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ZIMSEC A Level · N2006

Pure Mathematics Paper 1 November 2006

Questions
73
Total marks
99

Sit this paper online

Questions
73
Pass mark
44
Sit this paper

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

The questions

Question 501

[1 marks]sequences and functions
A sequence is defined by Un+1=11−UnU_{n+1} = \dfrac{1}{1-U_n}, with U1=aU_1 = a, where a≠1a \neq 1. What is U3U_3 in terms of aa?
  1. Aaa−1\dfrac{a}{a-1}
  2. B1−aa\dfrac{1-a}{a}
  3. Ca−1a\dfrac{a-1}{a}
  4. D11−a\dfrac{1}{1-a}

Question 502

[1 marks]sequences and functions
The function ff is defined by f:x↦ln⁡x−9f: x \mapsto \ln x - 9, where x∈R+x \in \mathbb{R}^+. What is f−1(x)f^{-1}(x)?
  1. Aex−9e^{x-9}
  2. Bex+9e^{x+9}
  3. Cex+9e^x + 9
  4. Dln⁡(x+9)\ln(x+9)

Question 503

[2 marks]sequences and functions
A sequence is defined by Un+1=11−UnU_{n+1}=\dfrac1{1-U_n}, with U1=aU_1=a, a≠1a\neq1, and U3=a−1aU_3=\dfrac{a-1}a. Continuing the sequence, state U4U_4 in terms of aa.

Answer this when you sit the paper.

Question 504

[1 marks]sequences and functions
The function ff is defined by f:x↦ln⁡x−9f:x\mapsto\ln x-9, x∈R+x\in\mathbb R^+, with inverse f−1(x)=ex+9f^{-1}(x)=e^{x+9}. State the domain of f−1f^{-1}.

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Question 505

[2 marks]sequences and functions
Given f−1(x)=ex+9f^{-1}(x)=e^{x+9} (the inverse of f:x↦ln⁡x−9f:x\mapsto\ln x-9), evaluate f−1(0)f^{-1}(0) in exact form.

Answer this when you sit the paper.

Question 601

[1 marks]trigonometry and iterative methods
The shape ABC is formed from a wire, where AB is an arc of a circle centre O radius 3 cm, and CA, CB are tangents to the circle each of length tt cm. This gives the iterative formula tn+1=5−3tan⁡−1 ⁣(tn3)t_{n+1} = 5 - 3\tan^{-1}\!\left(\dfrac{t_n}{3}\right), used with t1=3t_1 = 3. What is t2t_2, to 3 decimal places?
  1. A0.3280.328
  2. B2.6442.644
  3. C4.2154.215
  4. D7.3567.356

Question 602

[1 marks]trigonometry and iterative methods
Continuing the iteration tn+1=5−3tan⁡−1 ⁣(tn3)t_{n+1} = 5 - 3\tan^{-1}\!\left(\dfrac{t_n}{3}\right) from t1=3t_1 = 3, given that t2≈2.644t_2 \approx 2.644, what is t3t_3, to 3 decimal places?
  1. A2.6442.644
  2. B2.8332.833
  3. C4.2784.278
  4. D7.1667.166

Question 603

[1 marks]trigonometry and iterative methods
Continuing the iteration tn+1=5−3tan⁡−1 ⁣(tn3)t_{n+1} = 5 - 3\tan^{-1}\!\left(\dfrac{t_n}{3}\right), given that t3≈2.833t_3 \approx 2.833, what is t4t_4, to 3 decimal places?
  1. A2.6442.644
  2. B2.7322.732
  3. C4.2444.244
  4. D7.2687.268

Question 604

[1 marks]trigonometry and iterative methods
A wire of length 10 cm forms shape ABC, where AB is an arc of a circle centre O radius 3 cm subtending angle 2θ2\theta at the centre, and CA, CB are tangents to the circle, each of length tt cm. Since arc length == radius ×\times angle, arc AB =3(2θ)=6θ=3(2\theta)=6\theta. Using total wire length 10=10= arc AB +CA+CB+CA+CB, what is CA+CBCA+CB in terms of θ\theta?
  1. A10−3θ10-3\theta
  2. B10−6θ10-6\theta
  3. C5−6θ5-6\theta
  4. D5−3θ5-3\theta

Question 605

[1 marks]trigonometry and iterative methods
In the shape ABC (wire of length 10 cm, arc AB radius 3 cm subtending angle 2θ2\theta, tangents CA == CB =t=t cm), the right-angled triangle formed by the radius (3 cm), the tangent length tt, and the line from the centre to C has a right angle at the point of tangency. What is tan⁡θ\tan\theta in terms of tt?
  1. A6/t6/t
  2. B3/t3/t
  3. Ct/6t/6
  4. Dt/3t/3

Question 606

[2 marks]trigonometry and iterative methods
For the shape ABC (arc AB radius 3 cm, tangents CA == CB =t=t cm), using tan⁡θ=t/3\tan\theta=t/3 with the initial estimate t1=3t_1=3, state θ\theta in radians, exact form, in terms of pi.

Answer this when you sit the paper.

Question 701

[1 marks]integration and volumes of revolution
Evaluate exactly ∫142x+3x+1 dx\displaystyle\int_1^4 \dfrac{2x+3}{x+1}\,dx.
  1. A8+ln⁡ ⁣(52)8 + \ln\!\left(\dfrac{5}{2}\right)
  2. B6+ln⁡46 + \ln 4
  3. C6−ln⁡ ⁣(52)6 - \ln\!\left(\dfrac{5}{2}\right)
  4. D6+ln⁡ ⁣(52)6 + \ln\!\left(\dfrac{5}{2}\right)

Question 702

[1 marks]integration and volumes of revolution
The region R is bounded by the curve y=x2y=x^2, the yy-axis, and the lines y=1y=1 and y=92y=\dfrac{9}{2}. A cup is formed by rotating R through 4 right angles about the yy-axis. What is the volume of the cup, in terms of π\pi?
  1. A77π8\dfrac{77\pi}{8}
  2. B77π16\dfrac{77\pi}{16}
  3. C77π4\dfrac{77\pi}{4}
  4. D81π8\dfrac{81\pi}{8}

Question 703

[1 marks]integration and volumes of revolution
The region R is bounded by the curve y=x2y=x^2, the yy-axis, and the lines y=1y=1 and y=92y=\dfrac{9}{2}, and is rotated about the yy-axis to form a cup. Water is poured into the cup until the upper surface of the water has radius 3\sqrt3. What is the volume of water in the cup, in terms of π\pi?
  1. A9π2\dfrac{9\pi}{2}
  2. B77π8\dfrac{77\pi}{8}
  3. C4π4\pi
  4. Dπ\pi

Question 704

[1 marks]integration and volumes of revolution
Rewriting 2x+3x+1\dfrac{2x+3}{x+1} by polynomial division as A+Bx+1A+\dfrac{B}{x+1} (used before integrating ∫142x+3x+1 dx\displaystyle\int_1^4\dfrac{2x+3}{x+1}\,dx), state AA and BB as 'A, B'.

Answer this when you sit the paper.

Question 705

[2 marks]integration and volumes of revolution
Using 2x+3x+1=2+1x+1\dfrac{2x+3}{x+1}=2+\dfrac1{x+1}, evaluate ∫141x+1 dx\displaystyle\int_1^4\dfrac1{x+1}\,dx exactly (one part of the full integral).

Answer this when you sit the paper.

Question 706

[1 marks]integration and volumes of revolution
The region R, bounded by y=x2y=x^2, the yy-axis, and the lines y=1y=1 and y=92y=\dfrac92, is rotated about the yy-axis to form a cup. Water is poured in until the upper surface has radius 3\sqrt3. Since y=x2y=x^2, state the value of yy at this water level.

Answer this when you sit the paper.

Question 707

[1 marks]integration and volumes of revolution
Finding the volume of the cup (formed by rotating y=x2y=x^2 about the yy-axis between y=1y=1 and y=92y=\dfrac92) needs [y22]14.5\left[\dfrac{y^2}2\right]_1^{4.5}. State the value of y22\dfrac{y^2}2 at y=4.5y=4.5.

Answer this when you sit the paper.

Question 801

[1 marks]differentiation and tangents/normals
The point P(−2,6)(-2, 6) lies on the curve y=2−x2−x3y = 2 - x^2 - x^3. What is the equation of the tangent to the curve at P, in the form y=mx+cy = mx + c?
  1. Ay=−8x+6y = -8x + 6
  2. By=x8+254y = \dfrac{x}{8} + \dfrac{25}{4}
  3. Cy=−8x−10y = -8x - 10
  4. Dy=−16x−26y = -16x - 26

Question 802

[1 marks]differentiation and tangents/normals
The point P(−2,6)(-2, 6) lies on the curve y=2−x2−x3y = 2 - x^2 - x^3. What is the equation of the normal to the curve at P, in the form y=mx+cy = mx + c?
  1. Ay=8x+22y = 8x + 22
  2. By=−8x−10y = -8x - 10
  3. Cy=x8+254y = \dfrac{x}{8} + \dfrac{25}{4}
  4. Dy=−x8+234y = -\dfrac{x}{8} + \dfrac{23}{4}

Question 803

[1 marks]differentiation and tangents/normals
The point P(−2,6)(-2, 6) lies on the curve y=2−x2−x3y = 2 - x^2 - x^3. The tangent to the curve at P meets the yy-axis at A, and the normal to the curve at P meets the yy-axis at B. What is the length of AB?
  1. A16.2516.25 units
  2. B10.2510.25 units
  3. C3.753.75 units
  4. D21.2521.25 units

Question 804

[2 marks]differentiation and tangents/normals
The point P(−2,6)(-2,6) lies on the curve y=2−x2−x3y=2-x^2-x^3. Find dydx\dfrac{dy}{dx} and evaluate it at x=−2x=-2 (the gradient of the tangent at P).

Answer this when you sit the paper.

Question 805

[1 marks]differentiation and tangents/normals
The tangent to y=2−x2−x3y=2-x^2-x^3 at P(−2,6)(-2,6) has gradient −8-8. State the gradient of the normal at P.

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Question 806

[1 marks]differentiation and tangents/normals
The tangent to y=2−x2−x3y=2-x^2-x^3 at P(−2,6)(-2,6) has equation y=−8x−10y=-8x-10. State the yy-coordinate of point A, where this tangent meets the yy-axis.

Answer this when you sit the paper.

Question 807

[1 marks]differentiation and tangents/normals
The normal to y=2−x2−x3y=2-x^2-x^3 at P(−2,6)(-2,6) has equation y=x8+254y=\dfrac x8+\dfrac{25}4. State the yy-coordinate of point B, where this normal meets the yy-axis.

Answer this when you sit the paper.

Question 901

[1 marks]complex numbers and logarithms
The complex number z=x+iyz = x + iy satisfies zz+2=2−i\dfrac{z}{z+2} = 2 - i. What are the values of xx and yy?
  1. Ax=−1, y=−3x=-1,\ y=-3
  2. Bx=3, y=1x=3,\ y=1
  3. Cx=−3, y=1x=-3,\ y=1
  4. Dx=−3, y=−1x=-3,\ y=-1

Question 902

[1 marks]complex numbers and logarithms
Solve the simultaneous equations 2log⁡y=log⁡2+log⁡x2\log y = \log 2 + \log x and 2y=4x2^y = 4^x for xx and yy.
  1. Ax=18, y=12x=\tfrac18,\ y=\tfrac12
  2. Bx=1, y=2x=1,\ y=2
  3. Cx=12, y=1x=\tfrac12,\ y=1
  4. Dx=0, y=0x=0,\ y=0

Question 903

[2 marks]complex numbers and logarithms
The complex number z=x+iyz=x+iy satisfies zz+2=2−i\dfrac z{z+2}=2-i. Rearranging as z=(2−i)(z+2)z=(2-i)(z+2) and collecting all zz terms on one side gives z(a+bi)=4−2iz(a+bi)=4-2i for some complex coefficient. State this coefficient in the form 'a+bi'.

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Question 904

[1 marks]complex numbers and logarithms
Solving 2log⁡y=log⁡2+log⁡x2\log y=\log2+\log x without logarithms gives an equation in the form y2=kxy^2=kx. State kk.

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Question 905

[1 marks]complex numbers and logarithms
Rewriting 2y=4x2^y=4^x as 2y=22x2^y=2^{2x}, the exponents give y=cxy=cx. State cc.

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Question 906

[2 marks]complex numbers and logarithms
Substituting y=2xy=2x into y2=2xy^2=2x gives 4x2=2x4x^2=2x, which factorises as 2x(2x−c)=02x(2x-c)=0. State cc.

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Question 907

[1 marks]complex numbers and logarithms
Solving 2x(2x−1)=02x(2x-1)=0 for the simultaneous equations 2log⁡y=log⁡2+log⁡x2\log y=\log2+\log x and 2y=4x2^y=4^x gives x=0x=0 or x=12x=\tfrac12. Why is x=0x=0 rejected?
  1. A4x4^x is undefined when x=0x=0
  2. Bx=0x=0 gives an irrational value of yy
  3. Cx=0x=0 is not a solution of 2x(2x−1)=02x(2x-1)=0
  4. Dlog⁡x\log x is undefined at x=0x=0, so it lies outside the equation's domain

Question 1001

[1 marks]differentiation and related rates
A right circular metallic cylinder has radius rr cm and height hh cm, with volume 200 cm3200\text{ cm}^3. What is hh in terms of rr?
  1. Ah=200πr2h = \dfrac{200}{\pi r^2}
  2. Bh=200πrh = \dfrac{200}{\pi r}
  3. Ch=πr2200h = \dfrac{\pi r^2}{200}
  4. Dh=200πr2h = 200\pi r^2

Question 1002

[1 marks]differentiation and related rates
A right circular cylinder of radius rr cm and height hh cm always has volume 200 cm3200\text{ cm}^3. At the instant when r=4r=4 cm, hh is increasing at 0.40.4 cm/minute. What is the rate of change of rr at that instant, to 4 decimal places?
  1. A0.20110.2011 cm/min
  2. B−0.4021-0.4021 cm/min
  3. C−0.2011-0.2011 cm/min
  4. D−0.0040-0.0040 cm/min

Question 1003

[1 marks]differentiation and related rates
The total surface area of a cylinder of radius rr cm is S=2πr2+400rS = 2\pi r^2 + \dfrac{400}{r}. What is dSdr\dfrac{dS}{dr}?
  1. A4πr−400r4\pi r - \dfrac{400}{r}
  2. B4πr−400r24\pi r - \dfrac{400}{r^2}
  3. C2πr−400r22\pi r - \dfrac{400}{r^2}
  4. D4πr+400r24\pi r + \dfrac{400}{r^2}

Question 1004

[1 marks]differentiation and related rates
A cylinder's total surface area is S=2πr2+400rS = 2\pi r^2 + \dfrac{400}{r}, where rr is the radius in cm. At the instant when r=4r=4 cm and drdt≈−0.2011\dfrac{dr}{dt} \approx -0.2011 cm/min, what is the rate at which SS is changing, and is it increasing or decreasing?
  1. A≈−5.08 cm2/min\approx -5.08\text{ cm}^2/\text{min}, so SS is decreasing
  2. B≈−10.16 cm2/min\approx -10.16\text{ cm}^2/\text{min}, so SS is decreasing
  3. C≈+5.08 cm2/min\approx +5.08\text{ cm}^2/\text{min}, so SS is increasing
  4. D≈−2.54 cm2/min\approx -2.54\text{ cm}^2/\text{min}, so SS is decreasing

Question 1005

[2 marks]differentiation and related rates
A cylinder of volume 200 cm3200\text{ cm}^3 has h=200πr2h=\dfrac{200}{\pi r^2}. Evaluate hh when r=4r=4 cm, correct to 4 decimal places.

Answer this when you sit the paper.

Question 1006

[2 marks]differentiation and related rates
Differentiating V=πr2h=200V=\pi r^2h=200 implicitly with respect to tt and rearranging gives drdt=−r2hdhdt\dfrac{dr}{dt}=-\dfrac r{2h}\dfrac{dh}{dt}. Using r=4r=4 and h≈3.9789h\approx3.9789, evaluate the coefficient −r2h-\dfrac r{2h}, correct to 4 decimal places.

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Question 1007

[2 marks]differentiation and related rates
The total surface area of a cylinder of radius rr is S=2πr2+400rS=2\pi r^2+\dfrac{400}r, with dSdr=4πr−400r2\dfrac{dS}{dr}=4\pi r-\dfrac{400}{r^2}. Evaluate dSdr\dfrac{dS}{dr} at r=4r=4, correct to 3 decimal places.

Answer this when you sit the paper.

Question 1101

[1 marks]trigonometry and series expansions
Expanding cos⁡(2x+x)\cos(2x+x) as cos⁡2xcos⁡x−sin⁡2xsin⁡x\cos2x\cos x - \sin2x\sin x and using cos⁡2x=2cos⁡2x−1\cos2x = 2\cos^2x-1 and sin⁡2x=2sin⁡xcos⁡x\sin2x = 2\sin x\cos x, which of the following is the fully simplified result, in terms of cos⁡x\cos x only?
  1. A4cos⁡3x−3cos⁡x4\cos^3 x - 3\cos x
  2. B3cos⁡x−4cos⁡3x3\cos x - 4\cos^3 x
  3. C4cos⁡3x+3cos⁡x4\cos^3 x + 3\cos x
  4. D2cos⁡3x−cos⁡x2\cos^3 x - \cos x

Question 1102

[1 marks]trigonometry and series expansions
Using cos⁡3x=cos⁡3x+3cos⁡x4\cos^3x = \dfrac{\cos3x+3\cos x}{4} together with the series cos⁡x=1−x22+x424−x6720+…\cos x = 1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\dfrac{x^6}{720}+\ldots, what is the coefficient of x4x^4 in the resulting series for cos⁡3x\cos^3x?
  1. A78\dfrac{7}{8}
  2. B516\dfrac{5}{16}
  3. C2732\dfrac{27}{32}
  4. D72\dfrac{7}{2}

Question 1103

[1 marks]trigonometry and series expansions
Using the approximation cos⁡3x≈1−32x2+78x4−61240x6\cos^3x \approx 1-\dfrac{3}{2}x^2+\dfrac{7}{8}x^4-\dfrac{61}{240}x^6, what is the relative error in this approximation for cos⁡3x\cos^3x when x=13πx=\dfrac13\pi, to 1 significant figure?
  1. A0.30.3
  2. B0.40.4
  3. C0.50.5
  4. D0.60.6

Question 1104

[1 marks]trigonometry and series expansions
To prove cos⁡3x=4cos⁡3x−3cos⁡x\cos3x=4\cos^3x-3\cos x, the method begins by expanding cos⁡(2x+x)\cos(2x+x) using the addition formula cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A+B)=\cos A\cos B-\sin A\sin B. What is this expansion?
  1. Asin⁡2xcos⁡x−cos⁡2xsin⁡x\sin2x\cos x-\cos2x\sin x
  2. Bcos⁡2xsin⁡x−sin⁡2xcos⁡x\cos2x\sin x-\sin2x\cos x
  3. Ccos⁡2xcos⁡x−sin⁡2xsin⁡x\cos2x\cos x-\sin2x\sin x
  4. Dcos⁡2xcos⁡x+sin⁡2xsin⁡x\cos2x\cos x+\sin2x\sin x

Question 1105

[2 marks]trigonometry and series expansions
Expanding cos⁡(3x)\cos(3x) using the series cos⁡u=1−u22+u424−…\cos u=1-\dfrac{u^2}2+\dfrac{u^4}{24}-\ldots with u=3xu=3x, state the coefficient of x2x^2 in this series for cos⁡3x\cos3x alone (before combining with 3cos⁡x3\cos x or dividing by 4).

Answer this when you sit the paper.

Question 1106

[2 marks]trigonometry and series expansions
Expanding cos⁡(3x)\cos(3x) using the series cos⁡u=1−u22+u424−…\cos u=1-\dfrac{u^2}2+\dfrac{u^4}{24}-\ldots with u=3xu=3x, state the coefficient of x4x^4 in this series for cos⁡3x\cos3x alone (before combining with 3cos⁡x3\cos x or dividing by 4).

Answer this when you sit the paper.

Question 1107

[2 marks]trigonometry and series expansions
The relative error calculation for cos⁡3x\cos^3x at x=13πx=\dfrac13\pi needs the true value of cos⁡3x\cos^3x at that point. Using cos⁡(π/3)=12\cos(\pi/3)=\dfrac12, evaluate cos⁡3x\cos^3x exactly at x=π/3x=\pi/3.

Answer this when you sit the paper.

Question 1201

[1 marks]vectors
Points A, B and C have position vectors 6i+2j+6k6\mathbf{i}+2\mathbf{j}+6\mathbf{k}, 2i+3j+k2\mathbf{i}+3\mathbf{j}+\mathbf{k} and 14i+16k14\mathbf{i}+16\mathbf{k} respectively. What is the exact value of ∣AB→∣|\overrightarrow{AB}|?
  1. A50\sqrt{50}
  2. B42\sqrt{42}
  3. C41\sqrt{41}
  4. D59\sqrt{59}

Question 1202

[1 marks]vectors
Points A, B and C have position vectors 6i+2j+6k6\mathbf{i}+2\mathbf{j}+6\mathbf{k}, 2i+3j+k2\mathbf{i}+3\mathbf{j}+\mathbf{k} and 14i+16k14\mathbf{i}+16\mathbf{k} respectively, so that AB→=−4i+j−5k\overrightarrow{AB}=-4\mathbf{i}+\mathbf{j}-5\mathbf{k} and AC→=8i−2j+10k\overrightarrow{AC}=8\mathbf{i}-2\mathbf{j}+10\mathbf{k}. Which statement correctly describes the arrangement of A, B and C?
  1. AA, B and C are collinear, with B lying between A and C
  2. BA, B and C are collinear, with A lying between B and C
  3. CA, B and C are collinear, with C lying between A and B
  4. DA, B and C are not collinear, since AC→\overrightarrow{AC} is not a scalar multiple of AB→\overrightarrow{AB}

Question 1203

[1 marks]vectors
Point A has position vector 6i+2j+6k6\mathbf{i}+2\mathbf{j}+6\mathbf{k}, so that AB→=−4i+j−5k\overrightarrow{AB}=-4\mathbf{i}+\mathbf{j}-5\mathbf{k}. Point D has position vector i−dj+9k\mathbf{i}-d\mathbf{j}+9\mathbf{k}, and AD→\overrightarrow{AD} is perpendicular to AB→\overrightarrow{AB}. What is the value of dd?
  1. Ad=3d = 3
  2. Bd=−3d = -3
  3. Cd=5d = 5
  4. Dd=18d = 18

Question 1204

[1 marks]vectors
With A at 6i+2j+6k6\mathbf{i}+2\mathbf{j}+6\mathbf{k}, AB→=−4i+j−5k\overrightarrow{AB}=-4\mathbf{i}+\mathbf{j}-5\mathbf{k}, and D at i−3j+9k\mathbf{i}-3\mathbf{j}+9\mathbf{k} (so that AD→\overrightarrow{AD} is perpendicular to AB→\overrightarrow{AB}), what is the exact area of triangle ABD?
  1. A50.550.5
  2. B24782\dfrac{\sqrt{2478}}{2}
  3. C2478\sqrt{2478}
  4. D24784\dfrac{\sqrt{2478}}{4}

Question 1205

[2 marks]vectors
Points A, B and C have position vectors 6i+2j+6k6\mathbf i+2\mathbf j+6\mathbf k, 2i+3j+k2\mathbf i+3\mathbf j+\mathbf k and 14i+16k14\mathbf i+16\mathbf k respectively. State AC→=C−A\overrightarrow{AC}=C-A as 'x,y,z'.

Answer this when you sit the paper.

Question 1206

[1 marks]vectors
With AB→=−4i+j−5k\overrightarrow{AB}=-4\mathbf i+\mathbf j-5\mathbf k and AC→=8i−2j+10k\overrightarrow{AC}=8\mathbf i-2\mathbf j+10\mathbf k, state the scalar kk such that AC→=k⋅AB→\overrightarrow{AC}=k\cdot\overrightarrow{AB} (confirming A, B, C are collinear).

Answer this when you sit the paper.

Question 1207

[2 marks]vectors
Point D has position vector i−3j+9k\mathbf i-3\mathbf j+9\mathbf k (using d=3d=3), and A has position vector 6i+2j+6k6\mathbf i+2\mathbf j+6\mathbf k. State AD→=D−A\overrightarrow{AD}=D-A as 'x,y,z'.

Answer this when you sit the paper.

Question 1208

[2 marks]vectors
Given AD→=−5i−5j+3k\overrightarrow{AD}=-5\mathbf i-5\mathbf j+3\mathbf k, state ∣AD→∣|\overrightarrow{AD}| exactly.

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Question 1209

[1 marks]vectors
AD→\overrightarrow{AD} is perpendicular to AB→=−4i+j−5k\overrightarrow{AB}=-4\mathbf i+\mathbf j-5\mathbf k, where D has position vector i−dj+9k\mathbf i-d\mathbf j+9\mathbf k and A has 6i+2j+6k6\mathbf i+2\mathbf j+6\mathbf k. Before solving for dd, state AD→⋅AB→\overrightarrow{AD}\cdot\overrightarrow{AB} in terms of dd.

Answer this when you sit the paper.

Question 1301

[1 marks]series and sequences
Given that ∑r=1n(2r−3)=255\displaystyle\sum_{r=1}^{n}(2r-3) = 255, what is the value of nn?
  1. An=17n = 17
  2. Bn=15n = 15
  3. Cn=16n = 16
  4. Dn=19n = 19

Question 1302

[1 marks]series and sequences
An athlete runs a 40 km marathon, then each day after runs 80% of the distance run the previous day, starting with 32 km on the first day after the race. What distance does the athlete run on the tenth day after the marathon, to 3 decimal places?
  1. A5.3695.369 km
  2. B2.7492.749 km
  3. C3.4363.436 km
  4. D4.2954.295 km

Question 1303

[1 marks]series and sequences
An athlete runs a 40 km marathon, then each day after runs 80% of the distance run the previous day, starting with 32 km on the first day after the race. What is the first day after the marathon on which the athlete's total distance run (since the race) exceeds 155 km?
  1. Aday 13
  2. Bday 15
  3. Cday 17
  4. Dday 16

Question 1304

[2 marks]series and sequences
Simplifying ∑r=1n(2r−3)\displaystyle\sum_{r=1}^n(2r-3) gives a quadratic in nn of the form n2+bnn^2+bn (no constant term). State bb.

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Question 1305

[2 marks]series and sequences
Solving n2−2n−255=0n^2-2n-255=0 (from ∑r=1n(2r−3)=255\displaystyle\sum_{r=1}^n(2r-3)=255) gives two roots, one positive and one negative (the negative root is rejected since nn must be a positive integer). State the negative root.

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Question 1306

[1 marks]series and sequences
An athlete runs a 40 km marathon, then each day after runs 80% of the distance run the previous day, starting with 32 km on the first day after the race. State the common ratio of this geometric sequence of daily distances.

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Question 1307

[2 marks]series and sequences
An athlete runs 32 km on the first day after a marathon, then 80% of the previous day's distance each day after. The total distance after nn days is Sn=160(1−0.8n)S_n=160(1-0.8^n). Evaluate S15S_{15} (total distance after 15 days), correct to 3 decimal places.

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Question 1308

[2 marks]series and sequences
Using Sn=160(1−0.8n)S_n=160(1-0.8^n) for the athlete's total distance after nn days, evaluate S16S_{16}, correct to 3 decimal places.

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Question 1401

[1 marks]differential equations
Solve the differential equation dydx=x2y2\dfrac{dy}{dx} = x^2y^2 given that y=−12y=-\dfrac12 when x=3x=3, expressing yy in terms of xx.
  1. Ay=333−x3y = \dfrac{3}{33-x^3}
  2. By=321−x3y = \dfrac{3}{21-x^3}
  3. Cy=−3x3y = \dfrac{-3}{x^3}
  4. Dy=3x3−21y = \dfrac{3}{x^3-21}

Question 1402

[1 marks]differential equations
A liquid is heated in an oven kept at 200°C200°C. The rate of increase of the liquid's temperature θ°C\theta°C (θ<200\theta<200) at time tt minutes is proportional to (200−θ)(200-\theta). Which differential equation correctly models this, for some positive constant kk?
  1. Adθdt=200−kθ\dfrac{d\theta}{dt} = 200-k\theta
  2. Bdθdt=k(200−θ)\dfrac{d\theta}{dt} = k(200-\theta)
  3. Cdθdt=k(θ−200)\dfrac{d\theta}{dt} = k(\theta-200)
  4. Ddθdt=kθ(200−θ)\dfrac{d\theta}{dt} = k\theta(200-\theta)

Question 1403

[1 marks]differential equations
A liquid's temperature θ°C\theta°C satisfies θ=200−Ae−kt\theta = 200-Ae^{-kt}, where tt is time in minutes. The liquid starts at 0°C0°C and reaches 100°C100°C after 6 minutes. What is the temperature after 10 minutes, to the nearest degree?
  1. A168°C168°C
  2. B68°C68°C
  3. C100°C100°C
  4. D137°C137°C

Question 1404

[2 marks]differential equations
Separating variables in dydx=x2y2\dfrac{dy}{dx}=x^2y^2 gives ∫1y2 dy=∫x2 dx\displaystyle\int\dfrac1{y^2}\,dy=\displaystyle\int x^2\,dx. State the antiderivative of 1y2\dfrac1{y^2} with respect to yy (omit the constant of integration).

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Question 1405

[2 marks]differential equations
Solving dydx=x2y2\dfrac{dy}{dx}=x^2y^2 gives −1y=x33+C-\dfrac1y=\dfrac{x^3}3+C. Using y=−12y=-\dfrac12 when x=3x=3, evaluate CC.

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Question 1406

[2 marks]differential equations
A liquid's temperature θ\theta satisfies dθdt=k(200−θ)\dfrac{d\theta}{dt}=k(200-\theta). Separating variables, state the antiderivative of 1200−θ\dfrac1{200-\theta} with respect to θ\theta (omit the constant of integration).

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Question 1407

[1 marks]differential equations
A liquid's temperature satisfies θ=200−Ae−kt\theta=200-Ae^{-kt}, starting at θ=0\theta=0 when t=0t=0. State the value of AA.

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Question 1408

[2 marks]differential equations
A liquid's temperature satisfies θ=200−200e−kt\theta=200-200e^{-kt} and reaches 100°C100°C after 6 minutes. State e−6ke^{-6k} as a decimal.

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Question 1409

[2 marks]differential equations
Using e−6k=0.5e^{-6k}=0.5, evaluate kk, correct to 4 decimal places.

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Question 1410

[2 marks]differential equations
A liquid's temperature satisfies θ=200−200e−kt\theta=200-200e^{-kt} with k≈0.1155k\approx0.1155. Evaluate e−10ke^{-10k}, correct to 4 decimal places (needed to find θ\theta at t=10t=10).

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