2=02x+y-2=0.

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Question 1704

[2 marks]Coordinate geometry
Find the perpendicular distance from the point P(−2,6)P(-2,6) to the line 2y−x−5=02y-x-5=0, correct to 2 decimal places.

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Question 1705

[2 marks]Coordinate geometry
P is the point (−2,6)(-2,6), and the points Q(7;6)Q(7;6) and R(−7;−1)R(-7;-1) lie on the line 2y−x−5=02y-x-5=0, whose perpendicular distance from P is 95\dfrac{9}{\sqrt5}. Find the area of the triangle PQR.

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Question 1801

[1 marks]Newton-Raphson, root location
The graphs of y=e−xy=e^{-x} and y=sin⁡xy=\sin x are drawn on the same axes for 0≤x≤π0\leq x\leq\pi. State the number of roots of the equation e−x−sin⁡x=0e^{-x}-\sin x=0 in this interval.

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Question 1802

[1 marks]Newton-Raphson, root location
For f(x)=e−x−sin⁡xf(x)=e^{-x}-\sin x, write down f′(x)f'(x).

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Question 1803

[2 marks]Newton-Raphson, root location
Let f(x)=e−x−sin⁡xf(x)=e^{-x}-\sin x. Find f(π2)f\left(\dfrac{\pi}{2}\right) correct to 3 decimal places, and hence say what it shows when taken with f(0)=1f(0)=1.

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Question 1804

[3 marks]Newton-Raphson, root location
The Newton-Raphson method is applied to f(x)=e−x−sin⁡xf(x)=e^{-x}-\sin x starting with x0=0,8x_0=0,8. Find x1x_1 correct to 3 decimal places.

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Question 1805

[3 marks]Newton-Raphson, root location
Starting with x0=0,8x_0=0,8, use the Newton-Raphson method twice on f(x)=e−x−sin⁡xf(x)=e^{-x}-\sin x to estimate the smallest root of e−x−sin⁡x=0e^{-x}-\sin x=0, giving the answer correct to three decimal places.

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