Danho
ZIMSEC A Level · J2012

Pure Mathematics Paper 1 June 2012

Questions
66
Total marks
120

Sit this paper online

Questions
66
Pass mark
40
Sit this paper

Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[1 marks]modulus / inequalities
Given that −4<x<2-4 < x < 2 is the solution of the inequality ∣x+a∣<b|x + a| < b, find the values of aa and bb.
  1. Aa=1a = 1, b=3b = 3
  2. Ba=1a = 1, b=6b = 6
  3. Ca=−1a = -1, b=3b = 3
  4. Da=3a = 3, b=1b = 1

Question 102

[1 marks]modulus / inequalities
The inequality ∣x+a∣<b|x + a| < b (with b>0b > 0) is equivalent to which of the following?
  1. A−b−a<x<b−a-b - a < x < b - a
  2. B−b+a<x<b+a-b + a < x < b + a
  3. Cx<−b−ax < -b - a or x>b−ax > b - a
  4. Dx<b−ax < b - a only

Question 103

[1 marks]modulus / inequalities
Given −4<x<2-4 < x < 2 is the solution of ∣x+a∣<b|x + a| < b, matching gives −b−a=−4-b - a = -4 and b−a=2b - a = 2. Find the value of aa (by adding the two equations).

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Question 201

[1 marks]numerical integration / trapezium rule
Use the trapezium rule with 4 equal intervals to estimate ∫00.2cos⁡x2ex2 dx\displaystyle\int_0^{0.2} \dfrac{\cos x^2}{e^{x^2}}\,dx, correct to 4 decimal places.
  1. A0.12310.1231
  2. B0.19600.1960
  3. C0.19720.1972
  4. D0.39450.3945

Question 202

[3 marks]numerical integration / trapezium rule
Using the trapezium rule with 4 equal intervals over [0,0.2][0, 0.2] for ∫00.2cos⁡x2ex2 dx\int_0^{0.2}\frac{\cos x^2}{e^{x^2}}\,dx, the ordinates are y0=1y_0=1, y1=0.9975y_1=0.9975, y2=0.9904y_2=0.9904, y3=0.97775y_3=0.97775, y4=0.9229y_4=0.9229. Evaluate y0+y4+2(y1+y2+y3)y_0 + y_4 + 2(y_1+y_2+y_3).

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Question 301

[1 marks]small-angle approximations / trigonometry
In triangle ABDABD with height AD=hAD = h and angle ABD=16πABD = \tfrac{1}{6}\pi, find the length BDBD.
  1. Ah3h\sqrt{3}
  2. Bh3\dfrac{h}{\sqrt{3}}
  3. C2h2h
  4. Dh2\dfrac{h}{2}

Question 302

[1 marks]small-angle approximations / trigonometry
Given that xx is small enough for powers above x2x^2 to be neglected, the approximation for tan⁡(14π+x)\tan\left(\tfrac{1}{4}\pi + x\right) obtained from 1+tan⁡x1−tan⁡x\dfrac{1 + \tan x}{1 - \tan x} leads to DC≈DC \approx
  1. Ah(1+2x+2x2)h(1 + 2x + 2x^2)
  2. Bh(1−x+x2)h(1 - x + x^2)
  3. Ch(1−2x+2x2)h(1 - 2x + 2x^2)
  4. Dh(1−2x)h(1 - 2x)

Question 303

[1 marks]small-angle approximations / trigonometry
State the exact value of tan⁡16π\tan\frac{1}{6}\pi (used to find BDBD in the triangle).

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Question 304

[2 marks]small-angle approximations / trigonometry
Using the binomial expansion, write (1+x)−1(1+x)^{-1} up to and including the term in x2x^2.

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Question 401

[1 marks]small changes / implicit differentiation
Given that x3y2=1x^3 y^2 = 1, find the approximate percentage change in yy when xx increases by 0.5%0.5\%.
  1. Aa decrease of 0.75%0.75\%
  2. Ban increase of 0.75%0.75\%
  3. Ca decrease of 1.5%1.5\%
  4. Da decrease of 0.33%0.33\%

Question 402

[3 marks]small changes / implicit differentiation
Differentiate x3y2=1x^3y^2 = 1 implicitly with respect to xx to find dydx\frac{dy}{dx} in terms of xx and yy.

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Question 403

[1 marks]small changes / implicit differentiation
If xx increases by 0.5%0.5\%, express δx\delta x in terms of xx.

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Question 501

[1 marks]partial fractions
Express x4x4−1\dfrac{x^4}{x^4 - 1} in partial fractions.
  1. A1+12(x−1)−12(x+1)−1x2+11 + \dfrac{1}{2(x-1)} - \dfrac{1}{2(x+1)} - \dfrac{1}{x^2+1}
  2. B1+14(x−1)−14(x+1)−12(x2+1)1 + \dfrac{1}{4(x-1)} - \dfrac{1}{4(x+1)} - \dfrac{1}{2(x^2+1)}
  3. C14(x−1)−14(x+1)−12(x2+1)\dfrac{1}{4(x-1)} - \dfrac{1}{4(x+1)} - \dfrac{1}{2(x^2+1)}
  4. D1+14(x−1)+14(x+1)+12(x2+1)1 + \dfrac{1}{4(x-1)} + \dfrac{1}{4(x+1)} + \dfrac{1}{2(x^2+1)}

Question 502

[2 marks]partial fractions
Factorise x4−1x^4 - 1 completely over the real numbers.

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Question 503

[2 marks]partial fractions
Express 1(x2−1)(x2+1)\dfrac{1}{(x^2-1)(x^2+1)} in partial fractions with denominators (x2−1)(x^2-1) and (x2+1)(x^2+1).

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Question 601

[1 marks]binomial expansion
The expansions of (1+x)1/5(1+x)^{1/5} and 1+px1+qx\dfrac{1 + px}{1 + qx} in ascending powers of xx are identical up to and including the term in x2x^2. Find pp and qq.
  1. Ap=15p = \tfrac{1}{5}, q=−225q = -\tfrac{2}{25}
  2. Bp=35p = \tfrac{3}{5}, q=15q = \tfrac{1}{5}
  3. Cp=25p = \tfrac{2}{5}, q=35q = \tfrac{3}{5}
  4. Dp=35p = \tfrac{3}{5}, q=25q = \tfrac{2}{5}

Question 602

[3 marks]binomial expansion
Use the binomial series to expand (1+x)1/5(1+x)^{1/5} up to and including the term in x2x^2.

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Question 603

[3 marks]binomial expansion
Expand (1+px)(1+qx)−1(1+px)(1+qx)^{-1} up to and including the term in x2x^2, in terms of pp and qq.

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Question 701

[1 marks]arithmetic and geometric progressions
An arithmetic progression has first term aa and common difference d>0d > 0. Given that its first, second and fifth terms are consecutive terms of a geometric progression, express aa in terms of dd.
  1. Aa=da = d
  2. Ba=d4a = \tfrac{d}{4}
  3. Ca=d2a = \tfrac{d}{2}
  4. Da=2da = 2d

Question 702

[1 marks]arithmetic and geometric progressions
An arithmetic progression of 10 terms has a=d2a = \tfrac{d}{2} and the difference between its first and last terms is 36. Find the sum of all its terms.
  1. A180180
  2. B200200
  3. C220220
  4. D400400

Question 703

[3 marks]arithmetic and geometric progressions
The terms aa, a+da+d, a+4da+4d are consecutive terms of a geometric progression, so (a+d)2=a(a+4d)(a+d)^2 = a(a+4d). Expand and simplify this to a relation of the form d2=…d^2 = \ldots in terms of aa and dd.

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Question 704

[2 marks]arithmetic and geometric progressions
The arithmetic progression has 10 terms with d>0d>0, and the difference between its first and last (10th) terms is 36. Find the value of dd.

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Question 801

[1 marks]complex numbers
Express w=4+3i3−2iw = \dfrac{4 + 3i}{3 - 2i} in the form x+iyx + iy.
  1. A613+1713i\tfrac{6}{13} + \tfrac{17}{13}i
  2. B613−1713i\tfrac{6}{13} - \tfrac{17}{13}i
  3. C1813+113i\tfrac{18}{13} + \tfrac{1}{13}i
  4. D65+175i\tfrac{6}{5} + \tfrac{17}{5}i

Question 802

[1 marks]complex numbers
Find the argument of w=4+3i3−2iw = \dfrac{4 + 3i}{3 - 2i}, correct to the nearest 0.1°0.1°.
  1. A19.4°19.4°
  2. B70.6°70.6°
  3. C109.4°109.4°
  4. D−70.6°-70.6°

Question 803

[3 marks]complex numbers
Find the modulus of w=4+3i3−2iw = \dfrac{4+3i}{3-2i}.

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Question 804

[2 marks]complex numbers
Multiply out (4+3i)(3+2i)(4+3i)(3+2i) to find the numerator of w=4+3i3−2iw = \dfrac{4+3i}{3-2i} after rationalising, in the form p+qip + qi.

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Question 901

[1 marks]coordinate geometry / circles
Find the equation of the circle which passes through the origin and touches the line 4x−3y=54x - 3y = 5 at the point (2,1)(2, 1).
  1. A(x−2)2+(y−1)2=254(x - 2)^2 + (y - 1)^2 = \tfrac{25}{4}
  2. B(x−52)2+y2=254\left(x - \tfrac{5}{2}\right)^2 + y^2 = \tfrac{25}{4}
  3. Cx2+(y−52)2=25x^2 + \left(y - \tfrac{5}{2}\right)^2 = 25
  4. Dx2+(y−52)2=254x^2 + \left(y - \tfrac{5}{2}\right)^2 = \tfrac{25}{4}

Question 902

[3 marks]coordinate geometry / circles
Find the equation of the normal to the line 4x−3y=54x - 3y = 5 at the point (2,1)(2, 1), in the form ay=bx+cay = bx + c.

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Question 903

[3 marks]coordinate geometry / circles
Find the equation of the perpendicular bisector of the chord from (0,0)(0,0) to (2,1)(2,1), in the form y=mx+cy = mx + c.

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Question 1001

[1 marks]exponential and logarithmic equations
Solve the equation 3(7x)−2(7−x)=53(7^x) - 2(7^{-x}) = 5, giving your answer correct to 2 decimal places.
  1. Ax=0.36x = 0.36
  2. Bx=2.00x = 2.00
  3. Cx=0.56x = 0.56
  4. Dx=−0.56x = -0.56

Question 1002

[1 marks]exponential and logarithmic equations
Solve log⁡10(2x3+1)−3log⁡10x=1\log_{10}(2x^3 + 1) - 3\log_{10} x = 1.
  1. Ax=18x = \tfrac{1}{8}
  2. Bx=10x = 10
  3. Cx=2x = 2
  4. Dx=12x = \tfrac{1}{2}

Question 1003

[3 marks]exponential and logarithmic equations
Let y=7xy = 7^x. Rewrite 3(7x)−2(7−x)=53(7^x) - 2(7^{-x}) = 5 as a quadratic equation in yy.

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Question 1004

[3 marks]exponential and logarithmic equations
Rewrite log⁡10(2x3+1)−3log⁡10x=1\log_{10}(2x^3+1) - 3\log_{10}x = 1 without logarithms and hence find the value of x3x^3.

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Question 1101

[1 marks]vectors in 3D
In a cube of side 8 units with OO at the origin, M(0,4,4)M(0, 4, 4) is the centre of a face and P(4,8,8)P(4, 8, 8) is the midpoint of an edge. Find a unit vector parallel to MP→\overrightarrow{MP}.
  1. A14(i+j+k)\tfrac{1}{4}(\mathbf{i} + \mathbf{j} + \mathbf{k})
  2. B13(i+j+k)\tfrac{1}{\sqrt{3}}(\mathbf{i} + \mathbf{j} + \mathbf{k})
  3. C13(i−j+k)\tfrac{1}{\sqrt{3}}(\mathbf{i} - \mathbf{j} + \mathbf{k})
  4. D112(4i+4j+4k)\tfrac{1}{12}(4\mathbf{i} + 4\mathbf{j} + 4\mathbf{k})

Question 1102

[1 marks]vectors in 3D
In a cube of side 8 units, MP→=(4,4,4)\overrightarrow{MP} = (4, 4, 4) and MD→=(8,−4,4)\overrightarrow{MD} = (8, -4, 4). Find angle DMPDMP correct to the nearest 0.1°0.1°.
  1. A28.1°28.1°
  2. B45.0°45.0°
  3. C61.9°61.9°
  4. D70.5°70.5°

Question 1103

[3 marks]vectors in 3D
In the cube with OO at the origin, A=(8,0,0)A=(8,0,0), B=(0,8,0)B=(0,8,0), C=(0,0,8)C=(0,0,8), F=B+C=(0,8,8)F=B+C=(0,8,8), find the coordinates of MM, the point of intersection of OF→\overrightarrow{OF} and BC‾\overline{BC}.

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Question 1104

[3 marks]vectors in 3D
Given MP→=(4,4,4)\overrightarrow{MP} = (4,4,4), find ∣MP→∣|\overrightarrow{MP}|, leaving your answer in surd form.

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Question 1201

[1 marks]trigonometric identities and equations
Simplify csc⁡2x−cot⁡2x\csc 2x - \cot 2x.
  1. A2tan⁡x2\tan x
  2. Btan⁡x\tan x
  3. Ccot⁡x\cot x
  4. Dsec⁡x\sec x

Question 1202

[1 marks]trigonometric identities and equations
Using csc⁡2x−cot⁡2x≡tan⁡x\csc 2x - \cot 2x \equiv \tan x, solve csc⁡2x−cot⁡2x+5=3sec⁡2x\csc 2x - \cot 2x + 5 = 3\sec^2 x for 0<x≤360°0 < x \le 360°.
  1. Ax=45°, 225°x = 45°,\ 225° only
  2. Bx=45°, 146.3°, 225°, 326.3°x = 45°,\ 146.3°,\ 225°,\ 326.3°
  3. Cx=33.7°, 135°, 213.7°, 315°x = 33.7°,\ 135°,\ 213.7°,\ 315°
  4. Dx=45°, 135°, 225°, 315°x = 45°,\ 135°,\ 225°,\ 315°

Question 1203

[3 marks]trigonometric identities and equations
Substitute sec⁡2x=1+tan⁡2x\sec^2x = 1+\tan^2x into csc⁡2x−cot⁡2x+5=3sec⁡2x\csc 2x - \cot 2x + 5 = 3\sec^2x (using csc⁡2x−cot⁡2x=tan⁡x\csc2x-\cot2x=\tan x) and rearrange fully into the form 3tan⁡2x−tan⁡x−2=03\tan^2x - \tan x - 2 = 0. State this quadratic equation.

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Question 1204

[3 marks]trigonometric identities and equations
Solve 3tan⁡2x−tan⁡x−2=03\tan^2x - \tan x - 2 = 0 using the quadratic formula. State both values of tan⁡x\tan x.

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Question 1301

[1 marks]differentiation / optimisation
An open rectangular box has length 32x2\dfrac{32}{x^2} cm, width xx cm and volume 128128 cm3^3. Find its depth dd and total surface area AA in terms of xx.
  1. Ad=4xd = 4x, A=4x2+256xA = 4x^2 + \dfrac{256}{x}
  2. Bd=4xd = 4x, A=8x2+288xA = 8x^2 + \dfrac{288}{x}
  3. Cd=4xd = \dfrac{4}{x}, A=8x2+288xA = 8x^2 + \dfrac{288}{x}
  4. Dd=4xd = 4x, A=8x2+32xA = 8x^2 + \dfrac{32}{x}

Question 1302

[1 marks]differentiation / optimisation
The surface area of an open box is A=8x2+288xA = 8x^2 + \dfrac{288}{x} cm2^2. Find the value of xx that minimises AA, correct to 3 decimal places.
  1. Ax=3.000x = 3.000
  2. Bx=4.659x = 4.659
  3. Cx=10.483x = 10.483
  4. Dx=2.621x = 2.621

Question 1303

[3 marks]differentiation / optimisation
The open box has length 32x2\dfrac{32}{x^2} cm, width xx cm and depth d=4xd = 4x cm. Find the combined area of the two pairs of side faces (excluding the base), in terms of xx.

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Question 1304

[3 marks]differentiation / optimisation
Differentiate A=8x2+288xA = 8x^2 + \dfrac{288}{x} with respect to xx.

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Question 1305

[2 marks]differentiation / optimisation
Find d2Adx2\dfrac{d^2A}{dx^2} for A=8x2+288xA = 8x^2 + \dfrac{288}{x}, used to confirm the stationary point is a minimum.

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Question 1401

[1 marks]curve sketching / Newton-Raphson
Find the coordinates and nature of the turning points on the graph of y=x3−12x−12y = x^3 - 12x - 12.
  1. A(−2,−28)(-2, -28) minimum and (2,4)(2, 4) maximum
  2. B(−2,4)(-2, 4) minimum and (2,−28)(2, -28) maximum
  3. C(0,−12)(0, -12) maximum only
  4. D(−2,4)(-2, 4) maximum and (2,−28)(2, -28) minimum

Question 1402

[1 marks]curve sketching / Newton-Raphson
For which set of values of kk does the equation x3−12x−12−k=0x^3 - 12x - 12 - k = 0 have more than one real root?
  1. Ak>4k > 4
  2. Bk<−28k < -28
  3. C−28≤k≤4-28 \le k \le 4
  4. D−12≤k≤4-12 \le k \le 4

Question 1403

[1 marks]curve sketching / Newton-Raphson
Taking 3.93.9 as the first approximation, apply the Newton-Raphson method to x3−12x−12=0x^3 - 12x - 12 = 0 and give the root correct to 3 decimal places.
  1. A3.8643.864
  2. B3.8843.884
  3. C3.8853.885
  4. D3.9003.900

Question 1404

[3 marks]curve sketching / Newton-Raphson
Find dydx\dfrac{dy}{dx} for y=x3−12x−12y = x^3 - 12x - 12 and solve dydx=0\dfrac{dy}{dx}=0 for xx.

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Question 1405

[2 marks]curve sketching / Newton-Raphson
State the yy-intercept of the graph y=x3−12x−12y = x^3 - 12x - 12.

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Question 1406

[3 marks]curve sketching / Newton-Raphson
Using the Newton-Raphson method with f(x)=x3−12x−12f(x) = x^3 - 12x - 12 and first approximation x1=3.9x_1 = 3.9, find x2x_2, correct to 3 decimal places.

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Question 1501

[1 marks]quadratics / integration (area and volume)
Express 9x−x29x - x^2 in the form a+b(x+c)2a + b(x + c)^2 and hence state the coordinates of the turning point of y=9x−x2y = 9x - x^2.
  1. A812−(x−92)2\tfrac{81}{2} - \left(x - \tfrac{9}{2}\right)^2; turning point (92, 812)\left(\tfrac{9}{2},\ \tfrac{81}{2}\right)
  2. B9−(x−92)29 - \left(x - \tfrac{9}{2}\right)^2; turning point (92, 9)\left(\tfrac{9}{2},\ 9\right)
  3. C814−(x−92)2\tfrac{81}{4} - \left(x - \tfrac{9}{2}\right)^2; turning point (92, 814)\left(\tfrac{9}{2},\ \tfrac{81}{4}\right)
  4. D814+(x−92)2\tfrac{81}{4} + \left(x - \tfrac{9}{2}\right)^2; turning point (92, 814)\left(\tfrac{9}{2},\ \tfrac{81}{4}\right)

Question 1502

[1 marks]quadratics / integration (area and volume)
The region RR is bounded by the curve y=9x−x2y = 9x - x^2 and the xx-axis. Calculate the area of RR.
  1. A40.540.5 units2^2
  2. B81.081.0 units2^2
  3. C121.5121.5 units2^2
  4. D243.0243.0 units2^2

Question 1503

[1 marks]quadratics / integration (area and volume)
The region bounded by y=9x−x2y = 9x - x^2 and the xx-axis is rotated through four right angles about the xx-axis. Find the volume generated.
  1. A1968.3π1968.3\pi units3^3
  2. B984.15π984.15\pi units3^3
  3. C14762.25π14762.25\pi units3^3
  4. D121.5π121.5\pi units3^3

Question 1504

[1 marks]quadratics / integration (area and volume)
Find the roots of y=9x−x2y = 9x - x^2 (the xx-intercepts of the graph).

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Question 1505

[2 marks]quadratics / integration (area and volume)
Evaluate the indefinite integral ∫(9x−x2) dx\int(9x - x^2)\,dx.

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Question 1506

[3 marks]quadratics / integration (area and volume)
Expand (9x−x2)2(9x - x^2)^2 fully, in powers of xx.

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Question 1507

[3 marks]quadratics / integration (area and volume)
Evaluate ∫09(81x2−18x3+x4) dx\int_0^9 (81x^2 - 18x^3 + x^4)\,dx (the integral used before multiplying by π\pi to find the volume).

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Question 1601

[1 marks]differential equations
Solve the differential equation (4−x)dydx=y(4 - x)\dfrac{dy}{dx} = y given that y=4y = 4 when x=1x = 1.
  1. Ay=44−xy = \dfrac{4}{4 - x}
  2. By=4ex−1y = 4e^{x-1}
  3. Cy=124−xy = \dfrac{12}{4 - x}
  4. Dy=12(4−x)y = 12(4 - x)

Question 1602

[1 marks]differential equations
A liquid cools so that its rate of temperature decrease is proportional to θ\theta, its temperature in °°C. Given θ=60\theta = 60 at t=0t = 0 and θ=40\theta = 40 at t=5t = 5 minutes, find θ\theta in terms of tt.
  1. Aθ=40(23)t/5\theta = 40\left(\tfrac{2}{3}\right)^{t/5}
  2. Bθ=60(23)t/5\theta = 60\left(\tfrac{2}{3}\right)^{t/5}
  3. Cθ=60(32)t/5\theta = 60\left(\tfrac{3}{2}\right)^{t/5}
  4. Dθ=60e−t/5\theta = 60e^{-t/5}

Question 1603

[1 marks]differential equations
A liquid cools according to θ=60(23)t/5\theta = 60\left(\tfrac{2}{3}\right)^{t/5}, where tt is in minutes. Find its temperature 5 minutes after it has reached 40°40°C.
  1. A30.0°30.0°C
  2. B13.3°13.3°C
  3. C20.0°20.0°C
  4. D26.7°26.7°C

Question 1604

[3 marks]differential equations
Separate variables in (4−x)dydx=y(4-x)\dfrac{dy}{dx} = y and integrate both sides to express ln⁡y\ln y in terms of xx and an arbitrary constant cc.

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Question 1605

[2 marks]differential equations
Using y=4y = 4 when x=1x = 1 in ln⁡y=−ln⁡(4−x)+c\ln y = -\ln(4-x) + c, find the value of cc.

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Question 1606

[2 marks]differential equations
Write the differential equation relating θ\theta and tt, given that the rate of decrease of temperature is proportional to θ\theta (let k>0k>0 be the constant of proportionality).

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Question 1607

[3 marks]differential equations
Given θ=60\theta = 60 at t=0t=0 and θ=40\theta = 40 at t=5t=5, find the value of 5k5k.

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The answers, and why they are the answers

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