Danho
ZIMSEC A Level · J2009

Pure Mathematics Paper 1 June 2009

Questions
61
Total marks
92

Sit this paper online

Questions
61
Pass mark
37
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]algebra / factor theorem
Given that f(x)=x3+2x2−kx−1f(x) = x^3 + 2x^2 - kx - 1, find the value of kk for which (x+2)(x + 2) is a factor of f(x)f(x).
  1. Ak=1k = 1
  2. Bk=2k = 2
  3. Ck=12k = \frac{1}{2}
  4. Dk=−12k = -\frac{1}{2}

Question 102

[1 marks]algebra / factor theorem
With k=12k=\frac12, so that f(x)=x3+2x2−12x−1f(x) = x^3 + 2x^2 - \frac12 x - 1 and (x+2)(x+2) is a factor, which of the following is the complete factorisation of f(x)f(x)?
  1. A(x+2)(x−12)(x+12)(x+2)\left(x-\frac{1}{\sqrt2}\right)\left(x+\frac{1}{\sqrt2}\right)
  2. B(x+2)(x−12)(x+12)(x+2)\left(x-\frac12\right)\left(x+\frac12\right)
  3. C(x+2)(x2+12)(x+2)\left(x^2+\frac12\right)
  4. D(x−2)(x−12)(x+12)(x-2)\left(x-\frac{1}{\sqrt2}\right)\left(x+\frac{1}{\sqrt2}\right)

Question 103

[3 marks]algebra / factor theorem
Dividing f(x)=x3+2x2−12x−1f(x)=x^3+2x^2-\frac12x-1 by (x+2)(x+2) gives a quadratic quotient x2+bx+cx^2+bx+c. State bb and cc as 'b, c'.

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Question 201

[1 marks]coordinate geometry / simultaneous equations
The line y=2x+1y = 2x + 1 is substituted into the curve y2−xy=3y^2 - xy = 3. Which quadratic equation in xx results?
  1. A4x2+3x−2=04x^2+3x-2=0
  2. B2x2+3x−2=02x^2+3x-2=0
  3. C2x2+5x−2=02x^2+5x-2=0
  4. D2x2+3x+1=02x^2+3x+1=0

Question 202

[1 marks]coordinate geometry / simultaneous equations
The line y=2x+1y = 2x + 1 intersects the curve y2−xy=3y^2 - xy = 3 at points AA and BB. Find the coordinates of the mid-point of ABAB.
  1. A(−34,−12)\left(-\frac34,-\frac12\right)
  2. B(34,12)\left(\frac34,\frac12\right)
  3. C(−34,12)\left(-\frac34,\frac12\right)
  4. D(−12,−34)\left(-\frac12,-\frac34\right)

Question 203

[3 marks]coordinate geometry / simultaneous equations
Solving 2x2+3x−2=02x^2+3x-2=0 by factorising as (2x−1)(x+2)=0(2x-1)(x+2)=0, state both xx-values as 'x1, x2'.

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Question 301

[1 marks]functions / inverse functions
The function ff is defined by f:x↦x2−4xf: x \mapsto x^2 - 4x, x∈Rx \in \mathbb{R}, ∣x∣≤1|x| \leq 1. Which statement correctly explains why ff is one-one on this domain?
  1. AThe vertex of y=x2−4xy=x^2-4x is at x=−2x=-2, which lies outside [−1,1][-1,1]; since [−1,1][-1,1] lies to the right of this vertex, ff is strictly increasing there, so no two xx-values share an image.
  2. BThe vertex of y=x2−4xy=x^2-4x is at x=0x=0, inside [−1,1][-1,1], but the two branches on either side never take equal values within this domain.
  3. CThe vertex of y=x2−4xy=x^2-4x is at x=2x=2, which lies outside [−1,1][-1,1]; since [−1,1][-1,1] lies to the left of the vertex, ff is strictly decreasing there, so no two xx-values share an image.
  4. DThe vertex of y=x2−4xy=x^2-4x is at x=2x=2, and since [−1,1][-1,1] lies to the right of the vertex, ff is strictly increasing there, so no two xx-values share an image.

Question 302

[1 marks]functions / inverse functions
For f:x↦x2−4xf: x \mapsto x^2 - 4x with domain −1≤x≤1-1 \le x \le 1, find an expression for f−1(x)f^{-1}(x).
  1. Af−1(x)=x+4−2f^{-1}(x)=\sqrt{x+4}-2
  2. Bf−1(x)=2−x−4f^{-1}(x)=2-\sqrt{x-4}
  3. Cf−1(x)=2+x+4f^{-1}(x)=2+\sqrt{x+4}
  4. Df−1(x)=2−x+4f^{-1}(x)=2-\sqrt{x+4}

Question 303

[3 marks]functions / inverse functions
Complete the square: write f(x)=x2−4xf(x)=x^2-4x as (x−p)2+q(x-p)^2+q. State pp and qq as 'p, q'.

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Question 401

[1 marks]calculus / parametric differentiation
Given that y=ln⁡(t+2)y = \ln(t + 2) and x=t2x = t^2 for t>−2t > -2, find dydx\dfrac{dy}{dx} in terms of tt.
  1. A2t(t+2)2t(t+2)
  2. B2tt+2\dfrac{2t}{t+2}
  3. C12t(t+2)\dfrac{1}{2t(t+2)}
  4. D12(t+2)\dfrac{1}{2(t+2)}

Question 402

[1 marks]calculus / parametric differentiation
Given that y=ln⁡(t+2)y = \ln(t + 2) and x=t2x = t^2 for t>−2t > -2, for which values of tt does yy increase as xx increases?
  1. A−2<t<0-2<t<0
  2. Bt>0t>0
  3. Ct<−2t<-2
  4. Dt>−2t>-2

Question 403

[2 marks]calculus / parametric differentiation
For y=ln⁡(t+2)y=\ln(t+2), find dydt\dfrac{dy}{dt} in terms of tt.

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Question 404

[2 marks]calculus / parametric differentiation
For x=t2x=t^2, find dxdt\dfrac{dx}{dt} in terms of tt.

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Question 501

[1 marks]exponential equations / surds
Solve the inequality (0.2)2x−3>106(0.2)^{2x-3} > 10^6.
  1. Ax>−2.79x>-2.79
  2. Bx<2.79x<2.79
  3. Cx<−2.79x<-2.79
  4. Dx<−1.5x<-1.5

Question 502

[1 marks]exponential equations / surds
Solve the equation 3x−11=4x3\sqrt{x} - 11 = \dfrac{4}{\sqrt{x}} for x>0x>0.
  1. Ax=8x=8
  2. Bx=4x=4
  3. Cx=16x=16
  4. Dx=121x=121

Question 503

[2 marks]exponential equations / surds
Since 0.2=15=5−10.2=\dfrac15=5^{-1}, rewrite the left side of (0.2)2x−3>106(0.2)^{2x-3}>10^6 as 5k5^{k} for some expression kk in terms of xx. State kk.

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Question 504

[2 marks]exponential equations / surds
Evaluate log⁡510\log_5 10, correct to 4 decimal places (needed to solve (0.2)2x−3>106(0.2)^{2x-3}>10^6 using base 5).

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Question 505

[1 marks]exponential equations / surds
Solving 3x−11=4x3\sqrt{x}-11=\dfrac{4}{\sqrt{x}}: with u=xu=\sqrt{x}, this leads to 3u2−11u−4=03u^2-11u-4=0, which factorises as (3u+1)(u−c)=0(3u+1)(u-c)=0. State cc.

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Question 601

[1 marks]integration / inequalities
Find ∫xln⁡(2x) dx\int x\ln(2x)\,dx.
  1. Ax22ln⁡(2x)+x24+C\dfrac{x^2}{2}\ln(2x)+\dfrac{x^2}{4}+C
  2. Bx22ln⁡(2x)−x24+C\dfrac{x^2}{2}\ln(2x)-\dfrac{x^2}{4}+C
  3. Cx22ln⁡(2x)−x28+C\dfrac{x^2}{2}\ln(2x)-\dfrac{x^2}{8}+C
  4. Dx22ln⁡(2x)−x22+C\dfrac{x^2}{2}\ln(2x)-\dfrac{x^2}{2}+C

Question 602

[1 marks]integration / inequalities
Solve the inequality ∣4x−1∣≤∣2x+7∣|4x - 1| \leq |2x + 7|.
  1. A−4≤x≤1-4\le x\le1
  2. Bx≤−1 or x≥4x\le-1 \text{ or } x\ge4
  3. C−7≤x≤4-7\le x\le4
  4. D−1≤x≤4-1\le x\le4

Question 603

[2 marks]integration / inequalities
Integrating ∫xln⁡(2x) dx\int x\ln(2x)\,dx by parts with u=ln⁡(2x)u=\ln(2x), dv=x dxdv=x\,dx, state vv (the antiderivative of dvdv).

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Question 604

[1 marks]integration / inequalities
Using u=ln⁡(2x)u=\ln(2x), state dudx\dfrac{du}{dx}.

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Question 605

[2 marks]integration / inequalities
Squaring both sides of ∣4x−1∣≤∣2x+7∣|4x-1|\leq|2x+7| gives 16x2−8x+1≤4x2+28x+4916x^2-8x+1\leq4x^2+28x+49. Simplify this to the form x2−3x−c≤0x^2-3x-c\leq0. State cc.

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Question 701

[1 marks]complex numbers
The complex number p=3−5ip = 3 - 5i and q=4ipq = 4ip. Which of the following correctly states the relationship between ∣p∣|p| and ∣q∣|q|, and between arg⁡p\arg p and arg⁡q\arg q?
  1. A∣q∣=4∣p∣|q|=4|p| and arg⁡q=arg⁡p+π2\arg q=\arg p+\dfrac{\pi}{2}
  2. B∣q∣=4∣p∣|q|=4|p| and arg⁡q=arg⁡p−π2\arg q=\arg p-\dfrac{\pi}{2}
  3. C∣q∣=14∣p∣|q|=\dfrac14|p| and arg⁡q=arg⁡p+π2\arg q=\arg p+\dfrac{\pi}{2}
  4. D∣q∣=4∣p∣|q|=4|p| and arg⁡q=arg⁡p+π\arg q=\arg p+\pi

Question 702

[1 marks]complex numbers
Given p=3−5ip = 3 - 5i and q=4ipq = 4ip, find r=p+qr = p + q in the form a+bia + bi.
  1. Ar=17+7ir=17+7i
  2. Br=23−7ir=23-7i
  3. Cr=23+7ir=23+7i
  4. Dr=−17+7ir=-17+7i

Question 703

[1 marks]complex numbers
Points PP, QQ, RR in an Argand diagram represent p=3−5ip=3-5i, q=20+12iq=20+12i (where q=4ipq=4ip) and r=p+q=23+7ir=p+q=23+7i respectively, with OO the origin. What kind of quadrilateral is OPRQOPRQ, and what is its area?
  1. Arectangle, area =172=172
  2. Brhombus, area =136=136
  3. Csquare, area =136=136
  4. Drectangle, area =136=136

Question 704

[2 marks]complex numbers
Given p=3−5ip=3-5i and q=4ipq=4ip, compute qq in the form a+bia+bi.

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Question 705

[2 marks]complex numbers
Given pˉ=3+5i\bar p=3+5i (the conjugate of p=3−5ip=3-5i) and q=20+12iq=20+12i, compute pˉ q\bar p\,q in the form a+bia+bi (used to find the area of OPRQOPRQ).

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Question 801

[1 marks]calculus / differentiation of trig functions
Given that y=1sin⁡3x=csc⁡3xy = \dfrac{1}{\sin 3x} = \csc 3x, find dydx\dfrac{dy}{dx}.
  1. A−3csc⁡3xcot⁡3x-3\csc 3x\cot 3x
  2. B−3csc⁡23x-3\csc^2 3x
  3. C3csc⁡3xcot⁡3x3\csc 3x\cot 3x
  4. D−csc⁡3xcot⁡3x-\csc 3x\cot 3x

Question 802

[1 marks]calculus / differentiation of trig functions
Given that y=csc⁡3xy = \csc 3x, express d2ydx2+9y\dfrac{d^2y}{dx^2} + 9y in terms of yy.
  1. A00
  2. B18y318y^3
  3. C9y39y^3
  4. D18y3−18y18y^3-18y

Question 803

[2 marks]calculus / differentiation of trig functions
For y=csc⁡3xy=\csc3x, evaluate yy at x=π/18x=\pi/18 (i.e. csc⁡(3×π/18)=csc⁡(π/6)\csc(3\times\pi/18)=\csc(\pi/6)).

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Question 804

[2 marks]calculus / differentiation of trig functions
For dydx=−3cot⁡3xcsc⁡3x\dfrac{dy}{dx}=-3\cot3x\csc3x, evaluate dydx\dfrac{dy}{dx} at x=π/18x=\pi/18 (using cot⁡(π/6)=3\cot(\pi/6)=\sqrt3 and csc⁡(π/6)=2\csc(\pi/6)=2).

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Question 805

[2 marks]calculus / differentiation of trig functions
Given y=csc⁡3xy=\csc3x satisfies d2ydx2+9y=18y3\dfrac{d^2y}{dx^2}+9y=18y^3, evaluate d2ydx2\dfrac{d^2y}{dx^2} at x=π/18x=\pi/18, using y=2y=2 there.

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Question 901

[1 marks]trigonometry / sequences and series
Triangle ABCABC is right-angled at AA, with AC=2aAC=2a and angle ACB=θACB=\theta radians. A circular arc ADAD, with centre CC and radius 2a2a, divides the triangle into a sector PP (bounded by the arc, of angle θ\theta) and the remaining region QQ. Given that the area of PP is 5 times the area of QQ, which equation must θ\theta satisfy?
  1. A6θ=5θtan⁡θ6\theta=5\theta\tan\theta
  2. B5θ=6tan⁡θ5\theta=6\tan\theta
  3. C6θ=5tan⁡θ6\theta=5\tan\theta
  4. D6θ=5cot⁡θ6\theta=5\cot\theta

Question 902

[1 marks]trigonometry / sequences and series
The first two terms of a geometric progression are 4 and 6 respectively. Calculate the sum of the first 20 terms, giving your answer correct to the nearest integer.
  1. A1212
  2. B1772717727
  3. C2659426594
  4. D3989539895

Question 903

[1 marks]trigonometry / sequences and series
The first term of an arithmetic progression is 6 and the sum of the first 36 terms is 90. Find the common difference.
  1. Ad=15d=\frac15
  2. Bd=−16d=-\frac16
  3. Cd=−718d=-\frac{7}{18}
  4. Dd=−15d=-\frac15

Question 904

[2 marks]trigonometry / sequences and series
Triangle ABCABC is right-angled at AA with AC=2aAC=2a, angle ACB=θACB=\theta. State the area of triangle ABCABC in terms of aa and θ\theta (using AB=2atan⁡θAB=2a\tan\theta).

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Question 905

[2 marks]trigonometry / sequences and series
State the area of sector PP (radius 2a2a, angle θ\theta radians) cut from the triangle in the arc/sector problem.

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Question 906

[2 marks]trigonometry / sequences and series
A geometric progression has first two terms 4 and 6. State the common ratio rr as a fraction.

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Question 1001

[1 marks]differential equations / data analysis
Satellite pictures give the area AA km2^2 of a lake at times t=0,1,2,3,4t=0,1,2,3,4 weeks as 76,70,62,51,3176, 70, 62, 51, 31. What do the successive decreases in area (6,8,11,206, 8, 11, 20) show about the rate of decrease of the area?
  1. AThe rate of decrease is constant, since 6,8,116, 8, 11 and 2020 average to about 1111 km2^2 per week.
  2. BThe rate of decrease is not constant: it grows smaller each week, so the graph flattens out over time.
  3. CThe rate of decrease is constant at exactly 2020 km2^2 per week, matching the largest recorded drop.
  4. DThe rate of decrease is not constant: it grows larger each week, so the graph of AA against tt is a curve, not a straight line.

Question 1002

[1 marks]differential equations / data analysis
In a second lake, the rate of decrease of the area AA km2^2 is inversely proportional to the area at time tt weeks. Which differential equation correctly models this, where k>0k>0 is a constant?
  1. AdAdt=−kA\dfrac{dA}{dt}=-\dfrac{k}{A}
  2. BdAdt=−kA\dfrac{dA}{dt}=-kA
  3. CdAdt=−kA2\dfrac{dA}{dt}=-\dfrac{k}{A^2}
  4. DdAdt=kA\dfrac{dA}{dt}=\dfrac{k}{A}

Question 1003

[1 marks]differential equations / data analysis
Solving dAdt=−kA\dfrac{dA}{dt}=-\dfrac{k}{A} gives A2=−2kt+CA^2=-2kt+C. Given that A=8A=8 when t=1t=1 and A=6A=6 when t=5t=5, find the area when t=0t=0, correct to 2 decimal places.
  1. AA=8.37A=8.37
  2. BA=8.43A=8.43
  3. CA=8.00A=8.00
  4. DA=71A=71

Question 1004

[2 marks]differential equations / data analysis
Solving dAdt=−kA\dfrac{dA}{dt}=-\dfrac{k}{A} by separating variables gives A dA=−k dtA\,dA=-k\,dt. Integrate the left-hand side, ∫A dA\int A\,dA (omit the constant of integration).

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Question 1005

[2 marks]differential equations / data analysis
Using A2=−2kt+CA^2=-2kt+C with A=8A=8 at t=1t=1 and A=6A=6 at t=5t=5, forming two equations and subtracting to eliminate CC gives 28=nk28=nk. State nn.

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Question 1006

[2 marks]differential equations / data analysis
Using 28=8k28=8k, state kk as a decimal.

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Question 1007

[1 marks]differential equations / data analysis
Using k=3.5k=3.5 and 64=−2k+C64=-2k+C, state CC.

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Question 1101

[1 marks]vectors
Points AA, BB, CC have position vectors a=2i+j−ka=2i+j-k, b=3i+4j−2kb=3i+4j-2k and c=5i−j+2kc=5i-j+2k relative to the origin OO. Find the size of angle AB^CA\hat{B}C, giving your answer to the nearest 0.1°0.1°.
  1. A20.1°20.1°
  2. B40.2°40.2°
  3. C139.8°139.8°
  4. D49.8°49.8°

Question 1102

[1 marks]vectors
Points AA, BB, CC have position vectors a=2i+j−ka=2i+j-k, b=3i+4j−2kb=3i+4j-2k and c=5i−j+2kc=5i-j+2k. Given that ABCDABCD is a parallelogram, find the position vector of DD.
  1. A6i+2j+k6i+2j+k
  2. B0i+6j−5k0i+6j-5k
  3. C−4i+4j−3k-4i+4j-3k
  4. D4i−4j+3k4i-4j+3k

Question 1103

[1 marks]vectors
Points AA, BB have position vectors a=2i+j−ka=2i+j-k, b=3i+4j−2kb=3i+4j-2k, and D=4i−4j+3kD=4i-4j+3k is the fourth vertex of parallelogram ABCDABCD. Find the area of ABCDABCD, giving your answer in exact form.
  1. A206/2\sqrt{206}/2
  2. B206\sqrt{206}
  3. C206206
  4. D166\sqrt{166}

Question 1104

[2 marks]vectors
For A(2,1,−1)A(2,1,-1), B(3,4,−2)B(3,4,-2), C(5,−1,2)C(5,-1,2) (position vectors a,b,ca,b,c), compute a−ba-b as a vector, in the form 'x,y,z'.

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Question 1105

[2 marks]vectors
Compute c−bc-b as a vector, in the form 'x,y,z'.

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Question 1106

[2 marks]vectors
Evaluate the scalar product (a−b)⋅(c−b)(a-b)\cdot(c-b) using a−b=(−1,−3,1)a-b=(-1,-3,1) and c−b=(2,−5,4)c-b=(2,-5,4).

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Question 1107

[2 marks]vectors
State ∣a−b∣|a-b| as an exact surd, using a−b=(−1,−3,1)a-b=(-1,-3,1).

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Question 1201

[1 marks]logarithms / exponential growth
Express eln⁡x+ln⁡ye^{\ln x + \ln y} in its simplest form, for x,y>0x,y>0.
  1. Axyxy
  2. Bxyx^y
  3. Cx+yx+y
  4. Dln⁡(xy)\ln(xy)

Question 1202

[1 marks]logarithms / exponential growth
Express ln⁡e2x\ln e^{2x} in its simplest form.
  1. Ae2xe^{2x}
  2. B2ln⁡x2\ln x
  3. Cx2x^2
  4. D2x2x

Question 1203

[1 marks]logarithms / exponential growth
The number NN of bacteria in a culture at time tt hours is modelled by N=600ectN=600e^{ct}. Given that N=600N=600 when t=0t=0 and N=1800N=1800 when t=2t=2, find the exact value of cc.
  1. Ac=12ln⁡3c=\frac12\ln3
  2. Bc=ln⁡3c=\ln3
  3. Cc=12ln⁡1800c=\frac12\ln1800
  4. Dc=2ln⁡3c=2\ln3

Question 1204

[2 marks]logarithms / exponential growth
For N=600ectN=600e^{ct}, differentiate to find dNdt\dfrac{dN}{dt} in terms of tt and cc (before rewriting in terms of NN).

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Question 1205

[2 marks]logarithms / exponential growth
Using 600e2c=1800600e^{2c}=1800, divide both sides by 600 to isolate the exponential; state the resulting equation in the form e2c=me^{2c}=m. State mm.

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Question 1206

[2 marks]logarithms / exponential growth
From e2c=3e^{2c}=3, take natural logs of both sides; state the resulting expression for 2c2c.

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Question 1207

[2 marks]logarithms / exponential growth
Using N=600ectN=600e^{ct} with c=12ln⁡3c=\frac12\ln3, find NN at t=4t=4 hours (as an exact integer, using e2c=3e^{2c}=3).

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Question 1208

[1 marks]logarithms / exponential growth
For N=600ectN=600e^{ct}, state NN when t=0t=0 by direct substitution.

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The answers, and why they are the answers

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