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ZIMSEC A Level · N2013

Pure Mathematics Paper 1 November 2013

Questions
75
Total marks
120

Sit this paper online

Questions
75
Pass mark
45
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 5x−1+5x−2=305^{x-1} + 5^{x-2} = 30.
  1. Ax=125x = 125
  2. Bx=2x = 2
  3. Cx=3x = 3
  4. Dx=5x = 5

Question 102

[2 marks]indices / exponential equations
Writing 5x−1+5x−2=625⋅5x=305^{x-1}+5^{x-2} = \frac{6}{25}\cdot5^x = 30, find the value of 5x5^x (before solving for xx).

Answer this when you sit the paper.

Question 201

[1 marks]rational inequalities
Solve the inequality 3x+19−x2≥−1\dfrac{3x+1}{9-x^2} \ge -1.
  1. Ax<−3x < -3 or −2≤x<3-2 \le x < 3 or x≥5x \ge 5
  2. Bx≤−2x \le -2 or x≥5x \ge 5
  3. C−3<x<3-3 < x < 3
  4. D−2≤x≤5-2 \le x \le 5

Question 202

[2 marks]rational inequalities
Adding 1 to both sides of 3x+19−x2≥−1\dfrac{3x+1}{9-x^2}\ge-1 and simplifying gives (x−5)(x+2)(3−x)(3+x)≤0\dfrac{(x-5)(x+2)}{(3-x)(3+x)}\le0. State all four critical values, in ascending order.

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Question 203

[1 marks]rational inequalities
Which critical values from the inequality (x−5)(x+2)(3−x)(3+x)≤0\dfrac{(x-5)(x+2)}{(3-x)(3+x)}\le0 are excluded from the solution (where the denominator is zero)?

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Question 301

[1 marks]arithmetic progression / sectors
A circular plank is cut into 12 sectors whose areas are in arithmetic progression, the largest being twice the smallest. Find the angle between the straight edges of the smallest sector.
  1. Aπ9\dfrac{\pi}{9}
  2. Bπ12\dfrac{\pi}{12}
  3. Cπ6\dfrac{\pi}{6}
  4. D2π9\dfrac{2\pi}{9}

Question 302

[1 marks]arithmetic progression / sectors
The 12 sector angles are in AP with first term aa, summing to 2π2\pi, and the largest is twice the smallest. Find the common difference dd in terms of aa.

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Question 303

[2 marks]arithmetic progression / sectors
Using d=a11d=\frac{a}{11}, write the largest angle a+11da+11d in terms of aa alone.

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Question 401

[1 marks]vectors / scalar product
OM→=(2p−2)i+(1−p)j+(p−2)k\overrightarrow{OM} = (2p-2)\mathbf{i} + (1-p)\mathbf{j} + (p-2)\mathbf{k} and ON→=(p+2)i+pj+2pk\overrightarrow{ON} = (p+2)\mathbf{i} + p\mathbf{j} + 2p\mathbf{k}. Find pp such that ∣OM→∣=∣ON→∣|\overrightarrow{OM}| = |\overrightarrow{ON}|.
  1. Ap=−518p = -\tfrac{5}{18}
  2. Bp=43p = \tfrac{4}{3}
  3. Cp=12p = \tfrac{1}{2}
  4. Dp=518p = \tfrac{5}{18}

Question 402

[1 marks]vectors / scalar product
OM→=(2p−2)i+(1−p)j+(p−2)k\overrightarrow{OM} = (2p-2)\mathbf{i} + (1-p)\mathbf{j} + (p-2)\mathbf{k} and ON→=(p+2)i+pj+2pk\overrightarrow{ON} = (p+2)\mathbf{i} + p\mathbf{j} + 2p\mathbf{k}. Find the values of pp for which MO^N=90°M\hat{O}N = 90°.
  1. Ap=−43p = -\tfrac{4}{3} or p=1p = 1
  2. Bp=43p = \tfrac{4}{3} or p=−1p = -1
  3. Cp=518p = \tfrac{5}{18} only
  4. Dp=3p = 3 or p=−4p = -4

Question 403

[2 marks]vectors / scalar product
Expand and simplify ∣OM→∣2=(2p−2)2+(1−p)2+(p−2)2|\overrightarrow{OM}|^2 = (2p-2)^2+(1-p)^2+(p-2)^2 as a quadratic in pp.

Answer this when you sit the paper.

Question 404

[1 marks]vectors / scalar product
Expand and simplify ∣ON→∣2=(p+2)2+p2+(2p)2|\overrightarrow{ON}|^2 = (p+2)^2+p^2+(2p)^2 as a quadratic in pp.

Answer this when you sit the paper.

Question 501

[1 marks]factor theorem / polynomials
Given that (x+k)(x + k) is a factor of x3+2x2−3x−6x^3 + 2x^2 - 3x - 6, where k>0k > 0, find kk.
  1. Ak=6k = 6
  2. Bk=1k = 1
  3. Ck=3k = 3
  4. Dk=2k = 2

Question 502

[1 marks]factor theorem / polynomials
Find the exact roots of the equation x3+2x2−3x−6=0x^3 + 2x^2 - 3x - 6 = 0.
  1. Ax=−2x = -2 only
  2. Bx=−2x = -2, x=3x = \sqrt{3}, x=−3x = -\sqrt{3}
  3. Cx=2x = 2, x=3x = \sqrt{3}, x=−3x = -\sqrt{3}
  4. Dx=−2x = -2, x=3x = 3, x=−3x = -3

Question 503

[1 marks]factor theorem / polynomials
Evaluate f(−2)f(-2) for f(x)=x3+2x2−3x−6f(x)=x^3+2x^2-3x-6, used to confirm (x+2)(x+2) is a factor.

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Question 504

[2 marks]factor theorem / polynomials
Divide f(x)=x3+2x2−3x−6f(x)=x^3+2x^2-3x-6 by (x+2)(x+2) to find the quadratic quotient.

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Question 601

[1 marks]implicit differentiation / tangents
Given the curve x225+y216=1\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1, find dydx\dfrac{dy}{dx}.
  1. A16x25y\dfrac{16x}{25y}
  2. B−xy-\dfrac{x}{y}
  3. C−25y16x-\dfrac{25y}{16x}
  4. D−16x25y-\dfrac{16x}{25y}

Question 602

[1 marks]implicit differentiation / tangents
Find the equation of the tangent to x225+y216=1\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1 at the point (3,165)\left(3, \tfrac{16}{5}\right).
  1. A5x+3y=255x + 3y = 25
  2. B3x−5y=253x - 5y = 25
  3. C3x+5y=163x + 5y = 16
  4. D3x+5y=253x + 5y = 25

Question 603

[2 marks]implicit differentiation / tangents
Find the numeric value of dydx\dfrac{dy}{dx} at the point (3,165)\left(3,\frac{16}{5}\right) on x225+y216=1\dfrac{x^2}{25}+\dfrac{y^2}{16}=1.

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Question 604

[1 marks]implicit differentiation / tangents
State the yy-intercept of the tangent line 3x+5y=253x+5y=25.

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Question 701

[1 marks]differential equations / separation of variables
Find the general solution of dydx=1−y\dfrac{dy}{dx} = 1 - y in the form y=Ae−x+Cy = Ae^{-x} + C.
  1. Ay=Aex+1y = Ae^{x} + 1
  2. By=Ae−xy = Ae^{-x}
  3. Cy=Ae−x+1y = Ae^{-x} + 1
  4. Dy=Ae−x−1y = Ae^{-x} - 1

Question 702

[1 marks]differential equations / separation of variables
Find the particular solution of dydx=1−y\dfrac{dy}{dx} = 1 - y for which the yy-intercept is 3.
  1. Ay=3e−xy = 3e^{-x}
  2. By=1+2e−xy = 1 + 2e^{-x}
  3. Cy=1+3e−xy = 1 + 3e^{-x}
  4. Dy=2+e−xy = 2 + e^{-x}

Question 703

[2 marks]differential equations / separation of variables
For the general solution y=Ae−x+1y=Ae^{-x}+1, if instead the curve passes through (0,5)(0,5), find AA.

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Question 704

[2 marks]differential equations / separation of variables
For the particular solution y=1+2e−xy=1+2e^{-x}, find yy when x=ln⁡2x=\ln2.

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Question 801

[1 marks]complex numbers
Given z2=−1−3iz_2 = -1 - \sqrt{3}i, find its modulus and argument.
  1. A∣z2∣=2|z_2| = 2, arg⁡z2=2π3\arg z_2 = \tfrac{2\pi}{3}
  2. B∣z2∣=2|z_2| = 2, arg⁡z2=−2π3\arg z_2 = -\tfrac{2\pi}{3}
  3. C∣z2∣=2|z_2| = 2, arg⁡z2=−π3\arg z_2 = -\tfrac{\pi}{3}
  4. D∣z2∣=4|z_2| = 4, arg⁡z2=−π3\arg z_2 = -\tfrac{\pi}{3}

Question 802

[1 marks]complex numbers
Given z1=−1+iz_1 = -1 + i and z2=−1−3iz_2 = -1 - \sqrt{3}i, find z1z2z_1 z_2.
  1. A−1+3i-1 + \sqrt{3}i
  2. B(1+3)+(3−1)i(1+\sqrt{3}) + (\sqrt{3}-1)i
  3. C(1−3)+(3+1)i(1-\sqrt{3}) + (\sqrt{3}+1)i
  4. D(1+3)−(3−1)i(1+\sqrt{3}) - (\sqrt{3}-1)i

Question 803

[3 marks]complex numbers
Given z1=−1+iz_1=-1+i and z2=−1−3iz_2=-1-\sqrt3i, find z1z2\dfrac{z_1}{z_2} in the form a+iba+ib.

Answer this when you sit the paper.

Question 804

[1 marks]complex numbers
In which quadrant of the Argand diagram does z2=−1−3iz_2=-1-\sqrt3i lie?
  1. ASecond quadrant
  2. BThird quadrant
  3. CFourth quadrant
  4. DFirst quadrant

Question 901

[1 marks]trapezium rule / integration by substitution
Use the trapezium rule with 5 ordinates to evaluate ∫0141+x2 dx\displaystyle\int_0^1 \dfrac{4}{1+x^2}\,dx correct to 4 decimal places.
  1. A3.00003.0000
  2. B3.10003.1000
  3. C3.13123.1312
  4. D3.14163.1416

Question 902

[1 marks]trapezium rule / integration by substitution
By using the substitution x=tan⁡θx = \tan\theta, find the exact value of ∫0141+x2 dx\displaystyle\int_0^1 \dfrac{4}{1+x^2}\,dx.
  1. Aπ\pi
  2. Bπ4\tfrac{\pi}{4}
  3. C44
  4. D2π2\pi

Question 903

[1 marks]trapezium rule / integration by substitution
The trapezium rule gives 3.13123.1312 as an approximation to ∫0141+x2 dx=π\displaystyle\int_0^1 \dfrac{4}{1+x^2}\,dx = \pi. Find the percentage error correct to 2 decimal places.
  1. A0.33%0.33\%
  2. B0.01%0.01\%
  3. C3.31%3.31\%
  4. D1.04%1.04\%

Question 904

[2 marks]trapezium rule / integration by substitution
State the ordinate y2=f(0.5)y_2 = f(0.5) for f(x)=41+x2f(x)=\dfrac{4}{1+x^2}, used in the trapezium rule.

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Question 905

[2 marks]trapezium rule / integration by substitution
State the ordinate y4=f(1)y_4 = f(1) for f(x)=41+x2f(x)=\dfrac{4}{1+x^2}.

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Question 906

[2 marks]trapezium rule / integration by substitution
Using x=tan⁡θx=\tan\theta, find the upper limit of θ\theta (in terms of π\pi) corresponding to x=1x=1.

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Question 1001

[1 marks]implicit differentiation / Maclaurin series
If y=(1+3e−x)1/2y = \left(1 + 3e^{-x}\right)^{1/2}, find dydx\dfrac{dy}{dx} at x=0x = 0.
  1. A−3-3
  2. B−32-\tfrac{3}{2}
  3. C−34-\tfrac{3}{4}
  4. D34\tfrac{3}{4}

Question 1002

[1 marks]implicit differentiation / Maclaurin series
For y=(1+3e−x)1/2y = \left(1 + 3e^{-x}\right)^{1/2}, find the Maclaurin expansion of yy up to and including the term in x2x^2.
  1. A2−34x+1564x22 - \tfrac{3}{4}x + \tfrac{15}{64}x^2
  2. B1−34x+1564x21 - \tfrac{3}{4}x + \tfrac{15}{64}x^2
  3. C2−34x+1532x22 - \tfrac{3}{4}x + \tfrac{15}{32}x^2
  4. D2+34x+1564x22 + \tfrac{3}{4}x + \tfrac{15}{64}x^2

Question 1003

[1 marks]implicit differentiation / Maclaurin series
For y=(1+3e−x)1/2y=(1+3e^{-x})^{1/2}, state y(0)y(0).

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Question 1004

[2 marks]implicit differentiation / Maclaurin series
Differentiating y2=1+3e−xy^2=1+3e^{-x} implicitly gives 2ydydx=−3e−x2y\frac{dy}{dx} = -3e^{-x}. Evaluate the right side at x=0x=0.

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Question 1005

[3 marks]implicit differentiation / Maclaurin series
Find y′′(0)y''(0), the second derivative of y=(1+3e−x)1/2y=(1+3e^{-x})^{1/2} at x=0x=0 (used in the Maclaurin series).

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Question 1101

[1 marks]areas and volumes of revolution
The region RR is bounded by the curve f(x)=a−x2f(x) = a - x^2 and the axes, and has area 18. Find the value of aa.
  1. Aa=27a = 27
  2. Ba=6a = 6
  3. Ca=9a = 9
  4. Da=3a = 3

Question 1102

[1 marks]areas and volumes of revolution
The region bounded by y=9−x2y = 9 - x^2 and the axes (for −3≤x≤0-3 \le x \le 0) is rotated completely about the xx-axis. Find the exact volume generated.
  1. A648π5\dfrac{648\pi}{5} cubic units
  2. B324π5\dfrac{324\pi}{5} cubic units
  3. C1296π5\dfrac{1296\pi}{5} cubic units
  4. D81π81\pi cubic units

Question 1103

[2 marks]areas and volumes of revolution
The curve f(x)=a−x2f(x)=a-x^2 meets the xx-axis at x=−ax=-\sqrt a. Using a=9a=9, state this xx-intercept.

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Question 1104

[3 marks]areas and volumes of revolution
Evaluate −23a3/2-\dfrac{2}{3}a^{3/2} at a=9a=9 (the definite integral value used to find aa).

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Question 1105

[2 marks]areas and volumes of revolution
Evaluate 81x−6x3+x5581x-6x^3+\dfrac{x^5}{5} (part of the volume antiderivative) at x=−3x=-3.

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Question 1201

[1 marks]trigonometric identities and equations
Simplify (tan⁡θ+1)2(tan⁡θ−1)2\dfrac{(\tan\theta + 1)^2}{(\tan\theta - 1)^2}.
  1. A1+sin⁡2θ1−sin⁡2θ\dfrac{1 + \sin 2\theta}{1 - \sin 2\theta}
  2. B1+cos⁡2θ1−cos⁡2θ\dfrac{1 + \cos 2\theta}{1 - \cos 2\theta}
  3. Ctan⁡22θ\tan^2 2\theta
  4. D1−sin⁡2θ1+sin⁡2θ\dfrac{1 - \sin 2\theta}{1 + \sin 2\theta}

Question 1202

[1 marks]trigonometric identities and equations
Solve the equation tan⁡xcos⁡2x=sin⁡x\tan x \cos 2x = \sin x for 0°≤x≤360°0° \le x \le 360°.
  1. Ax=120°, 240°x = 120°,\ 240° only
  2. Bx=0°, 60°, 180°, 300°, 360°x = 0°,\ 60°,\ 180°,\ 300°,\ 360°
  3. Cx=0°, 120°, 180°, 240°, 360°x = 0°,\ 120°,\ 180°,\ 240°,\ 360°
  4. Dx=0°, 180°, 360°x = 0°,\ 180°,\ 360° only

Question 1203

[2 marks]trigonometric identities and equations
Expanding (sin⁡θ+cos⁡θ)2(\sin\theta+\cos\theta)^2 gives 1+2sin⁡θcos⁡θ1+2\sin\theta\cos\theta. Simplify 2sin⁡θcos⁡θ2\sin\theta\cos\theta to a single trig ratio of 2θ2\theta.

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Question 1204

[2 marks]trigonometric identities and equations
Solving tan⁡xcos⁡2x=sin⁡x\tan x\cos2x=\sin x by factoring sin⁡x(cos⁡2x−cos⁡x)=0\sin x(\cos2x-\cos x)=0, one branch is sin⁡x=0\sin x=0. State this branch's solutions for 0°≤x≤360°0°\le x\le360°.

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Question 1205

[2 marks]trigonometric identities and equations
The other branch requires cos⁡2x=cos⁡x\cos2x=\cos x, i.e. 2cos⁡2x−cos⁡x−1=02\cos^2x-\cos x-1=0, which factorises as (2cos⁡x+1)(cos⁡x−c)=0(2\cos x+1)(\cos x-c)=0. State cc.

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Question 1206

[2 marks]trigonometric identities and equations
From (2cos⁡x+1)(cos⁡x−1)=0(2\cos x+1)(\cos x-1)=0, state the value of cos⁡x\cos x from the first factor.

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Question 1301

[1 marks]coordinate geometry of the circle / radian measure
Find the centre and radius of the circle passing through P(0,0)P(0, 0), Q(1,7)Q(1, 7) and R(7,−1)R(7, -1).
  1. Acentre (4,3)(4, 3), radius 2525
  2. Bcentre (−4,−3)(-4, -3), radius 55
  3. Ccentre (3,4)(3, 4), radius 55
  4. Dcentre (4,3)(4, 3), radius 55

Question 1302

[1 marks]coordinate geometry of the circle / radian measure
A region PQSRPQSR is bounded by sectors APQAPQ and ARSARS with angle PAQ=2PAQ = 2 radians. Given that the perimeter of PQSRPQSR is 32 cm, calculate the length of ASAS.
  1. A1616 cm
  2. B44 cm
  3. C66 cm
  4. D88 cm

Question 1303

[1 marks]coordinate geometry of the circle / radian measure
The region PQSRPQSR between two sectors of angle 2 radians has AS=8AS = 8 cm and area 28 cm2^2. Find the length of AQAQ.
  1. A44 cm
  2. B55 cm
  3. C66 cm
  4. D3636 cm

Question 1304

[2 marks]coordinate geometry of the circle / radian measure
Using P(0,0)P(0,0) and Q(1,7)Q(1,7) equidistant from centre (g,f)(g,f) gives a linear equation 2g+14f=k2g+14f=k. State kk.

Answer this when you sit the paper.

Question 1305

[2 marks]coordinate geometry of the circle / radian measure
Using P(0,0)P(0,0) and R(7,−1)R(7,-1) equidistant from centre (g,f)(g,f) gives a linear equation 14g−2f=k14g-2f=k. State kk.

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Question 1306

[3 marks]coordinate geometry of the circle / radian measure
Area of PQSR=12(2)(r22−r12)=64−r12PQSR = \dfrac12(2)(r_2^2-r_1^2) = 64-r_1^2. Using area =28=28 cm2^2, find r12r_1^2 (before taking the square root to get AQAQ).

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Question 1401

[1 marks]functions / inverse functions / binomial expansion
Functions are defined by f:x→(x−2)(x+3)f : x \to (x-2)(x+3) for x≥−12x \ge -\tfrac{1}{2} and h:x→4x2+1h : x \to 4x^2 + 1. Find the exact values of xx for which fh(x)=0fh(x) = 0.
  1. Ax=±12x = \pm \tfrac{1}{2}
  2. Bx=2x = 2 or x=−3x = -3
  3. Cx=12x = \tfrac{1}{2} only
  4. Dx=±1x = \pm 1

Question 1402

[1 marks]functions / inverse functions / binomial expansion
For f:x→(x−2)(x+3)f : x \to (x-2)(x+3) with x≥−12x \ge -\tfrac{1}{2}, find f−1(x)f^{-1}(x) and its domain.
  1. Af−1(x)=x+6f^{-1}(x) = \sqrt{x + 6}, domain x≥−6x \ge -6
  2. Bf−1(x)=−12−x+254f^{-1}(x) = -\tfrac{1}{2} - \sqrt{x + \tfrac{25}{4}}, domain x≥−254x \ge -\tfrac{25}{4}
  3. Cf−1(x)=12+x+254f^{-1}(x) = \tfrac{1}{2} + \sqrt{x + \tfrac{25}{4}}, domain x≥0x \ge 0
  4. Df−1(x)=−12+x+254f^{-1}(x) = -\tfrac{1}{2} + \sqrt{x + \tfrac{25}{4}}, domain x≥−254x \ge -\tfrac{25}{4}

Question 1403

[1 marks]functions / inverse functions / binomial expansion
For which values of xx is the series expansion of (4−x)1/22x2−1\dfrac{(4-x)^{1/2}}{2x^2 - 1} valid?
  1. A∣x∣<4|x| < 4
  2. B∣x∣<2|x| < 2
  3. C∣x∣<1|x| < 1
  4. D−12<x<12-\tfrac{1}{\sqrt{2}} < x < \tfrac{1}{\sqrt{2}}

Question 1404

[3 marks]functions / inverse functions / binomial expansion
Find the series expansion of (4−x)1/2(4-x)^{1/2} up to and including the term in x2x^2.

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Question 1405

[2 marks]functions / inverse functions / binomial expansion
Find the series expansion of 12x2−1\dfrac{1}{2x^2-1} up to and including the term in x2x^2.

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Question 1406

[3 marks]functions / inverse functions / binomial expansion
Multiplying the two expansions of (4−x)1/2(4-x)^{1/2} and 12x2−1\dfrac{1}{2x^2-1} and simplifying, find the coefficient of x2x^2 in the full expansion.

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Question 1501

[1 marks]optimisation / implicit differentiation / small increments
A closed right circular cylinder has capacity 300 ml. Write down its total surface area AA in terms of the radius rr.
  1. AA=πr2+600rA = \pi r^2 + \dfrac{600}{r}
  2. BA=2πr2+600πrA = 2\pi r^2 + \dfrac{600}{\pi r}
  3. CA=2πr2+600rA = 2\pi r^2 + \dfrac{600}{r}
  4. DA=2πr2+300rA = 2\pi r^2 + \dfrac{300}{r}

Question 1502

[1 marks]optimisation / implicit differentiation / small increments
A closed cylindrical tin of capacity 300 ml has surface area A=2πr2+600rA = 2\pi r^2 + \dfrac{600}{r}. Find the radius and height that minimise AA, correct to 3 significant figures.
  1. Ar=7.26r = 7.26 cm, h=3.63h = 3.63 cm
  2. Br=3.63r = 3.63 cm, h=7.26h = 7.26 cm
  3. Cr=3.63r = 3.63 cm, h=3.63h = 3.63 cm
  4. Dr=4.57r = 4.57 cm, h=4.57h = 4.57 cm

Question 1503

[1 marks]optimisation / implicit differentiation / small increments
Given that x−yln⁡x=ln⁡yx - y\ln x = \ln y, find dydx\dfrac{dy}{dx} in terms of xx and yy.
  1. Ay(x−y)x(1+yln⁡x)\dfrac{y(x - y)}{x(1 + y\ln x)}
  2. Bx−y1+yln⁡x\dfrac{x - y}{1 + y\ln x}
  3. Cy(x+y)x(1−yln⁡x)\dfrac{y(x + y)}{x(1 - y\ln x)}
  4. Dx(1+yln⁡x)y(x−y)\dfrac{x(1 + y\ln x)}{y(x - y)}

Question 1504

[2 marks]optimisation / implicit differentiation / small increments
At x=1x=1, solve x−yln⁡x=ln⁡yx-y\ln x=\ln y for yy.

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Question 1505

[3 marks]optimisation / implicit differentiation / small increments
Evaluate dydx\dfrac{dy}{dx} at x=1x=1, y=ey=e, using dydx=y(x−y)x(1+yln⁡x)\dfrac{dy}{dx}=\dfrac{y(x-y)}{x(1+y\ln x)}.

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Question 1506

[3 marks]optimisation / implicit differentiation / small increments
Setting dAdr=4πr−600r2=0\dfrac{dA}{dr}=4\pi r-\dfrac{600}{r^2}=0, find r3r^3 in terms of π\pi (exact fraction).

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Question 1601

[1 marks]stationary points / Newton-Raphson method
Find the stationary points of y=2sin⁡2x+1y = 2\sin 2x + 1 for 0≤x≤π0 \le x \le \pi and determine their nature.
  1. A(π4,3)\left(\tfrac{\pi}{4}, 3\right) minimum and (3π4,−1)\left(\tfrac{3\pi}{4}, -1\right) maximum
  2. B(π2,1)\left(\tfrac{\pi}{2}, 1\right) maximum only
  3. C(π4,3)\left(\tfrac{\pi}{4}, 3\right) maximum and (3π4,−1)\left(\tfrac{3\pi}{4}, -1\right) minimum
  4. D(π4,2)\left(\tfrac{\pi}{4}, 2\right) maximum and (3π4,0)\left(\tfrac{3\pi}{4}, 0\right) minimum

Question 1602

[1 marks]stationary points / Newton-Raphson method
How many real roots does the equation ex(1+x)=2e^x(1 + x) = 2 have?
  1. Anone
  2. Bexactly one
  3. Cexactly two
  4. Dexactly three

Question 1603

[1 marks]stationary points / Newton-Raphson method
Taking x1=0.5x_1 = 0.5, apply the Newton-Raphson method twice to ex(1+x)=2e^x(1 + x) = 2 and give the root correct to 3 decimal places.
  1. A0.3620.362
  2. B0.3750.375
  3. C0.3850.385
  4. D0.5000.500

Question 1604

[2 marks]stationary points / Newton-Raphson method
For y=2sin⁡2x+1y=2\sin2x+1, find d2ydx2\dfrac{d^2y}{dx^2} (used to classify the stationary points).

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Question 1605

[2 marks]stationary points / Newton-Raphson method
Evaluate yy at x=π4x=\frac{\pi}{4} for y=2sin⁡2x+1y=2\sin2x+1 (the maximum value).

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Question 1606

[3 marks]stationary points / Newton-Raphson method
Using x1=0.5x_1=0.5, find the first Newton-Raphson iterate x2x_2, correct to 6 decimal places, for f(x)=ex(1+x)−2f(x)=e^x(1+x)-2.

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Question 1607

[2 marks]stationary points / Newton-Raphson method
Find f′(x)f'(x) for f(x)=ex(1+x)−2f(x)=e^x(1+x)-2, in terms of xx.

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