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ZIMSEC A Level · N2007

Pure Mathematics Paper 1 November 2007

Questions
50
Total marks
75

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Questions
50
Pass mark
30
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 / polynomial identity
In the identity 2x4+6x3+7x2+15x+5≡(x2+3x+1)(ax2+bx+c)2x^4+6x^3+7x^2+15x+5 \equiv (x^2+3x+1)(ax^2+bx+c), what is the value of bb?
  1. Ab=6b=6
  2. Bb=2b=2
  3. Cb=−6b=-6
  4. Db=0b=0

Question 102

[1 marks]algebra / polynomial identity
In the identity 2x4+6x3+7x2+15x+5≡(x2+3x+1)(ax2+bx+c)2x^4+6x^3+7x^2+15x+5 \equiv (x^2+3x+1)(ax^2+bx+c), what are the values of aa, bb, cc respectively?
  1. Aa=1, b=3, c=1a=1,\ b=3,\ c=1
  2. Ba=2, b=6, c=5a=2,\ b=6,\ c=5
  3. Ca=2, b=0, c=5a=2,\ b=0,\ c=5
  4. Da=2, b=0, c=−5a=2,\ b=0,\ c=-5

Question 103

[1 marks]algebra / polynomial identity
Matching the coefficient of x2x^2 in 2x4+6x3+7x2+15x+5≡(x2+3x+1)(ax2+bx+c)2x^4+6x^3+7x^2+15x+5 \equiv (x^2+3x+1)(ax^2+bx+c) gives the equation c+3b+a=kc+3b+a=k. State kk.

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Question 201

[1 marks]differential equations
Solving x2dydx=yx^2\dfrac{dy}{dx}=y by separating variables gives ln⁡y=−1x+C\ln y = -\dfrac{1}{x} + C. Given y=e2y=e^2 when x=1x=1, what is the value of CC?
  1. AC=2C=2
  2. BC=3C=3
  3. CC=−3C=-3
  4. DC=1C=1

Question 202

[1 marks]differential equations
What is the particular solution yy, in terms of xx, of x2dydx=yx^2\dfrac{dy}{dx}=y given that y=e2y=e^2 when x=1x=1?
  1. Ay=e−1/x+3y=e^{-1/x}+3
  2. By=3e−1/xy=3e^{-1/x}
  3. Cy=e3−1/xy=e^{3-1/x}
  4. Dy=e3+1/xy=e^{3+1/x}

Question 203

[2 marks]differential equations
Separating variables in x2dydx=yx^2\dfrac{dy}{dx}=y gives dyy=dxx2\dfrac{dy}{y}=\dfrac{dx}{x^2}. Integrate the right-hand side, ∫dxx2\int \dfrac{dx}{x^2} (omit the constant of integration).

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Question 301

[1 marks]inequalities / absolute value
What is the solution set of ∣2x−3∣<x+1|2x-3| < x+1?
  1. A23<x<32\frac{2}{3} < x < \frac{3}{2}
  2. B23<x<4\frac{2}{3} < x < 4
  3. C32<x<4\frac{3}{2} < x < 4
  4. D−4<x<23-4 < x < \frac{2}{3}

Question 302

[1 marks]inequalities / absolute value
When solving ∣2x−3∣<x+1|2x-3|<x+1 for the branch x<32x<\frac32, the inequality becomes −(2x−3)<x+1-(2x-3)<x+1. At what value of xx do the two sides become equal (the boundary of this branch)?
  1. Ax=32x=\frac32
  2. Bx=4x=4
  3. Cx=23x=\frac23
  4. Dx=−23x=-\frac23

Question 303

[2 marks]inequalities / absolute value
For the branch x≥32x\geq\frac32 of ∣2x−3∣<x+1|2x-3|<x+1, solve 2x−3=x+12x-3=x+1 to find the boundary value of xx for this branch.

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Question 401

[1 marks]complex numbers
Given 3+2i2+ai=x+iy\dfrac{3+2i}{2+ai}=x+iy with x,yx,y real and x=yx=y, what is the value of aa?
  1. Aa=25a=\frac{2}{5}
  2. Ba=43a=\frac{4}{3}
  3. Ca=−23a=-\frac{2}{3}
  4. Da=−25a=-\frac{2}{5}

Question 402

[2 marks]complex numbers
Expanding (3+2i)(2−ai)(3+2i)(2-ai), state the real part of the result, in terms of aa.

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Question 403

[2 marks]complex numbers
Expanding (3+2i)(2−ai)(3+2i)(2-ai), state the coefficient of ii (the imaginary part), in terms of aa.

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Question 501

[1 marks]functions / inverse functions
For f:x↦x2+4x+1f:x\mapsto x^2+4x+1, x≥−2x\geq -2, what is the range of ff?
  1. Af(x)≥−2f(x)\geq -2
  2. Bf(x)≤−3f(x)\leq -3
  3. Cf(x)≥−3f(x)\geq -3
  4. Df(x)≥1f(x)\geq 1

Question 502

[1 marks]functions / inverse functions
For the same function f:x↦x2+4x+1f:x\mapsto x^2+4x+1, x≥−2x\geq -2, which correctly gives f−1(x)f^{-1}(x) together with its domain?
  1. Af−1(x)=x+3−2f^{-1}(x)=\sqrt{x+3}-2, domain x≥−3x\geq -3
  2. Bf−1(x)=x+3−2f^{-1}(x)=\sqrt{x+3}-2, domain x≥−2x\geq -2
  3. Cf−1(x)=x−3−2f^{-1}(x)=\sqrt{x-3}-2, domain x≥−3x\geq -3
  4. Df−1(x)=−x+3−2f^{-1}(x)=-\sqrt{x+3}-2, domain x≥−3x\geq -3

Question 503

[2 marks]functions / inverse functions
Write f(x)=x2+4x+1f(x)=x^2+4x+1 in completed-square form (x+p)2+q(x+p)^2+q. State pp and qq as 'p, q'.

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Question 504

[2 marks]functions / inverse functions
Given y=(x+2)2−3y=(x+2)^2-3, make x+2x+2 the subject (taking the positive square root since x≥−2x\geq-2). State the resulting expression for x+2x+2 in terms of yy.

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Question 601

[1 marks]calculus / volumes of revolution
The region bounded by x2=y−1x^2=y-1, x=0x=0, y=0y=0 and x=1x=1 is rotated completely about the xx-axis. Which integral gives the volume of the solid formed?
  1. Aπ∫01(x2−1)2 dx\pi\int_0^1 (x^2-1)^2\,dx
  2. Bπ∫01(x2+1) dx\pi\int_0^1 (x^2+1)\,dx
  3. Cπ∫01(x2+1)2 dx\pi\int_0^1 (x^2+1)^2\,dx
  4. D2π∫01x(x2+1) dx2\pi\int_0^1 x(x^2+1)\,dx

Question 602

[1 marks]calculus / volumes of revolution
What is the volume of the solid generated when the region bounded by x2=y−1x^2=y-1, x=0x=0, y=0y=0 and x=1x=1 is rotated completely about the xx-axis?
  1. A56π15\frac{56\pi}{15}
  2. B13π15\frac{13\pi}{15}
  3. C4π3\frac{4\pi}{3}
  4. D28π15\frac{28\pi}{15}

Question 603

[2 marks]calculus / volumes of revolution
Expand (x2+1)2(x^2+1)^2 as a polynomial in xx.

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Question 604

[2 marks]calculus / volumes of revolution
Evaluate ∫01(x4+2x2+1) dx\int_0^1 (x^4+2x^2+1)\,dx as a single fraction (before multiplying by π\pi).

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Question 701

[1 marks]calculus / small changes and errors
For a cylindrical wire, an error δx\delta x in the measured diameter produces an error δV\delta V in the calculated volume. Approximately how many times the relative error in the diameter is the relative error in the volume?

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Question 702

[1 marks]calculus / small changes and errors
A cylindrical wire of fixed length ll and diameter xx has volume V=πlx24V=\dfrac{\pi l x^2}{4}. Which of these correctly relates the error δV\delta V to an error δx\delta x in the diameter?
  1. AδV≈πlx2δx\delta V \approx \dfrac{\pi l x}{2}\delta x
  2. BδV≈πlx24δx\delta V \approx \dfrac{\pi l x^2}{4}\delta x
  3. CδV≈πlx4δx\delta V \approx \dfrac{\pi l x}{4}\delta x
  4. DδV≈πlx δx\delta V \approx \pi l x\,\delta x

Question 703

[2 marks]calculus / small changes and errors
For V=πlx24V=\dfrac{\pi l x^2}{4} (volume of a cylindrical wire in terms of diameter xx), find dVdx\dfrac{dV}{dx} in terms of ll and xx.

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Question 704

[2 marks]calculus / small changes and errors
Simplify the ratio πlx/2πlx2/4\dfrac{\pi l x/2}{\pi l x^2/4} to an expression in xx only.

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Question 1301

[1 marks]arithmetic series / sequences
Ten oranges O1,…,O10O_1,\ldots,O_{10} are placed in a line 6 m apart, with O1O_1 15 m from the starting line. A contestant starts at the line, runs to fetch each orange one at a time in order, returning it to a box at the start before fetching the next. What is the total distance covered to collect all 10 oranges?
  1. A540 m
  2. B840 m
  3. C420 m
  4. D960 m

Question 1302

[1 marks]arithmetic series / sequences
Using the same set-up (oranges 6 m apart, O1O_1 15 m from the start, each fetched and returned to the box before the next), what round-trip distance, in metres, does the contestant cover to fetch O5O_5 alone?

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Question 1303

[2 marks]arithmetic series / sequences
Ten oranges are placed 6 m apart with O1O_1 15 m from the start. The round-trip distance to collect orange OnO_n alone is 2(15+6(n−1))2(15+6(n-1)) m. Evaluate this for n=10n=10 (the last orange).

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Question 1401

[1 marks]differential equations / exponential decay
The mass mm of mealie-meal satisfies dmdt=−k5m\dfrac{dm}{dt}=-\dfrac{k}{5}m, with m=m0m=m_0 at t=0t=0. What is the general solution?
  1. Am=m0e−kt/5m=m_0e^{-kt/5}
  2. Bm=m0e−ktm=m_0e^{-kt}
  3. Cm=m0−k5tm=m_0-\dfrac{k}{5}t
  4. Dm=m0e−5ktm=m_0e^{-5kt}

Question 1402

[1 marks]differential equations / exponential decay
Modelling mealie-meal loss by m=m0e−kt/5m=m_0e^{-kt/5}, if 10% is lost in the first hour (so m=0.9m0m=0.9m_0 at t=1t=1), what percentage of the mealie-meal remains after the girl's full 2-hour journey?
  1. A72.9%
  2. B80%
  3. C81%
  4. D90%

Question 1403

[2 marks]differential equations / exponential decay
Separating variables in dmdt=−k5m\dfrac{dm}{dt}=-\dfrac{k}{5}m gives dmm=−k5dt\dfrac{dm}{m}=-\dfrac{k}{5}dt. Integrate the left-hand side, ∫dmm\int\dfrac{dm}{m} (omit the constant of integration).

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Question 1404

[2 marks]differential equations / exponential decay
Given that after t=1t=1 hour, 10% of the mealie-meal is lost (so m=0.9m0m=0.9m_0), and m=m0e−kt/5m=m_0e^{-kt/5}, state the value of e−k/5e^{-k/5}.

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Question 1405

[2 marks]differential equations / exponential decay
Given e−k/5=0.9e^{-k/5}=0.9 and that after 2 hours the mass is m0(e−k/5)2m_0(e^{-k/5})^2, what percentage of the mealie-meal is LOST during the full 2-hour journey (not the percentage remaining)?

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Question 1406

[2 marks]differential equations / exponential decay
Using ln⁡m=−k5t+C\ln m = -\dfrac{k}{5}t + C and m=m0m=m_0 at t=0t=0, find CC in terms of m0m_0.

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Question 1407

[2 marks]differential equations / exponential decay
Given e−k/5=0.9e^{-k/5}=0.9, taking natural logs gives −k5=ln⁡0.9-\dfrac{k}{5}=\ln0.9, so k=mln⁡0.9k=m\ln0.9 for some constant mm. State mm.

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Question 1501

[1 marks]logarithms / series expansion
Which expression is equal to (1−1n2)−1\left(1-\dfrac{1}{n^2}\right)^{-1}?
  1. An2n2−1\dfrac{n^2}{n^2-1}
  2. Bn2−1n2\dfrac{n^2-1}{n^2}
  3. Cn2n2+1\dfrac{n^2}{n^2+1}
  4. D1n2−1\dfrac{1}{n^2-1}

Question 1502

[1 marks]logarithms / series expansion
What are the first three terms of the series expansion of ln⁡(1−x)\ln(1-x) in ascending powers of xx?
  1. A−x−x22−x33-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}
  2. Bx−x22+x33x-\dfrac{x^2}{2}+\dfrac{x^3}{3}
  3. C−x−x22−x36-x-\dfrac{x^2}{2}-\dfrac{x^3}{6}
  4. D−x+x22−x33-x+\dfrac{x^2}{2}-\dfrac{x^3}{3}

Question 1503

[1 marks]logarithms / series expansion
Using ln⁡10=2.3025851\ln 10=2.3025851, ln⁡3=1.0986123\ln 3=1.0986123, and n=10n=10 in ln⁡(n2n2−1)≈1n2+12n4+13n6\ln\left(\dfrac{n^2}{n^2-1}\right)\approx \dfrac{1}{n^2}+\dfrac{1}{2n^4}+\dfrac{1}{3n^6} together with ln⁡(n2n2−1)=2ln⁡n−ln⁡(n+1)−ln⁡(n−1)\ln\left(\dfrac{n^2}{n^2-1}\right)=2\ln n-\ln(n+1)-\ln(n-1), what is ln⁡11\ln 11 to six decimal places?
  1. A2.397895
  2. B2.397946
  3. C2.407946
  4. D2.417996

Question 1504

[2 marks]logarithms / series expansion
Substituting x=1/n2x=1/n^2 into ln⁡(1−x)≈−x−x22−x33\ln(1-x)\approx -x-\dfrac{x^2}{2}-\dfrac{x^3}{3}, state the first (leading) term of the expansion of ln⁡(1−1n2)\ln\left(1-\dfrac1{n^2}\right), in terms of nn.

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Question 1505

[2 marks]logarithms / series expansion
The expansion of ln⁡(1−1n2)−1\ln\left(1-\frac1{n^2}\right)^{-1} is the negative of the expansion of ln⁡(1−1n2)\ln\left(1-\frac1{n^2}\right). State its leading (first) term, in terms of nn.

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Question 1506

[2 marks]logarithms / series expansion
Using ln⁡10=2.3025851\ln 10=2.3025851, evaluate 2ln⁡102\ln 10, correct to 7 decimal places.

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Question 1507

[2 marks]logarithms / series expansion
Using ln⁡3=1.0986123\ln 3=1.0986123, evaluate 2ln⁡32\ln 3 (needed since ln⁡9=2ln⁡3\ln 9 = 2\ln 3), correct to 7 decimal places.

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Question 1508

[1 marks]logarithms / series expansion
Using n=10n=10, evaluate ln⁡(10099)\ln\left(\dfrac{100}{99}\right) approximately using 1n2+12n4+13n6\dfrac1{n^2}+\dfrac1{2n^4}+\dfrac1{3n^6}, correct to 6 decimal places.

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Question 1601

[1 marks]coordinate geometry / circles
The circle through (−2,−4)(-2,-4), (3,1)(3,1) and (−2,0)(-2,0) has centre (1,−2)(1,-2). Using the chord joining (−2,−4)(-2,-4) and (−2,0)(-2,0), which is vertical, what is the equation of its perpendicular bisector (the locus on which the centre must lie)?
  1. Ay=−4y=-4
  2. By=−2y=-2
  3. Cx=1x=1
  4. Dx=−2x=-2

Question 1602

[1 marks]coordinate geometry / circles
The circle (x−1)2+(y+2)2=13(x-1)^2+(y+2)^2=13 has centre (1,−2)(1,-2). What is the equation, in the form ax+by+c=0ax+by+c=0, of the diameter through the point (−2,0)(-2,0)?
  1. A3x+2y+4=03x+2y+4=0
  2. B2x+3y−4=02x+3y-4=0
  3. C2x−3y+4=02x-3y+4=0
  4. D2x+3y+4=02x+3y+4=0

Question 1603

[1 marks]coordinate geometry / circles
What is the gradient of the tangent to the circle (x−1)2+(y+2)2=13(x-1)^2+(y+2)^2=13 at the point (3,1)(3,1)?
  1. A−32-\dfrac32
  2. B−23-\dfrac23
  3. C32\dfrac32
  4. D23\dfrac23

Question 1604

[2 marks]coordinate geometry / circles
The circle (x−1)2+(y+2)2=13(x-1)^2+(y+2)^2=13 has centre (1,−2)(1,-2) and passes through (3,1)(3,1). State the gradient of the radius from the centre to (3,1)(3,1).

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Question 1605

[2 marks]coordinate geometry / circles
The diameter of the circle (x−1)2+(y+2)2=13(x-1)^2+(y+2)^2=13 (centre (1,−2)(1,-2)) passes through the point (−2,0)(-2,0). State its gradient.

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Question 1606

[2 marks]coordinate geometry / circles
Find the distance between the two given points (−2,−4)(-2,-4) and (−2,0)(-2,0) (a chord of the circle (x−1)2+(y+2)2=13(x-1)^2+(y+2)^2=13).

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Question 1607

[2 marks]coordinate geometry / circles
State the gradient of the chord joining the two given points (−2,−4)(-2,-4) and (3,1)(3,1) on the circle (not a diameter).

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Question 1608

[2 marks]coordinate geometry / circles
State the midpoint of the chord joining (−2,−4)(-2,-4) and (3,1)(3,1), as 'x, y'.

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