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ZIMSEC A Level · N2014

Pure Mathematics Paper 1 November 2014

Questions
87
Total marks
120

Sit this paper online

Questions
87
Pass mark
53
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]surds / rationalisation
Express 63+5+14\dfrac{6}{3 + \sqrt{5} + \sqrt{14}} in the form a+bc+dea + b\sqrt{c} + d\sqrt{e}.
  1. A3+5−143 + \sqrt{5} - \sqrt{14}
  2. B1+355+15701 + \tfrac{3}{5}\sqrt{5} + \tfrac{1}{5}\sqrt{70}
  3. C1+355−15701 + \tfrac{3}{5}\sqrt{5} - \tfrac{1}{5}\sqrt{70}
  4. D1−355+15701 - \tfrac{3}{5}\sqrt{5} + \tfrac{1}{5}\sqrt{70}

Question 102

[2 marks]surds / rationalisation
To rationalise 63+5+14\dfrac{6}{3+\sqrt5+\sqrt{14}}, the first step multiplies top and bottom by 3+5−143+\sqrt5-\sqrt{14}. What does the new denominator simplify to?
  1. A−65-6\sqrt5
  2. B656\sqrt5
  3. C6146\sqrt{14}
  4. D14−6514-6\sqrt5

Question 201

[1 marks]rational inequalities
Solve the inequality 10−2xx−2>x+1\dfrac{10-2x}{x-2} > x + 1.
  1. Ax<−4x < -4 or x>3x > 3
  2. B−4<x<2-4 < x < 2 or x>3x > 3
  3. Cx<−4x < -4 or 2<x<32 < x < 3
  4. D2<x<32 < x < 3 only

Question 202

[2 marks]rational inequalities
Clearing denominators (multiplying both sides of 10−2xx−2>x+1\dfrac{10-2x}{x-2}>x+1 by (x−2)2(x-2)^2, or combining into one fraction) gives 10−2x−(x−2)(x+1)x−2>0\dfrac{10-2x-(x-2)(x+1)}{x-2}>0. Which single fraction does this simplify to?
  1. A−x2+x+12x−2<0\dfrac{-x^2+x+12}{x-2}<0
  2. Bx2+x−12x−2>0\dfrac{x^2+x-12}{x-2}>0
  3. C−x2+x+12x−2>0\dfrac{-x^2+x+12}{x-2}>0
  4. Dx2−x−12x−2>0\dfrac{x^2-x-12}{x-2}>0

Question 203

[1 marks]rational inequalities
Rewriting −x2+x+12x−2>0\dfrac{-x^2+x+12}{x-2}>0 as (x+4)(x−3)x−2<0\dfrac{(x+4)(x-3)}{x-2}<0, what are the three critical values of xx?
  1. Ax=4, −2, −3x=4,\,-2,\,-3
  2. Bx=−4, 3x=-4,\,3 only
  3. Cx=−4, 2, −3x=-4,\,2,\,-3
  4. Dx=−4, 2, 3x=-4,\,2,\,3

Question 301

[1 marks]small changes / percentage change
Given that F=kS2F = kS^2, where kk is a constant, find the percentage change in FF when SS increases by 5%5\%.
  1. Aan increase of 5%5\%
  2. Ban increase of 2.5%2.5\%
  3. Can increase of 25%25\%
  4. Dan increase of 10%10\%

Question 302

[1 marks]small changes / percentage change
Given F=kS2F=kS^2, what is dFdS\dfrac{dF}{dS}?
  1. A2kS2kS
  2. B2k2k
  3. CkSkS
  4. DkS2kS^2

Question 303

[2 marks]small changes / percentage change
Using δF≈dFdSδS\delta F \approx \dfrac{dF}{dS}\delta S with F=kS2F=kS^2, which expression gives the fractional change δFF\dfrac{\delta F}{F} in terms of δSS\dfrac{\delta S}{S}?
  1. AδFF=12δSS\dfrac{\delta F}{F}=\dfrac12\dfrac{\delta S}{S}
  2. BδFF=(δSS)2\dfrac{\delta F}{F}=\left(\dfrac{\delta S}{S}\right)^2
  3. CδFF=δSS\dfrac{\delta F}{F}=\dfrac{\delta S}{S}
  4. DδFF=2δSS\dfrac{\delta F}{F}=2\dfrac{\delta S}{S}

Question 401

[1 marks]trapezium rule
Use the trapezium rule with four trapezia of equal width to estimate ∫121xex/3 dx\displaystyle\int_1^2 \dfrac{1}{x}e^{x/3}\,dx, correct to 4 decimal places.
  1. A0.56520.5652
  2. B1.13041.1304
  3. C1.18301.1830
  4. D1.21361.2136

Question 402

[1 marks]trapezium rule
For ∫121xex/3 dx\displaystyle\int_1^2 \dfrac{1}{x}e^{x/3}\,dx, has the trapezium rule underestimated or overestimated the integral, and why?
  1. AUnderestimated, because the chords lie below the curve
  2. BOverestimated, because the chords lie above the convex curve
  3. CUnderestimated, because the function is decreasing
  4. DNeither: the rule is exact for this function

Question 403

[1 marks]trapezium rule
Using the trapezium rule with four trapezia to estimate ∫121xex/3dx\int_1^2 \frac1x e^{x/3}dx on [1,2][1,2], what strip width hh is used?

Answer this when you sit the paper.

Question 404

[2 marks]trapezium rule
For f(x)=1xex/3f(x)=\dfrac1x e^{x/3}, what is f(1.5)f(1.5), correct to 6 decimal places?

Answer this when you sit the paper.

Question 501

[1 marks]related rates / radian measure
PP is fixed on a circle of radius 5 cm centred at OO, and QQ moves round the circumference at 33 cm s−1^{-1}. If angle POQ=θPOQ = \theta radians, find dθdt\dfrac{d\theta}{dt}.
  1. A1515 rad s−1^{-1}
  2. B35\tfrac{3}{5} rad s−1^{-1}
  3. C33 rad s−1^{-1}
  4. D53\tfrac{5}{3} rad s−1^{-1}

Question 502

[1 marks]related rates / radian measure
A point QQ moves at 33 cm s−1^{-1} round a circle of radius 5 cm, so that dθdt=35\dfrac{d\theta}{dt} = \tfrac{3}{5} rad s−1^{-1}. Find the rate of increase of the area of sector POQPOQ.
  1. A1515 cm2^2 s−1^{-1}
  2. B252\tfrac{25}{2} cm2^2 s−1^{-1}
  3. C152\tfrac{15}{2} cm2^2 s−1^{-1}
  4. D35\tfrac{3}{5} cm2^2 s−1^{-1}

Question 503

[2 marks]related rates / radian measure
PP is fixed on a circle of radius 5 cm; QQ moves round the circumference at 3 cm s−1^{-1}, and arc PQ=x=5θPQ=x=5\theta. Applying the chain rule dxdt=dxdθ⋅dθdt\dfrac{dx}{dt}=\dfrac{dx}{d\theta}\cdot\dfrac{d\theta}{dt}, which equation results?
  1. A3=5dθdt3=5\dfrac{d\theta}{dt}
  2. B5=3dθdt5=3\dfrac{d\theta}{dt}
  3. C3=θdθdt3=\theta\dfrac{d\theta}{dt}
  4. D15=dθdt15=\dfrac{d\theta}{dt}

Question 504

[2 marks]related rates / radian measure
The area of sector POQPOQ is A=12r2θ=252θA=\dfrac12 r^2\theta=\dfrac{25}{2}\theta. What is dAdθ\dfrac{dA}{d\theta}?
  1. A252\dfrac{25}{2}
  2. B25
  3. C52\dfrac{5}{2}
  4. D152\dfrac{15}{2}

Question 601

[1 marks]partial fractions
Express 2x3−17x−1(x−2)(x2+5)\dfrac{2x^3 - 17x - 1}{(x-2)(x^2+5)} in partial fractions.
  1. A2−199(x−2)+55x+1339(x2+5)2 - \dfrac{19}{9(x-2)} + \dfrac{55x+133}{9(x^2+5)}
  2. B−199(x−2)+55x−1339(x2+5)\dfrac{-19}{9(x-2)} + \dfrac{55x-133}{9(x^2+5)}
  3. C2−199(x−2)+55x−1339(x2+5)2 - \dfrac{19}{9(x-2)} + \dfrac{55x-133}{9(x^2+5)}
  4. D2+199(x−2)−55x−1339(x2+5)2 + \dfrac{19}{9(x-2)} - \dfrac{55x-133}{9(x^2+5)}

Question 602

[2 marks]partial fractions
Dividing 2x3−17x−12x^3-17x-1 by (x−2)(x2+5)=x3−2x2+5x−10(x-2)(x^2+5)=x^3-2x^2+5x-10, the quotient is 2 with a remainder. Which polynomial is the remainder?
  1. A2x2−17x−12x^2-17x-1
  2. B4x2−27x−194x^2-27x-19
  3. C4x2−27x+194x^2-27x+19
  4. D4x2+27x+194x^2+27x+19

Question 603

[2 marks]partial fractions
Writing 4x2−27x+19(x−2)(x2+5)=Ax−2+Bx+Cx2+5\dfrac{4x^2-27x+19}{(x-2)(x^2+5)}=\dfrac{A}{x-2}+\dfrac{Bx+C}{x^2+5} and substituting x=2x=2 gives 16−54+19=9A16-54+19=9A. What is AA?
  1. AA=−19A=-19
  2. BA=−919A=-\dfrac{9}{19}
  3. CA=−199A=-\dfrac{19}{9}
  4. DA=199A=\dfrac{19}{9}

Question 604

[1 marks]partial fractions
With A=−199A=-\dfrac{19}{9}, substituting x=0x=0 into 4x2−27x+19=A(x2+5)+(Bx+C)(x−2)4x^2-27x+19=A(x^2+5)+(Bx+C)(x-2) gives 19=5A−2C19=5A-2C. What is CC?

Answer this when you sit the paper.

Question 701

[1 marks]logarithms / linearising relationships
Variables pp and qq satisfy log⁡p=blog⁡q+log⁡a\log p = b\log q + \log a. The graph of log⁡p\log p against log⁡q\log q is a straight line through (0.30,0.55)(0.30, 0.55) and (0.70,−0.05)(0.70, -0.05). Find aa and bb.
  1. Aa=10a = 10, b=1.5b = 1.5
  2. Ba=10a = 10, b=−1.5b = -1.5
  3. Ca=1a = 1, b=−1.5b = -1.5
  4. Da=−1.5a = -1.5, b=10b = 10

Question 702

[2 marks]logarithms / linearising relationships
The table gives q=4, p=1.26q=4,\,p=1.26. To 2 decimal places, what are log⁡10q\log_{10}q and log⁡10p\log_{10}p for this pair?
  1. Alog⁡q=0.60, log⁡p=0.26\log q=0.60,\ \log p=0.26
  2. Blog⁡q=1.26, log⁡p=0.60\log q=1.26,\ \log p=0.60
  3. Clog⁡q=0.60, log⁡p=0.10\log q=0.60,\ \log p=0.10
  4. Dlog⁡q=0.40, log⁡p=1.26\log q=0.40,\ \log p=1.26

Question 703

[2 marks]logarithms / linearising relationships
The line through the plotted points (log⁡q,log⁡p)=(0.30,0.55)(\log q,\log p)=(0.30,0.55) and (0.70,−0.05)(0.70,-0.05) has gradient b=−1.5b=-1.5. Using log⁡a=0.55−b(0.30)\log a=0.55-b(0.30), what is the value of aa?

Answer this when you sit the paper.

Question 704

[1 marks]logarithms / linearising relationships
In the relation log⁡p=blog⁡q+log⁡a\log p=b\log q+\log a, plotting log⁡p\log p against log⁡q\log q gives a straight line. What does the gradient of that line equal?
  1. Aaa
  2. Bbb
  3. Clog⁡a\log a
  4. Dpp

Question 801

[1 marks]quadratics / remainder theorem
Given f(x)=px2+5x−4f(x) = px^2 + 5x - 4, find the set of values of pp for which f(x)=0f(x) = 0 has no real roots.
  1. Ap<−2516p < -\tfrac{25}{16}
  2. B−2516<p<0-\tfrac{25}{16} < p < 0
  3. Cp>0p > 0
  4. Dp>2516p > \tfrac{25}{16}

Question 802

[1 marks]quadratics / remainder theorem
The remainder when 4x4−5x2−13x+34x^4 - 5x^2 - 13x + 3 is divided by x+bx + b equals the square of the remainder when 2x2−32x^2 - 3 is divided by x+bx + b. Find the possible values of bb, correct to 2 decimal places.
  1. Ab=0.38b = 0.38 or b=−2.24b = -2.24
  2. Bb=−0.38b = -0.38 or b=2.24b = 2.24
  3. Cb=0.38b = 0.38 only
  4. Db=1.86b = 1.86 or b=−1.86b = -1.86

Question 803

[1 marks]quadratics / remainder theorem
For f(x)=px2+5x−4f(x)=px^2+5x-4 to be positive for all real xx (no real roots), what must the discriminant 25+16p25+16p satisfy?
  1. A25+16p<025+16p<0
  2. B25+16p>025+16p>0
  3. C25+16p=025+16p=0
  4. D25+16p≤025+16p\le0

Question 804

[2 marks]quadratics / remainder theorem
By the remainder theorem, the remainder when 4x4−5x2−13x+34x^4-5x^2-13x+3 is divided by x+bx+b is found by substituting x=−bx=-b. Which expression gives this remainder?
  1. A4b4−5b2−13b+34b^4-5b^2-13b+3
  2. B4b4−5b2+13b−34b^4-5b^2+13b-3
  3. C4b4−5b2+13b+34b^4-5b^2+13b+3
  4. D4b4+5b2+13b+34b^4+5b^2+13b+3

Question 805

[1 marks]quadratics / remainder theorem
Equating 4b4−5b2+13b+34b^4-5b^2+13b+3 to the square of the remainder of 2x2−32x^2-3 divided by x+bx+b (which is 2b2−32b^2-3) and simplifying, which quadratic in bb results?
  1. A7b2+13b+6=07b^2+13b+6=0
  2. B7b2+13b−6=07b^2+13b-6=0
  3. C7b2−13b−6=07b^2-13b-6=0
  4. D4b2+13b−6=04b^2+13b-6=0

Question 901

[1 marks]vectors / scalar product
In a triangular prism, PS→=3i+9j−4k\overrightarrow{PS} = 3\mathbf{i} + 9\mathbf{j} - 4\mathbf{k}. Find the unit vector in the direction of PS→\overrightarrow{PS}.
  1. A1106(−3i+9j−4k)\tfrac{1}{\sqrt{106}}(-3\mathbf{i} + 9\mathbf{j} - 4\mathbf{k})
  2. B1106(3i+9j−4k)\tfrac{1}{106}(3\mathbf{i} + 9\mathbf{j} - 4\mathbf{k})
  3. C1106(3i+9j−4k)\tfrac{1}{\sqrt{106}}(3\mathbf{i} + 9\mathbf{j} - 4\mathbf{k})
  4. D194(3i+9j−4k)\tfrac{1}{\sqrt{94}}(3\mathbf{i} + 9\mathbf{j} - 4\mathbf{k})

Question 902

[1 marks]vectors / scalar product
In a triangular prism, PS→=3i+9j−4k\overrightarrow{PS} = 3\mathbf{i} + 9\mathbf{j} - 4\mathbf{k} and PQ→=−3i+9j−4k\overrightarrow{PQ} = -3\mathbf{i} + 9\mathbf{j} - 4\mathbf{k}. Find angle SPQSPQ correct to the nearest 0.1°0.1°.
  1. A146.1°146.1°
  2. B33.9°33.9°
  3. C45.0°45.0°
  4. D56.1°56.1°

Question 903

[2 marks]vectors / scalar product
In the triangular prism, PS→=3i+9j−4k\overrightarrow{PS}=3\mathbf i+9\mathbf j-4\mathbf k. What is ∣PS→∣|\overrightarrow{PS}|?
  1. A1010
  2. B94\sqrt{94}
  3. C106\sqrt{106}
  4. D106106

Question 904

[2 marks]vectors / scalar product
In the triangular prism, PS→=3i+9j−4k\overrightarrow{PS}=3\mathbf i+9\mathbf j-4\mathbf k and PQ→=−3i+9j−4k\overrightarrow{PQ}=-3\mathbf i+9\mathbf j-4\mathbf k. What is PS→⋅PQ→\overrightarrow{PS}\cdot\overrightarrow{PQ}?

Answer this when you sit the paper.

Question 1001

[1 marks]complex numbers
The complex number zz satisfies z+2zˉ=13−2+3iz + 2\bar{z} = \dfrac{13}{-2+3i}. Find zz in the form x+iyx + iy.
  1. A23+3i\tfrac{2}{3} + 3i
  2. B−23+3i-\tfrac{2}{3} + 3i
  3. C−2−3i-2 - 3i
  4. D−23−3i-\tfrac{2}{3} - 3i

Question 1002

[1 marks]complex numbers
Given z=−23+3iz = -\tfrac{2}{3} + 3i, find the modulus of 1z\dfrac{1}{z}, correct to 3 decimal places.
  1. A0.1080.108
  2. B0.3250.325
  3. C3.0733.073
  4. D9.2229.222

Question 1003

[2 marks]complex numbers
To simplify 13−2+3i\dfrac{13}{-2+3i}, the numerator and denominator are multiplied by −2−3i-2-3i. Given (−2+3i)(−2−3i)=13(-2+3i)(-2-3i)=13, what does 13−2+3i\dfrac{13}{-2+3i} equal in the form x+iyx+iy?
  1. A2+3i2+3i
  2. B−2+3i-2+3i
  3. C−3−2i-3-2i
  4. D−2−3i-2-3i

Question 1004

[1 marks]complex numbers
For z=−23+3iz=-\dfrac23+3i, what is x2+y2x^2+y^2 (that is, ∣z∣2|z|^2) as a fraction?

Answer this when you sit the paper.

Question 1005

[2 marks]complex numbers
For z=−23+3iz=-\dfrac23+3i, which lies in the second quadrant, arg⁡z=180°−tan⁡−1 ⁣(32/3)=102.5°\arg z=180°-\tan^{-1}\!\left(\dfrac{3}{2/3}\right)=102.5°. What is arg⁡(1z)\arg\left(\dfrac1z\right)?
  1. A102.5°102.5°
  2. B−102.5°-102.5°
  3. C77.5°77.5°
  4. D−77.5°-77.5°

Question 1101

[1 marks]geometric series
In a geometric series with r>0r > 0, the ratio of the sum of the first four terms to the sum of the first two terms is 17:1617 : 16. Find rr.
  1. Ar=14r = \tfrac{1}{4}
  2. Br=12r = \tfrac{1}{2}
  3. Cr=4r = 4
  4. Dr=116r = \tfrac{1}{16}

Question 1102

[1 marks]geometric series
A geometric series has common ratio 14\tfrac{1}{4} and third term 2342\tfrac{3}{4}. Find its sum to infinity.
  1. A113\tfrac{11}{3}
  2. B1763\tfrac{176}{3}
  3. C443\tfrac{44}{3}
  4. D4444

Question 1103

[2 marks]geometric series
In a geometric series with first term aa and ratio rr, the condition a(1+r+r2+r3)a(1+r)=1716\dfrac{a(1+r+r^2+r^3)}{a(1+r)}=\dfrac{17}{16} reduces, using 1+r+r2+r3=(1+r)(1+r2)1+r+r^2+r^3=(1+r)(1+r^2), to which cubic equation in rr?
  1. A16r3+16r2+r−1=016r^3+16r^2+r-1=0
  2. B16r3+16r2−r−1=016r^3+16r^2-r-1=0
  3. C16r3−16r2+r+1=016r^3-16r^2+r+1=0
  4. D16r2+16r−1=016r^2+16r-1=0

Question 1104

[1 marks]geometric series
The cubic 16r3+16r2−r−1=016r^3+16r^2-r-1=0 has r=14r=\tfrac14 as one root. What are the other two roots?
  1. Ar=1r=1 and r=14r=\tfrac14
  2. Br=−1r=-1 and r=14r=\tfrac14
  3. Cr=−4r=-4 and r=−1r=-1
  4. Dr=−1r=-1 and r=−14r=-\tfrac14

Question 1105

[2 marks]geometric series
Given r=14r=\dfrac14 and third term ar2=234ar^2=2\dfrac34, what is the first term aa?

Answer this when you sit the paper.

Question 1201

[1 marks]logarithmic series / Maclaurin expansion
Given e−3y(1−4x)=7+3xe^{-3y}(1-4x) = 7+3x, express yy in terms of xx.
  1. Ay=−13[ln⁡(7+3x)−ln⁡(1−4x)]y = -\tfrac{1}{3}\left[\ln(7+3x) - \ln(1-4x)\right]
  2. By=−3[ln⁡(7+3x)−ln⁡(1−4x)]y = -3\left[\ln(7+3x) - \ln(1-4x)\right]
  3. Cy=−13[ln⁡(7+3x)+ln⁡(1−4x)]y = -\tfrac{1}{3}\left[\ln(7+3x) + \ln(1-4x)\right]
  4. Dy=13[ln⁡(7+3x)−ln⁡(1−4x)]y = \tfrac{1}{3}\left[\ln(7+3x) - \ln(1-4x)\right]

Question 1202

[1 marks]logarithmic series / Maclaurin expansion
In the Maclaurin expansion of y=−13[ln⁡(7+3x)−ln⁡(1−4x)]y = -\tfrac{1}{3}\left[\ln(7+3x) - \ln(1-4x)\right], what is the coefficient of xx?
  1. A3121\tfrac{31}{21}
  2. B−317-\tfrac{31}{7}
  3. C−17-\tfrac{1}{7}
  4. D−3121-\tfrac{31}{21}

Question 1203

[1 marks]logarithmic series / Maclaurin expansion
Given e−3y(1−4x)=7+3xe^{-3y}(1-4x)=7+3x, making e−3ye^{-3y} the subject gives which expression?
  1. Ae−3y=7+3x−1+4xe^{-3y}=7+3x-1+4x
  2. Be−3y=7+3x1−4xe^{-3y}=\dfrac{7+3x}{1-4x}
  3. Ce−3y=1−4x7+3xe^{-3y}=\dfrac{1-4x}{7+3x}
  4. De−3y=(7+3x)(1−4x)e^{-3y}=(7+3x)(1-4x)

Question 1204

[2 marks]logarithmic series / Maclaurin expansion
What is the Maclaurin expansion of ln⁡(7+3x)\ln(7+3x) up to and including the term in x2x^2?
  1. Aln⁡7+3x7+9x298\ln7+\dfrac{3x}{7}+\dfrac{9x^2}{98}
  2. Bln⁡7−3x7−9x298\ln7-\dfrac{3x}{7}-\dfrac{9x^2}{98}
  3. Cln⁡7+x7−x298\ln7+\dfrac{x}{7}-\dfrac{x^2}{98}
  4. Dln⁡7+3x7−9x298\ln7+\dfrac{3x}{7}-\dfrac{9x^2}{98}

Question 1205

[2 marks]logarithmic series / Maclaurin expansion
What is the Maclaurin expansion of ln⁡(1−4x)\ln(1-4x) up to and including the term in x2x^2?
  1. A4x−8x24x-8x^2
  2. B−4x+8x2-4x+8x^2
  3. C−4x−16x2-4x-16x^2
  4. D−4x−8x2-4x-8x^2

Question 1301

[1 marks]binomial expansion
Use the binomial expansion to simplify (x+2)7−(x−2)7(x+2)^7 - (x-2)^7.
  1. A28x6+560x4+1344x228x^6 + 560x^4 + 1344x^2
  2. B14x6+280x4+672x2+12814x^6 + 280x^4 + 672x^2 + 128
  3. C28x6+560x4+1344x2+25628x^6 + 560x^4 + 1344x^2 + 256
  4. D2x7+168x5+1344x22x^7 + 168x^5 + 1344x^2

Question 1302

[1 marks]binomial expansion
Find the exact value of (5+2)7−(5−2)7\left(\sqrt{5}+2\right)^7 - \left(\sqrt{5}-2\right)^7.
  1. A35003500
  2. B1223812238
  3. C2198021980
  4. D2447624476

Question 1303

[2 marks]binomial expansion
In the binomial expansion of (x+2)7(x+2)^7, what is the term in x4x^4?
  1. A672x4672x^4
  2. B448x4448x^4
  3. C560x4560x^4
  4. D280x4280x^4

Question 1304

[2 marks]binomial expansion
When (x−2)7(x-2)^7 is subtracted from (x+2)7(x+2)^7, which terms cancel?
  1. Athe terms with odd powers of xx, since they carry opposite signs in the two expansions
  2. Bno terms cancel; all 8 terms remain
  3. Conly the constant terms cancel
  4. Dthe terms with even powers of xx, since both expansions share the same sign there

Question 1305

[1 marks]binomial expansion
In the simplified result (x+2)7−(x−2)7=28x6+560x4+1344x2+256(x+2)^7-(x-2)^7=28x^6+560x^4+1344x^2+256, what is the coefficient of x2x^2?

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Question 1306

[1 marks]binomial expansion
To evaluate (5+2)7−(5−2)7(\sqrt5+2)^7-(\sqrt5-2)^7 using the result for (x+2)7−(x−2)7(x+2)^7-(x-2)^7, what value of xx is substituted?

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Question 1401

[1 marks]trigonometric identities / integration by substitution
Simplify sec⁡2θ+tan⁡2θ\sec 2\theta + \tan 2\theta.
  1. A1−sin⁡2θcos⁡2θ\dfrac{1 - \sin 2\theta}{\cos 2\theta}
  2. Btan⁡θ+1\tan\theta + 1
  3. Ccos⁡θ−sin⁡θsin⁡θ+cos⁡θ\dfrac{\cos\theta - \sin\theta}{\sin\theta + \cos\theta}
  4. Dsin⁡θ+cos⁡θcos⁡θ−sin⁡θ\dfrac{\sin\theta + \cos\theta}{\cos\theta - \sin\theta}

Question 1402

[1 marks]trigonometric identities / integration by substitution
By using the substitution u=sin⁡xu = \sin x, find the exact value of ∫0π/2cos⁡x3+cos⁡2x dx\displaystyle\int_0^{\pi/2} \dfrac{\cos x}{3 + \cos^2 x}\,dx.
  1. A12ln⁡3\tfrac{1}{2}\ln 3
  2. B14ln⁡3\tfrac{1}{4}\ln 3
  3. C14ln⁡13\tfrac{1}{4}\ln \tfrac{1}{3}
  4. Dln⁡3\ln 3

Question 1403

[2 marks]trigonometric identities / integration by substitution
To prove sec⁡2θ+tan⁡2θ=sin⁡θ+cos⁡θcos⁡θ−sin⁡θ\sec2\theta+\tan2\theta=\dfrac{\sin\theta+\cos\theta}{\cos\theta-\sin\theta}, the left side is first written over a common denominator. Which expression results?
  1. A1−sin⁡2θcos⁡2θ\dfrac{1-\sin2\theta}{\cos2\theta}
  2. Bsin⁡2θ1+cos⁡2θ\dfrac{\sin2\theta}{1+\cos2\theta}
  3. C1+cos⁡2θsin⁡2θ\dfrac{1+\cos2\theta}{\sin2\theta}
  4. D1+sin⁡2θcos⁡2θ\dfrac{1+\sin2\theta}{\cos2\theta}

Question 1404

[1 marks]trigonometric identities / integration by substitution
In the identity cos⁡2θ=cos⁡2θ−sin⁡2θ\cos2\theta=\cos^2\theta-\sin^2\theta, which alternative single-term form is also equal to cos⁡2θ\cos2\theta?
  1. A1−2sin⁡2θ1-2\sin^2\theta
  2. B2sin⁡2θ−12\sin^2\theta-1
  3. C1−2cos⁡2θ1-2\cos^2\theta
  4. Dsin⁡2θ−cos⁡2θ\sin^2\theta-\cos^2\theta

Question 1405

[1 marks]trigonometric identities / integration by substitution
Using sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1, the expression 3+cos⁡2x3+\cos^2x can be rewritten entirely in terms of sin⁡x\sin x as:
  1. A4+sin⁡2x4+\sin^2x
  2. B3−sin⁡2x3-\sin^2x
  3. C2−sin⁡2x2-\sin^2x
  4. D4−sin⁡2x4-\sin^2x

Question 1406

[1 marks]trigonometric identities / integration by substitution
Using the substitution u=sin⁡xu=\sin x in ∫0π/2cos⁡x3+cos⁡2xdx\displaystyle\int_0^{\pi/2}\dfrac{\cos x}{3+\cos^2x}dx, the limits x=0x=0 and x=π2x=\dfrac{\pi}{2} become which values of uu?
  1. Au=0u=0 to u=1u=1
  2. Bu=1u=1 to u=0u=0
  3. Cu=0u=0 to u=π/2u=\pi/2
  4. Du=−1u=-1 to u=1u=1

Question 1407

[2 marks]trigonometric identities / integration by substitution
With u=sin⁡xu=\sin x, the integral becomes ∫01du4−u2=∫01du(2−u)(2+u)\displaystyle\int_0^1\dfrac{du}{4-u^2}=\int_0^1\dfrac{du}{(2-u)(2+u)}. What is the partial-fraction form of 1(2−u)(2+u)\dfrac{1}{(2-u)(2+u)}?
  1. A14(2−u)+14(2+u)\dfrac{1}{4(2-u)}+\dfrac{1}{4(2+u)}
  2. B14(2−u)−14(2+u)\dfrac{1}{4(2-u)}-\dfrac{1}{4(2+u)}
  3. C12−u+12+u\dfrac{1}{2-u}+\dfrac{1}{2+u}
  4. D14(2+u)−14(2−u)\dfrac{1}{4(2+u)}-\dfrac{1}{4(2-u)}

Question 1501

[1 marks]graphical solution / Newton-Raphson method
Let f(x)=sin⁡x+1−4x2f(x) = \sin x + 1 - 4x^2. Which calculation verifies that the positive root of f(x)=0f(x) = 0 lies between x=0.6x = 0.6 and x=0.7x = 0.7?
  1. Af(0.6)=0.12464f(0.6) = 0.12464 and f(0.7)=−0.31578f(0.7) = -0.31578
  2. Bf(0.6)=−0.12464f(0.6) = -0.12464 and f(0.7)=0.31578f(0.7) = 0.31578
  3. Cf(0.6)=0.12464f(0.6) = 0.12464 and f(0.7)=0.31578f(0.7) = 0.31578
  4. Df(0.6)=1.5646f(0.6) = 1.5646 and f(0.7)=1.6442f(0.7) = 1.6442

Question 1502

[1 marks]graphical solution / Newton-Raphson method
Taking x0=0.6x_0 = 0.6, apply the Newton-Raphson method twice to sin⁡x+1−4x2=0\sin x + 1 - 4x^2 = 0 and give the positive root correct to 5 decimal places.
  1. A0.600000.60000
  2. B0.630370.63037
  3. C0.631360.63136
  4. D0.700000.70000

Question 1503

[2 marks]graphical solution / Newton-Raphson method
To show that sin⁡x+1−4x2=0\sin x+1-4x^2=0 (for −2π≤x≤2π-2\pi\le x\le2\pi) has two real roots by sketching, which pair of graphs should be drawn on the same axes?
  1. Ay=cos⁡xy=\cos x and y=4x2−1y=4x^2-1
  2. By=sin⁡xy=\sin x and y=4x2−1y=4x^2-1
  3. Cy=sin⁡xy=\sin x and y=1−4x2y=1-4x^2
  4. Dy=sin⁡x+1y=\sin x+1 and y=4x2y=4x^2

Question 1504

[1 marks]graphical solution / Newton-Raphson method
For f(x)=sin⁡x+1−4x2f(x)=\sin x+1-4x^2, what is f(0.7)f(0.7), correct to 5 decimal places?

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Question 1505

[1 marks]graphical solution / Newton-Raphson method
Given f(0.6)=0.12464>0f(0.6)=0.12464>0 and f(0.7)=−0.31578<0f(0.7)=-0.31578<0, what does this sign change confirm?
  1. Af(x)f(x) is increasing throughout [0.6,0.7][0.6,0.7]
  2. Bf(x)f(x) has a turning point at x=0.65x=0.65
  3. Cf(x)=0f(x)=0 has no root in [0.6,0.7][0.6,0.7]
  4. Da root of f(x)=0f(x)=0 lies strictly between x=0.6x=0.6 and x=0.7x=0.7

Question 1506

[1 marks]graphical solution / Newton-Raphson method
For f(x)=sin⁡x+1−4x2f(x)=\sin x+1-4x^2, needed to apply the Newton-Raphson method, what is f′(x)f'(x)?
  1. A−cos⁡x−8x-\cos x-8x
  2. Bcos⁡x−4x\cos x-4x
  3. Ccos⁡x+8x\cos x+8x
  4. Dcos⁡x−8x\cos x-8x

Question 1507

[2 marks]graphical solution / Newton-Raphson method
Taking x0=0.6x_0=0.6, one Newton-Raphson iteration x1=x0−f(x0)f′(x0)x_1=x_0-\dfrac{f(x_0)}{f'(x_0)} on f(x)=sin⁡x+1−4x2f(x)=\sin x+1-4x^2 gives which value of x1x_1, correct to 5 decimal places?

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Question 1601

[1 marks]coordinate geometry / circles
Find the perpendicular distance of the point P(−3,2)P(-3, 2) from the line 3x−5y=73x - 5y = 7, correct to 2 decimal places.
  1. A2.232.23
  2. B3.093.09
  3. C4.464.46
  4. D5.835.83

Question 1602

[1 marks]coordinate geometry / circles
Find the equation of the circle which passes through (5,4)(5, 4) and touches the yy-axis at (0,2)(0, 2).
  1. A(x−2910)2+(y−2)2=4\left(x - \tfrac{29}{10}\right)^2 + (y-2)^2 = 4
  2. B(x−2910)2+(y−2)2=(2910)2\left(x - \tfrac{29}{10}\right)^2 + (y-2)^2 = \left(\tfrac{29}{10}\right)^2
  3. C(x−52)2+(y−3)2=(2910)2\left(x - \tfrac{5}{2}\right)^2 + (y-3)^2 = \left(\tfrac{29}{10}\right)^2
  4. Dx2+(y−2)2=(2910)2x^2 + (y-2)^2 = \left(\tfrac{29}{10}\right)^2

Question 1603

[1 marks]coordinate geometry / circles
What is the gradient of a line perpendicular to 3x−5y=73x-5y=7 (which has gradient 35\dfrac35)?

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Question 1604

[2 marks]coordinate geometry / circles
What is the equation of the line through P(−3,2)P(-3,2) with gradient −53-\dfrac53?
  1. A5x+3y=−95x+3y=-9
  2. B5x+3y=95x+3y=9
  3. C3x+5y=−93x+5y=-9
  4. D5x−3y=−95x-3y=-9

Question 1605

[2 marks]coordinate geometry / circles
Solving 5x+3y=−95x+3y=-9 together with 3x−5y=73x-5y=7, what is the point of intersection?
  1. A(−1217,−3117)\left(-\dfrac{12}{17},-\dfrac{31}{17}\right)
  2. B(1217,3117)\left(\dfrac{12}{17},\dfrac{31}{17}\right)
  3. C(−1217,3117)\left(-\dfrac{12}{17},\dfrac{31}{17}\right)
  4. D(−3117,−1217)\left(-\dfrac{31}{17},-\dfrac{12}{17}\right)

Question 1606

[1 marks]coordinate geometry / circles
What is the midpoint of the chord joining (5,4)(5,4) and (0,2)(0,2)?

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Question 1607

[1 marks]coordinate geometry / circles
What is the gradient of the chord joining (5,4)(5,4) and (0,2)(0,2)?

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Question 1608

[2 marks]coordinate geometry / circles
The perpendicular bisector of the chord joining (5,4)(5,4) and (0,2)(0,2) passes through (2.5,3)(2.5,3) with gradient −52-\dfrac52. What is its equation?
  1. A10x−4y=3710x-4y=37
  2. B4x+10y=374x+10y=37
  3. C10x+4y=1310x+4y=13
  4. D10x+4y=3710x+4y=37

Question 1609

[1 marks]coordinate geometry / circles
The circle's centre lies on both 10x+4y=3710x+4y=37 and y=2y=2 (since it touches the y-axis at (0,2)(0,2)). What is the x-coordinate of the centre?

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Question 1701

[1 marks]integration / differential equations
A curve has gradient function dydx=15(3x+4)2\dfrac{dy}{dx} = \dfrac{15}{(3x+4)^2} and passes through M(2,6)M(2, 6). Find the equation of the curve.
  1. Ay=53x+4+112y = \dfrac{5}{3x+4} + \dfrac{11}{2}
  2. By=−153x+4+152y = -\dfrac{15}{3x+4} + \dfrac{15}{2}
  3. Cy=−53x+4+132y = -\dfrac{5}{3x+4} + \dfrac{13}{2}
  4. Dy=−5(3x+4)3+6y = -\dfrac{5}{(3x+4)^3} + 6

Question 1702

[1 marks]integration / differential equations
The number xx of laptop owners grows at a rate proportional to x(N−x)x(N - x), with x=110Nx = \tfrac{1}{10}N initially and x=14Nx = \tfrac{1}{4}N after one week, so that xN−x=3t−2\dfrac{x}{N-x} = 3^{t-2}. Find tt when x=34Nx = \tfrac{3}{4}N.
  1. At=1t = 1 week
  2. Bt=3t = 3 weeks
  3. Ct=2t = 2 weeks
  4. Dt=4t = 4 weeks

Question 1703

[1 marks]integration / differential equations
For a population in which xN−x=3t−2\dfrac{x}{N-x} = 3^{t-2}, what is the shape of the graph of xx against tt?
  1. AA parabola with maximum at t=2t = 2
  2. BA straight line of positive gradient
  3. CAn S-shaped (logistic) curve rising from N10\tfrac{N}{10} towards the asymptote x=Nx = N
  4. DAn exponential decay curve towards x=0x = 0

Question 1704

[2 marks]integration / differential equations
What is ∫15(3x+4)2 dx\displaystyle\int\dfrac{15}{(3x+4)^2}\,dx (before applying any boundary condition)?
  1. A−53x+4+k-\dfrac{5}{3x+4}+k
  2. B53x+4+k\dfrac{5}{3x+4}+k
  3. C−5(3x+4)3+k-\dfrac{5}{(3x+4)^3}+k
  4. D−153x+4+k-\dfrac{15}{3x+4}+k

Question 1705

[1 marks]integration / differential equations
Given y=−53x+4+ky=-\dfrac{5}{3x+4}+k passes through M(2,6)M(2,6), what is kk?

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Question 1706

[1 marks]integration / differential equations
The number of laptop owners xx increases at a rate proportional to the product of xx and N−xN-x. Which differential equation models this?
  1. Adxdt=kx(N−x)\dfrac{dx}{dt}=kx(N-x)
  2. Bdxdt=k(N−x)\dfrac{dx}{dt}=k(N-x)
  3. Cdxdt=kx+N−x\dfrac{dx}{dt}=kx+N-x
  4. Ddxdt=kx(N−x)\dfrac{dx}{dt}=\dfrac{k}{x(N-x)}

Question 1707

[2 marks]integration / differential equations
To separate variables in dxdt=kx(N−x)\dfrac{dx}{dt}=kx(N-x), 1x(N−x)\dfrac{1}{x(N-x)} is split into partial fractions. Which decomposition is correct?
  1. A1x+1N−x\dfrac1x+\dfrac{1}{N-x}
  2. B1N(1N−x−1x)\dfrac{1}{N}\left(\dfrac{1}{N-x}-\dfrac1x\right)
  3. C1N(1x+1N−x)\dfrac{1}{N}\left(\dfrac1x+\dfrac{1}{N-x}\right)
  4. D1N(1x−1N−x)\dfrac{1}{N}\left(\dfrac1x-\dfrac{1}{N-x}\right)

Question 1708

[2 marks]integration / differential equations
Integrating 1N(1x+1N−x)dx=k dt\dfrac{1}{N}\left(\dfrac1x+\dfrac{1}{N-x}\right)dx=k\,dt gives which relation (with constant CC)?
  1. A1Nln⁡ ⁣(N−xx)=kt+C\dfrac{1}{N}\ln\!\left(\dfrac{N-x}{x}\right)=kt+C
  2. B1Nln⁡ ⁣(xN−x)=kt+C\dfrac{1}{N}\ln\!\left(\dfrac{x}{N-x}\right)=kt+C
  3. C1Nln⁡(x(N−x))=kt+C\dfrac{1}{N}\ln(x(N-x))=kt+C
  4. DNln⁡ ⁣(xN−x)=kt+CN\ln\!\left(\dfrac{x}{N-x}\right)=kt+C

Question 1709

[2 marks]integration / differential equations
At t=0t=0, x=N10x=\dfrac{N}{10}. Substituting into 1Nln⁡ ⁣(xN−x)=kt+C\dfrac{1}{N}\ln\!\left(\dfrac{x}{N-x}\right)=kt+C, what is CC?
  1. AC=1Nln⁡9C=\dfrac{1}{N}\ln9
  2. BC=1Nln⁡110C=\dfrac{1}{N}\ln\dfrac{1}{10}
  3. CC=Nln⁡19C=N\ln\dfrac19
  4. DC=1Nln⁡19C=\dfrac{1}{N}\ln\dfrac19

Question 1710

[2 marks]integration / differential equations
At t=1t=1, x=N4x=\dfrac{N}{4}. Using C=1Nln⁡19C=\dfrac1N\ln\dfrac19, what is kk?
  1. Ak=Nln⁡3k=N\ln3
  2. Bk=1Nln⁡27k=\dfrac{1}{N}\ln27
  3. Ck=1Nln⁡3k=\dfrac{1}{N}\ln3
  4. Dk=1Nln⁡13k=\dfrac{1}{N}\ln\dfrac13

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