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ZIMSEC A Level · J2020

Pure Mathematics Paper 1 June 2020

Questions
89
Total marks
120

Sit this paper online

Questions
89
Pass mark
54
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]Indices and quadratic substitution
Using the substitution y=x1/3y = x^{1/3}, find the values of xx for which x1/3−3x−1/3=2x^{1/3} - 3x^{-1/3} = 2.
  1. Ax=27x = 27 or x=−1x = -1
  2. Bx=27x = 27 or x=1x = 1
  3. Cx=9x = 9 or x=−1x = -1
  4. Dx=3x = 3 or x=−1x = -1

Question 102

[2 marks]Indices and quadratic substitution
Using the substitution y=x1/3y=x^{1/3}, rewrite x1/3−3x−1/3=2x^{1/3}-3x^{-1/3}=2 as a quadratic equation in yy, in the form y2+by+c=0y^2+by+c=0.

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Question 103

[1 marks]Indices and quadratic substitution
Solve y2−2y−3=0y^2-2y-3=0 by factorisation. State both values of yy.

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Question 201

[1 marks]Coordinate geometry (parallel and perpendicular lines)
Find the value of kk for which the line kx+(k−2)y+10=0kx + (k-2)y + 10 = 0 is parallel to 3x+2y−16=03x + 2y - 16 = 0.
  1. Ak=3k = 3
  2. Bk=−6k = -6
  3. Ck=2k = 2
  4. Dk=6k = 6

Question 202

[1 marks]Coordinate geometry (parallel and perpendicular lines)
Find the gradient of any line perpendicular to 3x+2y−16=03x + 2y - 16 = 0.
  1. A32\tfrac{3}{2}
  2. B23\tfrac{2}{3}
  3. C−32-\tfrac{3}{2}
  4. D−23-\tfrac{2}{3}

Question 203

[2 marks]Coordinate geometry (parallel and perpendicular lines)
Substitute k=6k=6 into kx+(k−2)y+10=0kx+(k-2)y+10=0 to find the resulting equation, simplified to the form ax+by+c=0ax+by+c=0 with integer coefficients.

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Question 301

[1 marks]Completing the square and quadratic inequalities
Express 6x2−24x−256x^2 - 24x - 25 in the form A(x+B)2+CA(x+B)^2 + C.
  1. A6(x−4)2−496(x-4)^2 - 49
  2. B6(x−2)2−496(x-2)^2 - 49
  3. C6(x−2)2−256(x-2)^2 - 25
  4. D6(x+2)2−496(x+2)^2 - 49

Question 302

[1 marks]Completing the square and quadratic inequalities
Solve exactly the inequality 6x2−24x−25>06x^2 - 24x - 25 > 0.
  1. Ax>12+766x > \dfrac{12+7\sqrt{6}}{6} only
  2. Bx<2−496x < 2 - \dfrac{49}{6} or x>2+496x > 2 + \dfrac{49}{6}
  3. C12−766<x<12+766\dfrac{12-7\sqrt{6}}{6} < x < \dfrac{12+7\sqrt{6}}{6}
  4. Dx<12−766x < \dfrac{12-7\sqrt{6}}{6} or x>12+766x > \dfrac{12+7\sqrt{6}}{6}

Question 303

[1 marks]Completing the square and quadratic inequalities
Completing the square, 6x2−24x−25=6(x−2)2+C6x^2-24x-25=6(x-2)^2+C. Find CC.

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Question 304

[2 marks]Completing the square and quadratic inequalities
Solve (x−2)2>496(x-2)^2>\tfrac{49}{6} for x−2x-2, giving the positive critical value in exact surd form.

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Question 401

[1 marks]Modulus inequalities
Solve the inequality ∣2x+2∣>1−4x|2x+2| > 1 - 4x.
  1. Ax>32x > \tfrac{3}{2}
  2. Bx>−16x > -\tfrac{1}{6}
  3. Cx<−16x < -\tfrac{1}{6}
  4. D−16<x<32-\tfrac{1}{6} < x < \tfrac{3}{2}

Question 402

[2 marks]Modulus inequalities
Expanding and simplifying (2x+2)2>(1−4x)2(2x+2)^2>(1-4x)^2 gives 12x2−16x−3<012x^2-16x-3<0. Factorise 12x2−16x−312x^2-16x-3.

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Question 403

[1 marks]Modulus inequalities
State the smaller root of 12x2−16x−3=012x^2-16x-3=0.

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Question 404

[1 marks]Modulus inequalities
State the larger root of 12x2−16x−3=012x^2-16x-3=0.

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Question 501

[1 marks]Binomial series expansion
Expand (p−x)−2(p-x)^{-2} in ascending powers of xx up to the term in x3x^3, where p>0p > 0.
  1. A1p2+2px+3p2x2+4p3x3\dfrac{1}{p^2} + \dfrac{2}{p}x + \dfrac{3}{p^2}x^2 + \dfrac{4}{p^3}x^3
  2. Bp−2+2p−3x+6p−4x2+24p−5x3p^{-2} + 2p^{-3}x + 6p^{-4}x^2 + 24p^{-5}x^3
  3. C1p2−2p3x+3p4x2−4p5x3\dfrac{1}{p^2} - \dfrac{2}{p^3}x + \dfrac{3}{p^4}x^2 - \dfrac{4}{p^5}x^3
  4. D1p2+2p3x+3p4x2+4p5x3\dfrac{1}{p^2} + \dfrac{2}{p^3}x + \dfrac{3}{p^4}x^2 + \dfrac{4}{p^5}x^3

Question 502

[1 marks]Binomial series expansion
In the expansion of (p−x)−2(p-x)^{-2} the coefficient of x2x^2 is 316\tfrac{3}{16}. Find pp and the set of values of xx for which the expansion is valid.
  1. Ap=16p = 16, valid for −16<x<16-16 < x < 16
  2. Bp=2p = 2, valid for −2<x<2-2 < x < 2
  3. Cp=2p = 2, valid for −12<x<12-\tfrac{1}{2} < x < \tfrac{1}{2}
  4. Dp=4p = 4, valid for −4<x<4-4 < x < 4

Question 503

[1 marks]Binomial series expansion
Find the coefficient of x2x^2 in the expansion of (p−x)−2(p-x)^{-2}, in terms of pp.

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Question 504

[1 marks]Binomial series expansion
Given 3p4=316\dfrac{3}{p^4}=\dfrac{3}{16}, find p4p^4.

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Question 505

[3 marks]Binomial series expansion
Find the coefficient of x3x^3 in the expansion of (p−x)−2(p-x)^{-2}, in terms of pp.

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Question 506

[1 marks]Binomial series expansion
With p=2p=2, evaluate the coefficient of x3x^3, i.e. 4p5\dfrac{4}{p^5}.

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Question 601

[1 marks]Vectors
Points AA and BB have position vectors i−2j+pk\mathbf{i} - 2\mathbf{j} + p\mathbf{k} and qi+5j+6kq\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}. Given AB→=−5i+7j+3k\overrightarrow{AB} = -5\mathbf{i} + 7\mathbf{j} + 3\mathbf{k}, find pp and qq.
  1. Ap=9p = 9, q=−4q = -4
  2. Bp=3p = 3, q=6q = 6
  3. Cp=3p = 3, q=−4q = -4
  4. Dp=−3p = -3, q=4q = 4

Question 602

[1 marks]Vectors
With a=i−2j+3k\mathbf{a} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} and c=5i+7j\mathbf{c} = 5\mathbf{i} + 7\mathbf{j}, find the exact length of ACAC.
  1. A106\sqrt{106}
  2. B106106
  3. C74\sqrt{74}
  4. D83\sqrt{83}

Question 603

[1 marks]Vectors
Given AB→=−5i+7j+3k\overrightarrow{AB} = -5\mathbf{i} + 7\mathbf{j} + 3\mathbf{k} and AC→=4i+9j−3k\overrightarrow{AC} = 4\mathbf{i} + 9\mathbf{j} - 3\mathbf{k}, find the acute angle BA^CB\hat{A}C to the nearest degree.
  1. A111°111°
  2. B21°21°
  3. C45°45°
  4. D69°69°

Question 604

[1 marks]Vectors
Find ∣AB→∣|\overrightarrow{AB}| for AB→=−5i+7j+3k\overrightarrow{AB}=-5\mathbf{i}+7\mathbf{j}+3\mathbf{k}, in surd form.

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Question 605

[2 marks]Vectors
Find AB→⋅AC→\overrightarrow{AB}\cdot\overrightarrow{AC} for AB→=−5i+7j+3k\overrightarrow{AB}=-5\mathbf{i}+7\mathbf{j}+3\mathbf{k} and AC→=4i+9j−3k\overrightarrow{AC}=4\mathbf{i}+9\mathbf{j}-3\mathbf{k}.

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Question 606

[1 marks]Vectors
State cos⁡BA^C\cos B\hat{A}C as an exact fraction, using AB→⋅AC→=34\overrightarrow{AB}\cdot\overrightarrow{AC}=34, ∣AB→∣=83|\overrightarrow{AB}|=\sqrt{83}, ∣AC→∣=106|\overrightarrow{AC}|=\sqrt{106} (before evaluating the angle).

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Question 701

[1 marks]Geometric growth
A city's population is 500 000 and grows at 5%5\% per year. Write down the population at the end of the nnth year.
  1. A500 000+25 000n500\,000 + 25\,000n
  2. B500 000(0.05)n500\,000(0.05)^n
  3. C500 000(1.5)n500\,000(1.5)^n
  4. D500 000(1.05)n500\,000(1.05)^n

Question 702

[1 marks]Geometric growth
A population of 500 000 grows at 5%5\% per year. Find the population after ten years, to the nearest thousand.
  1. A814 000814\,000
  2. B1 000 0001\,000\,000
  3. C525 000525\,000
  4. D750 000750\,000

Question 703

[1 marks]Geometric growth
A population of 500 000 grows at 5%5\% per year. In which year does it first exceed 1 000 000?
  1. Ayear 14
  2. Byear 20
  3. Cyear 15
  4. Dyear 10

Question 704

[2 marks]Geometric growth
Evaluate (1,05)10(1{,}05)^{10}, correct to 4 decimal places (used to find the population after ten years).

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Question 705

[2 marks]Geometric growth
Solving (1,05)n>2(1{,}05)^n>2 using logarithms, find n=ln⁡2ln⁡1,05n=\dfrac{\ln2}{\ln1{,}05}, correct to 1 decimal place.

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Question 801

[1 marks]Functions and inverses
The function f(x)=2−1xf(x) = 2 - \dfrac{1}{x} is defined for x>0x > 0. State its range.
  1. Af(x)>0f(x) > 0
  2. Bf(x)>2f(x) > 2
  3. C0<f(x)<20 < f(x) < 2
  4. Df(x)<2f(x) < 2

Question 802

[1 marks]Functions and inverses
Find f−1(x)f^{-1}(x) for f(x)=2−1xf(x) = 2 - \dfrac{1}{x}, x>0x > 0.
  1. Af−1(x)=12−xf^{-1}(x) = \dfrac{1}{2-x}
  2. Bf−1(x)=2−1xf^{-1}(x) = 2 - \dfrac{1}{x}
  3. Cf−1(x)=1x−2f^{-1}(x) = \dfrac{1}{x-2}
  4. Df−1(x)=x2x−1f^{-1}(x) = \dfrac{x}{2x-1}

Question 803

[1 marks]Functions and inverses
For f(x)=2−1xf(x) = 2 - \dfrac{1}{x}, x>0x > 0, calculate the value of xx for which f(x)=f−1(x)f(x) = f^{-1}(x).
  1. Ax=12x = \tfrac{1}{2}
  2. Bx=1x = 1
  3. Cx=0x = 0
  4. Dx=2x = 2

Question 804

[1 marks]Functions and inverses
For f(x)=2−1xf(x)=2-\dfrac1x, x>0x>0, find the xx-intercept (where f(x)=0f(x)=0).

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Question 805

[1 marks]Functions and inverses
To invert ff, write x=2−1yx=2-\dfrac1y. Rearrange to express 1y\dfrac1y in terms of xx.

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Question 806

[1 marks]Functions and inverses
State the vertical asymptote of f−1(x)=12−xf^{-1}(x)=\dfrac{1}{2-x}.

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Question 807

[2 marks]Functions and inverses
Verify that x=1x=1 solves f(x)=f−1(x)f(x)=f^{-1}(x) by evaluating both f(1)f(1) and f−1(1)f^{-1}(1). State their common value.

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Question 901

[1 marks]Coordinate geometry of the circle
A circle with centre (2,−5)(2, -5) touches the line x+6y−9=0x + 6y - 9 = 0. Find the point of contact.
  1. A(−1,2)(-1, 2)
  2. B(3,1)(3, 1)
  3. C(9,0)(9, 0)
  4. D(2,−5)(2, -5)

Question 902

[1 marks]Coordinate geometry of the circle
Find the equation of the circle with centre (2,−5)(2, -5) which touches the line x+6y−9=0x + 6y - 9 = 0.
  1. A(x−2)2+(y+5)2=37(x-2)^2 + (y+5)^2 = \sqrt{37}
  2. B(x−3)2+(y−1)2=37(x-3)^2 + (y-1)^2 = 37
  3. C(x−2)2+(y+5)2=37(x-2)^2 + (y+5)^2 = 37
  4. D(x−2)2+(y−5)2=37(x-2)^2 + (y-5)^2 = 37

Question 903

[3 marks]Coordinate geometry of the circle
The tangent x+6y−9=0x+6y-9=0 has gradient −16-\tfrac16, so the radius (perpendicular to it) has gradient 66. Find the equation of the radius through the centre (2,−5)(2,-5), in the form y=6x+cy=6x+c.

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Question 904

[2 marks]Coordinate geometry of the circle
Given the point of contact (3,1)(3,1) and centre (2,−5)(2,-5), find the exact radius rr.

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Question 1001

[1 marks]Complex numbers
Given u=1+2iu = 1 + 2i and v=−2−iv = -2 - i, find w=5uvw = \dfrac{5u}{v} in the form a+iba + ib.
  1. A−4−3i-4 - 3i
  2. B−20−15i-20 - 15i
  3. C4+3i4 + 3i
  4. D−4+3i-4 + 3i

Question 1002

[1 marks]Complex numbers
Given w=−4−3iw = -4 - 3i, find ∣w∣|w| and arg⁡w\arg w.
  1. A∣w∣=5|w| = 5, arg⁡w≈−2.5\arg w \approx -2.5 rad
  2. B∣w∣=5|w| = 5, arg⁡w≈2.5\arg w \approx 2.5 rad
  3. C∣w∣=25|w| = 25, arg⁡w≈−2.5\arg w \approx -2.5 rad
  4. D∣w∣=7|w| = 7, arg⁡w≈−0.64\arg w \approx -0.64 rad

Question 1003

[1 marks]Complex numbers
To rationalise w=5uv=5+10i−2−iw=\dfrac{5u}{v}=\dfrac{5+10i}{-2-i}, multiply numerator and denominator by −2+i-2+i. Find the denominator (−2−i)(−2+i)(-2-i)(-2+i).

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Question 1004

[2 marks]Complex numbers
Find the numerator (5+10i)(−2+i)(5+10i)(-2+i), in the form p+qip+qi, before dividing by 5 to get ww.

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Question 1005

[1 marks]Complex numbers
Given u=1+2iu=1+2i, find 5u5u, in the form p+qip+qi.

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Question 1006

[2 marks]Complex numbers
For w=−4−3iw=-4-3i, find the reference angle tan⁡−1(34)\tan^{-1}\left(\tfrac34\right), correct to 1 decimal place (degrees), used to find arg⁡w\arg w.

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Question 1101

[1 marks]Remainder and factor theorems
The polynomial 2x3−11x2+ax+b2x^3 - 11x^2 + ax + b is exactly divisible by (x−2)(x-2) and leaves remainder −36-36 when divided by (x+1)(x+1). Find aa and bb.
  1. Aa=−17a = -17, b=6b = 6
  2. Ba=6a = 6, b=−17b = -17
  3. Ca=17a = 17, b=6b = 6
  4. Da=17a = 17, b=−6b = -6

Question 1102

[1 marks]Remainder and factor theorems
Factorise 2x3−11x2+17x−62x^3 - 11x^2 + 17x - 6 completely.
  1. A(x−2)(2x−1)(x−3)(x-2)(2x-1)(x-3)
  2. B(x−2)(x−1)(2x−3)(x-2)(x-1)(2x-3)
  3. C(x−2)(2x+1)(x−3)(x-2)(2x+1)(x-3)
  4. D(x+2)(2x−1)(x−3)(x+2)(2x-1)(x-3)

Question 1103

[1 marks]Remainder and factor theorems
Evaluate f(2)=16−44+2a+bf(2)=16-44+2a+b for f(x)=2x3−11x2+ax+bf(x)=2x^3-11x^2+ax+b (which equals 0 since (x−2)(x-2) is a factor), and state the resulting equation in the form 2a+b=k2a+b=k. Find kk.

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Question 1104

[1 marks]Remainder and factor theorems
Evaluate f(−1)=−2−11−a+bf(-1)=-2-11-a+b (which equals −36-36, the given remainder), and state the resulting equation in the form −a+b=k-a+b=k. Find kk.

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Question 1105

[2 marks]Remainder and factor theorems
Using 2a+b=282a+b=28 and −a+b=−23-a+b=-23, solve simultaneously for aa.

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Question 1106

[2 marks]Remainder and factor theorems
Dividing 2x3−11x2+17x−62x^3-11x^2+17x-6 by (x−2)(x-2) gives a quadratic quotient. Find this quotient, in the form ax2+bx+cax^2+bx+c.

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Question 1107

[1 marks]Remainder and factor theorems
Factorise the quadratic 2x2−7x+32x^2-7x+3, stating both linear factors.

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Question 1201

[1 marks]Improper algebraic fractions and integration
Express f(x)=x3x2−5x+6f(x) = \dfrac{x^3}{x^2 - 5x + 6} in the form Ax+B+Cx−2+Dx−3Ax + B + \dfrac{C}{x-2} + \dfrac{D}{x-3}.
  1. AA=1A = 1, B=5B = 5, C=8C = 8, D=−27D = -27
  2. BA=1A = 1, B=5B = 5, C=−8C = -8, D=27D = 27
  3. CA=1A = 1, B=−5B = -5, C=−8C = -8, D=27D = 27
  4. DA=1A = 1, B=5B = 5, C=−27C = -27, D=8D = 8

Question 1202

[1 marks]Improper algebraic fractions and integration
Find the exact value of ∫46(x+5−8x−2+27x−3)dx\displaystyle\int_4^6 \left(x + 5 - \dfrac{8}{x-2} + \dfrac{27}{x-3}\right)dx.
  1. A20−8ln⁡4+27ln⁡320 - 8\ln 4 + 27\ln 3
  2. B20+8ln⁡2−27ln⁡320 + 8\ln 2 - 27\ln 3
  3. C20−8ln⁡2+27ln⁡320 - 8\ln 2 + 27\ln 3
  4. D48−8ln⁡4+27ln⁡348 - 8\ln 4 + 27\ln 3

Question 1203

[2 marks]Improper algebraic fractions and integration
By polynomial division, find the quotient obtained when x3x^3 is divided by x2−5x+6x^2-5x+6, in the form Ax+BAx+B.

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Question 1204

[1 marks]Improper algebraic fractions and integration
Substituting x=2x=2 into the partial-fraction identity for f(x)=x+5+Cx−2+Dx−3f(x)=x+5+\dfrac{C}{x-2}+\dfrac{D}{x-3}, find CC.

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Question 1205

[1 marks]Improper algebraic fractions and integration
Substituting x=3x=3 into the partial-fraction identity, find DD.

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Question 1206

[3 marks]Improper algebraic fractions and integration
Find the antiderivative of x+5−8x−2+27x−3x+5-\dfrac{8}{x-2}+\dfrac{27}{x-3} with respect to xx (ignore the constant of integration).

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Question 1301

[1 marks]R-form and trigonometric equations
Express 2cos⁡x−5sin⁡x2\cos x - 5\sin x in the form Rcos⁡(x+θ)R\cos(x + \theta), where R>0R > 0 and 0°<θ<90°0° < \theta < 90°.
  1. A7cos⁡(x+68°)7\cos(x + 68°)
  2. B29cos⁡(x+22°)\sqrt{29}\cos(x + 22°)
  3. C29cos⁡(x+68°)\sqrt{29}\cos(x + 68°)
  4. D29cos⁡(x−68°)\sqrt{29}\cos(x - 68°)

Question 1302

[1 marks]R-form and trigonometric equations
Solve 2cos⁡2x−5sin⁡2x=2.52\cos 2x - 5\sin 2x = 2.5 for 0°≤x≤360°0° \le x \le 360°, giving answers to the nearest degree.
  1. Ax=115°, 177°, 295°, 357°x = 115°,\ 177°,\ 295°,\ 357°
  2. Bx=62°, 118°, 242°, 298°x = 62°,\ 118°,\ 242°,\ 298°
  3. Cx=115°, 295°x = 115°,\ 295° only
  4. Dx=56°, 146°, 236°, 326°x = 56°,\ 146°,\ 236°,\ 326°

Question 1303

[1 marks]R-form and trigonometric equations
At which values of xx between 0°0° and 360°360° does 2cos⁡2x−5sin⁡2x2\cos 2x - 5\sin 2x take its maximum value?
  1. Ax=0°x = 0° and x=180°x = 180°
  2. Bx=56°x = 56° and x=236°x = 236°
  3. Cx=146°x = 146° and x=326°x = 326°
  4. Dx=68°x = 68° and x=248°x = 248°

Question 1304

[2 marks]R-form and trigonometric equations
For 2cos⁡x−5sin⁡x=Rcos⁡(x+θ)2\cos x-5\sin x=R\cos(x+\theta) with Rcos⁡θ=2R\cos\theta=2, Rsin⁡θ=5R\sin\theta=5, find the exact value of RR.

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Question 1305

[2 marks]R-form and trigonometric equations
State the values of xx between 0∘0^\circ and 360∘360^\circ at which 2cos⁡2x−5sin⁡2x2\cos2x-5\sin2x takes its minimum value.

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Question 1306

[2 marks]R-form and trigonometric equations
Using R=29R=\sqrt{29}, θ=68∘\theta=68^\circ, solve cos⁡(2x+68∘)=2,529\cos(2x+68^\circ)=\dfrac{2{,}5}{\sqrt{29}} for the principal value of 2x+68∘2x+68^\circ, correct to the nearest degree.

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Question 1307

[1 marks]R-form and trigonometric equations
State cos⁡(2x+68∘)\cos(2x+68^\circ) as a decimal, correct to 4 decimal places (i.e. 2,529\dfrac{2{,}5}{\sqrt{29}}).

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Question 1401

[1 marks]Trapezium rule, areas and volumes of revolution
Use the trapezium rule with 4 ordinates to evaluate ∫01.5x3sin⁡2x dx\displaystyle\int_0^{1.5} x^3\sin^2 x\,dx, correct to 3 significant figures.
  1. A0.6050.605
  2. B1.211.21
  3. C1.471.47
  4. D2.422.42

Question 1402

[1 marks]Trapezium rule, areas and volumes of revolution
Find the area of the region bounded by the curve y2=4xy^2 = 4x and the line y=xy = x.
  1. A83\tfrac{8}{3} units2^2
  2. B163\tfrac{16}{3} units2^2
  3. C323\tfrac{32}{3} units2^2
  4. D88 units2^2

Question 1403

[1 marks]Trapezium rule, areas and volumes of revolution
The region bounded by y2=4xy^2 = 4x and y=xy = x is rotated through 360°360° about the yy-axis. Find the volume generated.
  1. A25615π\dfrac{256}{15}\pi units3^3
  2. B6415π\dfrac{64}{15}\pi units3^3
  3. C12815π\dfrac{128}{15}\pi units3^3
  4. D83π\dfrac{8}{3}\pi units3^3

Question 1404

[2 marks]Trapezium rule, areas and volumes of revolution
For the trapezium rule with h=0,5h=0{,}5 over x=0,0,5,1,1,5x=0, 0{,}5, 1, 1{,}5 applied to x3sin⁡2xx^3\sin^2x, state the ordinate at x=0,5x=0{,}5, correct to 4 decimal places.

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Question 1405

[2 marks]Trapezium rule, areas and volumes of revolution
State the ordinate at x=1x=1 (i.e. 13sin⁡2(1)1^3\sin^2(1)), correct to 4 decimal places.

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Question 1406

[2 marks]Trapezium rule, areas and volumes of revolution
State the ordinate at x=1,5x=1{,}5 (i.e. 1,53sin⁡2(1,5)1{,}5^3\sin^2(1{,}5)), correct to 4 decimal places.

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Question 1407

[1 marks]Trapezium rule, areas and volumes of revolution
Find the xx-coordinate (other than the origin) where y2=4xy^2=4x and y=xy=x intersect.

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Question 1408

[2 marks]Trapezium rule, areas and volumes of revolution
Find the antiderivative of 4x−x\sqrt{4x}-x with respect to xx (ignore the constant of integration).

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Question 1409

[1 marks]Trapezium rule, areas and volumes of revolution
For the volume integral V=π∫04(y2−y416)dyV=\pi\int_0^4\left(y^2-\dfrac{y^4}{16}\right)dy, find the antiderivative of y2−y416y^2-\dfrac{y^4}{16} with respect to yy (ignore π\pi and the constant of integration).

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Question 1501

[1 marks]Maclaurin series and laws in linear form
Given y=11+cos⁡xy = \dfrac{1}{1+\cos x}, find d2ydx2\dfrac{d^2y}{dx^2} when x=0x = 0.
  1. A12\tfrac{1}{2}
  2. B14\tfrac{1}{4}
  3. C18\tfrac{1}{8}
  4. D00

Question 1502

[1 marks]Maclaurin series and laws in linear form
Find the Maclaurin series of y=11+cos⁡xy = \dfrac{1}{1+\cos x} up to and including the term in x2x^2.
  1. A12+14x2\tfrac{1}{2} + \tfrac{1}{4}x^2
  2. B1+18x21 + \tfrac{1}{8}x^2
  3. C12+12x+18x2\tfrac{1}{2} + \tfrac{1}{2}x + \tfrac{1}{8}x^2
  4. D12+18x2\tfrac{1}{2} + \tfrac{1}{8}x^2

Question 1503

[1 marks]Maclaurin series and laws in linear form
Variables xx and yy satisfy y=abxy = ab^x. Taking logarithms to base 10, the graph of lg⁡y\lg y against xx is a straight line whose gradient and intercept are
  1. Agradient lg⁡a\lg a, intercept lg⁡b\lg b
  2. Bgradient bb, intercept aa
  3. Cgradient lg⁡b\lg b, intercept lg⁡a\lg a
  4. Dgradient lg⁡(ab)\lg(ab), intercept 00

Question 1504

[2 marks]Maclaurin series and laws in linear form
For y=11+cos⁡xy=\dfrac{1}{1+\cos x}, find dydx\dfrac{dy}{dx}, in terms of xx.

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Question 1505

[1 marks]Maclaurin series and laws in linear form
Evaluate dydx=sin⁡x(1+cos⁡x)2\dfrac{dy}{dx}=\dfrac{\sin x}{(1+\cos x)^2} at x=0x=0.

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Question 1506

[1 marks]Maclaurin series and laws in linear form
Evaluate y=11+cos⁡xy=\dfrac{1}{1+\cos x} at x=0x=0.

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Question 1507

[1 marks]Maclaurin series and laws in linear form
Taking logarithms of y=abxy=ab^x, express lg⁡y\lg y as a linear function of xx, in the form lg⁡y=mx+c\lg y=mx+c, in terms of lg⁡a\lg a and lg⁡b\lg b.

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Question 1508

[1 marks]Maclaurin series and laws in linear form
Given the gradient of the linear graph is 0,70{,}7 (i.e. lg⁡b=0,7\lg b=0{,}7), find bb, correct to 1 decimal place.

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Question 1509

[1 marks]Maclaurin series and laws in linear form
Given the intercept of the linear graph is 2,32{,}3 (i.e. lg⁡a=2,3\lg a=2{,}3), find aa, correct to 1 decimal place.

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Question 1510

[3 marks]Maclaurin series and laws in linear form
Differentiate dydx=sin⁡x(1+cos⁡x)2\dfrac{dy}{dx}=\dfrac{\sin x}{(1+\cos x)^2} again (quotient rule) and simplify to find d2ydx2\dfrac{d^2y}{dx^2}, as a function of xx.

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Question 1511

[1 marks]Maclaurin series and laws in linear form
Evaluate the numerator 2+cos⁡x−cos⁡2x2+\cos x-\cos^2x at x=0x=0 (used in d2ydx2\dfrac{d^2y}{dx^2}).

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Question 1512

[1 marks]Maclaurin series and laws in linear form
Evaluate the denominator (1+cos⁡x)3(1+\cos x)^3 at x=0x=0 (used in d2ydx2\dfrac{d^2y}{dx^2}).

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