Danho
ZIMSEC A Level · J2011

Pure Mathematics Paper 1 June 2011

Questions
85
Total marks
120

Sit this paper online

Questions
85
Pass mark
51
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]inequalities / indices
Solve the inequality ∣3−3x−54∣<27|3^{-3x} - 54| < 27.
  1. A1<x<431 < x < \tfrac{4}{3}
  2. B−43<x<43-\tfrac{4}{3} < x < \tfrac{4}{3}
  3. Cx<−43x < -\tfrac{4}{3} or x>−1x > -1
  4. D−43<x<−1-\tfrac{4}{3} < x < -1

Question 102

[1 marks]inequalities / indices
Removing the modulus from ∣3−3x−54∣<27|3^{-3x} - 54| < 27 reduces it to which inequality in 3−3x3^{-3x}?
  1. A27<3−3x<8127 < 3^{-3x} < 81
  2. B0<3−3x<270 < 3^{-3x} < 27
  3. C−27<3−3x<27-27 < 3^{-3x} < 27
  4. D3−3x>813^{-3x} > 81

Question 103

[2 marks]inequalities / indices
Rewrite 27<3−3x27 < 3^{-3x} using base 3 (i.e. 33<3−3x3^3 < 3^{-3x}), then solve for xx, stating only this one bound as an inequality in xx.

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Question 201

[1 marks]binomial expansion
Find the expansion of 4−3x2\sqrt{4 - 3x^2} up to and including the term in x4x^4.
  1. A2−34x2−964x42 - \tfrac{3}{4}x^2 - \tfrac{9}{64}x^4
  2. B2−32x2−916x42 - \tfrac{3}{2}x^2 - \tfrac{9}{16}x^4
  3. C4−32x2−932x44 - \tfrac{3}{2}x^2 - \tfrac{9}{32}x^4
  4. D2−34x2+964x42 - \tfrac{3}{4}x^2 + \tfrac{9}{64}x^4

Question 202

[1 marks]binomial expansion
In the expansion of 4−3x2\sqrt{4 - 3x^2} in ascending powers of xx, what is the coefficient of x4x^4?
  1. A−916-\tfrac{9}{16}
  2. B−34-\tfrac{3}{4}
  3. C−964-\tfrac{9}{64}
  4. D964\tfrac{9}{64}

Question 203

[2 marks]binomial expansion
Write 4−3x2\sqrt{4-3x^2} as 2(1+u)1/22(1+u)^{1/2}, stating uu in terms of xx.

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Question 301

[1 marks]coordinate geometry
Point QQ lies on the line y−3x=0y - 3x = 0 and on the perpendicular bisector of the line joining M(6,3)M(6, 3) and N(2,1)N(2, 1). Find the coordinates of QQ.
  1. A(6,18)(6, 18)
  2. B(2,6)(2, 6)
  3. C(−2,−6)(-2, -6)
  4. D(4,2)(4, 2)

Question 302

[1 marks]coordinate geometry
What is the gradient of the perpendicular bisector of the line joining M(6,3)M(6, 3) and N(2,1)N(2, 1)?
  1. A−12-\tfrac12
  2. B−2-2
  3. C12\tfrac12
  4. D22

Question 303

[1 marks]coordinate geometry
Find the midpoint of M(6,3)M(6,3) and N(2,1)N(2,1).

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Question 304

[2 marks]coordinate geometry
Hence write the equation of the perpendicular bisector of MN, in the form y=mx+cy=mx+c.

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Question 401

[1 marks]indices / equations
Find the value of xx which satisfies 3(22x+1)−21−2x=−53(2^{2x+1}) - 2^{1-2x} = -5.
  1. Ax=1x = 1
  2. Bx=−2x = -2
  3. Cx=14x = \tfrac14
  4. Dx=−1x = -1

Question 402

[1 marks]indices / equations
Which substitution turns 3(22x+1)−21−2x=−53(2^{2x+1}) - 2^{1-2x} = -5 into a quadratic equation?
  1. Au=log⁡2xu = \log_2 x
  2. Bu=2xu = 2^{x}
  3. Cu=22xu = 2^{2x}
  4. Du=2x+1u = 2x + 1

Question 403

[2 marks]indices / equations
Solve (4u−1)(3u+2)=0(4u-1)(3u+2)=0 and state the positive root uu.

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Question 404

[1 marks]indices / equations
Why is u=−23u=-\frac23 rejected as a solution, given u=22xu=2^{2x}?
  1. Auu must be an integer since 2x2x is an exponent.
  2. Bu=−23u=-\frac23 gives a non-real value of xx when substituted back.
  3. Cu=−23u=-\frac23 makes (4u−1)(3u+2)=0(4u-1)(3u+2)=0 false, so it isn't an algebraic root.
  4. D22x2^{2x} is always positive, so uu cannot be negative.

Question 501

[1 marks]numerical integration / trapezium rule
Using the trapezium rule with 5 trapezia, find the approximate value of ∫01x2ex dx\displaystyle\int_0^1 x^2 e^x \, dx, correct to 4 decimal places.
  1. A0.64870.6487
  2. B0.71830.7183
  3. C0.74540.7454
  4. D1.49081.4908

Question 502

[1 marks]numerical integration / trapezium rule
The trapezium rule with 5 trapezia estimates ∫01x2ex dx\displaystyle\int_0^1 x^2 e^x \, dx as 0.74540.7454. Given that the exact value is e−2e - 2, the percentage error in the approximation is closest to
  1. A0.3%0.3\%
  2. B3.8%3.8\%
  3. C7.5%7.5\%
  4. D26.6%26.6\%

Question 503

[1 marks]numerical integration / trapezium rule
For the trapezium rule with 5 strips over [0,1][0,1], what is the strip width hh?

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Question 504

[2 marks]numerical integration / trapezium rule
Calculate y3=(0.6)2e0.6y_3=(0.6)^2e^{0.6}, to 4 decimal places.

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Question 505

[1 marks]numerical integration / trapezium rule
State the exact value of the integral, e−2e-2, to 4 decimal places.

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Question 601

[1 marks]trigonometry
In triangle ABCABC the sides BCBC, CACA and ABAB are aa, bb and cc respectively, and cot⁡C=accosec⁡B−cot⁡B\cot C = \dfrac{a}{c}\operatorname{cosec} B - \cot B. Given a=14.7a = 14.7 cm, c=17.3c = 17.3 cm and B=64.2°B = 64.2°, find angle CC correct to the nearest 0.1°0.1°.
  1. A41.5°41.5°
  2. B50.5°50.5°
  3. C65.3°65.3°
  4. D24.7°24.7°

Question 602

[1 marks]trigonometry
In triangle ABCABC, a=14.7a = 14.7 cm, c=17.3c = 17.3 cm, B=64.2°B = 64.2° and C=65.3°C = 65.3°. Find angle AA correct to the nearest 0.1°0.1°.
  1. A50.5°50.5°
  2. B65.3°65.3°
  3. C115.5°115.5°
  4. D25.8°25.8°

Question 603

[1 marks]trigonometry
In right triangle ABD (D the foot of the perpendicular from A), express AD=xAD=x in terms of cc and angle BB.

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Question 604

[1 marks]trigonometry
In right triangle ABD, express BDBD in terms of cc and angle BB.

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Question 605

[2 marks]trigonometry
Using a=14.7a=14.7, c=17.3c=17.3, B=64.2°B=64.2°, evaluate cot⁡C=accosec⁡B−cot⁡B\cot C=\dfrac{a}{c}\operatorname{cosec}B-\cot B to 4 decimal places.

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Question 701

[1 marks]graph transformations
The diagram shows the graph of y=f(x)y = f(x), passing through A(−1,0)A(-1, 0), B(0,1)B(0, 1) and C(1,0)C(1, 0). On the graph of y=f(−x)+3y = f(-x) + 3, the image of BB is
  1. A(0,3)(0, 3)
  2. B(0,−2)(0, -2)
  3. C(0,4)(0, 4)
  4. D(3,1)(3, 1)

Question 702

[1 marks]graph transformations
The diagram shows the graph of y=f(x)y = f(x), passing through A(−1,0)A(-1, 0), B(0,1)B(0, 1) and C(1,0)C(1, 0). On the graph of y=2f(x+1)y = 2f(x + 1), the image of CC is
  1. A(0,2)(0, 2)
  2. B(0,0)(0, 0)
  3. C(2,0)(2, 0)
  4. D(1,0)(1, 0)

Question 703

[1 marks]graph transformations
Under y=f(2x)y=f(2x) (horizontal stretch factor 12\frac12), find the image of A(−1,0)A(-1,0).

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Question 704

[1 marks]graph transformations
Under y=f(2x)y=f(2x), find the image of C(1,0)C(1,0).

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Question 705

[1 marks]graph transformations
Under y=f(−x)+3y=f(-x)+3, find the image of A(−1,0)A(-1,0).

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Question 706

[1 marks]graph transformations
Under y=2f(x+1)y=2f(x+1), find the image of B(0,1)B(0,1).

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Question 801

[1 marks]differentiation
Given that y=e2xsin⁡5xy = e^{2x}\sin 5x, which differential equation does yy satisfy?
  1. Ad2ydx2−4dydx−21y=0\dfrac{d^2y}{dx^2} - 4\dfrac{dy}{dx} - 21y = 0
  2. Bd2ydx2−2dydx+25y=0\dfrac{d^2y}{dx^2} - 2\dfrac{dy}{dx} + 25y = 0
  3. Cd2ydx2−4dydx+29y=0\dfrac{d^2y}{dx^2} - 4\dfrac{dy}{dx} + 29y = 0
  4. Dd2ydx2+4dydx+29y=0\dfrac{d^2y}{dx^2} + 4\dfrac{dy}{dx} + 29y = 0

Question 802

[1 marks]differentiation
If y=e2xsin⁡5xy = e^{2x}\sin 5x, then dydx=\dfrac{dy}{dx} =
  1. A2e2xsin⁡5x−5e2xcos⁡5x2e^{2x}\sin 5x - 5e^{2x}\cos 5x
  2. B10e2xsin⁡5xcos⁡5x10e^{2x}\sin 5x\cos 5x
  3. C2e2xcos⁡5x2e^{2x}\cos 5x
  4. D2e2xsin⁡5x+5e2xcos⁡5x2e^{2x}\sin 5x + 5e^{2x}\cos 5x

Question 803

[2 marks]differentiation
Differentiate 5e2xcos⁡5x5e^{2x}\cos5x (one term of dy/dxdy/dx) using the product rule; give the result as a sum of two terms before simplifying.

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Question 804

[1 marks]differentiation
Differentiate 2e2xsin⁡5x2e^{2x}\sin5x (the other term of dy/dxdy/dx); give the result as a sum of two terms.

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Question 805

[1 marks]differentiation
From dydx=5e2xcos⁡5x+2y\frac{dy}{dx}=5e^{2x}\cos5x+2y, make 5e2xcos⁡5x5e^{2x}\cos5x the subject.

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Question 901

[1 marks]complex numbers
Given z1=2−4iz_1 = 2 - 4i and z2=6−2iz_2 = 6 - 2i, express z1z2z_1 z_2 in the form a+iba + ib.
  1. A4−28i4 - 28i
  2. B20−28i20 - 28i
  3. C12−28i12 - 28i
  4. D12+8i12 + 8i

Question 902

[1 marks]complex numbers
Given z1=2−4iz_1 = 2 - 4i and w=1z1w = \dfrac{1}{z_1}, find the exact value of ∣w∣|w|.
  1. A510\tfrac{\sqrt{5}}{10}
  2. B20\sqrt{20}
  3. C110\tfrac{1}{10}
  4. D55\tfrac{\sqrt{5}}{5}

Question 903

[1 marks]complex numbers
Find z1−z2z_1-z_2 for z1=2−4iz_1=2-4i, z2=6−2iz_2=6-2i, in the form a+bia+bi.

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Question 904

[2 marks]complex numbers
Find w=1z1w=\dfrac{1}{z_1} for z1=2−4iz_1=2-4i, in the form a+bia+bi (rationalise using the conjugate).

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Question 905

[2 marks]complex numbers
Find arg⁡(w)\arg(w) for w=110+15iw=\dfrac{1}{10}+\dfrac{1}{5}i, to 1 decimal place (in degrees).

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Question 1001

[1 marks]parametric differentiation
A curve has parametric equations x=t−ln⁡(2t+1)x = t - \ln(2t+1) and y=t+ln⁡(2t+1)y = t + \ln(2t+1), t>−0.5t > -0.5. Find the equation of the tangent to the curve at t=1t = 1.
  1. Ay=15x+1+ln⁡3y = \tfrac15 x + 1 + \ln 3
  2. By=5x+4−6ln⁡3y = 5x + 4 - 6\ln 3
  3. Cy=5x−4−6ln⁡3y = 5x - 4 - 6\ln 3
  4. Dy=5x−4+6ln⁡3y = 5x - 4 + 6\ln 3

Question 1002

[1 marks]parametric differentiation
For the curve x=t−ln⁡(2t+1)x = t - \ln(2t+1), y=t+ln⁡(2t+1)y = t + \ln(2t+1) with t>−0.5t > -0.5, it can be shown that dydx=2t+32t−1\dfrac{dy}{dx} = \dfrac{2t+3}{2t-1}. What is the gradient of the curve at t=1t = 1?
  1. A55
  2. B15\tfrac15
  3. C53\tfrac53
  4. D−5-5

Question 1003

[1 marks]parametric differentiation
Find dxdt\dfrac{dx}{dt} for x=t−ln⁡(2t+1)x=t-\ln(2t+1).

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Question 1004

[1 marks]parametric differentiation
Find dydt\dfrac{dy}{dt} for y=t+ln⁡(2t+1)y=t+\ln(2t+1).

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Question 1005

[3 marks]parametric differentiation
Find the coordinates (x,y)(x,y) of the point on the curve at t=1t=1, giving exact values involving ln⁡3\ln3.

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Question 1101

[1 marks]trigonometric equations / calculus
Solve the equation cos⁡2x=cos⁡x\cos 2x = \cos x for 0≤x≤2π0 \le x \le 2\pi.
  1. Ax=π3,5π3x = \tfrac{\pi}{3}, \tfrac{5\pi}{3} only
  2. Bx=0,2π3,4π3,2πx = 0, \tfrac{2\pi}{3}, \tfrac{4\pi}{3}, 2\pi
  3. Cx=0,π3,5π3,2πx = 0, \tfrac{\pi}{3}, \tfrac{5\pi}{3}, 2\pi
  4. Dx=0,π,2πx = 0, \pi, 2\pi

Question 1102

[1 marks]trigonometric equations / calculus
Given f(x)=cos⁡2x−cos⁡xf(x) = \cos 2x - \cos x, the derivative f′(x)f'(x) factorises as
  1. A−2sin⁡2xcos⁡x-2\sin 2x \cos x
  2. Bsin⁡x(4cos⁡x−1)\sin x(4\cos x - 1)
  3. Ccos⁡x(1−4sin⁡x)\cos x(1 - 4\sin x)
  4. Dsin⁡x(1−4cos⁡x)\sin x(1 - 4\cos x)

Question 1103

[2 marks]trigonometric equations / calculus
Solve cos⁡x=14\cos x=\frac14 for 0≤x≤2π0\le x\le2\pi, giving the first (smaller) solution in radians, to 3 d.p.

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Question 1104

[2 marks]trigonometric equations / calculus
Solve cos⁡x=14\cos x=\frac14 for 0≤x≤2π0\le x\le2\pi, giving the second (larger) solution in radians, to 3 d.p.

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Question 1105

[2 marks]trigonometric equations / calculus
Using cos⁡2x=2cos⁡2x−1\cos2x=2\cos^2x-1, rewrite cos⁡2x=cos⁡x\cos2x=\cos x as a quadratic in cos⁡x\cos x. State the quadratic expression, in the form acos⁡2x+bcos⁡x+ca\cos^2x+b\cos x+c.

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Question 1201

[1 marks]arithmetic series
A teacher earns $x\$x in his first year and his salary increases each year by 10% of his first year's salary. His total salary after nn years is
  1. Anx(n+19)20\dfrac{nx(n+19)}{20}
  2. Bnx(n+9)10\dfrac{nx(n+9)}{10}
  3. Cx(1.1)nx(1.1)^{n}
  4. Dnx(n−1)20\dfrac{nx(n-1)}{20}

Question 1202

[1 marks]arithmetic series
A teacher's total salary after nn years is nx(n+19)20\dfrac{nx(n+19)}{20}, where $x\$x is his first year's salary. Find the least value of nn for which his total salary exceeds 100 times his first salary.
  1. An=45n = 45
  2. Bn=100n = 100
  3. Cn=37n = 37
  4. Dn=36n = 36

Question 1203

[1 marks]arithmetic series
What is the common difference of the salary arithmetic progression, in terms of xx?

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Question 1204

[2 marks]arithmetic series
From nx(n+19)20>100x\dfrac{nx(n+19)}{20}>100x, cancel xx and simplify to a quadratic inequality in nn, in the form n2+bn+c>0n^2+bn+c>0.

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Question 1205

[3 marks]arithmetic series
Using the quadratic formula on n2+19n−2000=0n^2+19n-2000=0, evaluate the positive root, to 1 decimal place.

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Question 1301

[1 marks]vectors
PQRSPQRS is a square with PR→=(−3312)\overrightarrow{PR} = \begin{pmatrix} -3 \\ 3 \\ 12 \end{pmatrix} and SQ→=(5−114)\overrightarrow{SQ} = \begin{pmatrix} 5 \\ -11 \\ 4 \end{pmatrix}. Find PQ→\overrightarrow{PQ}.
  1. A(1−48)\begin{pmatrix} 1 \\ -4 \\ 8 \end{pmatrix}
  2. B(−474)\begin{pmatrix} -4 \\ 7 \\ 4 \end{pmatrix}
  3. C(−14−8)\begin{pmatrix} -1 \\ 4 \\ -8 \end{pmatrix}
  4. D(2−816)\begin{pmatrix} 2 \\ -8 \\ 16 \end{pmatrix}

Question 1302

[1 marks]vectors
PQRSPQRS is a square in which PQ→=(1−48)\overrightarrow{PQ} = \begin{pmatrix} 1 \\ -4 \\ 8 \end{pmatrix}. Calculate the area of the square.
  1. A99 square units
  2. B1818 square units
  3. C40.540.5 square units
  4. D8181 square units

Question 1303

[2 marks]vectors
Find OP→=OQ→−PQ→\overrightarrow{OP}=\overrightarrow{OQ}-\overrightarrow{PQ}, given OQ→=(3,−1,2)\overrightarrow{OQ}=(3,-1,2) and PQ→=(1,−4,8)\overrightarrow{PQ}=(1,-4,8).

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Question 1304

[2 marks]vectors
Find OR→=OP→+PR→\overrightarrow{OR}=\overrightarrow{OP}+\overrightarrow{PR}, given OP→=(2,3,−6)\overrightarrow{OP}=(2,3,-6) and PR→=(−3,3,12)\overrightarrow{PR}=(-3,3,12).

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Question 1305

[2 marks]vectors
Find OS→=OQ→−SQ→\overrightarrow{OS}=\overrightarrow{OQ}-\overrightarrow{SQ}, given OQ→=(3,−1,2)\overrightarrow{OQ}=(3,-1,2) and SQ→=(5,−11,4)\overrightarrow{SQ}=(5,-11,4).

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Question 1306

[1 marks]vectors
Find ∣PQ→∣|\overrightarrow{PQ}|, the side length of the square, given PQ→=(1,−4,8)\overrightarrow{PQ}=(1,-4,8).

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Question 1401

[1 marks]integration (substitution and by parts)
Using the substitution y=xy = \sqrt{x}, find the exact value of ∫041x+1 dx\displaystyle\int_0^4 \dfrac{1}{\sqrt{x}+1}\,dx.
  1. A2ln⁡32\ln 3
  2. B4−2ln⁡34 - 2\ln 3
  3. C2−2ln⁡32 - 2\ln 3
  4. D4+2ln⁡34 + 2\ln 3

Question 1402

[1 marks]integration (substitution and by parts)
Given that ∫01dxx+1=2−2ln⁡2\displaystyle\int_0^1 \dfrac{dx}{\sqrt{x}+1} = 2 - 2\ln 2, find the exact value of ∫01ln⁡(x+1)x dx\displaystyle\int_0^1 \dfrac{\ln(\sqrt{x}+1)}{\sqrt{x}}\,dx.
  1. A2ln⁡2−22\ln 2 - 2
  2. B2−2ln⁡22 - 2\ln 2
  3. C4ln⁡2+24\ln 2 + 2
  4. D4ln⁡2−24\ln 2 - 2

Question 1403

[1 marks]integration (substitution and by parts)
With y=xy=\sqrt{x}, find dxdx in terms of yy and dydy.

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Question 1404

[2 marks]integration (substitution and by parts)
Rewrite 2yy+1\dfrac{2y}{y+1} by polynomial division, in the form 2−2y+12-\dfrac{2}{y+1}. State the simplified integrand.

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Question 1405

[2 marks]integration (substitution and by parts)
Evaluate ∫01dxx+1\displaystyle\int_0^1\dfrac{dx}{\sqrt x+1} (needed in part (ii)), exact value.

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Question 1406

[2 marks]integration (substitution and by parts)
In the integration by parts for ∫01ln⁡(x+1)xdx\displaystyle\int_0^1\dfrac{\ln(\sqrt x+1)}{\sqrt x}dx, with u=ln⁡(x+1)u=\ln(\sqrt x+1) and dv=dxxdv=\dfrac{dx}{\sqrt x}, find vv.

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Question 1407

[3 marks]integration (substitution and by parts)
Evaluate [2xln⁡(x+1)]01[2\sqrt x\ln(\sqrt x+1)]_0^1, the boundary term from integration by parts.

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Question 1501

[1 marks]exponentials, area and volume of revolution
The region RR is bounded by the curves y=exy = e^x and y=e−xy = e^{-x} and the line x=1x = 1. Find the exact area of RR.
  1. Ae+1ee + \dfrac{1}{e}
  2. Be+1e−2e + \dfrac{1}{e} - 2
  3. Ce−1e−2e - \dfrac{1}{e} - 2
  4. De−1ee - \dfrac{1}{e}

Question 1502

[1 marks]exponentials, area and volume of revolution
The region RR bounded by y=exy = e^x, y=e−xy = e^{-x} and x=1x = 1 is rotated through 2π2\pi radians about the xx-axis. Find the exact volume generated.
  1. Aπ(e+1e−2)2\pi\left(e + \dfrac{1}{e} - 2\right)^2
  2. Bπ(e2+1e2−2)\pi\left(e^2 + \dfrac{1}{e^2} - 2\right)
  3. Cπ2(e2+1e2−2)\dfrac{\pi}{2}\left(e^2 + \dfrac{1}{e^2} - 2\right)
  4. Dπ2(e2−1e2)\dfrac{\pi}{2}\left(e^2 - \dfrac{1}{e^2}\right)

Question 1503

[1 marks]exponentials, area and volume of revolution
Solve ex−e−x=0e^x-e^{-x}=0 for xx.

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Question 1504

[1 marks]exponentials, area and volume of revolution
Find the point of intersection of y=exy=e^x and y=e−xy=e^{-x}.

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Question 1505

[2 marks]exponentials, area and volume of revolution
Find the antiderivative of ex−e−xe^x-e^{-x} (ignore the constant of integration).

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Question 1506

[2 marks]exponentials, area and volume of revolution
Expand (ex)2−(e−x)2(e^x)^2-(e^{-x})^2 in terms of e2xe^{2x} and e−2xe^{-2x}.

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Question 1507

[3 marks]exponentials, area and volume of revolution
Find the antiderivative of e2x−e−2xe^{2x}-e^{-2x} (used for the volume integral), ignoring the constant.

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Question 1508

[2 marks]exponentials, area and volume of revolution
Evaluate e+1ee+\dfrac1e to 4 decimal places (used in the area calculation).

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Question 1601

[1 marks]differential equations (Newton's law of cooling)
Bread cools in a box at 20°20°C so that dθdt=−k(θ−20)\dfrac{d\theta}{dt} = -k(\theta - 20), where θ °\theta\,°C is its temperature at time tt minutes. Given that θ=100\theta = 100 when t=0t = 0 and θ=80\theta = 80 when t=10t = 10, the particular solution is
  1. Aθ=20(34)t/10+80\theta = 20\left(\tfrac34\right)^{t/10} + 80
  2. Bθ=80(34)t/10+20\theta = 80\left(\tfrac34\right)^{t/10} + 20
  3. Cθ=100(45)t/10\theta = 100\left(\tfrac45\right)^{t/10}
  4. Dθ=80e−t/10+20\theta = 80e^{-t/10} + 20

Question 1602

[1 marks]differential equations (Newton's law of cooling)
The temperature of cooling bread is θ=80(34)t/10+20\theta = 80\left(\tfrac34\right)^{t/10} + 20, where tt is in minutes. Find its temperature when t=20t = 20.
  1. A45°45°C
  2. B60°60°C
  3. C65°65°C
  4. D80°80°C

Question 1603

[1 marks]differential equations (Newton's law of cooling)
The temperature of cooling bread is θ=80(34)t/10+20\theta = 80\left(\tfrac34\right)^{t/10} + 20, where tt is in minutes. To the nearest minute, when has the temperature dropped by 40°40°C?
  1. A2424 minutes
  2. B2929 minutes
  3. C1414 minutes
  4. D2020 minutes

Question 1604

[1 marks]differential equations (Newton's law of cooling)
Write the right-hand side of the differential equation dθdt=…\dfrac{d\theta}{dt}=\ldots, given the cooling rate is proportional to the difference between θ\theta and 20°C20°C, with constant kk.

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Question 1605

[1 marks]differential equations (Newton's law of cooling)
Using θ=100\theta=100 at t=0t=0 in θ=Be−kt+20\theta=Be^{-kt}+20, find BB.

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Question 1606

[2 marks]differential equations (Newton's law of cooling)
Using θ=80\theta=80 at t=10t=10, write and solve 60=80e−10k60=80e^{-10k} for e−10ke^{-10k}.

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Question 1607

[2 marks]differential equations (Newton's law of cooling)
Hence evaluate k=110ln⁡(43)k=\dfrac{1}{10}\ln\left(\dfrac{4}{3}\right), to 4 decimal places.

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Question 1608

[2 marks]differential equations (Newton's law of cooling)
Solve (34)t/10=0.5\left(\frac34\right)^{t/10}=0.5 for t10=ln⁡0.5ln⁡0.75\dfrac{t}{10}=\dfrac{\ln0.5}{\ln0.75}, to 3 decimal places.

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Question 1609

[3 marks]differential equations (Newton's law of cooling)
Separate variables in dθdt=−k(θ−20)\dfrac{d\theta}{dt}=-k(\theta-20) to write ∫dθθ−20=−k∫dt\int\frac{d\theta}{\theta-20}=-k\int dt, then integrate the left side (state the antiderivative, ignoring the constant).

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