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ZIMSEC A Level · 6042/1 · N2022

Pure Mathematics Paper 1 November 2022

Questions
54
Total marks
120
Syllabus code
6042/1

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Questions
54
Pass mark
33
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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]Binomial expansion
Expand (4−3x)−12(4-3x)^{-\frac{1}{2}} in ascending powers of xx. Give the coefficient of xx as an exact fraction.

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Question 102

[2 marks]Binomial expansion
Expand (4−3x)−12(4-3x)^{-\frac{1}{2}} in ascending powers of xx. Give the coefficient of x2x^2 as an exact fraction.

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Question 103

[1 marks]Binomial expansion
The binomial expansion of (4−3x)−12(4-3x)^{-\frac{1}{2}} in ascending powers of xx is valid for ∣x∣<a|x|<a. State the value of aa.

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Question 201

[1 marks]Functions
The function gg is defined by g:x→xg:x\rightarrow\sqrt{x}, x≥px\geq p. State the least value of pp for which gg is defined.

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Question 202

[2 marks]Functions
The functions hh and gg are defined by h:x→x3h:x\rightarrow x^3, x∈Rx\in\mathbb{R} and g:x→xg:x\rightarrow\sqrt{x}, x≥0x\geq0. Find gh(x)gh(x) in the form xnx^n, where nn is rational. State nn.

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Question 203

[2 marks]Functions
The function gg is defined by g:x→xg:x\rightarrow\sqrt{x}, x≥0x\geq0. Find g2(x)g^2(x), that is g(g(x))g(g(x)), in the form xnx^n where nn is rational. State nn.

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Question 301

[2 marks]Parametric equations
A curve CC has parametric equations x=3+4cos⁡θx=3+4\cos\theta and y=1+4sin⁡θy=1+4\sin\theta. Eliminate θ\theta to find the cartesian equation of CC.

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Question 302

[2 marks]Parametric equations
A curve CC has parametric equations x=3+4cos⁡θx=3+4\cos\theta and y=1+4sin⁡θy=1+4\sin\theta, and its cartesian equation is (x−3)2+(y−1)2=16(x-3)^2+(y-1)^2=16. State the coordinates of the centre of this circle.

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Question 303

[1 marks]Parametric equations
A circle has equation (x−3)2+(y−1)2=16(x-3)^2+(y-1)^2=16. State its radius.

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Question 401

[2 marks]Graph transformations
The curve y=f(x)y=f(x), where f(x)=ln⁡(x+1)f(x)=\ln(x+1), has vertical asymptote x=−1x=-1 and passes through the origin. The graph of y=∣f(x)∣y=|f(x)| is sketched. State the coordinates of the point where y=∣f(x)∣y=|f(x)| touches the xx-axis.

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Question 402

[2 marks]Graph transformations
The curve y=f(x)y=f(x), where f(x)=ln⁡(x+1)f(x)=\ln(x+1), has vertical asymptote x=−1x=-1. State the equation of the vertical asymptote of the curve y=f(x−2)y=f(x-2).

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Question 403

[2 marks]Graph transformations
The curve y=f(x)y=f(x) is defined by f(x)=ln⁡(x+1)f(x)=\ln(x+1). Find the yy-intercept of the curve y=1−f(x)y=1-f(x).

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Question 501

[1 marks]Linear laws
In an experiment the variables xx and yy satisfy y=ax2+by=ax^2+b, where aa and bb are constants. To obtain a straight line graph, yy is plotted on the vertical axis. State what must be plotted on the horizontal axis.

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Question 502

[1 marks]Linear laws
In an experiment the variables xx and yy satisfy y=ax2+by=ax^2+b, and yy is plotted against x2x^2. One reading is x=3.8x=3.8. Calculate the value of x2x^2 that is plotted for this reading.

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Question 503

[2 marks]Linear laws
In an experiment the variables xx and yy satisfy y=ax2+by=ax^2+b. Two of the readings are x=1.6x=1.6, y=5.4y=5.4 and x=5.0x=5.0, y=32.3y=32.3. Plotting yy against x2x^2 gives a straight line of gradient aa. Use these two readings to calculate aa, correct to 2 significant figures.

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Question 504

[2 marks]Linear laws
In an experiment the variables xx and yy satisfy y=ax2+by=ax^2+b, and the gradient of the graph of yy against x2x^2 gives a=1.2a=1.2. One reading is x=1.6x=1.6, y=5.4y=5.4. Find the value of bb, correct to 2 significant figures.

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Question 601

[3 marks]Differential equations
Find the particular solution of the differential equation (x+1)2dydx+2=0(x+1)^2\dfrac{dy}{dx}+2=0, given that y=1y=1 when x=0x=0. Express yy in terms of xx.

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Question 602

[2 marks]Differential equations
The curve y=2x+1−1y=\dfrac{2}{x+1}-1 is sketched. State the equation of its vertical asymptote.

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Question 603

[2 marks]Differential equations
Find the xx-coordinate of the point where the curve y=2x+1−1y=\dfrac{2}{x+1}-1 crosses the xx-axis.

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Question 701

[1 marks]Variation
The total daily takings CC of a cooperative partly varies as the number of chickens sold NN, and partly varies as the number of dozens of eggs sold DD. Write down an expression for CC in terms of NN, DD and the constants hh and kk.

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Question 702

[3 marks]Variation
The total daily takings of a cooperative are given by C=hN+kDC=hN+kD, where NN is the number of chickens sold and DD the number of dozens of eggs sold. A total of $120\$120 was collected when 15 chickens and 15 dozens of eggs were sold, and $90\$90 was collected when 10 chickens and 20 dozens of eggs were sold. Find the value of hh and the value of kk.

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Question 703

[1 marks]Variation
The total daily takings of a cooperative are C=hN+kDC=hN+kD, where NN is the number of chickens sold, DD the number of dozens of eggs sold, and it is found that h=7h=7 and k=1k=1. State, in dollars, the amount taken for one dozen eggs.

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Question 704

[2 marks]Variation
The total daily takings of a cooperative are C=7N+DC=7N+D, where NN is the number of chickens sold and DD the number of dozens of eggs sold. Find the total daily takings, in dollars, if 20 chickens and 30 dozen eggs were sold.

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Question 801

[1 marks]Mathematical induction
The statement ∑r=1n(8r−1)=4n2+3n\displaystyle\sum_{r=1}^{n}(8r-1)=4n^2+3n is to be proved by induction. Evaluate the left hand side when n=1n=1.

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Question 802

[2 marks]Mathematical induction
In proving ∑r=1n(8r−1)=4n2+3n\displaystyle\sum_{r=1}^{n}(8r-1)=4n^2+3n by induction, it is assumed that ∑r=1k(8r−1)=4k2+3k\displaystyle\sum_{r=1}^{k}(8r-1)=4k^2+3k. Write down, in terms of kk and in its simplest form, the single term that must be added to reach ∑r=1k+1(8r−1)\displaystyle\sum_{r=1}^{k+1}(8r-1).

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Question 803

[2 marks]Mathematical induction
Given that ∑r=1n(8r−1)=4n2+3n\displaystyle\sum_{r=1}^{n}(8r-1)=4n^2+3n, evaluate ∑r=110(8r−1)\displaystyle\sum_{r=1}^{10}(8r-1).

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Question 901

[3 marks]Partial fractions
Express 24r2−1\dfrac{2}{4r^2-1} in partial fractions.

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Question 902

[3 marks]Partial fractions
Given that 24r2−1=12r−1−12r+1\dfrac{2}{4r^2-1}=\dfrac{1}{2r-1}-\dfrac{1}{2r+1}, find ∑r=1n24r2−1\displaystyle\sum_{r=1}^{n}\dfrac{2}{4r^2-1} as a single fraction in terms of nn.

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Question 903

[2 marks]Partial fractions
Given that ∑r=1n24r2−1=2n2n+1\displaystyle\sum_{r=1}^{n}\dfrac{2}{4r^2-1}=\dfrac{2n}{2n+1}, evaluate ∑r=1524r2−1\displaystyle\sum_{r=1}^{5}\dfrac{2}{4r^2-1}, giving the answer as a fraction in its lowest terms.

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Question 1001

[1 marks]Integration
The curve y=x(4−x)y=x(4-x) meets the xx-axis at two points. State the two values of xx at which it does so.

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Question 1002

[2 marks]Integration
Find ∫x(4−x) dx\displaystyle\int x(4-x)\,dx.

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Question 1003

[2 marks]Integration
The region RR is bounded by the curve y=x(4−x)y=x(4-x) and the xx-axis between x=0x=0 and x=4x=4. Calculate the area of RR, giving the answer as an exact fraction.

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Question 1004

[3 marks]Integration
The region bounded by the curve y=x(4−x)y=x(4-x) and the xx-axis between x=0x=0 and x=4x=4 is rotated through 360∘360^\circ about the xx-axis. Calculate the exact volume of the solid generated, in terms of π\pi.

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Question 1101

[2 marks]Quadratic and cubic equations
For the equation x2+(3k−2)x+k+2=0x^2+(3k-2)x+k+2=0, the condition for no real roots reduces to 9k2−16k−4<09k^2-16k-4<0. Solve the equation 9k2−16k−4=09k^2-16k-4=0.

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Question 1102

[3 marks]Quadratic and cubic equations
Find the range of values of kk for which the equation x2+(3k−2)x+k+2=0x^2+(3k-2)x+k+2=0 has no real roots.

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Question 1103

[2 marks]Quadratic and cubic equations
Show that x=2x=2 is a root of 2x3−9x2+7x+6=02x^3-9x^2+7x+6=0 by evaluating 2x3−9x2+7x+62x^3-9x^2+7x+6 at x=2x=2.

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Question 1104

[3 marks]Quadratic and cubic equations
Solve the equation 2x3−9x2+7x+6=02x^3-9x^2+7x+6=0.

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Question 1201

[2 marks]Remainder and factor theorem
f(x)=ax4−9x3+bx2−3x+2f(x)=ax^4-9x^3+bx^2-3x+2 is exactly divisible by x+2x+2. Using the factor theorem, this reduces to an equation of the form 4a+b=c4a+b=c. Find the value of cc.

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Question 1202

[3 marks]Remainder and factor theorem
f(x)=ax4−9x3+bx2−3x+2f(x)=ax^4-9x^3+bx^2-3x+2 is exactly divisible by x+2x+2 and leaves a remainder −12-12 when divided by x−1x-1. Find the values of aa and bb.

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Question 1203

[2 marks]Remainder and factor theorem
Verify that 1−2x1-2x is a factor of f(x)=−6x4−9x3+4x2−3x+2f(x)=-6x^4-9x^3+4x^2-3x+2 by evaluating f(12)f\left(\dfrac{1}{2}\right).

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Question 1204

[3 marks]Remainder and factor theorem
The quartic f(x)=−6x4−9x3+4x2−3x+2f(x)=-6x^4-9x^3+4x^2-3x+2 factorises as (x+2)(1−2x)(3x2+1)(x+2)(1-2x)\left(3x^2+1\right). State all the real roots of f(x)=0f(x)=0.

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Question 1301

[2 marks]Numerical methods
To show that the root of cot⁡x=1+x2\cot x=1+x^2 lies between 0,6 and 0,9, the value of cot⁡x\cot x is needed at each end. Evaluate cot⁡(0,6)\cot(0,6), with xx in radians, correct to 3 decimal places.

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Question 1302

[2 marks]Numerical methods
The iterative formula xn+1=tan⁡−1(11+xn2)x_{n+1}=\tan^{-1}\left(\dfrac{1}{1+x_n^2}\right), with xx in radians, is used to find the root of cot⁡x=1+x2\cot x=1+x^2. Taking x0=0.6x_0=0.6, find x1x_1 correct to 4 decimal places.

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Question 1303

[2 marks]Numerical methods
The iterative formula xn+1=tan⁡−1(11+xn2)x_{n+1}=\tan^{-1}\left(\dfrac{1}{1+x_n^2}\right), with xx in radians, is used to find the root of cot⁡x=1+x2\cot x=1+x^2. Given that x1=0.6340x_1=0.6340, find x2x_2 correct to 4 decimal places.

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Question 1304

[2 marks]Numerical methods
The iterative formula xn+1=tan⁡−1(11+xn2)x_{n+1}=\tan^{-1}\left(\dfrac{1}{1+x_n^2}\right), with xx in radians, is used to find the root of cot⁡x=1+x2\cot x=1+x^2. Given that x2=0.6196x_2=0.6196, find x3x_3 correct to 4 decimal places.

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Question 1401

[2 marks]Matrices
Find the determinant of A=(23−1121332)A=\begin{pmatrix}2 & 3 & -1\\ 1 & 2 & 1\\ 3 & 3 & 2\end{pmatrix}.

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Question 1402

[2 marks]Matrices
The simultaneous equations 2x+3y−z=02x+3y-z=0, x+2y+z=2x+2y+z=2 and 3x+3y+2z=43x+3y+2z=4 are written in the matrix form AX=BA\mathbf{X}=\mathbf{B}, where A=(23−1121332)A=\begin{pmatrix}2 & 3 & -1\\ 1 & 2 & 1\\ 3 & 3 & 2\end{pmatrix} and X=(xyz)\mathbf{X}=\begin{pmatrix}x\\y\\z\end{pmatrix}. Write down the column vector B\mathbf{B}.

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Question 1403

[2 marks]Matrices
For A=(23−1121332)A=\begin{pmatrix}2 & 3 & -1\\ 1 & 2 & 1\\ 3 & 3 & 2\end{pmatrix}, whose determinant is 8, the adjugate is (1−9517−3−331)\begin{pmatrix}1 & -9 & 5\\ 1 & 7 & -3\\ -3 & 3 & 1\end{pmatrix}. Find the element of A−1A^{-1} in the first row and second column.

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Question 1404

[3 marks]Matrices
Given that A−1=18(1−9517−3−331)A^{-1}=\dfrac18\begin{pmatrix}1 & -9 & 5\\ 1 & 7 & -3\\ -3 & 3 & 1\end{pmatrix}, solve the simultaneous equations 2x+3y−z=02x+3y-z=0, x+2y+z=2x+2y+z=2, 3x+3y+2z=43x+3y+2z=4. Give the values of xx, yy and zz.

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Question 1501

[2 marks]Vectors
The points AA and BB have position vectors a=3i+3j+4k\mathbf{a}=3\mathbf{i}+3\mathbf{j}+4\mathbf{k} and b=2i+5j+3k\mathbf{b}=2\mathbf{i}+5\mathbf{j}+3\mathbf{k}. Find the vector BA→\overrightarrow{BA}.

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Question 1502

[3 marks]Vectors
The points AA, BB and CC have position vectors a=3i+3j+4k\mathbf{a}=3\mathbf{i}+3\mathbf{j}+4\mathbf{k}, b=2i+5j+3k\mathbf{b}=2\mathbf{i}+5\mathbf{j}+3\mathbf{k} and c=3i+4j+3k\mathbf{c}=3\mathbf{i}+4\mathbf{j}+3\mathbf{k}. Find the angle AB^CA\hat{B}C, in degrees.

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Question 1503

[3 marks]Vectors
The points AA, BB and CC have position vectors a=3i+3j+4k\mathbf{a}=3\mathbf{i}+3\mathbf{j}+4\mathbf{k}, b=2i+5j+3k\mathbf{b}=2\mathbf{i}+5\mathbf{j}+3\mathbf{k} and c=3i+4j+3k\mathbf{c}=3\mathbf{i}+4\mathbf{j}+3\mathbf{k}, and the angle AB^CA\hat{B}C is 30∘30^\circ. Find the area of triangle ABCABC in exact form.

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Question 1504

[2 marks]Vectors
The points AA and BB have position vectors a=3i+3j+4k\mathbf{a}=3\mathbf{i}+3\mathbf{j}+4\mathbf{k} and b=2i+5j+3k\mathbf{b}=2\mathbf{i}+5\mathbf{j}+3\mathbf{k}. Find the position vector of FF, the midpoint of ABAB.

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Question 1505

[2 marks]Vectors
The points AA, BB and CC have position vectors a=3i+3j+4k\mathbf{a}=3\mathbf{i}+3\mathbf{j}+4\mathbf{k}, b=2i+5j+3k\mathbf{b}=2\mathbf{i}+5\mathbf{j}+3\mathbf{k} and c=3i+4j+3k\mathbf{c}=3\mathbf{i}+4\mathbf{j}+3\mathbf{k}, and FF is the midpoint of ABAB with position vector 52i+4j+72k\dfrac52\mathbf{i}+4\mathbf{j}+\dfrac72\mathbf{k}. Evaluate the scalar product CF→⋅AB→\overrightarrow{CF}\cdot\overrightarrow{AB}.

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