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ZIMSEC A Level · J2014

Pure Mathematics Paper 1 June 2014

Questions
83
Total marks
120

Sit this paper online

Questions
83
Pass mark
50
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 / exponential equations
Solve the equation 274x34=9(x−1)\dfrac{27^{4x}}{3^4} = 9^{(x-1)} for xx.
  1. Ax=15x=\dfrac15
  2. Bx=−15x=-\dfrac15
  3. Cx=310x=\dfrac{3}{10}
  4. Dx=−12x=-\dfrac12

Question 102

[1 marks]indices / exponential equations
Express 274x27^{4x} as 33 raised to what power, in terms of xx?

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Question 103

[1 marks]indices / exponential equations
Express 9x−19^{x-1} as 33 raised to what power, in terms of xx?

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Question 201

[1 marks]binomial expansion
Find the coefficient of x2x^2 in the expansion of (x−25x)6\left(x-\dfrac{2}{5x}\right)^6.
  1. A−125-\dfrac{12}{5}
  2. B125\dfrac{12}{5}
  3. C65\dfrac{6}{5}
  4. D245\dfrac{24}{5}

Question 202

[1 marks]binomial expansion
In the expansion of (x−25x)6\left(x-\dfrac{2}{5x}\right)^6, the general term is (6r)x6−r(−25x)r=(6r)(−1)r2r5rx6−2r\binom{6}{r}x^{6-r}\left(-\dfrac{2}{5x}\right)^r=\binom{6}{r}(-1)^r\dfrac{2^r}{5^r}x^{6-2r}. What value of rr gives the term in x2x^2?
  1. Ar=1r=1
  2. Br=4r=4
  3. Cr=3r=3
  4. Dr=2r=2

Question 203

[1 marks]binomial expansion
Evaluate (62)\binom{6}{2}, the binomial coefficient used in finding the x2x^2 term of (x−25x)6\left(x-\frac{2}{5x}\right)^6.

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Question 301

[1 marks]small changes / differentiation
Given that y=12x−1y=\dfrac{1}{\sqrt{2x-1}}, find the exact value of dydx\dfrac{dy}{dx} at x=5x=5.
  1. A−154-\dfrac{1}{54}
  2. B−13-\dfrac13
  3. C−127-\dfrac{1}{27}
  4. D127\dfrac{1}{27}

Question 302

[1 marks]small changes / differentiation
Given that y=12x−1y=\dfrac{1}{\sqrt{2x-1}}, find the approximate percentage change in yy when x=5x=5 is decreased by 5%.
  1. A1.39% increase
  2. B2.78% increase
  3. C2.78% decrease
  4. D5.56% increase

Question 303

[1 marks]small changes / differentiation
Find δx\delta x, the change in xx, when x=5x=5 is decreased by 5%5\%.

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Question 304

[1 marks]small changes / differentiation
State the value of y=12x−1y=\dfrac{1}{\sqrt{2x-1}} at x=5x=5.

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Question 401

[1 marks]composite functions
The functions ff and gg are defined for x∈Rx\in\mathbb{R} by f:x→x2+4x+1f:x\to x^2+4x+1 and g:x→ax+bg:x\to ax+b. Given that fg(2)=−2fg(2)=-2 and gf(0)=−3gf(0)=-3, find the values of aa and bb.
  1. Aa=2, b=−5a=2,\ b=-5
  2. Ba=4, b=−7a=4,\ b=-7
  3. Ca=−2, b=−1a=-2,\ b=-1
  4. Da=−5, b=2a=-5,\ b=2

Question 402

[1 marks]composite functions
Evaluate f(0)f(0) for f(x)=x2+4x+1f(x)=x^2+4x+1 (used to find gf(0)gf(0)).

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Question 403

[2 marks]composite functions
Given a+b=−3a+b=-3 (from gf(0)=−3gf(0)=-3), using fg(2)=−2fg(2)=-2 leads to (2a+b)2+4(2a+b)+3=0(2a+b)^2+4(2a+b)+3=0, which factorises as (2a+b+1)(2a+b+3)=0(2a+b+1)(2a+b+3)=0. State the two possible values of 2a+b2a+b.

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Question 404

[1 marks]composite functions
Using a+b=−3a+b=-3 and 2a+b=−12a+b=-1 (the valid branch), find aa.

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Question 501

[1 marks]polynomial remainder theorem / inequalities
The polynomials P(x)=x3−x2+4xP(x)=x^3-x^2+4x and Q(x)=x3+6x−10Q(x)=x^3+6x-10 leave the same remainder when divided by x−ax-a. Find the possible values of aa.
  1. Aa=−2±211a=-2\pm2\sqrt{11}
  2. Ba=−1±11a=-1\pm\sqrt{11}
  3. Ca=−1±14a=-1\pm\sqrt{14}
  4. Da=1±11a=1\pm\sqrt{11}

Question 502

[1 marks]polynomial remainder theorem / inequalities
The polynomials P(x)=x3−x2+4xP(x)=x^3-x^2+4x and Q(x)=x3+6x−10Q(x)=x^3+6x-10 leave the same remainder when divided by x−ax-a, where a=−1±11a=-1\pm\sqrt{11}. Solve the inequality P(x)>Q(x)P(x)>Q(x).
  1. A−1−11<x<1+11-1-\sqrt{11}<x<1+\sqrt{11}
  2. B−1−11<x<−1+11-1-\sqrt{11}<x<-1+\sqrt{11}
  3. Cx<−1−11x<-1-\sqrt{11} or x>−1+11x>-1+\sqrt{11}
  4. D1−11<x<1+111-\sqrt{11}<x<1+\sqrt{11}

Question 503

[2 marks]polynomial remainder theorem / inequalities
Setting P(a)=Q(a)P(a)=Q(a) for P(x)=x3−x2+4xP(x)=x^3-x^2+4x and Q(x)=x3+6x−10Q(x)=x^3+6x-10 leads to a2+2a−10=0a^2+2a-10=0. State the discriminant b2−4acb^2-4ac used in the quadratic formula.

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Question 504

[1 marks]polynomial remainder theorem / inequalities
P(x)−Q(x)P(x)-Q(x) simplifies to −x2−2x+10-x^2-2x+10, so P(x)>Q(x)P(x)>Q(x) becomes x2+2x−10<0x^2+2x-10<0. Evaluate x2+2x−10x^2+2x-10 at x=0x=0 (to confirm 00 lies in the solution interval).

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Question 601

[1 marks]optimisation / differentiation
An open rectangular tin (no lid) has a square base of side xx cm and holds 500 cm3^3 of liquid when full. Which expression gives the total area, AA cm2^2, of sheet metal used?
  1. AA=2x2+2000xA=2x^2+\dfrac{2000}{x}
  2. BA=x2+1000xA=x^2+\dfrac{1000}{x}
  3. CA=x2+500xA=x^2+\dfrac{500}{x}
  4. DA=x2+2000xA=x^2+\dfrac{2000}{x}

Question 602

[1 marks]optimisation / differentiation
An open rectangular tin (no lid) has a square base of side xx cm and holds 500 cm3^3 of liquid when full, so its sheet-metal area is A=x2+2000xA=x^2+\dfrac{2000}{x} cm2^2. Find the minimum value of AA.
  1. A300 cm2300\ \text{cm}^2
  2. B400 cm2400\ \text{cm}^2
  3. C150 cm2150\ \text{cm}^2
  4. D200 cm2200\ \text{cm}^2

Question 603

[2 marks]optimisation / differentiation
For the open tin with square base of side xx, find hh (the height) in terms of xx, given x2h=500x^2h=500.

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Question 604

[2 marks]optimisation / differentiation
Setting dAdx=2x−2000x2=0\dfrac{dA}{dx}=2x-\dfrac{2000}{x^2}=0, solve x3=1000x^3=1000 for xx.

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Question 701

[1 marks]trigonometry / areas
In triangle ABCABC, AB=5AB=5 cm, BC=7BC=7 cm and CA=8CA=8 cm. Find angle BACBAC.
  1. Aπ3\dfrac{\pi}{3}
  2. Bπ4\dfrac{\pi}{4}
  3. C2π3\dfrac{2\pi}{3}
  4. Dπ6\dfrac{\pi}{6}

Question 702

[1 marks]trigonometry / areas
In triangle ABCABC, AB=5AB=5 cm, BC=7BC=7 cm, CA=8CA=8 cm and angle BAC=π3BAC=\dfrac{\pi}{3}. Find sin⁡B\sin B.
  1. A5314\dfrac{5\sqrt3}{14}
  2. B47\dfrac{4}{7}
  3. C437\dfrac{4\sqrt3}{7}
  4. D5316\dfrac{5\sqrt3}{16}

Question 703

[1 marks]trigonometry / areas
Triangle ABCABC has AB=5AB=5 cm, BC=7BC=7 cm, CA=8CA=8 cm and angle BAC=π3BAC=\dfrac{\pi}{3}, with sin⁡B=437\sin B=\dfrac{4\sqrt3}{7}. A circular arc, centred at AA, is drawn touching line BCBC at RR, so its radius is AR=ABsin⁡BAR=AB\sin B. Find the exact area of the region between the triangle and the sector swept out by the arc between the two sides from AA.
  1. A103−25π610\sqrt3-\dfrac{25\pi}{6}
  2. B103−200π4910\sqrt3-\dfrac{200\pi}{49}
  3. C103+200π4910\sqrt3+\dfrac{200\pi}{49}
  4. D10310\sqrt3

Question 704

[2 marks]trigonometry / areas
Find the exact area of triangle ABCABC, given AB=5AB=5 cm, CA=8CA=8 cm and angle BAC=π3BAC=\frac{\pi}{3}.

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Question 705

[2 marks]trigonometry / areas
Find the exact radius r=ARr=AR of the arc, given AR=ABsin⁡BAR=AB\sin B with AB=5AB=5 and sin⁡B=437\sin B=\frac{4\sqrt3}{7}.

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Question 801

[1 marks]numerical methods / Newton-Raphson
For the equation x−3+3tan⁡x=0x-3+3\tan x=0 (with xx in radians), let h(x)=x−3+3tan⁡xh(x)=x-3+3\tan x. Evaluate h(0.6)h(0.6) correct to 2 decimal places.
  1. A−2.37-2.37
  2. B−1.72-1.72
  3. C−0.35-0.35
  4. D0.350.35

Question 802

[1 marks]numerical methods / Newton-Raphson
Taking x0=0.6x_0=0.6 as the first approximation to the root of x−3+3tan⁡x=0x-3+3\tan x=0 (with xx in radians), apply the Newton-Raphson method to find the root correct to 3 decimal places.
  1. A0.60.6
  2. B0.6520.652
  3. C0.6620.662
  4. D0.6640.664

Question 803

[2 marks]numerical methods / Newton-Raphson
Evaluate h(0.7)=0.7−3+3tan⁡(0.7)h(0.7)=0.7-3+3\tan(0.7), correct to 2 decimal places (used to confirm the root lies between 0.6 and 0.7).

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Question 804

[2 marks]numerical methods / Newton-Raphson
Given h′(x)=1+3sec⁡2xh'(x)=1+3\sec^2x, evaluate h′(0.6)h'(0.6), correct to 3 significant figures (used in the Newton-Raphson formula).

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Question 805

[2 marks]numerical methods / Newton-Raphson
Using x0=0,6x_0=0,6, h(0,6)=−0,3476h(0,6)=-0,3476 and h′(0,6)=5,404h'(0,6)=5,404, find the first Newton-Raphson iterate x1=x0−h(x0)h′(x0)x_1=x_0-\dfrac{h(x_0)}{h'(x_0)}, correct to 4 decimal places.

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Question 901

[1 marks]differential equations / separation of variables
A rumour spreads through a school so that the proportion xx who have heard it satisfies dxdt=kx(1−x)\dfrac{dx}{dt}=kx(1-x), where initially (at t=0t=0) a proportion cc of the population had heard it. Which expression correctly gives xx in terms of tt?
  1. Ax=cc+(1−c)ektx=\dfrac{c}{c+(1-c)e^{kt}}
  2. Bx=ce−ktc+(1−c)e−ktx=\dfrac{ce^{-kt}}{c+(1-c)e^{-kt}}
  3. Cx=1−c(1−c)+ce−ktx=\dfrac{1-c}{(1-c)+ce^{-kt}}
  4. Dx=cc+(1−c)e−ktx=\dfrac{c}{c+(1-c)e^{-kt}}

Question 902

[2 marks]differential equations / separation of variables
Express 1x(1−x)\dfrac{1}{x(1-x)} in partial fractions, in the form 1x+1A−x\dfrac{1}{x}+\dfrac{1}{A-x}. State AA.

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Question 903

[2 marks]differential equations / separation of variables
Integrating gives ln⁡x−ln⁡(1−x)=kt+C\ln x-\ln(1-x)=kt+C, i.e. x1−x=Aekt\dfrac{x}{1-x}=Ae^{kt}. Using x=cx=c at t=0t=0, find AA in terms of cc.

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Question 904

[3 marks]differential equations / separation of variables
From x1−x=c1−cekt\dfrac{x}{1-x}=\dfrac{c}{1-c}e^{kt}, cross-multiplying and collecting xx terms gives x[(1−c)+cekt]=x[(1-c)+ce^{kt}] = [blank]. State the blank (the right-hand side).

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Question 1001

[1 marks]coordinate geometry / circles
The line l1l_1 has equation x+3y−33=0x+3y-33=0. The line l2l_2 is parallel to l1l_1 and passes through P(3,0)P(3,0). Find the equation of l2l_2.
  1. Ax+3y+3=0x+3y+3=0
  2. Bx+3y−33=0x+3y-33=0
  3. Cx+3y−3=0x+3y-3=0
  4. D3x−y−9=03x-y-9=0

Question 1002

[1 marks]coordinate geometry / circles
The line l1l_1 has equation x+3y−33=0x+3y-33=0. Find the product of the gradient of l1l_1 and the gradient of the line joining P(3,0)P(3,0) to Q(6,9)Q(6,9).
  1. A33
  2. B11
  3. C−1-1
  4. D−19-\dfrac19

Question 1003

[1 marks]coordinate geometry / circles
The line l1l_1 has equation x+3y−33=0x+3y-33=0, and l2l_2 (parallel to l1l_1) passes through P(3,0)P(3,0). Given that R(9,−2)R(9,-2) lies on l2l_2 and Q(6,9)Q(6,9) lies on l1l_1, find the equation of the circle passing through PP, QQ and RR.
  1. A(x−7.5)2+(y−3.5)2=68.5(x-7.5)^2+(y-3.5)^2=68.5
  2. B(x−3.5)2+(y−7.5)2=32.5(x-3.5)^2+(y-7.5)^2=32.5
  3. C(x−7.5)2+(y−3.5)2=32.5(x-7.5)^2+(y-3.5)^2=32.5
  4. D(x+7.5)2+(y+3.5)2=32.5(x+7.5)^2+(y+3.5)^2=32.5

Question 1004

[2 marks]coordinate geometry / circles
Find the gradient of line l1:x+3y−33=0l_1: x+3y-33=0.

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Question 1005

[1 marks]coordinate geometry / circles
Find the gradient of the line joining P(3,0)P(3,0) to Q(6,9)Q(6,9).

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Question 1006

[2 marks]coordinate geometry / circles
Find the midpoint of PQPQ, where P(3,0)P(3,0) and Q(6,9)Q(6,9).

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Question 1101

[1 marks]vectors
Points PP, QQ, RR have position vectors i−2j+k\mathbf{i}-2\mathbf{j}+\mathbf{k}, 3i−j−k3\mathbf{i}-\mathbf{j}-\mathbf{k} and 7i+j−5k7\mathbf{i}+\mathbf{j}-5\mathbf{k} respectively. Find the angle between OQ→\overrightarrow{OQ} and PR→\overrightarrow{PR}, correct to 1 decimal place.
  1. A44.7°44.7°
  2. B38.9°38.9°
  3. C45.3°45.3°
  4. D72.5°72.5°

Question 1102

[1 marks]vectors
Points PP and QQ have position vectors i−2j+k\mathbf{i}-2\mathbf{j}+\mathbf{k} and 3i−j−k3\mathbf{i}-\mathbf{j}-\mathbf{k}, and RR has position vector 7i+j−5k7\mathbf{i}+\mathbf{j}-5\mathbf{k}. Find the value of kk such that PR→=kPQ→\overrightarrow{PR}=k\overrightarrow{PQ}, hence showing PP, QQ, RR are collinear.
  1. Ak=3k=3
  2. Bk=13k=\dfrac13
  3. Ck=2k=2
  4. Dk=−3k=-3

Question 1103

[1 marks]vectors
Point PP has position vector i−2j+k\mathbf{i}-2\mathbf{j}+\mathbf{k}. Given that the vector λi+3j−4k\lambda\mathbf{i}+3\mathbf{j}-4\mathbf{k} is perpendicular to OP→\overrightarrow{OP}, find the value of λ\lambda.
  1. Aλ=−2\lambda=-2
  2. Bλ=10\lambda=10
  3. Cλ=2\lambda=2
  4. Dλ=4\lambda=4

Question 1104

[2 marks]vectors
Find vector PR→=OR→−OP→\overrightarrow{PR}=\overrightarrow{OR}-\overrightarrow{OP}, for P=(1,−2,1)P=(1,-2,1) and R=(7,1,−5)R=(7,1,-5).

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Question 1105

[2 marks]vectors
Find ∣OQ→∣|\overrightarrow{OQ}|, correct to 3 significant figures, for OQ→=3i−j−k\overrightarrow{OQ}=3\mathbf{i}-\mathbf{j}-\mathbf{k}.

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Question 1106

[2 marks]vectors
Find ∣PR→∣|\overrightarrow{PR}|, for PR→=6i+3j−6k\overrightarrow{PR}=6\mathbf{i}+3\mathbf{j}-6\mathbf{k}.

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Question 1201

[1 marks]sequences and series
The numbers pp, 10, qq are three consecutive terms of an arithmetic progression, and pp, 6, qq are three consecutive terms of a geometric progression. Which pair of equations correctly expresses these two conditions?
  1. Ap−q=20, pq=36p-q=20,\ pq=36
  2. Bp+q=20, pq=6p+q=20,\ pq=6
  3. Cp+q=10, pq=36p+q=10,\ pq=36
  4. Dp+q=20, pq=36p+q=20,\ pq=36

Question 1202

[1 marks]sequences and series
The numbers pp, 10, qq are consecutive terms of an arithmetic progression, and pp, 6, qq are consecutive terms of a geometric progression, so p+q=20p+q=20 and pq=36pq=36, giving p2−20p+36=0p^2-20p+36=0. Find the values of pp, qq and the common ratio rr for which the geometric progression converges.
  1. Ap=18, q=2, r=13p=18,\ q=2,\ r=\tfrac13
  2. Bp=2, q=18, r=3p=2,\ q=18,\ r=3
  3. Cp=18, q=2, r=3p=18,\ q=2,\ r=3
  4. Dp=2, q=18, r=13p=2,\ q=18,\ r=\tfrac13

Question 1203

[1 marks]sequences and series
A woman measures her child's height at birth and then at monthly intervals. The height increases by 5% each month. Find how many measurements (including the one at birth) she will have made by the time the child's height is first at least double its height at birth.
  1. A1414
  2. B1515
  3. C1616
  4. D1717

Question 1204

[2 marks]sequences and series
From p+q=20p+q=20 and pq=36pq=36, substituting q=20−pq=20-p gives p2−20p+36=0p^2-20p+36=0. State the discriminant 400−144400-144.

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Question 1205

[2 marks]sequences and series
Solve p2−20p+36=0p^2-20p+36=0 using 256=16\sqrt{256}=16. State the two roots for pp, in ascending order.

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Question 1206

[2 marks]sequences and series
For the child's height increasing 5%5\% per month, find nn from (1.05)n=2(1.05)^n=2 using logarithms, correct to 1 decimal place (before rounding up to the next whole measurement).

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Question 1301

[1 marks]complex numbers
Given that a=2+ia=2+i and b=1+3ib=1+3i, find abab.
  1. A−1+7i-1+7i
  2. B−1−7i-1-7i
  3. C−7+i-7+i
  4. D5+7i5+7i

Question 1302

[1 marks]complex numbers
Given that a=2+ia=2+i and b=1+3ib=1+3i, find ab\dfrac{a}{b}.
  1. A−12+12i-\tfrac12+\tfrac12i
  2. B12+12i\tfrac12+\tfrac12i
  3. C54−54i\tfrac54-\tfrac54i
  4. D12−12i\tfrac12-\tfrac12i

Question 1303

[1 marks]complex numbers
Given that a=2+ia=2+i and b=1+3ib=1+3i, so that ab=−1+7iab=-1+7i, find the modulus and argument of abab.
  1. Amodulus 525\sqrt2, argument ≈81.9°\approx81.9°
  2. Bmodulus 222\sqrt2, argument ≈98.1°\approx98.1°
  3. Cmodulus 525\sqrt2, argument ≈−81.9°\approx-81.9°
  4. Dmodulus 525\sqrt2, argument ≈98.1°\approx98.1°

Question 1304

[1 marks]complex numbers
Given that a=2+ia=2+i and b=1+3ib=1+3i, so that ab=12−12i\dfrac{a}{b}=\tfrac12-\tfrac12i, find the modulus and argument of ab\dfrac{a}{b}.
  1. Amodulus 12\tfrac12, argument −45°-45°
  2. Bmodulus 22\tfrac{\sqrt2}{2}, argument 135°135°
  3. Cmodulus 22\tfrac{\sqrt2}{2}, argument −45°-45°
  4. Dmodulus 22\tfrac{\sqrt2}{2}, argument 45°45°

Question 1305

[2 marks]complex numbers
Given a=2+ia=2+i and b=1+3ib=1+3i, evaluate i(1+3i)i(1+3i) in the form c+dic+di (part of expanding abab).

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Question 1306

[2 marks]complex numbers
Rationalising ab=2+i1+3i\dfrac{a}{b}=\dfrac{2+i}{1+3i} by multiplying by the conjugate 1−3i1-3i, find the resulting denominator (1+3i)(1−3i)(1+3i)(1-3i).

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Question 1307

[2 marks]complex numbers
Given a=2+ia=2+i, b=1+3ib=1+3i, so ab=−1+7iab=-1+7i, in which quadrant of the Argand diagram does abab lie?
  1. AThird quadrant
  2. BFourth quadrant
  3. CFirst quadrant
  4. DSecond quadrant

Question 1401

[1 marks]Maclaurin series
Given that f(x)=e3xsin⁡xf(x)=e^{3x}\sin x, f′(x)=e3xcos⁡x+3e3xsin⁡xf'(x)=e^{3x}\cos x+3e^{3x}\sin x, f′′(x)=e3x(8sin⁡x+6cos⁡x)f''(x)=e^{3x}(8\sin x+6\cos x) and f′′′(x)=e3x(18sin⁡x+26cos⁡x)f'''(x)=e^{3x}(18\sin x+26\cos x), write down the Maclaurin series of f(x)f(x) up to and including the term in x3x^3.
  1. Ax+3x2+133x3x+3x^2+\dfrac{13}{3}x^3
  2. Bx+3x2+263x3x+3x^2+\dfrac{26}{3}x^3
  3. Cx+3x2−133x3x+3x^2-\dfrac{13}{3}x^3
  4. Dx+6x2+26x3x+6x^2+26x^3

Question 1402

[1 marks]Maclaurin series
Given that f(x)=e3xsin⁡xf(x)=e^{3x}\sin x, evaluate f(0.5)f(0.5) correct to 4 decimal places.
  1. A0.03910.0391
  2. B1.64461.6446
  3. C2.14862.1486
  4. D2.14902.1490

Question 1403

[1 marks]Maclaurin series
Given that f(x)=e3xsin⁡xf(x)=e^{3x}\sin x, with exact value f(0.5)≈2.1486f(0.5)\approx2.1486 and Maclaurin approximation f(x)≈x+3x2+133x3f(x)\approx x+3x^2+\dfrac{13}{3}x^3, state the absolute error in using the Maclaurin series to estimate f(0.5)f(0.5).
  1. A0.35700.3570
  2. B0.35730.3573
  3. C0.54170.5417
  4. D1.79171.7917

Question 1404

[2 marks]Maclaurin series
Given f′′(x)=e3x(8sin⁡x+6cos⁡x)f''(x)=e^{3x}(8\sin x+6\cos x), evaluate f′′(0)f''(0).

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Question 1405

[2 marks]Maclaurin series
Given f′′′(x)=e3x(18sin⁡x+26cos⁡x)f'''(x)=e^{3x}(18\sin x+26\cos x), evaluate f′′′(0)f'''(0).

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Question 1406

[3 marks]Maclaurin series
Using the Maclaurin series f(x)≈x+3x2+133x3f(x)\approx x+3x^2+\frac{13}{3}x^3, evaluate the series estimate at x=0.5x=0.5 (before comparing to the exact value).

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Question 1501

[1 marks]integration / area and volume
A curve has equation f(x)=−x3+3x+2f(x)=-x^3+3x+2. The point A(0,2)A(0,2) and point B(1,4)B(1,4) both lie on the curve, and the line ABAB has equation y=2x+2y=2x+2. Find the area of the finite region bounded by the curve and the line ABAB.
  1. A12\dfrac12
  2. B14\dfrac14
  3. C−14-\dfrac14
  4. D134\dfrac{13}{4}

Question 1502

[1 marks]integration / area and volume
A curve has equation f(x)=−x3+3x+2f(x)=-x^3+3x+2, and the line ABAB from A(0,2)A(0,2) to B(1,4)B(1,4) has equation y=2x+2y=2x+2. Find the volume of the solid formed when the finite region bounded by the curve and the line ABAB is rotated completely about the xx-axis, correct to 2 decimal places.
  1. A0.240.24
  2. B1.611.61
  3. C5.065.06
  4. D34.3834.38

Question 1503

[2 marks]integration / area and volume
The curve f(x)=−x3+3x+2f(x)=-x^3+3x+2 and line ABAB: y=2x+2y=2x+2 intersect where f(x)−(2x+2)=−x3+x=0f(x)-(2x+2)=-x^3+x=0. Factorise −x3+x-x^3+x completely.

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Question 1504

[2 marks]integration / area and volume
State the yy-coordinate of point BB on the curve f(x)=−x3+3x+2f(x)=-x^3+3x+2, where x=1x=1.

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Question 1505

[2 marks]integration / area and volume
Find the gradient of line ABAB, from A(0,2)A(0,2) to B(1,4)B(1,4).

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Question 1506

[3 marks]integration / area and volume
Expanding (−x3+3x+2)2=x6−6x4−4x3+9x2+12x+4(-x^3+3x+2)^2 = x^6-6x^4-4x^3+9x^2+12x+4, state the coefficient of x4x^4 (part of the volume integrand).

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Question 1601

[1 marks]trigonometry
In quadrilateral RPQSRPQS, PQ=8PQ=8 cm, RS=12RS=12 cm, angle QPR=60°QPR=60°, angle QSR=90°QSR=90°, diagonal QR=13QR=13 cm, and angle PRQ=θPRQ=\theta. Find sin⁡θ\sin\theta and cos⁡θ\cos\theta.
  1. Asin⁡θ=413, cos⁡θ=31713\sin\theta=\tfrac{4}{13},\ \cos\theta=\tfrac{3\sqrt{17}}{13}
  2. Bsin⁡θ=4313, cos⁡θ=−1113\sin\theta=\tfrac{4\sqrt3}{13},\ \cos\theta=-\tfrac{11}{13}
  3. Csin⁡θ=13316, cos⁡θ=1113\sin\theta=\tfrac{13\sqrt3}{16},\ \cos\theta=\tfrac{11}{13}
  4. Dsin⁡θ=4313, cos⁡θ=1113\sin\theta=\tfrac{4\sqrt3}{13},\ \cos\theta=\tfrac{11}{13}

Question 1602

[1 marks]trigonometry
In quadrilateral RPQSRPQS with PQ=8PQ=8 cm, RS=12RS=12 cm, QR=13QR=13 cm, angle QPR=60°QPR=60° and angle QSR=90°QSR=90°, it is given that sin⁡θ=4313\sin\theta=\tfrac{4\sqrt3}{13}, cos⁡θ=1113\cos\theta=\tfrac{11}{13} (where θ=∠PRQ\theta=\angle PRQ), and cos⁡(∠QRS)=1213\cos(\angle QRS)=\tfrac{12}{13}, sin⁡(∠QRS)=513\sin(\angle QRS)=\tfrac{5}{13}. Find cos⁡(∠PRS)\cos(\angle PRS), where ∠PRS=θ+∠QRS\angle PRS=\theta+\angle QRS.
  1. A4(53−33)169\dfrac{4(5\sqrt3-33)}{169}
  2. B4(33−53)169\dfrac{4(33-5\sqrt3)}{169}
  3. C4(33+53)169\dfrac{4(33+5\sqrt3)}{169}
  4. D132−20313\dfrac{132-20\sqrt3}{13}

Question 1603

[1 marks]trigonometry
The graph of y=sin⁡xy=\sin x is drawn for 0≤x≤3π0\le x\le3\pi. Point AA has coordinates (α,sin⁡α)(\alpha,\sin\alpha), where 0<α<π20<\alpha<\dfrac{\pi}{2}. A horizontal line through AA cuts the graph again at points BB, CC and DD (in order of increasing xx). Write down the coordinates of BB, CC and DD.
  1. AB=(π−α,sin⁡α), C=(2π+α,sin⁡α), D=(3π−α,sin⁡α)B=(\pi-\alpha,\sin\alpha),\ C=(2\pi+\alpha,\sin\alpha),\ D=(3\pi-\alpha,\sin\alpha)
  2. BB=(π+α,sin⁡α), C=(2π−α,sin⁡α), D=(3π+α,sin⁡α)B=(\pi+\alpha,\sin\alpha),\ C=(2\pi-\alpha,\sin\alpha),\ D=(3\pi+\alpha,\sin\alpha)
  3. CB=(π−α,sin⁡α), C=(2π−α,sin⁡α), D=(3π−α,sin⁡α)B=(\pi-\alpha,\sin\alpha),\ C=(2\pi-\alpha,\sin\alpha),\ D=(3\pi-\alpha,\sin\alpha)
  4. DB=(2π+α,sin⁡α), C=(π−α,sin⁡α), D=(3π−α,sin⁡α)B=(2\pi+\alpha,\sin\alpha),\ C=(\pi-\alpha,\sin\alpha),\ D=(3\pi-\alpha,\sin\alpha)

Question 1604

[1 marks]trigonometry
Solve the equation esin⁡x=2e^{\sin x}=2 for 0≤x≤3π0\le x\le3\pi, giving your answers correct to 2 decimal places.
  1. Ax≈2.38, 7.05x\approx2.38,\ 7.05
  2. Bx≈0.77, 2.38x\approx0.77,\ 2.38
  3. Cx≈0.77, 2.38, 7.05, 8.66x\approx0.77,\ 2.38,\ 7.05,\ 8.66
  4. Dx≈0.76, 2.38, 7.04, 8.66x\approx0.76,\ 2.38,\ 7.04,\ 8.66

Question 1605

[2 marks]trigonometry
In right triangle QRSQRS (right angle at SS), QR=13QR=13 cm and RS=12RS=12 cm. Find QSQS.

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Question 1606

[1 marks]trigonometry
State sin⁡(∠QRS)\sin(\angle QRS), given QS=5QS=5 cm and QR=13QR=13 cm.

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Question 1607

[1 marks]trigonometry
State cos⁡(∠QRS)\cos(\angle QRS), given RS=12RS=12 cm and QR=13QR=13 cm.

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Question 1608

[2 marks]trigonometry
For 0≤x≤3π0\le x\le3\pi, write down the solution of sin⁡x>sin⁡α\sin x>\sin\alpha (in terms of α\alpha) for the first interval only, between AA and BB.

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Question 1609

[2 marks]trigonometry
Using cos⁡θ=1113\cos\theta=\frac{11}{13} and cos⁡(∠QRS)=1213\cos(\angle QRS)=\frac{12}{13}, evaluate cos⁡θcos⁡(∠QRS)\cos\theta\cos(\angle QRS) (the first term of cos⁡(∠PRS)\cos(\angle PRS)).

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Question 1610

[2 marks]trigonometry
Using sin⁡θ=4313\sin\theta=\frac{4\sqrt3}{13} and sin⁡(∠QRS)=513\sin(\angle QRS)=\frac{5}{13}, evaluate sin⁡θsin⁡(∠QRS)\sin\theta\sin(\angle QRS) (the second term of cos⁡(∠PRS)\cos(\angle PRS)).

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