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ZIMSEC A Level · N2017

Pure Mathematics Paper 1 November 2017

Questions
77
Total marks
120

Sit this paper online

Questions
77
Pass mark
47
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]Complex numbers
Given w=1+2iw = 1 + 2i and u=3−iu = 3 - i, express uwuw in the form a+iba + ib.
  1. A5−5i5 - 5i
  2. B5+5i5 + 5i
  3. C3−2i3 - 2i
  4. D1+5i1 + 5i

Question 102

[1 marks]Complex numbers
Given w=1+2iw = 1 + 2i and u=3−iu = 3 - i, find arg⁡(uw)\arg(uw).
  1. A3π4\tfrac{3\pi}{4}
  2. Bπ4\tfrac{\pi}{4}
  3. Cπ2\tfrac{\pi}{2}
  4. D−π4-\tfrac{\pi}{4}

Question 103

[2 marks]Complex numbers
Given w=1+2iw=1+2i and u=3−iu=3-i, find u+wu+w in the form a+iba+ib.

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Question 201

[1 marks]Composite functions
Given f(x)=3x−1f(x) = 3x - 1 and h(x)=2x+5h(x) = 2x + 5, find the value of xx for which fh(x)=2hf(x)fh(x) = 2hf(x).
  1. Ax=2x = 2
  2. Bx=43x = \tfrac{4}{3}
  3. Cx=34x = \tfrac{3}{4}
  4. Dx=−43x = -\tfrac{4}{3}

Question 202

[2 marks]Composite functions
Which pair correctly gives fh(x)fh(x) and hf(x)hf(x) for f(x)=3x−1f(x)=3x-1, h(x)=2x+5h(x)=2x+5?
  1. Afh(x)=6x+14fh(x)=6x+14, hf(x)=6x+3hf(x)=6x+3
  2. Bfh(x)=6x+3fh(x)=6x+3, hf(x)=6x+14hf(x)=6x+14
  3. Cfh(x)=6x+9fh(x)=6x+9, hf(x)=6x+4hf(x)=6x+4
  4. Dfh(x)=5x+14fh(x)=5x+14, hf(x)=5x+3hf(x)=5x+3

Question 203

[1 marks]Composite functions
Find fh(0)fh(0) where f(x)=3x−1f(x)=3x-1, h(x)=2x+5h(x)=2x+5.

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Question 301

[1 marks]Completing the square
Express 2x2−3x+72x^2 - 3x + 7 in the form p(x+q)2+rp(x+q)^2 + r.
  1. A2(x+34)2+4782\left(x + \tfrac{3}{4}\right)^2 + \tfrac{47}{8}
  2. B2(x−32)2+522\left(x - \tfrac{3}{2}\right)^2 + \tfrac{5}{2}
  3. C2(x−34)2+4782\left(x - \tfrac{3}{4}\right)^2 + \tfrac{47}{8}
  4. D2(x−34)2+4742\left(x - \tfrac{3}{4}\right)^2 + \tfrac{47}{4}

Question 302

[1 marks]Completing the square
Write down the coordinates of the turning point of y=2x2−3x+7y = 2x^2 - 3x + 7.
  1. A(34, 478)\left(\tfrac{3}{4},\ \tfrac{47}{8}\right)
  2. B(34, 7)\left(\tfrac{3}{4},\ 7\right)
  3. C(32, 478)\left(\tfrac{3}{2},\ \tfrac{47}{8}\right)
  4. D(−34, 478)\left(-\tfrac{3}{4},\ \tfrac{47}{8}\right)

Question 303

[2 marks]Completing the square
Verify the identity 2x2−3x+7=2(x−34)2+4782x^2-3x+7=2\left(x-\tfrac34\right)^2+\tfrac{47}{8} by evaluating 2x2−3x+72x^2-3x+7 at x=2x=2.

Answer this when you sit the paper.

Question 401

[1 marks]Partial fractions
Express 3x+8(2x+1)(x2+3)\dfrac{3x+8}{(2x+1)(x^2+3)} in partial fractions.
  1. A22x+1+2−xx2+3\dfrac{2}{2x+1} + \dfrac{2-x}{x^2+3}
  2. B22x+1−2−xx2+3\dfrac{2}{2x+1} - \dfrac{2-x}{x^2+3}
  3. C22x+1+x−2x2+3\dfrac{2}{2x+1} + \dfrac{x-2}{x^2+3}
  4. D12x+1+2−xx2+3\dfrac{1}{2x+1} + \dfrac{2-x}{x^2+3}

Question 402

[2 marks]Partial fractions
Substituting x=−12x=-\tfrac12 into 3x+8=A(x2+3)+(Bx+C)(2x+1)3x+8=A(x^2+3)+(Bx+C)(2x+1) isolates AA (the (Bx+C)(2x+1)(Bx+C)(2x+1) term vanishes). Find AA.

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Question 403

[2 marks]Partial fractions
Comparing coefficients of x2x^2 in 3x+8=A(x2+3)+(Bx+C)(2x+1)3x+8=A(x^2+3)+(Bx+C)(2x+1) gives A+2B=0A+2B=0. Using A=2A=2, find BB.

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Question 501

[1 marks]Small changes / rates
Quantities yy and xx satisfy y2x3=cy^2 x^3 = c, where cc is constant. Find the percentage decrease in yy when xx increases by 0.5%0.5\%.
  1. A1.5%1.5\%
  2. B0.33%0.33\%
  3. C0.5%0.5\%
  4. D0.75%0.75\%

Question 502

[2 marks]Small changes / rates
Implicitly differentiating y2x3=cy^2x^3=c gives 2yx3dydx+3y2x2=02yx^3\dfrac{dy}{dx}+3y^2x^2=0. Solve for dydx\dfrac{dy}{dx} in terms of xx and yy.

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Question 503

[2 marks]Small changes / rates
Hence find the constant kk in δyy=kδxx\dfrac{\delta y}{y}=k\dfrac{\delta x}{x}, using dydx=−3y2x\dfrac{dy}{dx}=-\dfrac{3y}{2x}.

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Question 601

[1 marks]Geometric progression
A geometric progression has first term aa and common ratio rr with 0<r<10 < r < 1. Given that the sum of the first four terms is half the sum to infinity, find the exact value of rr.
  1. Ar=12r = \tfrac{1}{\sqrt{2}}
  2. Br=124r = \dfrac{1}{\sqrt[4]{2}}
  3. Cr=12r = \tfrac{1}{2}
  4. Dr=116r = \tfrac{1}{16}

Question 602

[1 marks]Geometric progression
A geometric progression has a=2a = 2 and r=(12)1/4r = \left(\tfrac{1}{2}\right)^{1/4}. Find its 9th term.
  1. A12\tfrac{1}{2}
  2. B14\tfrac{1}{4}
  3. C11
  4. D18\tfrac{1}{8}

Question 603

[2 marks]Geometric progression
The condition 'sum of the first four terms is half the sum to infinity' gives 1−r4=121-r^4=\tfrac12. Solve for r4r^4.

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Question 604

[1 marks]Geometric progression
With a=2a=2 and r4=12r^4=\tfrac12, find the 5th term ar4ar^4.

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Question 701

[1 marks]Differentiation / normals
Find the equation of the normal to the curve y=3e−2x+x+3y = 3e^{-2x} + x + 3 at the point where x=0x = 0.
  1. Ay=15x+3y = \tfrac{1}{5}x + 3
  2. By=−5x+6y = -5x + 6
  3. Cy=15x+6y = \tfrac{1}{5}x + 6
  4. Dy=−15x+6y = -\tfrac{1}{5}x + 6

Question 702

[2 marks]Differentiation / normals
Find dydx\dfrac{dy}{dx} for y=3e−2x+x+3y=3e^{-2x}+x+3.

Answer this when you sit the paper.

Question 703

[2 marks]Differentiation / normals
At x=0x=0, what are the gradient of the tangent and the value of yy for y=3e−2x+x+3y=3e^{-2x}+x+3?
  1. Agradient −5-5, y=3y=3
  2. Bgradient 55, y=6y=6
  3. Cgradient −4-4, y=6y=6
  4. Dgradient −5-5, y=6y=6

Question 704

[1 marks]Differentiation / normals
The gradient of the tangent at x=0x=0 is −5-5. Find the gradient of the normal.

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Question 801

[1 marks]Numerical methods / Newton-Raphson
Taking x1=1x_1 = 1, apply the Newton-Raphson method twice to e−x−2x+3=0e^{-x} - 2x + 3 = 0 and give the root correct to 3 decimal places.
  1. A1.0001.000
  2. B1.5001.500
  3. C1.5781.578
  4. D1.6011.601

Question 802

[2 marks]Numerical methods / Newton-Raphson
Find f′(x)f'(x) for f(x)=e−x−2x+3f(x)=e^{-x}-2x+3.

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Question 803

[2 marks]Numerical methods / Newton-Raphson
Taking x1=1x_1=1, perform one Newton-Raphson iteration on f(x)=e−x−2x+3f(x)=e^{-x}-2x+3 to find x2x_2, correct to 4 decimal places.

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Question 804

[2 marks]Numerical methods / Newton-Raphson
Why does sketching y=e−xy=e^{-x} and y=2x−3y=2x-3 on the same axes show that e−x−2x+3=0e^{-x}-2x+3=0 has only one real root?
  1. Athe curves touch at the origin only
  2. Bthe curves intersect exactly once
  3. Cthe curves intersect exactly three times
  4. Dthe curves are parallel everywhere

Question 901

[1 marks]Vectors in 3D
Points AA and BB have position vectors i+2j−3k\mathbf{i}+2\mathbf{j}-3\mathbf{k} and 3i−2j+5k3\mathbf{i}-2\mathbf{j}+5\mathbf{k}. Find a unit vector in the direction of AB→\overrightarrow{AB}.
  1. A121(−i+2j−4k)\tfrac{1}{\sqrt{21}}(-\mathbf{i} + 2\mathbf{j} - 4\mathbf{k})
  2. B121(i−2j+4k)\tfrac{1}{\sqrt{21}}(\mathbf{i} - 2\mathbf{j} + 4\mathbf{k})
  3. C184(i−2j+4k)\tfrac{1}{\sqrt{84}}(\mathbf{i} - 2\mathbf{j} + 4\mathbf{k})
  4. D121(2i−4j+8k)\tfrac{1}{21}(2\mathbf{i} - 4\mathbf{j} + 8\mathbf{k})

Question 902

[1 marks]Vectors in 3D
Points AA and BB have position vectors i+2j−3k\mathbf{i}+2\mathbf{j}-3\mathbf{k} and 3i−2j+5k3\mathbf{i}-2\mathbf{j}+5\mathbf{k} relative to the origin OO. Find angle AO^BA\hat{O}B to the nearest degree.
  1. A116°116°
  2. B134°134°
  3. C46°46°
  4. D90°90°

Question 903

[1 marks]Vectors in 3D
OB→=3i−2j+5k\overrightarrow{OB} = 3\mathbf{i} - 2\mathbf{j} + 5\mathbf{k} and OC→=pi−pj+(p−1)k\overrightarrow{OC} = p\mathbf{i} - p\mathbf{j} + (p-1)\mathbf{k}. Find pp for which OB→\overrightarrow{OB} is perpendicular to OC→\overrightarrow{OC}.
  1. Ap=5p = 5
  2. Bp=2p = 2
  3. Cp=−12p = -\tfrac{1}{2}
  4. Dp=12p = \tfrac{1}{2}

Question 904

[2 marks]Vectors in 3D
Given A=(1,2,−3)A=(1,2,-3) and B=(3,−2,5)B=(3,-2,5), find the vector AB⃗\vec{AB} (before finding its magnitude).

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Question 905

[2 marks]Vectors in 3D
Find ∣AB⃗∣|\vec{AB}|, where AB⃗=(2,−4,8)\vec{AB}=(2,-4,8), in simplified surd form.

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Question 906

[1 marks]Vectors in 3D
Find OA⃗⋅OB⃗\vec{OA}\cdot\vec{OB} (the dot product) for OA⃗=(1,2,−3)\vec{OA}=(1,2,-3), OB⃗=(3,−2,5)\vec{OB}=(3,-2,5).

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Question 1001

[1 marks]Optimisation / stationary points
A rectangular block has base 3x3x m by 2x2x m, height hh m and volume 144144 m3^3. Express the total surface area AA in terms of xx.
  1. AA=6x2+240xA = 6x^2 + \dfrac{240}{x}
  2. BA=12x2+240xA = 12x^2 + \dfrac{240}{x}
  3. CA=12x2+24x2A = 12x^2 + \dfrac{24}{x^2}
  4. DA=12x2+144xA = 12x^2 + \dfrac{144}{x}

Question 1002

[1 marks]Optimisation / stationary points
The surface area of a block is A=12x2+240xA = 12x^2 + \dfrac{240}{x}. Find the stationary value of xx and the nature of the stationary point.
  1. Ax=103x = \sqrt[3]{10}, a maximum
  2. Bx=203x = \sqrt[3]{20}, a minimum
  3. Cx=10x = 10, a minimum
  4. Dx=103x = \sqrt[3]{10}, a minimum

Question 1003

[2 marks]Optimisation / stationary points
Given 6x2h=1446x^2h=144, express the height hh in terms of xx.

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Question 1004

[2 marks]Optimisation / stationary points
Find dAdx\dfrac{dA}{dx} for A=12x2+240xA=12x^2+\dfrac{240}{x}.

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Question 1005

[2 marks]Optimisation / stationary points
Find d2Adx2\dfrac{d^2A}{dx^2} for A=12x2+240xA=12x^2+\dfrac{240}{x}.

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Question 1101

[1 marks]Polynomials / factor theorem
Given that g(x)=3x4+bx3+cx2−7x−4g(x) = 3x^4 + bx^3 + cx^2 - 7x - 4 has factors (x+1)(x+1) and (x−1)(x-1), find bb and cc.
  1. Ab=1b = 1, c=7c = 7
  2. Bb=−7b = -7, c=1c = 1
  3. Cb=7b = 7, c=1c = 1
  4. Db=7b = 7, c=−1c = -1

Question 1102

[1 marks]Polynomials / factor theorem
Factorise g(x)=3x4+7x3+x2−7x−4g(x) = 3x^4 + 7x^3 + x^2 - 7x - 4 completely.
  1. A(x+1)(x−1)2(3x+4)(x+1)(x-1)^2(3x+4)
  2. B(x+1)2(x−1)(3x−4)(x+1)^2(x-1)(3x-4)
  3. C(x+1)(x−1)(3x2−7x+4)(x+1)(x-1)(3x^2 - 7x + 4)
  4. D(x+1)2(x−1)(3x+4)(x+1)^2(x-1)(3x+4)

Question 1103

[2 marks]Polynomials / factor theorem
Evaluate g(1)=3+b+c−7−4g(1)=3+b+c-7-4 using g(1)=0g(1)=0. State the resulting equation in the form b+c=kb+c=k: give kk.

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Question 1104

[2 marks]Polynomials / factor theorem
Evaluate g(−1)=3−b+c+7−4g(-1)=3-b+c+7-4 using g(−1)=0g(-1)=0. State the resulting equation in the form −b+c=k-b+c=k: give kk.

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Question 1105

[2 marks]Polynomials / factor theorem
Solving b+c=8b+c=8 and −b+c=−6-b+c=-6 simultaneously, find cc.

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Question 1201

[1 marks]Trigonometric identities and equations
Simplify cot⁡θ−cosec⁡2θ\cot\theta - \operatorname{cosec} 2\theta.
  1. Acot⁡θ\cot\theta
  2. Btan⁡2θ\tan 2\theta
  3. Ccot⁡2θ\cot 2\theta
  4. Dcosec⁡θ\operatorname{cosec}\theta

Question 1202

[1 marks]Trigonometric identities and equations
Solve cot⁡θ−cosec⁡2θ=32\cot\theta - \operatorname{cosec} 2\theta = \dfrac{\sqrt{3}}{2} for 0°≤θ≤360°0° \le \theta \le 360°.
  1. Aθ=24.55°, 114.55°, 204.55°, 294.55°\theta = 24.55°,\ 114.55°,\ 204.55°,\ 294.55°
  2. Bθ=24.55°, 204.55°\theta = 24.55°,\ 204.55°
  3. Cθ=49.1°, 229.1°\theta = 49.1°,\ 229.1°
  4. Dθ=30°, 120°, 210°, 300°\theta = 30°,\ 120°,\ 210°,\ 300°

Question 1203

[2 marks]Trigonometric identities and equations
Verify the identity cot⁡θ−cosec⁡2θ≡cot⁡2θ\cot\theta-\operatorname{cosec}2\theta\equiv\cot2\theta at θ=45°\theta=45°: evaluate cot⁡45°−cosec⁡90°\cot45°-\operatorname{cosec}90°.

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Question 1204

[2 marks]Trigonometric identities and equations
Continuing the check at θ=45°\theta=45°, evaluate cot⁡(2×45°)\cot(2\times45°).

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Question 1205

[2 marks]Trigonometric identities and equations
How many solutions does cot⁡θ−cosec⁡2θ=32\cot\theta-\operatorname{cosec}2\theta=\dfrac{\sqrt3}{2} have for 0°≤θ≤360°0°\le\theta\le360°?
  1. A1
  2. B2
  3. C4
  4. D8

Question 1206

[1 marks]Trigonometric identities and equations
State the period, in degrees, of cot⁡2θ\cot2\theta.

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Question 1301

[1 marks]Areas and volumes of revolution
The region SS is bounded by the curve y=x+1y = \sqrt{x+1}, the line y=x−1y = x - 1 and the xx-axis. Find the exact area of SS.
  1. A163\tfrac{16}{3}
  2. B103\tfrac{10}{3}
  3. C43\tfrac{4}{3}
  4. D223\tfrac{22}{3}

Question 1302

[1 marks]Areas and volumes of revolution
The region bounded by y=x+1y = \sqrt{x+1}, the line y=x−1y = x - 1 and the xx-axis is rotated completely about the xx-axis. Find the exact volume generated.
  1. A8π8\pi
  2. B8π3\tfrac{8\pi}{3}
  3. C10π3\tfrac{10\pi}{3}
  4. D16π3\tfrac{16\pi}{3}

Question 1303

[2 marks]Areas and volumes of revolution
The curve y=x+1y=\sqrt{x+1} meets the line y=x−1y=x-1 where x+1=(x−1)2x+1=(x-1)^2. Solve for the positive value of xx where they meet.

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Question 1304

[2 marks]Areas and volumes of revolution
Find the exact value of ∫−13x+1 dx\displaystyle\int_{-1}^{3}\sqrt{x+1}\,dx (before subtracting the triangle).

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Question 1305

[2 marks]Areas and volumes of revolution
The triangle subtracted from the area under the curve has base 2 and height 2 (between x=1x=1 and x=3x=3 along y=x−1y=x-1). Find its area.

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Question 1306

[2 marks]Areas and volumes of revolution
For the volume, find π∫−13(x+1) dx\pi\displaystyle\int_{-1}^{3}(x+1)\,dx (before subtracting the cone).

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Question 1401

[1 marks]Indices / modulus inequalities
Solve the equation 21+2x−9(2x)=−42^{1+2x} - 9(2^x) = -4.
  1. Ax=−1x = -1 or x=2x = 2
  2. Bx=1x = 1 or x=4x = 4
  3. Cx=12x = \tfrac{1}{2} or x=4x = 4
  4. Dx=−2x = -2 or x=1x = 1

Question 1402

[1 marks]Indices / modulus inequalities
Solve the inequality ∣2x−3∣<x2|2x - 3| < x^2.
  1. Ax>1x > 1 only
  2. B−1<x<3-1 < x < 3
  3. C−3<x<1-3 < x < 1
  4. Dx<−3x < -3 or x>1x > 1

Question 1403

[2 marks]Indices / modulus inequalities
Letting u=2xu=2^x, the equation 2(2x)2−9(2x)+4=02(2^x)^2-9(2^x)+4=0 becomes 2u2−9u+4=02u^2-9u+4=0. Solve for the larger value of uu.

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Question 1404

[2 marks]Indices / modulus inequalities
Solving x2+2x−3=0x^2+2x-3=0 (from x2=3−2xx^2=3-2x) gives two values of xx. State the larger one.

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Question 1405

[2 marks]Indices / modulus inequalities
Why is the branch x2=2x−3x^2=2x-3 (from the other case of ∣2x−3∣<x2|2x-3|<x^2) not used to find critical values?
  1. Ait duplicates the other branch
  2. Bit only applies for x<0x<0
  3. Cit has no real solutions
  4. Dit gives negative xx only

Question 1406

[2 marks]Indices / modulus inequalities
Confirm x2−2x+3=0x^2-2x+3=0 has no real roots by evaluating its discriminant b2−4acb^2-4ac.

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Question 1501

[1 marks]Integration by parts / trapezium rule
Find the exact value of ∫012x2ex dx\displaystyle\int_0^1 2x^2 e^x\,dx.
  1. A2e−42e - 4
  2. B2e−22e - 2
  3. C2e2e
  4. D4−2e4 - 2e

Question 1502

[1 marks]Integration by parts / trapezium rule
Use the trapezium rule with 6 ordinates to evaluate ∫012x2ex dx\displaystyle\int_0^1 2x^2 e^x\,dx, correct to 4 decimal places.
  1. A1.38001.3800
  2. B1.43661.4366
  3. C1.49081.4908
  4. D1.54361.5436

Question 1503

[1 marks]Integration by parts / trapezium rule
The trapezium rule gives 1.49081.4908 as an estimate of ∫012x2ex dx=2e−4\displaystyle\int_0^1 2x^2 e^x\,dx = 2e - 4. Find the percentage error correct to 3 decimal places.
  1. A0.378%0.378\%
  2. B37.760%37.760\%
  3. C5.424%5.424\%
  4. D3.776%3.776\%

Question 1504

[2 marks]Integration by parts / trapezium rule
Using integration by parts twice, ∫2x2ex dx=2x2ex−4xex+4ex+C\int2x^2e^x\,dx=2x^2e^x-4xe^x+4e^x+C. Evaluate this antiderivative at x=1x=1 (omit CC), in terms of ee.

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Question 1505

[2 marks]Integration by parts / trapezium rule
For the trapezium rule with h=0.2h=0.2, find the ordinate y3=2(0.6)2e0.6y_3=2(0.6)^2e^{0.6}, to 5 decimal places.

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Question 1506

[1 marks]Integration by parts / trapezium rule
State the trapezium-rule strip width hh used for 6 ordinates over [0,1][0,1].

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Question 1507

[3 marks]Integration by parts / trapezium rule
Find the ordinate y4=2(0.8)2e0.8y_4=2(0.8)^2e^{0.8}, correct to 4 decimal places.

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Question 1601

[1 marks]Successive differentiation / Maclaurin series / differential equations
Given y=e2xsin⁡xy = e^{2x}\sin x with dydx=e2xcos⁡x+2e2xsin⁡x\dfrac{dy}{dx} = e^{2x}\cos x + 2e^{2x}\sin x, find d2ydx2\dfrac{d^2y}{dx^2}.
  1. Ae2x(2sin⁡x+11cos⁡x)e^{2x}(2\sin x + 11\cos x)
  2. Be2x(3sin⁡x−4cos⁡x)e^{2x}(3\sin x - 4\cos x)
  3. Ce2x(3sin⁡x+4cos⁡x)e^{2x}(3\sin x + 4\cos x)
  4. De2x(4sin⁡x+3cos⁡x)e^{2x}(4\sin x + 3\cos x)

Question 1602

[1 marks]Successive differentiation / Maclaurin series / differential equations
Obtain the Maclaurin series for y=e2xsin⁡xy = e^{2x}\sin x up to and including the term in x3x^3.
  1. Ax+2x2+116x3x + 2x^2 + \tfrac{11}{6}x^3
  2. Bx+4x2+11x3x + 4x^2 + 11x^3
  3. C1+x+2x2+116x31 + x + 2x^2 + \tfrac{11}{6}x^3
  4. Dx+2x2+113x3x + 2x^2 + \tfrac{11}{3}x^3

Question 1603

[1 marks]Successive differentiation / Maclaurin series / differential equations
A hot iron bar cools so that dθdt=−kθ\dfrac{d\theta}{dt} = -\dfrac{k}{\theta}. Given that its temperature falls from 80°80°C to 70°70°C in 20 minutes, find its temperature after a further 20 minutes.
  1. A58.3°58.3°C
  2. B60.0°60.0°C
  3. C65.0°65.0°C
  4. D50.0°50.0°C

Question 1604

[2 marks]Successive differentiation / Maclaurin series / differential equations
Find d3ydx3\dfrac{d^3y}{dx^3} for y=e2xsin⁡xy=e^{2x}\sin x, given d2ydx2=e2x(3sin⁡x+4cos⁡x)\dfrac{d^2y}{dx^2}=e^{2x}(3\sin x+4\cos x).

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Question 1605

[2 marks]Successive differentiation / Maclaurin series / differential equations
Which triple correctly gives y(0)y(0), y′(0)y'(0), y′′(0)y''(0) for y=e2xsin⁡xy=e^{2x}\sin x?
  1. A(0,1,11)(0,1,11)
  2. B(1,0,4)(1,0,4)
  3. C(0,1,4)(0,1,4)
  4. D(0,4,1)(0,4,1)

Question 1606

[1 marks]Successive differentiation / Maclaurin series / differential equations
State y′′′(0)y'''(0) for y=e2xsin⁡xy=e^{2x}\sin x (the coefficient basis for the Maclaurin x3x^3 term).

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Question 1607

[2 marks]Successive differentiation / Maclaurin series / differential equations
The rate of cooling is inversely proportional to temperature, with θ\theta decreasing over time. Which differential equation correctly captures this (constant k>0k>0)?
  1. Adθdt=kθ\dfrac{d\theta}{dt}=\dfrac{k}{\theta}
  2. Bdθdt=−kθ\dfrac{d\theta}{dt}=-k\theta
  3. Cdθdt=−kθ\dfrac{d\theta}{dt}=-\dfrac{k}{\theta}
  4. Dθdθdt=k\theta\dfrac{d\theta}{dt}=k

Question 1608

[2 marks]Successive differentiation / Maclaurin series / differential equations
Solving θ dθ=−k dt\theta\,d\theta=-k\,dt by integrating both sides gives θ2=−2kt+c\theta^2=-2kt+c. Using θ=80\theta=80 at t=0t=0, find cc.

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Question 1609

[2 marks]Successive differentiation / Maclaurin series / differential equations
Using θ=70\theta=70 at t=20t=20 and c=6400c=6400 in θ2=−2kt+c\theta^2=-2kt+c, find kk.

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