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ZIMSEC A Level · J2013

Pure Mathematics Paper 1 June 2013

Questions
79
Total marks
120

Sit this paper online

Questions
79
Pass mark
48
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]indices / simultaneous equations
Solve the simultaneous equations 2x+3y=52^x + 3^y = 5 and 2x+2−3y+1=132^{x+2} - 3^{y+1} = 13.
  1. Ax=0x = 0, y=2y = 2
  2. Bx=4x = 4, y=1y = 1
  3. Cx=1x = 1, y=2y = 2
  4. Dx=2x = 2, y=0y = 0

Question 102

[1 marks]indices / simultaneous equations
Which substitution reduces 2x+3y=52^x + 3^y = 5 and 2x+2−3y+1=132^{x+2} - 3^{y+1} = 13 to a pair of linear equations?
  1. Ap=2xp = 2x, q=3yq = 3y
  2. Bp=2xp = 2^x, q=3yq = 3^y
  3. Cp=x2p = x^2, q=y3q = y^3
  4. Dp=log⁡2xp = \log_2 x, q=log⁡3yq = \log_3 y

Question 103

[2 marks]indices / simultaneous equations
Substituting p=2xp = 2^x, q=3yq = 3^y turns the equations into p+q=5p + q = 5 and 4p−3q=134p - 3q = 13. Solve for pp.

Answer this when you sit the paper.

Question 201

[1 marks]modulus / inequalities
Solve the inequality ∣2x+1∣<3x+2|2x + 1| < 3x + 2.
  1. A−1<x<−35-1 < x < -\tfrac{3}{5}
  2. Bx>−23x > -\tfrac{2}{3}
  3. Cx>−35x > -\tfrac{3}{5}
  4. Dx<−35x < -\tfrac{3}{5}

Question 202

[2 marks]modulus / inequalities
Squaring both sides of ∣2x+1∣<3x+2|2x+1| < 3x+2 gives 5x2+8x+3>05x^2+8x+3 > 0. Factorise the left side as (5x+3)(x+c)(5x+3)(x+c) and state cc.

Answer this when you sit the paper.

Question 203

[1 marks]modulus / inequalities
Since ∣2x+1∣≥0|2x+1| \ge 0, squaring is only valid when 3x+2>03x+2 > 0. State this condition on xx.

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Question 301

[1 marks]logarithms / linearising relationships
Two quantities satisfy y=abxy = ab^x. Plotting ln⁡y\ln y against xx gives a straight line of gradient 0.80.8 with ln⁡y\ln y-intercept 3.33.3. Find aa and bb, each correct to one decimal place.
  1. Aa=2.2a = 2.2, b=27.1b = 27.1
  2. Ba=27.1a = 27.1, b=2.2b = 2.2
  3. Ca=0.8a = 0.8, b=3.3b = 3.3
  4. Da=3.3a = 3.3, b=0.8b = 0.8

Question 302

[2 marks]logarithms / linearising relationships
Given y=abxy=ab^x and ln⁡b=0.8\ln b = 0.8 (the gradient), find bb correct to one decimal place.

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Question 303

[1 marks]logarithms / linearising relationships
Given y=abxy=ab^x and ln⁡a=3.3\ln a = 3.3 (the intercept), find aa correct to one decimal place.

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Question 401

[1 marks]small changes / percentage error
A 1%1\% error is made in measuring the diameter of a sphere. Find the approximate resulting percentage error in the volume.
  1. A2%2\%
  2. B3%3\%
  3. C13%\tfrac{1}{3}\%
  4. D1%1\%

Question 402

[1 marks]small changes / percentage error
A 1%1\% error is made in measuring the diameter of a sphere. Find the approximate resulting percentage error in the surface area.
  1. A1%1\%
  2. B4%4\%
  3. C2%2\%
  4. D3%3\%

Question 403

[2 marks]small changes / percentage error
Express the volume VV of a sphere in terms of its diameter DD (substituting r=D/2r=D/2 into V=43πr3V=\frac{4}{3}\pi r^3).

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Question 501

[1 marks]polynomials / remainder and factor theorems
When f(x)=x3+ax2+bx+cf(x) = x^3 + ax^2 + bx + c is divided by x2−4x^2 - 4 the remainder is 2x+112x + 11, and x+1x + 1 is a factor of f(x)f(x). Find aa, bb and cc.
  1. Aa=−4a = -4, b=−2b = -2, c=5c = 5
  2. Ba=4a = 4, b=−2b = -2, c=−5c = -5
  3. Ca=2a = 2, b=−4b = -4, c=11c = 11
  4. Da=4a = 4, b=2b = 2, c=−5c = -5

Question 502

[2 marks]polynomials / remainder and factor theorems
Long division of f(x)=x3+ax2+bx+cf(x)=x^3+ax^2+bx+c by x2−4x^2-4 gives remainder (b+4)x+(4a+c)(b+4)x + (4a+c). Matching to 2x+112x+11, find bb.

Answer this when you sit the paper.

Question 503

[2 marks]polynomials / remainder and factor theorems
Using b=−2b=-2, the factor condition f(−1)=0f(-1)=0 gives a−b+c=1a-b+c=1, and matching constants gives 4a+c=114a+c=11. Find aa.

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Question 601

[1 marks]partial fractions / binomial expansion
Express 4x+6(x+2)(x+1)(x+3)\dfrac{4x + 6}{(x+2)(x+1)(x+3)} in partial fractions.
  1. A2x+2+1x+1−3x+3\dfrac{2}{x+2} + \dfrac{1}{x+1} - \dfrac{3}{x+3}
  2. B2x+2−1x+1+3x+3\dfrac{2}{x+2} - \dfrac{1}{x+1} + \dfrac{3}{x+3}
  3. C1x+2+2x+1−3x+3\dfrac{1}{x+2} + \dfrac{2}{x+1} - \dfrac{3}{x+3}
  4. D2x+2+1x+1+3x+3\dfrac{2}{x+2} + \dfrac{1}{x+1} + \dfrac{3}{x+3}

Question 602

[1 marks]partial fractions / binomial expansion
Using series expansions up to and including the term in x2x^2, 4x+6(x+2)(x+1)(x+3)\dfrac{4x + 6}{(x+2)(x+1)(x+3)} reduces to
  1. A1−76x+4136x21 - \tfrac{7}{6}x + \tfrac{41}{36}x^2
  2. B1+76x+4136x21 + \tfrac{7}{6}x + \tfrac{41}{36}x^2
  3. C1−76x−4136x21 - \tfrac{7}{6}x - \tfrac{41}{36}x^2
  4. D76−x+4136x2\tfrac{7}{6} - x + \tfrac{41}{36}x^2

Question 603

[2 marks]partial fractions / binomial expansion
In 4x+6(x+2)(x+1)(x+3)=Ax+2+Bx+1+Cx+3\dfrac{4x+6}{(x+2)(x+1)(x+3)} = \dfrac{A}{x+2}+\dfrac{B}{x+1}+\dfrac{C}{x+3}, find BB (the coefficient for x+1x+1).

Answer this when you sit the paper.

Question 604

[1 marks]partial fractions / binomial expansion
In the same partial-fraction decomposition, find CC (the coefficient for x+3x+3).

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Question 605

[2 marks]partial fractions / binomial expansion
Expand (1+x2)−1\left(1+\frac{x}{2}\right)^{-1} up to and including the term in x2x^2.

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Question 701

[1 marks]trigonometry / R-formula
Express 4sin⁡θ−3cos⁡θ4\sin\theta - 3\cos\theta in the form Rsin⁡(θ−α)R\sin(\theta - \alpha), where R>0R > 0 and α\alpha is acute.
  1. A25sin⁡(θ−36.9°)25\sin(\theta - 36.9°)
  2. B7sin⁡(θ−36.9°)7\sin(\theta - 36.9°)
  3. C5sin⁡(θ−53.1°)5\sin(\theta - 53.1°)
  4. D5sin⁡(θ−36.9°)5\sin(\theta - 36.9°)

Question 702

[1 marks]trigonometry / R-formula
Solve 4sin⁡θ−3cos⁡θ=34\sin\theta - 3\cos\theta = 3 for 0°≤θ≤360°0° \le \theta \le 360°.
  1. Aθ=36.9°\theta = 36.9° and θ=143.1°\theta = 143.1°
  2. Bθ=73.8°\theta = 73.8° and θ=180°\theta = 180°
  3. Cθ=73.8°\theta = 73.8° and θ=143.1°\theta = 143.1°
  4. Dθ=0°\theta = 0° and θ=180°\theta = 180°

Question 703

[1 marks]trigonometry / R-formula
For what value of θ\theta in 0°≤θ≤360°0° \le \theta \le 360° is 4sin⁡θ−3cos⁡θ4\sin\theta - 3\cos\theta a maximum?
  1. Aθ=126.9°\theta = 126.9°
  2. Bθ=36.9°\theta = 36.9°
  3. Cθ=306.9°\theta = 306.9°
  4. Dθ=90°\theta = 90°

Question 704

[2 marks]trigonometry / R-formula
State the value of α\alpha, in degrees correct to 1 d.p., in 4sin⁡θ−3cos⁡θ=5sin⁡(θ−α)4\sin\theta - 3\cos\theta = 5\sin(\theta - \alpha).

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Question 705

[2 marks]trigonometry / R-formula
For 0°≤θ≤360°0° \le \theta \le 360°, find the value of θ\theta at which 4sin⁡θ−3cos⁡θ4\sin\theta - 3\cos\theta is a minimum.

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Question 801

[1 marks]parametric differentiation / turning points
A curve has parametric equations x=1+t2x = 1 + t^2 and y=4t−t3y = 4t - t^3, t>0t > 0. Find dydx\dfrac{dy}{dx} in terms of tt.
  1. A2t4−3t2\dfrac{2t}{4 - 3t^2}
  2. B4−3t24 - 3t^2
  3. C4−t22t\dfrac{4 - t^2}{2t}
  4. D4−3t22t\dfrac{4 - 3t^2}{2t}

Question 802

[1 marks]parametric differentiation / turning points
For the curve x=1+t2x = 1 + t^2, y=4t−t3y = 4t - t^3 with t>0t > 0, find the coordinates of the turning point.
  1. A(73, 1633)\left(\tfrac{7}{3},\ \tfrac{16}{3\sqrt{3}}\right)
  2. B(1, 0)\left(1,\ 0\right)
  3. C(43, 1633)\left(\tfrac{4}{3},\ \tfrac{16}{3\sqrt{3}}\right)
  4. D(73, 83)\left(\tfrac{7}{3},\ \tfrac{8}{\sqrt{3}}\right)

Question 803

[2 marks]parametric differentiation / turning points
Setting dydx=0\frac{dy}{dx}=0 for x=1+t2x=1+t^2, y=4t−t3y=4t-t^3 (t>0t>0), find the exact value of tt at the turning point.

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Question 804

[1 marks]parametric differentiation / turning points
For y=4t−t3y=4t-t^3, find dydt\dfrac{dy}{dt}.

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Question 805

[1 marks]parametric differentiation / turning points
For x=1+t2x=1+t^2, find dxdt\dfrac{dx}{dt}.

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Question 901

[1 marks]circular measure / sectors
In a school logo, OO is the centre of sector OACBOACB with angle AOB=π3AOB = \tfrac{\pi}{3}, and XX is a point with OX=OA=OB=rOX = OA = OB = r. Find angle AXBAXB.
  1. Aπ6\tfrac{\pi}{6}
  2. B2π3\tfrac{2\pi}{3}
  3. Cπ12\tfrac{\pi}{12}
  4. Dπ3\tfrac{\pi}{3}

Question 902

[1 marks]circular measure / sectors
In the school-logo figure with OX=OA=OB=rOX = OA = OB = r and angle AOB=π3AOB = \tfrac{\pi}{3}, the exact area of the shaded region is
  1. Ar22+πr26\dfrac{r^2}{2} + \dfrac{\pi r^2}{6}
  2. B3πr212\dfrac{\sqrt{3}\pi r^2}{12}
  3. Cr212(6−3π)\dfrac{r^2}{12}\left(6 - \sqrt{3}\pi\right)
  4. Dr212(6+3π)\dfrac{r^2}{12}\left(6 + \sqrt{3}\pi\right)

Question 903

[2 marks]circular measure / sectors
In triangle AOXAOX, angle AOX=5π6AOX = \frac{5\pi}{6} and OA=OX=rOA=OX=r. Find the area of triangle AOXAOX in terms of rr (exact form).

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Question 904

[2 marks]circular measure / sectors
Using the cosine rule in triangle AOXAOX (angle 5π6\frac{5\pi}{6}, sides rr and rr), find AX2AX^2 in terms of rr (exact form).

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Question 905

[2 marks]circular measure / sectors
Find the exact area of sector OACBOACB (angle π3\frac{\pi}{3}, radius rr).

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Question 1001

[1 marks]vectors in 3D
Points AA and BB have position vectors 4i−9j−k4\mathbf{i} - 9\mathbf{j} - \mathbf{k} and i+3j+5k\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}. Find the unit vector parallel to AB→\overrightarrow{AB}.
  1. A1189(3i−12j−6k)\tfrac{1}{\sqrt{189}}(3\mathbf{i} - 12\mathbf{j} - 6\mathbf{k})
  2. B198(−3i+12j+6k)\tfrac{1}{\sqrt{98}}(-3\mathbf{i} + 12\mathbf{j} + 6\mathbf{k})
  3. C1189(−3i+12j+6k)\tfrac{1}{\sqrt{189}}(-3\mathbf{i} + 12\mathbf{j} + 6\mathbf{k})
  4. D1189(−3i+12j+6k)\tfrac{1}{189}(-3\mathbf{i} + 12\mathbf{j} + 6\mathbf{k})

Question 1002

[1 marks]vectors in 3D
Points AA, BB, CC have position vectors 4i−9j−k4\mathbf{i} - 9\mathbf{j} - \mathbf{k}, i+3j+5k\mathbf{i} + 3\mathbf{j} + 5\mathbf{k} and λi−j+3k\lambda\mathbf{i} - \mathbf{j} + 3\mathbf{k}. Find λ\lambda so that AA, BB and CC are collinear.
  1. Aλ=−2\lambda = -2
  2. Bλ=4\lambda = 4
  3. Cλ=1\lambda = 1
  4. Dλ=2\lambda = 2

Question 1003

[1 marks]vectors in 3D
Points AA and BB have position vectors 4i−9j−k4\mathbf{i} - 9\mathbf{j} - \mathbf{k} and i+3j+5k\mathbf{i} + 3\mathbf{j} + 5\mathbf{k} relative to the origin OO. Calculate angle AOBAOB correct to the nearest 0.1°0.1°.
  1. A151.4°151.4°
  2. B28.6°28.6°
  3. C61.4°61.4°
  4. D118.6°118.6°

Question 1004

[2 marks]vectors in 3D
Find OA→⋅OB→\overrightarrow{OA}\cdot\overrightarrow{OB} for OA→=4i−9j−k\overrightarrow{OA}=4\mathbf{i}-9\mathbf{j}-\mathbf{k} and OB→=i+3j+5k\overrightarrow{OB}=\mathbf{i}+3\mathbf{j}+5\mathbf{k}.

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Question 1005

[2 marks]vectors in 3D
Find ∣OA→∣|\overrightarrow{OA}|, correct to 3 significant figures, for OA→=4i−9j−k\overrightarrow{OA}=4\mathbf{i}-9\mathbf{j}-\mathbf{k}.

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Question 1006

[1 marks]vectors in 3D
Find ∣OB→∣|\overrightarrow{OB}|, correct to 3 significant figures, for OB→=i+3j+5k\overrightarrow{OB}=\mathbf{i}+3\mathbf{j}+5\mathbf{k}.

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Question 1101

[1 marks]complex numbers
Given p=5+ip = 5 + i and q=−2+3iq = -2 + 3i, find pqpq.
  1. A−10+3i-10 + 3i
  2. B−7−17i-7 - 17i
  3. C−13+13i-13 + 13i
  4. D−13−13i-13 - 13i

Question 1102

[1 marks]complex numbers
Given p=5+ip = 5 + i and q=−2+3iq = -2 + 3i, express pq\dfrac{p}{q} in the form a+iba + ib.
  1. A−7+17i13\dfrac{-7 + 17i}{13}
  2. B−13+13i13\dfrac{-13 + 13i}{13}
  3. C−7−17i26\dfrac{-7 - 17i}{26}
  4. D−7−17i13\dfrac{-7 - 17i}{13}

Question 1103

[1 marks]complex numbers
Describe the geometrical transformation which maps ipip onto pp on an Argand diagram.
  1. AAn enlargement of scale factor ii about the origin
  2. BA rotation of 90°90° clockwise about the origin
  3. CA rotation of 90°90° anticlockwise about the origin
  4. DA reflection in the real axis

Question 1104

[1 marks]complex numbers
Given p=5+ip = 5+i, find ipip in the form a+iba+ib.

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Question 1105

[1 marks]complex numbers
Given p=5+ip=5+i and q=−2+3iq=-2+3i, find p+qp+q in the form a+iba+ib.

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Question 1106

[2 marks]complex numbers
Find the modulus of p=5+ip=5+i, correct to 3 significant figures.

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Question 1107

[1 marks]complex numbers
Find the argument of p=5+ip=5+i, in degrees correct to 1 decimal place.

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Question 1201

[1 marks]differential equations
Compounds PP and QQ have masses xx and yy with x+y=10x + y = 10, and xx increases at a rate proportional to the product of the two masses. The differential equation satisfied by xx is
  1. Adxdt=k(10−x)\dfrac{dx}{dt} = k(10 - x)
  2. Bdxdt=kx+10\dfrac{dx}{dt} = kx + 10
  3. Cdxdt=10kx\dfrac{dx}{dt} = 10kx
  4. Ddxdt=kx(10−x)\dfrac{dx}{dt} = kx(10 - x)

Question 1202

[1 marks]differential equations
Given dxdt=kx(10−x)\dfrac{dx}{dt} = kx(10 - x) with x=2x = 2 at t=0t = 0 and x=5x = 5 at t=ln⁡2t = \ln 2, find tt when x=9.9x = 9.9, correct to 3 significant figures.
  1. A1.501.50
  2. B2.992.99
  3. C4.604.60
  4. D5.985.98

Question 1203

[2 marks]differential equations
Using x=2x=2 at t=0t=0 in 110ln⁡∣x10−x∣=kt+A\frac{1}{10}\ln\left|\frac{x}{10-x}\right| = kt+A, find the constant AA, correct to 3 decimal places.

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Question 1204

[2 marks]differential equations
Using x=5x=5 at t=ln⁡2t=\ln2 in 110ln⁡∣x10−x∣=kt+A\frac{1}{10}\ln\left|\frac{x}{10-x}\right| = kt+A, find kk.

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Question 1205

[2 marks]differential equations
Using k=15k=\frac15, find dxdt\dfrac{dx}{dt} when x=5x=5.

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Question 1301

[1 marks]parametric curves / coordinate geometry
A curve CC has parametric equations y=aty = at and x=atx = \dfrac{a}{t}, t>0t > 0. Write down the cartesian equation of CC.
  1. Ay=a2xy = \dfrac{a^2}{x}
  2. By=axy = ax
  3. Cy=axy = \dfrac{a}{x}
  4. Dxy=axy = a

Question 1302

[1 marks]parametric curves / coordinate geometry
For the curve y=a2xy = \dfrac{a^2}{x}, find the equation of the normal at the point where t=2t = 2, i.e. at P(a2, 2a)P\left(\tfrac{a}{2},\ 2a\right).
  1. A8y=−2x+15a8y = -2x + 15a
  2. By=4x+15ay = 4x + 15a
  3. C8y=2x+15a8y = 2x + 15a
  4. D2y=8x+15a2y = 8x + 15a

Question 1303

[1 marks]parametric curves / coordinate geometry
The normal to y=a2xy = \dfrac{a^2}{x} at P(a2, 2a)P\left(\tfrac{a}{2},\ 2a\right) meets the curve again at Q(−8a, −a8)Q\left(-8a,\ -\tfrac{a}{8}\right). Find the centre of the circle having PQPQ as diameter.
  1. A(−15a8, 15a32)\left(-\tfrac{15a}{8},\ \tfrac{15a}{32}\right)
  2. B(−15a4, −15a16)\left(-\tfrac{15a}{4},\ -\tfrac{15a}{16}\right)
  3. C(15a4, 15a16)\left(\tfrac{15a}{4},\ \tfrac{15a}{16}\right)
  4. D(−15a4, 15a16)\left(-\tfrac{15a}{4},\ \tfrac{15a}{16}\right)

Question 1304

[2 marks]parametric curves / coordinate geometry
For the curve y=aty=at, x=atx=\frac{a}{t}, find the coordinates of point PP at t=2t=2 (in terms of aa).

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Question 1305

[2 marks]parametric curves / coordinate geometry
For the curve y=a2xy=\dfrac{a^2}{x}, find dydx\dfrac{dy}{dx} in terms of xx and aa.

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Question 1306

[3 marks]parametric curves / coordinate geometry
Find r2r^2, the square of the circle's radius (with PQPQ as diameter), in terms of aa, as an exact fraction.

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Question 1401

[1 marks]integration (area and volume)
The region in the first quadrant bounded by y=x2−xy = \dfrac{x}{2-x}, the line x=1x = 1 and the xx-axis has area AA. Find AA in exact form.
  1. A2ln⁡22\ln 2
  2. Bln⁡2−1\ln 2 - 1
  3. C2ln⁡2−12\ln 2 - 1
  4. D1−2ln⁡21 - 2\ln 2

Question 1402

[1 marks]integration (area and volume)
The region bounded by y=x2−xy = \dfrac{x}{2-x}, the line x=1x = 1 and the xx-axis is rotated completely about the xx-axis. Find the exact volume generated.
  1. Aπ(3−4ln⁡2)\pi(3 - 4\ln 2)
  2. Bπ(2ln⁡2−1)\pi(2\ln 2 - 1)
  3. Cπ(5−4ln⁡2)\pi(5 - 4\ln 2)
  4. Dπ(4ln⁡2−3)\pi(4\ln 2 - 3)

Question 1403

[2 marks]integration (area and volume)
Rewrite x2−x\dfrac{x}{2-x} in the form −1+k2−x-1+\dfrac{k}{2-x} by algebraic division. State kk.

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Question 1404

[2 marks]integration (area and volume)
Evaluate the antiderivative −x−2ln⁡∣2−x∣-x-2\ln|2-x| at x=1x=1 (the upper limit, before subtracting the lower-limit value).

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Question 1405

[2 marks]integration (area and volume)
Evaluate the antiderivative −x−2ln⁡∣2−x∣-x-2\ln|2-x| at x=0x=0 (the lower limit).

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Question 1406

[1 marks]integration (area and volume)
Evaluate x+4ln⁡∣2−x∣+42−xx+4\ln|2-x|+\dfrac{4}{2-x} at x=1x=1 (part of the volume antiderivative, upper limit).

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Question 1501

[1 marks]numerical methods
Use the trapezium rule with 4 ordinates to find an approximate value of ∫03xe−x dx\displaystyle\int_0^3 xe^{-x}\,dx, correct to 3 decimal places.
  1. A0.3560.356
  2. B0.7130.713
  3. C0.8010.801
  4. D1.4261.426

Question 1502

[1 marks]numerical methods
How many roots does the equation ex/2=x+2e^{x/2} = x + 2 have?
  1. Anone
  2. Bexactly one
  3. Cexactly two
  4. Dinfinitely many

Question 1503

[1 marks]numerical methods
Taking 44 as the first approximation to a root of ex/2=x+2e^{x/2} = x + 2, use the Newton-Raphson method to find the second approximation, correct to 3 significant figures.
  1. A3.003.00
  2. B3.483.48
  3. C3.623.62
  4. D4.284.28

Question 1504

[2 marks]numerical methods
Evaluate f(3)=e1.5−5f(3)=e^{1.5}-5, correct to 3 significant figures (used to show a root lies between 3 and 4).

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Question 1505

[2 marks]numerical methods
Evaluate f(4)=e2−6f(4)=e^2-6, correct to 3 significant figures.

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Question 1506

[2 marks]numerical methods
Find f′(4)=12e2−1f'(4) = \frac12 e^2-1, correct to 3 significant figures (used in the Newton-Raphson formula).

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Question 1507

[1 marks]numerical methods
State the interval width hh used in the trapezium rule with 4 ordinates over [0,3][0,3].

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Question 1601

[1 marks]geometric series / summation of series
The sum of the first 5 terms of a geometric series is 5 and the sum of the fifth to the ninth terms is 80. Find the possible values of the first term aa.
  1. Aa=531a = \tfrac{5}{31} or a=511a = \tfrac{5}{11}
  2. Ba=5a = 5 or a=80a = 80
  3. Ca=516a = \tfrac{5}{16} or a=511a = \tfrac{5}{11}
  4. Da=531a = \tfrac{5}{31} only

Question 1602

[1 marks]geometric series / summation of series
Express 4x3−6x2+2x4x^3 - 6x^2 + 2x in the form Ax(x+1)(x+2)+Bx(x+1)+CxAx(x+1)(x+2) + Bx(x+1) + Cx.
  1. AA=1A = 1, B=−18B = -18, C=12C = 12
  2. BA=4A = 4, B=−18B = -18, C=12C = 12
  3. CA=4A = 4, B=−6B = -6, C=2C = 2
  4. DA=4A = 4, B=18B = 18, C=−12C = -12

Question 1603

[1 marks]geometric series / summation of series
Given ∑x=1nx(x+1)(x+2)=n4(n+1)(n+2)(n+3)\sum_{x=1}^{n} x(x+1)(x+2) = \tfrac{n}{4}(n+1)(n+2)(n+3) and ∑x=1nx(x+1)=n3(n+1)(n+2)\sum_{x=1}^{n} x(x+1) = \tfrac{n}{3}(n+1)(n+2), evaluate ∑x=1n(4x3−6x2+2x)\sum_{x=1}^{n}\left(4x^3 - 6x^2 + 2x\right).
  1. An(n2−1)n(n^2 - 1)
  2. Bn2(n2−1)n^2(n^2 - 1)
  3. Cn2(n2+1)n^2(n^2 + 1)
  4. Dn2(n+1)2n^2(n+1)^2

Question 1604

[2 marks]geometric series / summation of series
Dividing ar4(1+r+r2+r3+r4)=80ar^4(1+r+r^2+r^3+r^4)=80 by a(1+r+r2+r3+r4)=5a(1+r+r^2+r^3+r^4)=5, find r4r^4.

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Question 1605

[2 marks]geometric series / summation of series
For r=2r=2, evaluate 1+r+r2+r3+r41+r+r^2+r^3+r^4 (used to find aa).

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Question 1606

[2 marks]geometric series / summation of series
For r=−2r=-2, evaluate 1+r+r2+r3+r41+r+r^2+r^3+r^4 (used to find the other value of aa).

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Question 1607

[2 marks]geometric series / summation of series
Verify the identity (n+2)(n+3)−6(n+2)+6=n2−n(n+2)(n+3)-6(n+2)+6 = n^2-n by evaluating the left side at n=2n=2.

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