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ZIMSEC A Level · J2019

Pure Mathematics Paper 1 June 2019

Questions
92
Total marks
120

Sit this paper online

Questions
92
Pass mark
56
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]Coordinate geometry
Find, in terms of kk, the equation of the perpendicular bisector of the line joining A(2k,k)A(2k, k) and B(4k,9k)B(4k, 9k).
  1. A4x+y=17k4x + y = 17k
  2. Bx−4y=−17kx - 4y = -17k
  3. Cx+4y=23kx + 4y = 23k
  4. D4x−y=7k4x - y = 7k

Question 102

[1 marks]Coordinate geometry
Find the midpoint of A(2k,k)A(2k,k) and B(4k,9k)B(4k,9k), in terms of kk.

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Question 103

[1 marks]Coordinate geometry
Find the gradient of the line ABAB joining A(2k,k)A(2k,k) and B(4k,9k)B(4k,9k).

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Question 201

[1 marks]Binomial expansion
Find the first three terms, in ascending powers of tt, of the expansion of (1−2t)−4(1-2t)^{-4}.
  1. A1+8t+40t21 + 8t + 40t^2
  2. B1+8t+20t21 + 8t + 20t^2
  3. C1+4t+10t21 + 4t + 10t^2
  4. D1−8t+40t21 - 8t + 40t^2

Question 202

[1 marks]Binomial expansion
Find the coefficient of tt in the expansion of (1−2t)−4(1-2t)^{-4}.

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Question 203

[1 marks]Binomial expansion
Find the coefficient of t2t^2 in the expansion of (1−2t)−4(1-2t)^{-4}.

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Question 301

[1 marks]Modulus inequalities
Solve the inequality ∣x−1∣<2x−4|x-1| < 2x - 4.
  1. Ax>2x > 2
  2. Bx>3x > 3
  3. Cx<3x < 3
  4. D53<x<3\tfrac{5}{3} < x < 3

Question 302

[1 marks]Modulus inequalities
Solve the boundary equation x−1=2x−4x-1=2x-4 for xx, the point where the two sides of ∣x−1∣<2x−4|x-1|<2x-4 are equal.

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Question 303

[1 marks]Modulus inequalities
Solving ∣x−1∣<2x−4|x-1| < 2x-4 for the case x<1x<1 leads to 1−x<2x−41-x<2x-4. Solve this inequality for xx, stating the boundary value.

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Question 304

[1 marks]Modulus inequalities
Since ∣x−1∣≥0|x-1|\ge0, the right-hand side 2x−42x-4 must also be positive. Solve 2x−4=02x-4=0 to find this boundary value of xx.

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Question 401

[1 marks]Factorisation of polynomials
Factorise completely 4+5x−7x2−2x34 + 5x - 7x^2 - 2x^3.
  1. A(1−x)(2x+1)(x+4)(1-x)(2x+1)(x+4)
  2. B(x−1)(2x+1)(x+4)(x-1)(2x+1)(x+4)
  3. C(1−x)(2x−1)(x+4)(1-x)(2x-1)(x+4)
  4. D(1−x)(2x+1)(x−4)(1-x)(2x+1)(x-4)

Question 402

[1 marks]Factorisation of polynomials
For f(x)=4+5x−7x2−2x3f(x)=4+5x-7x^2-2x^3, evaluate f(1)f(1) to confirm (x−1)(x-1) is a factor.

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Question 403

[2 marks]Factorisation of polynomials
Divide 4+5x−7x2−2x34+5x-7x^2-2x^3 by (x−1)(x-1) to find the quotient, in the form ax2+bx+cax^2+bx+c.

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Question 501

[1 marks]Exponential equations
The equation 2(2−x)+2x+2=102(2^{-x}) + 2^{x+2} = 10 becomes which quadratic in y=2xy = 2^x?
  1. A2y2−10y+4=02y^2 - 10y + 4 = 0
  2. By2−10y+4=0y^2 - 10y + 4 = 0
  3. C4y2+10y−2=04y^2 + 10y - 2 = 0
  4. D4y2−10y+2=04y^2 - 10y + 2 = 0

Question 502

[1 marks]Exponential equations
Solve 2(2−x)+2x+2=102(2^{-x}) + 2^{x+2} = 10, giving your answers correct to 2 significant figures.
  1. Ax=1.2x = 1.2 or x=2.2x = 2.2
  2. Bx=1.2x = 1.2 or x=−2.2x = -2.2
  3. Cx=−1.2x = -1.2 or x=2.2x = 2.2
  4. Dx=0.6x = 0.6 or x=−1.1x = -1.1

Question 503

[2 marks]Exponential equations
Solve 4y2−10y+2=04y^2-10y+2=0 for yy using the quadratic formula. State both values, correct to 2 decimal places.

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Question 504

[1 marks]Exponential equations
State the discriminant of the quadratic 4y2−10y+2=04y^2-10y+2=0.

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Question 601

[1 marks]Differentiation (quotient rule)
Given y=sin⁡θ2−cos⁡θy = \dfrac{\sin\theta}{2-\cos\theta} for 0≤θ≤2π0 \le \theta \le 2\pi, find the values of θ\theta for which yy is stationary.
  1. Aθ=π3\theta = \tfrac{\pi}{3} and θ=5π3\theta = \tfrac{5\pi}{3}
  2. Bθ=2π3\theta = \tfrac{2\pi}{3} and θ=4π3\theta = \tfrac{4\pi}{3}
  3. Cθ=π6\theta = \tfrac{\pi}{6} and θ=11π6\theta = \tfrac{11\pi}{6}
  4. Dθ=0\theta = 0 and θ=π\theta = \pi

Question 602

[2 marks]Differentiation (quotient rule)
Given y=sin⁡θ2−cos⁡θy=\dfrac{\sin\theta}{2-\cos\theta}, use the quotient rule to find dydθ\dfrac{dy}{d\theta}, in terms of θ\theta.

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Question 603

[1 marks]Differentiation (quotient rule)
Setting the numerator of dydθ\dfrac{dy}{d\theta} to zero, solve 2cos⁡θ−1=02\cos\theta-1=0 for cos⁡θ\cos\theta.

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Question 604

[1 marks]Differentiation (quotient rule)
Evaluate (2−cos⁡θ)2(2-\cos\theta)^2 when cos⁡θ=12\cos\theta=\tfrac12, the denominator of dydθ\dfrac{dy}{d\theta} at the stationary point.

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Question 701

[1 marks]Functions and inverses
Given f:x↦exf: x \mapsto e^x and h:x↦x+2h: x \mapsto x + 2, describe the graph of fh(x)fh(x).
  1. Ay=exy = e^x translated 2 units in the negative xx-direction
  2. By=exy = e^x translated 2 units in the positive xx-direction
  3. Cy=exy = e^x translated 2 units upwards
  4. Dy=exy = e^x stretched vertically by factor 2

Question 702

[1 marks]Functions and inverses
The function g:x↦x2−2xg: x \mapsto x^2 - 2x is defined for 0≤x≤10 \le x \le 1. Find g−1(x)g^{-1}(x).
  1. Ag−1(x)=x+1−1g^{-1}(x) = \sqrt{x+1} - 1
  2. Bg−1(x)=1−x+1g^{-1}(x) = 1 - \sqrt{x+1}
  3. Cg−1(x)=1+x+1g^{-1}(x) = 1 + \sqrt{x+1}
  4. Dg−1(x)=x+1g^{-1}(x) = \sqrt{x} + 1

Question 703

[1 marks]Functions and inverses
Write g(x)=x2−2xg(x)=x^2-2x in completed-square form a(x−h)2+ka(x-h)^2+k.

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Question 704

[1 marks]Functions and inverses
The inverse g−1(x)=1−x+1g^{-1}(x)=1-\sqrt{x+1} has domain x≥x\ge what value?

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Question 705

[1 marks]Functions and inverses
Given f:x↦exf:x\mapsto e^x and h:x↦x+2h:x\mapsto x+2, find fh(x)fh(x) as a single expression.

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Question 706

[1 marks]Functions and inverses
Find the value of fh(x)=ex+2fh(x)=e^{x+2} at x=−2x=-2.

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Question 801

[1 marks]Arithmetic and geometric progressions
Given that sin⁡2x\sin^2 x, cos⁡2x\cos^2 x and 5cos⁡2x−35\cos^2 x - 3 are consecutive terms of an arithmetic progression, find cos⁡2x\cos^2 x.
  1. A12\tfrac{1}{2}
  2. B00
  3. C34\tfrac{3}{4}
  4. D11

Question 802

[1 marks]Arithmetic and geometric progressions
A geometric progression has sum to infinity 500 and common ratio 0.80.8. Find the first term.
  1. A8080
  2. B100100
  3. C400400
  4. D625625

Question 803

[1 marks]Arithmetic and geometric progressions
A geometric progression has first term 100 and common ratio 0.80.8. Find the least number of terms whose sum exceeds 499.
  1. A2727
  2. B2828
  3. C2929
  4. D3030

Question 804

[1 marks]Arithmetic and geometric progressions
Given cos⁡2x=1\cos^2x=1, evaluate the third AP term 5cos⁡2x−35\cos^2x-3.

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Question 805

[1 marks]Arithmetic and geometric progressions
Given cos⁡2x=1\cos^2x=1 (so sin⁡2x=0\sin^2x=0), find the common difference dd of the arithmetic progression sin⁡2x, cos⁡2x, 5cos⁡2x−3\sin^2x,\ \cos^2x,\ 5\cos^2x-3.

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Question 806

[1 marks]Arithmetic and geometric progressions
State the formula for the sum to infinity S∞S_\infty of a geometric progression with first term aa and common ratio rr (where ∣r∣<1|r|<1).

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Question 807

[2 marks]Arithmetic and geometric progressions
For the geometric progression with first term 100 and common ratio 0,8, find the sum of the first 10 terms, correct to the nearest whole number.

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Question 901

[1 marks]Integration by parts and the trapezium rule
Evaluate ∫14xln⁡x dx\displaystyle\int_1^4 x\ln x\,dx.
  1. A8ln⁡4−48\ln 4 - 4
  2. B4ln⁡4−1544\ln 4 - \tfrac{15}{4}
  3. C8ln⁡4−1548\ln 4 - \tfrac{15}{4}
  4. D8ln⁡4+1548\ln 4 + \tfrac{15}{4}

Question 902

[1 marks]Integration by parts and the trapezium rule
Use the trapezium rule with 4 ordinates to approximate ∫14xln⁡x dx\displaystyle\int_1^4 x\ln x\,dx, correct to 2 significant figures.
  1. A6.36.3
  2. B7.37.3
  3. C7.57.5
  4. D8.38.3

Question 903

[2 marks]Integration by parts and the trapezium rule
Integrating by parts with u=ln⁡xu=\ln x, v′=xv'=x, find the antiderivative of xln⁡xx\ln x (ignore the constant of integration).

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Question 904

[2 marks]Integration by parts and the trapezium rule
For the trapezium rule with h=1h=1 over x=1,2,3,4x=1,2,3,4 applied to xln⁡xx\ln x, find y1+y2y_1+y_2 (the sum of the two middle ordinates 2ln⁡2+3ln⁡32\ln2+3\ln3), correct to 4 decimal places.

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Question 905

[1 marks]Integration by parts and the trapezium rule
State the ordinate y4=4ln⁡4y_4=4\ln4 at x=4x=4 for the trapezium rule applied to xln⁡xx\ln x, correct to 4 decimal places.

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Question 906

[2 marks]Integration by parts and the trapezium rule
The trapezium rule gives ≈7.4547\approx7.4547 for ∫14xln⁡x dx\int_1^4x\ln x\,dx, whose exact value is 8ln⁡4−154≈7.34048\ln4-\tfrac{15}{4}\approx7.3404. Find the relative error, correct to 3 decimal places.

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Question 1001

[1 marks]Partial fractions and integration
Express f(x)=x3−x−2(x−1)(x2+1)f(x) = \dfrac{x^3 - x - 2}{(x-1)(x^2+1)} in the form A+Bx−1+Cx+Dx2+1A + \dfrac{B}{x-1} + \dfrac{Cx+D}{x^2+1}.
  1. AA=0A = 0, B=−1B = -1, C=2C = 2, D=1D = 1
  2. BA=1A = 1, B=−1B = -1, C=2C = 2, D=0D = 0
  3. CA=1A = 1, B=1B = 1, C=2C = 2, D=0D = 0
  4. DA=1A = 1, B=−1B = -1, C=−2C = -2, D=0D = 0

Question 1002

[1 marks]Partial fractions and integration
Evaluate ∫23(1−1x−1+2xx2+1)dx\displaystyle\int_2^3 \left(1 - \dfrac{1}{x-1} + \dfrac{2x}{x^2+1}\right)dx.
  1. A11
  2. B1−ln⁡21 - \ln 2
  3. Cln⁡105\ln\tfrac{10}{5}
  4. D1+ln⁡21 + \ln 2

Question 1003

[2 marks]Partial fractions and integration
Find the indefinite integral of 1−1x−1+2xx2+11-\dfrac{1}{x-1}+\dfrac{2x}{x^2+1} with respect to xx (ignore the constant of integration).

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Question 1004

[2 marks]Partial fractions and integration
Evaluate the antiderivative x−ln⁡(x−1)+ln⁡(x2+1)x-\ln(x-1)+\ln(x^2+1) at x=3x=3, correct to 4 decimal places.

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Question 1005

[2 marks]Partial fractions and integration
Evaluate the antiderivative x−ln⁡(x−1)+ln⁡(x2+1)x-\ln(x-1)+\ln(x^2+1) at x=2x=2, correct to 4 decimal places.

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Question 1006

[1 marks]Partial fractions and integration
State the value of ln⁡10\ln10 (used when evaluating the antiderivative at x=3x=3), correct to 4 decimal places.

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Question 1101

[1 marks]Numerical methods (iteration)
The iterative formula xn+1=cos⁡−1 ⁣(13−xn2)x_{n+1} = \cos^{-1}\!\left(\dfrac{1}{3 - x_n^2}\right) converges to a root of which equation?
  1. Asec⁡x=3−x2\sec x = 3 - x^2
  2. Bsec⁡x=3+x2\sec x = 3 + x^2
  3. Ctan⁡x=3−x2\tan x = 3 - x^2
  4. Dcos⁡x=3−x2\cos x = 3 - x^2

Question 1102

[1 marks]Numerical methods (iteration)
Using x0=1.024627x_0 = 1.024627 in the iteration xn+1=cos⁡−1 ⁣(13−xn2)x_{n+1} = \cos^{-1}\!\left(\dfrac{1}{3 - x_n^2}\right), determine the root correct to 2 decimal places.
  1. A1.001.00
  2. B1.021.02
  3. C1.031.03
  4. D1.051.05

Question 1103

[1 marks]Numerical methods (iteration)
Starting from x=cos⁡−1 ⁣(13−x2)x=\cos^{-1}\!\left(\dfrac{1}{3-x^2}\right), take the cosine of both sides. State the resulting expression for cos⁡x\cos x, in terms of xx.

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Question 1104

[2 marks]Numerical methods (iteration)
Using the iteration xn+1=cos⁡−1 ⁣(13−xn2)x_{n+1}=\cos^{-1}\!\left(\dfrac{1}{3-x_n^2}\right) with x0=1,024627x_0=1{,}024627, find x1x_1, correct to 6 decimal places.

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Question 1105

[2 marks]Numerical methods (iteration)
Continuing the iteration xn+1=cos⁡−1 ⁣(13−xn2)x_{n+1}=\cos^{-1}\!\left(\dfrac{1}{3-x_n^2}\right) from x1=1,032372x_1=1{,}032372, find x2x_2, correct to 6 decimal places.

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Question 1106

[2 marks]Numerical methods (iteration)
Continuing the iteration xn+1=cos⁡−1 ⁣(13−xn2)x_{n+1}=\cos^{-1}\!\left(\dfrac{1}{3-x_n^2}\right) from x2=1,027445x_2=1{,}027445, find x3x_3, correct to 6 decimal places.

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Question 1201

[1 marks]Differential equations
The rate at which a quantity xx decreases is proportional to the product of xx and 1−x1 - x. Which differential equation models this?
  1. Adxdt=kx(1−x)\dfrac{dx}{dt} = kx(1-x)
  2. Bdxdt=−k(1−x)\dfrac{dx}{dt} = -k(1-x)
  3. Cdxdt=−kx(1−x)\dfrac{dx}{dt} = -\dfrac{k}{x(1-x)}
  4. Ddxdt=−kx(1−x)\dfrac{dx}{dt} = -kx(1-x)

Question 1202

[1 marks]Differential equations
Solve dxdt=−kx(1−x)\dfrac{dx}{dt} = -kx(1-x) given that x=0.2x = 0.2 initially and the rate of decrease is then 1.61.6 units per unit time.
  1. Ax=4e10t+1x = \dfrac{4}{e^{10t} + 1}
  2. Bx=1e10t+4x = \dfrac{1}{e^{10t} + 4}
  3. Cx=14e10t+1x = \dfrac{1}{4e^{10t} + 1}
  4. Dx=14e−10t+1x = \dfrac{1}{4e^{-10t} + 1}

Question 1203

[2 marks]Differential equations
Separate variables in dxdt=−kx(1−x)\dfrac{dx}{dt}=-kx(1-x) and integrate the left side ∫dxx(1−x)\int\dfrac{dx}{x(1-x)} using 1x(1−x)=1x+11−x\dfrac1{x(1-x)}=\dfrac1x+\dfrac1{1-x}. State the result (ignore the constant of integration).

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Question 1204

[2 marks]Differential equations
Given x=0,2x=0,2 at t=0t=0, find the constant of integration cc in ln⁡x1−x=−kt+c\ln\dfrac{x}{1-x}=-kt+c, correct to 4 decimal places.

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Question 1205

[2 marks]Differential equations
Given that x=0,2x=0,2 initially and the rate of decrease is then 1,61,6 units per unit time, use dxdt=−kx(1−x)\dfrac{dx}{dt}=-kx(1-x) to find kk.

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Question 1206

[1 marks]Differential equations
State the value of x(1−x)x(1-x) at x=0,2x=0,2, used to find kk from the initial rate.

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Question 1301

[1 marks]R-form, trigonometric equations and series expansion
Express 2sin⁡x−4cos⁡x2\sin x - 4\cos x in the form Rsin⁡(x−α)R\sin(x - \alpha), where R>0R > 0 and α\alpha is acute.
  1. A20sin⁡(x−63.4°)\sqrt{20}\sin(x - 63.4°)
  2. B6sin⁡(x−63.4°)6\sin(x - 63.4°)
  3. C25sin⁡(x+63.4°)2\sqrt{5}\sin(x + 63.4°)
  4. D25sin⁡(x−26.6°)2\sqrt{5}\sin(x - 26.6°)

Question 1302

[1 marks]R-form, trigonometric equations and series expansion
Solve 2sin⁡x−4cos⁡x=52\sin x - 4\cos x = \sqrt{5} for 0°≤x≤360°0° \le x \le 360°, giving answers to the nearest degree.
  1. Ax=63°x = 63° and x=243°x = 243°
  2. Bx=93°x = 93° and x=273°x = 273°
  3. Cx=30°x = 30° and x=150°x = 150°
  4. Dx=93°x = 93° and x=213°x = 213°

Question 1303

[1 marks]R-form, trigonometric equations and series expansion
In the series expansion of y=ea−xy = e^{a-x}, the coefficient of xx is −4-4. Find the exact value of aa.
  1. Aa=−ln⁡4a = -\ln 4
  2. Ba=ln⁡4a = \ln 4
  3. Ca=e4a = e^4
  4. Da=4a = 4

Question 1304

[1 marks]R-form, trigonometric equations and series expansion
For 2sin⁡x−4cos⁡x=Rsin⁡(x−α)2\sin x-4\cos x=R\sin(x-\alpha), state the value of tan⁡α\tan\alpha.

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Question 1305

[2 marks]R-form, trigonometric equations and series expansion
Given tan⁡α=2\tan\alpha=2, find α\alpha correct to 1 decimal place (degrees).

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Question 1306

[1 marks]R-form, trigonometric equations and series expansion
State the exact value of RR for 2sin⁡x−4cos⁡x=Rsin⁡(x−α)2\sin x-4\cos x=R\sin(x-\alpha).

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Question 1307

[2 marks]R-form, trigonometric equations and series expansion
Find the first three terms, in ascending powers of xx, of the series expansion of y=ea−x=eae−xy=e^{a-x}=e^ae^{-x}, in terms of eae^a.

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Question 1308

[1 marks]R-form, trigonometric equations and series expansion
The coefficient of xx in the series expansion of ea−xe^{a-x} is −ea-e^a. Given this coefficient is −4-4, find eae^a.

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Question 1309

[1 marks]R-form, trigonometric equations and series expansion
Given 25sin⁡(x−63,4∘)=52\sqrt5\sin(x-63{,}4^\circ)=\sqrt5, find sin⁡(x−63,4∘)\sin(x-63{,}4^\circ).

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Question 1401

[1 marks]Parametric differentiation and circular measure
A curve has parametric equations x=cos⁡2θx = \cos^2\theta and y=1−2sin⁡2θy = 1 - 2\sin^2\theta. Find dydx\dfrac{dy}{dx}.
  1. A−2-2
  2. B12\tfrac{1}{2}
  3. C2cos⁡2θ2\cos 2\theta
  4. D22

Question 1402

[1 marks]Parametric differentiation and circular measure
A sector AOBAOB of a circle of radius rr has arc ABAB of length 12r\tfrac{1}{2}r. Find angle AOBAOB in degrees, correct to the nearest degree.
  1. A14°14°
  2. B29°29°
  3. C30°30°
  4. D57°57°

Question 1403

[1 marks]Parametric differentiation and circular measure
In a sector of angle 1 radian, the segment cut off by the chord is removed. Express the segment area as a percentage of the remaining triangular area, correct to 2 significant figures.
  1. A27%27\%
  2. B16%16\%
  3. C19%19\%
  4. D84%84\%

Question 1404

[1 marks]Parametric differentiation and circular measure
For x=cos⁡2θx=\cos^2\theta, find dxdθ\dfrac{dx}{d\theta}.

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Question 1405

[1 marks]Parametric differentiation and circular measure
For y=1−2sin⁡2θy=1-2\sin^2\theta, find dydθ\dfrac{dy}{d\theta}.

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Question 1406

[2 marks]Parametric differentiation and circular measure
Using cos⁡3θ=cos⁡2θcos⁡θ−sin⁡2θsin⁡θ\cos3\theta=\cos2\theta\cos\theta-\sin2\theta\sin\theta, substitute cos⁡2θ=2cos⁡2θ−1\cos2\theta=2\cos^2\theta-1 and sin⁡2θ=2sin⁡θcos⁡θ\sin2\theta=2\sin\theta\cos\theta, and simplify to express cos⁡3θ\cos3\theta in terms of sin⁡θ\sin\theta and cos⁡θ\cos\theta only (as cos⁡θ\cos\theta minus a term).

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Question 1407

[1 marks]Parametric differentiation and circular measure
Arc ABAB of the sector has length 12r\tfrac12r. Using arc length =rθ=r\theta, find angle AOBAOB in radians.

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Question 1408

[2 marks]Parametric differentiation and circular measure
BB is the midpoint of arc ACAC, so the full sector angle AOC=2×0,5=1AOC=2\times0,5=1 radian. Find the area of triangle AOCAOC, in terms of rr (using 12r2sin⁡θ\tfrac12r^2\sin\theta).

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Question 1409

[1 marks]Parametric differentiation and circular measure
Find the area of the circular sector AOCAOC (angle 1 radian), in terms of rr (using 12r2θ\tfrac12r^2\theta).

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Question 1501

[1 marks]Areas and volumes of revolution
Find the xx-coordinate of the point of intersection PP of the curves y=e2xy = e^{2x} and y=e3−xy = e^{3-x}.
  1. Ax=3x = 3
  2. Bx=32x = \tfrac{3}{2}
  3. Cx=1x = 1
  4. Dx=ln⁡3x = \ln 3

Question 1502

[1 marks]Areas and volumes of revolution
Find the exact area of the region bounded by y=e2xy = e^{2x}, y=e3−xy = e^{3-x} and the yy-axis.
  1. Ae3+32e2−12e^3 + \tfrac{3}{2}e^2 - \tfrac{1}{2}
  2. B32e2−e3+12\tfrac{3}{2}e^2 - e^3 + \tfrac{1}{2}
  3. Ce3−e2+12e^3 - e^2 + \tfrac{1}{2}
  4. De3−32e2+12e^3 - \tfrac{3}{2}e^2 + \tfrac{1}{2}

Question 1503

[1 marks]Areas and volumes of revolution
The region bounded by y=e2xy = e^{2x}, y=e3−xy = e^{3-x} and the yy-axis is rotated through 360°360° about the xx-axis. Find the exact volume generated.
  1. A(e6−34e4+14)π\left(e^6 - \tfrac{3}{4}e^4 + \tfrac{1}{4}\right)\pi
  2. B(e62−34e4+14)π\left(\tfrac{e^6}{2} - \tfrac{3}{4}e^4 + \tfrac{1}{4}\right)\pi
  3. C(e62−14e4+34)π\left(\tfrac{e^6}{2} - \tfrac{1}{4}e^4 + \tfrac{3}{4}\right)\pi
  4. D(e62+34e4−14)π\left(\tfrac{e^6}{2} + \tfrac{3}{4}e^4 - \tfrac{1}{4}\right)\pi

Question 1504

[2 marks]Areas and volumes of revolution
Find the antiderivative of e3−x−e2xe^{3-x}-e^{2x} with respect to xx (ignore the constant of integration).

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Question 1505

[2 marks]Areas and volumes of revolution
Evaluate the antiderivative −e3−x−12e2x-e^{3-x}-\tfrac12e^{2x} at x=1x=1, in terms of ee.

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Question 1506

[2 marks]Areas and volumes of revolution
Evaluate the antiderivative −e3−x−12e2x-e^{3-x}-\tfrac12e^{2x} at x=0x=0, in terms of ee.

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Question 1507

[2 marks]Areas and volumes of revolution
For the volume of revolution V=π∫01(e6−2x−e4x) dxV=\pi\displaystyle\int_0^1(e^{6-2x}-e^{4x})\,dx, find the antiderivative of e6−2x−e4xe^{6-2x}-e^{4x} with respect to xx (ignore π\pi and the constant of integration).

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Question 1601

[1 marks]Vectors and complex numbers
A building has floor OABCOABC with OA=4OA = 4 m and OC=8OC = 8 m, and apex VV vertically 55 m above MM, the centre of the floor. Taking OO as origin, write down OB→\overrightarrow{OB} and OV→\overrightarrow{OV}.
  1. AOB→=4i+8j\overrightarrow{OB} = 4\mathbf{i} + 8\mathbf{j}, OV→=4i+8j+5k\overrightarrow{OV} = 4\mathbf{i} + 8\mathbf{j} + 5\mathbf{k}
  2. BOB→=8i+4j\overrightarrow{OB} = 8\mathbf{i} + 4\mathbf{j}, OV→=4i+2j+5k\overrightarrow{OV} = 4\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}
  3. COB→=4i+8j\overrightarrow{OB} = 4\mathbf{i} + 8\mathbf{j}, OV→=2i+4j+5k\overrightarrow{OV} = 2\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}
  4. DOB→=2i+4j\overrightarrow{OB} = 2\mathbf{i} + 4\mathbf{j}, OV→=2i+4j+5k\overrightarrow{OV} = 2\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}

Question 1602

[1 marks]Vectors and complex numbers
Given OV→=2i+4j+5k\overrightarrow{OV} = 2\mathbf{i} + 4\mathbf{j} + 5\mathbf{k} and OB→=4i+8j\overrightarrow{OB} = 4\mathbf{i} + 8\mathbf{j}, find the exact value of cos⁡θ\cos\theta, where θ\theta is the angle between them.
  1. A12\tfrac{1}{2}
  2. B34\tfrac{3}{4}
  3. C23\tfrac{2}{3}
  4. D4045\tfrac{40}{45}

Question 1603

[1 marks]Vectors and complex numbers
Express u=4−8iiu = \dfrac{4-8i}{i} in the form x+iyx + iy and state its modulus.
  1. Au=8+4iu = 8 + 4i, ∣u∣=45|u| = 4\sqrt{5}
  2. Bu=−8+4iu = -8 + 4i, ∣u∣=45|u| = 4\sqrt{5}
  3. Cu=−8−4iu = -8 - 4i, ∣u∣=80|u| = 80
  4. Du=−8−4iu = -8 - 4i, ∣u∣=45|u| = 4\sqrt{5}

Question 1604

[2 marks]Vectors and complex numbers
Find the argument of u=−8−4iu=-8-4i, correct to 2 decimal places (radians).

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Question 1605

[1 marks]Vectors and complex numbers
Find ∣OB→∣|\overrightarrow{OB}| for OB→=4i+8j\overrightarrow{OB}=4\mathbf{i}+8\mathbf{j}, in surd form.

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Question 1606

[1 marks]Vectors and complex numbers
Find ∣OV→∣|\overrightarrow{OV}| for OV→=2i+4j+5k\overrightarrow{OV}=2\mathbf{i}+4\mathbf{j}+5\mathbf{k}, in surd form.

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Question 1607

[1 marks]Vectors and complex numbers
Find OV→⋅OB→\overrightarrow{OV}\cdot\overrightarrow{OB} for OV→=2i+4j+5k\overrightarrow{OV}=2\mathbf{i}+4\mathbf{j}+5\mathbf{k} and OB→=4i+8j\overrightarrow{OB}=4\mathbf{i}+8\mathbf{j}.

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Question 1608

[3 marks]Vectors and complex numbers
To rationalise u=4−8iiu=\dfrac{4-8i}{i}, multiply numerator and denominator by ii. Simplify the numerator (4−8i)(i)(4-8i)(i), using i2=−1i^2=-1, in the form p+qip+qi.

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Question 1609

[2 marks]Vectors and complex numbers
Given u=−8−4iu=-8-4i, find u2u^2, in the form p+qip+qi.

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