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ZIMSEC A Level · N2016

Pure Mathematics Paper 1 November 2016

Questions
80
Total marks
120

Sit this paper online

Questions
80
Pass mark
48
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]complex numbers
The complex number u=1−2+iu = \dfrac{1}{-2+i}. Find ∣u∣|u|.
  1. A15\dfrac{1}{5}
  2. B25\dfrac{2}{5}
  3. C5\sqrt{5}
  4. D55\dfrac{\sqrt{5}}{5}

Question 102

[1 marks]complex numbers
The complex number u=1−2+iu = \dfrac{1}{-2+i}. Find arg⁡u\arg u correct to the nearest 0.1°0.1°.
  1. A153.4°153.4°
  2. B−153.4°-153.4°
  3. C−26.6°-26.6°
  4. D26.6°26.6°

Question 103

[2 marks]complex numbers
Express u=1−2+iu=\dfrac1{-2+i} in the form x+iyx+iy.

Answer this when you sit the paper.

Question 201

[1 marks]binomial expansion
Find the series expansion of (9+2x)−3/2(9+2x)^{-3/2} up to and including the term in x2x^2.
  1. A127−x81+51458x2\tfrac{1}{27} - \tfrac{x}{81} + \tfrac{5}{1458}x^2
  2. B127−x9+5162x2\tfrac{1}{27} - \tfrac{x}{9} + \tfrac{5}{162}x^2
  3. C13−x81+51458x2\tfrac{1}{3} - \tfrac{x}{81} + \tfrac{5}{1458}x^2
  4. D127+x81+51458x2\tfrac{1}{27} + \tfrac{x}{81} + \tfrac{5}{1458}x^2

Question 202

[1 marks]binomial expansion
Factoring (9+2x)−3/2=9−3/2(1+2x9)−3/2(9+2x)^{-3/2}=9^{-3/2}\left(1+\dfrac{2x}9\right)^{-3/2}, state 9−3/29^{-3/2} as a fraction.

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Question 203

[2 marks]binomial expansion
In the binomial expansion (1+u)n≈1+nu+…(1+u)^n\approx1+nu+\ldots with n=−32n=-\dfrac32, u=2x9u=\dfrac{2x}9 (used for (1+2x9)−3/2\left(1+\frac{2x}9\right)^{-3/2}, before multiplying by 9−3/29^{-3/2}), state the coefficient of xx.

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Question 301

[1 marks]circles / tangency / discriminant
The line y=2mx+cy = 2mx + c is a tangent to the circle x2+y2=4x^2 + y^2 = 4. Which condition must mm and cc satisfy?
  1. A16m2−c2−4=016m^2 - c^2 - 4 = 0
  2. B16m2−c2+4=016m^2 - c^2 + 4 = 0
  3. C4m2−c2+4=04m^2 - c^2 + 4 = 0
  4. D16m2+c2−4=016m^2 + c^2 - 4 = 0

Question 302

[1 marks]circles / tangency / discriminant
Substituting y=2mx+cy=2mx+c into x2+y2=4x^2+y^2=4 gives (1+4m2)x2+4mcx+(c2−4)=0(1+4m^2)x^2+4mcx+(c^2-4)=0. State the coefficient of xx in this quadratic, in terms of mm and cc.

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Question 303

[2 marks]circles / tangency / discriminant
For the tangency condition (4mc)2−4(1+4m2)(c2−4)=0(4mc)^2-4(1+4m^2)(c^2-4)=0 (from substituting y=2mx+cy=2mx+c into x2+y2=4x^2+y^2=4 and setting the discriminant to zero), state (4mc)2(4mc)^2 in simplified form.

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Question 401

[1 marks]modulus functions / inequalities
Given f(x)=2−∣x+3∣f(x) = 2 - |x+3|, solve the inequality ∣f(x)∣>1|f(x)| > 1.
  1. Ax<−6x < -6 or x>0x > 0 only
  2. Bx<−6x < -6 or −4<x<−2-4 < x < -2 or x>0x > 0
  3. C−6<x<0-6 < x < 0
  4. D−4<x<−2-4 < x < -2 only

Question 402

[2 marks]modulus functions / inequalities
For f(x)=2−∣x+3∣f(x)=2-|x+3|, state the coordinates of its vertex (maximum point), as 'x,y'.

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Question 403

[1 marks]modulus functions / inequalities
For f(x)=2−∣x+3∣f(x)=2-|x+3|, state the two values of xx where f(x)=0f(x)=0 (where y=∣f(x)∣y=|f(x)| touches the xx-axis), as 'a, b' in ascending order.

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Question 501

[1 marks]linearising relationships / graphs
Variables xx and yy satisfy y=Pxy = \dfrac{P}{\sqrt{x}}. Which graph is a straight line, and what does its gradient give?
  1. Ayy against x\sqrt{x}: gradient PP
  2. Bln⁡y\ln y against ln⁡x\ln x: gradient PP
  3. Cyy against xx: gradient PP
  4. Dyy against 1x\dfrac{1}{\sqrt{x}}: gradient PP

Question 502

[2 marks]linearising relationships / graphs
For the data y=Pxy=\dfrac P{\sqrt x} plotted against 1x\dfrac1{\sqrt x}, evaluate 1x\dfrac1{\sqrt x} at x=20x=20, correct to 4 decimal places.

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Question 503

[2 marks]linearising relationships / graphs
For the same data, evaluate 1x\dfrac1{\sqrt x} at x=75x=75, correct to 4 decimal places.

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Question 601

[1 marks]related rates
An aeroplane flies horizontally at height 5 500 m; its angle of elevation from an observer xx m from the point directly beneath it is θ\theta. Which expression gives dθdx\dfrac{d\theta}{dx}?
  1. A5500x2sec⁡2θ\dfrac{5500}{x^2\sec^2\theta}
  2. B−5500x2-\dfrac{5500}{x^2}
  3. C−5500x2sec⁡2θ-\dfrac{5500}{x^2\sec^2\theta}
  4. D−x2sec⁡2θ5500-\dfrac{x^2\sec^2\theta}{5500}

Question 602

[1 marks]related rates
An aeroplane flies at a constant height of 5 500 m at 550 000550\,000 m/hr. When the angle of elevation from the observer is 25°25°, the horizontal distance xx from the observer to the point below the plane is
  1. A5500tan⁡25°\dfrac{5500}{\tan 25°} m
  2. B5500tan⁡25°5500\tan 25° m
  3. C5500sin⁡25°\dfrac{5500}{\sin 25°} m
  4. D5500cos⁡25°5500\cos 25° m

Question 603

[2 marks]related rates
For the aeroplane problem, evaluate sec⁡225°\sec^225°, correct to 4 decimal places (needed in dθdx=−5500x2sec⁡2θ\dfrac{d\theta}{dx}=-\dfrac{5500}{x^2\sec^2\theta} at θ=25°\theta=25°).

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Question 604

[1 marks]related rates
For the aeroplane flying at 550 000550\,000 m/hr, state dxdt\dfrac{dx}{dt} in m/hr (the rate at which the horizontal distance xx to the point below the plane changes, since xx decreases as the plane approaches).

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Question 701

[1 marks]partial fractions / integration
Express x−3x2−1\dfrac{x-3}{x^2-1} in partial fractions.
  1. A−2x+1+1x−1-\dfrac{2}{x+1} + \dfrac{1}{x-1}
  2. B2x+1−1x−1\dfrac{2}{x+1} - \dfrac{1}{x-1}
  3. C1x+1−2x−1\dfrac{1}{x+1} - \dfrac{2}{x-1}
  4. D2x+1+1x−1\dfrac{2}{x+1} + \dfrac{1}{x-1}

Question 702

[1 marks]partial fractions / integration
Find ∫23x−3x2−1 dx\displaystyle\int_2^3 \dfrac{x-3}{x^2-1}\,dx, giving your answer as a single logarithm.
  1. Aln⁡98\ln\tfrac{9}{8}
  2. Bln⁡169\ln\tfrac{16}{9}
  3. Cln⁡89\ln\tfrac{8}{9}
  4. D2ln⁡432\ln\tfrac{4}{3}

Question 703

[2 marks]partial fractions / integration
Using x−3x2−1=2x+1−1x−1\dfrac{x-3}{x^2-1}=\dfrac2{x+1}-\dfrac1{x-1}, evaluate [2ln⁡(x+1)−ln⁡(x−1)][2\ln(x+1)-\ln(x-1)] at x=3x=3, in the form 2ln⁡4−ln⁡p2\ln4-\ln p. State pp.

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Question 704

[2 marks]partial fractions / integration
Using x−3x2−1=2x+1−1x−1\dfrac{x-3}{x^2-1}=\dfrac2{x+1}-\dfrac1{x-1}, evaluate [2ln⁡(x+1)−ln⁡(x−1)][2\ln(x+1)-\ln(x-1)] at x=2x=2, in the form 2ln⁡3−ln⁡q2\ln3-\ln q. State qq.

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Question 801

[1 marks]compound angle identities
The equation cos⁡(45°+θ)=2sin⁡(45°−θ)\cos(45° + \theta) = 2\sin(45° - \theta) reduces to which simple equation?
  1. Atan⁡θ=1\tan\theta = 1
  2. Btan⁡θ=2\tan\theta = 2
  3. Ctan⁡θ=12\tan\theta = \tfrac{1}{2}
  4. Dtan⁡θ=−1\tan\theta = -1

Question 802

[1 marks]compound angle identities
Solve cos⁡(45°+θ)=2sin⁡(45°−θ)\cos(45° + \theta) = 2\sin(45° - \theta) for 0°≤θ≤360°0° \le \theta \le 360°.
  1. Aθ=45°\theta = 45° only
  2. Bθ=135°\theta = 135° or θ=315°\theta = 315°
  3. Cθ=45°\theta = 45° or θ=225°\theta = 225°
  4. Dθ=45°\theta = 45° or θ=135°\theta = 135°

Question 803

[2 marks]compound angle identities
Expanding 2sin⁡(45°−θ)=2sin⁡45°cos⁡θ−2cos⁡45°sin⁡θ2\sin(45°-\theta)=2\sin45°\cos\theta-2\cos45°\sin\theta (used to reduce cos⁡(45°+θ)=2sin⁡(45°−θ)\cos(45°+\theta)=2\sin(45°-\theta)), evaluate 2sin⁡45°2\sin45° exactly.

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Question 804

[2 marks]compound angle identities
Expanding cos⁡(45°+θ)=cos⁡45°cos⁡θ−sin⁡45°sin⁡θ\cos(45°+\theta)=\cos45°\cos\theta-\sin45°\sin\theta, evaluate cos⁡45°\cos45° exactly.

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Question 901

[1 marks]geometric and arithmetic series
A country's population is 27 million now and is projected to be 38 million in 36 years. Which equation determines the annual growth factor rr under a geometric model?
  1. A38=27r3638 = 27r^{36}
  2. B38=27+36r38 = 27 + 36r
  3. C27=38r3627 = 38r^{36}
  4. D38=27r1/3638 = 27r^{1/36}

Question 902

[1 marks]geometric and arithmetic series
Find the sum S=∑n=115(2n+12)S = \displaystyle\sum_{n=1}^{15}\left(2n + \tfrac{1}{2}\right).
  1. A127.5127.5
  2. B240240
  3. C247.5247.5
  4. D255255

Question 903

[2 marks]geometric and arithmetic series
A country's population grows from 27 million to 38 million in 36 years, following 38=27r3638=27r^{36}. Evaluate rr, correct to 4 decimal places.

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Question 904

[1 marks]geometric and arithmetic series
Given r≈1.0095r\approx1.0095 for the population growth, state the annual growth rate as a percentage, correct to 2 decimal places.

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Question 905

[1 marks]geometric and arithmetic series
For S=∑n=115(2n+12)S=\displaystyle\sum_{n=1}^{15}\left(2n+\dfrac12\right), state the first term (at n=1n=1).

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Question 1001

[1 marks]circle geometry / radian measure
Points PP, QQ, RR lie on a circle of centre OO radius rr, and SS is on OQOQ produced with QS=rQS = r and PS=RSPS = RS. If angle POS=θPOS = \theta radians, the shaded area PQRSPQRS is
  1. Ar2sin⁡θ−12r2θr^2\sin\theta - \tfrac{1}{2}r^2\theta
  2. B2r2sin⁡θ+r2θ2r^2\sin\theta + r^2\theta
  3. C2r2sin⁡θ−r2θ2r^2\sin\theta - r^2\theta
  4. Dr2θ−2r2sin⁡θr^2\theta - 2r^2\sin\theta

Question 1002

[1 marks]circle geometry / radian measure
In a circle of radius rr centre OO, SS lies on OQOQ produced with OS=2rOS = 2r and angle POS=θPOS = \theta. Find PSPS.
  1. Ar5+4cos⁡θr\sqrt{5 + 4\cos\theta}
  2. Br(5−4cos⁡θ)r\left(5 - 4\cos\theta\right)
  3. C2rsin⁡θ2r\sin\theta
  4. Dr5−4cos⁡θr\sqrt{5 - 4\cos\theta}

Question 1003

[3 marks]circle geometry / radian measure
In the circle centre OO radius rr with points P,Q,R,TP,Q,R,T on it and angle POS=θPOS=\theta (SS on OQOQ produced), PP and RR are symmetric about line OSOS, so the reflex arc PTRPTR (not through QQ) subtends angle 2π−2θ2\pi-2\theta at OO. State the arc length of PTRPTR, in terms of rr and θ\theta.

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Question 1004

[2 marks]circle geometry / radian measure
For the circle problem, PS=r5−4cos⁡θPS=r\sqrt{5-4\cos\theta}. Evaluate 5−4cos⁡θ\sqrt{5-4\cos\theta} when θ=π3\theta=\dfrac{\pi}3, correct to 3 decimal places.

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Question 1005

[2 marks]circle geometry / radian measure
The shaded area is PQRS=2r2sin⁡θ−r2θPQRS=2r^2\sin\theta-r^2\theta. Evaluate this area when θ=π3\theta=\dfrac{\pi}3 and r=2r=2, correct to 3 decimal places.

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Question 1101

[1 marks]factor and remainder theorems
For f(x)=2x3+ax2−bx+12f(x) = 2x^3 + ax^2 - bx + 12, (x−1)(x-1) is a factor of f(x)f(x) and f′(x)f'(x) leaves remainder −5-5 when divided by (x−1)(x-1). Find aa and bb.
  1. Aa=3a = 3, b=17b = 17
  2. Ba=3a = 3, b=−11b = -11
  3. Ca=−3a = -3, b=−17b = -17
  4. Da=17a = 17, b=3b = 3

Question 1102

[1 marks]factor and remainder theorems
Solve the equation 2x3+3x2−17x+12=02x^3 + 3x^2 - 17x + 12 = 0.
  1. Ax=1, x=−4, x=32x = 1,\ x = -4,\ x = \tfrac{3}{2}
  2. Bx=−1, x=4, x=32x = -1,\ x = 4,\ x = \tfrac{3}{2}
  3. Cx=1, x=−4, x=−32x = 1,\ x = -4,\ x = -\tfrac{3}{2}
  4. Dx=1, x=4, x=−32x = 1,\ x = 4,\ x = -\tfrac{3}{2}

Question 1103

[2 marks]factor and remainder theorems
For f(x)=2x3+ax2−bx+12f(x)=2x^3+ax^2-bx+12, the factor theorem (f(1)=0f(1)=0, since (x−1)(x-1) is a factor) gives an equation relating aa and bb, in the form a−b=ka-b=k. State kk.

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Question 1104

[2 marks]factor and remainder theorems
For f′(x)=6x2+2ax−bf'(x)=6x^2+2ax-b, the remainder theorem (f′(1)=−5f'(1)=-5) gives an equation, in the form 2a−b=k2a-b=k. State kk.

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Question 1105

[3 marks]factor and remainder theorems
With a=3a=3, b=17b=17, f(x)=2x3+3x2−17x+12=(x−1)(2x2+px+q)f(x)=2x^3+3x^2-17x+12=(x-1)(2x^2+px+q). State pp and qq as 'p, q'.

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Question 1201

[1 marks]vectors
Given OB→=−i+j−k\overrightarrow{OB} = -\mathbf{i} + \mathbf{j} - \mathbf{k}, find the unit vector in the direction of OB→\overrightarrow{OB}.
  1. A−i+j−k-\mathbf{i} + \mathbf{j} - \mathbf{k}
  2. B13(−i+j−k)\tfrac{1}{\sqrt{3}}(-\mathbf{i} + \mathbf{j} - \mathbf{k})
  3. C13(−i+j−k)\tfrac{1}{3}(-\mathbf{i} + \mathbf{j} - \mathbf{k})
  4. D13(i−j+k)\tfrac{1}{\sqrt{3}}(\mathbf{i} - \mathbf{j} + \mathbf{k})

Question 1202

[1 marks]vectors
Given OA→=3i+4j+(2−b)k\overrightarrow{OA} = 3\mathbf{i} + 4\mathbf{j} + (2-b)\mathbf{k} and OB→=−i+j−k\overrightarrow{OB} = -\mathbf{i} + \mathbf{j} - \mathbf{k}, find bb such that BO^A=90°B\hat{O}A = 90°.
  1. Ab=2b = 2
  2. Bb=1b = 1
  3. Cb=−1b = -1
  4. Db=3b = 3

Question 1203

[1 marks]vectors
With AB→=−4i−3j+(b−3)k\overrightarrow{AB} = -4\mathbf{i} - 3\mathbf{j} + (b-3)\mathbf{k}, find the value of bb for which ∣AB→∣2|\overrightarrow{AB}|^2 is a minimum.
  1. Ab=0b = 0
  2. Bb=1b = 1
  3. Cb=3b = 3
  4. Db=6b = 6

Question 1204

[2 marks]vectors
With OA→=3i+4j+(2−b)k\overrightarrow{OA}=3\mathbf i+4\mathbf j+(2-b)\mathbf k and OB→=−i+j−k\overrightarrow{OB}=-\mathbf i+\mathbf j-\mathbf k, state AB→=OB→−OA→\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA} as 'x,y,z', in terms of bb.

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Question 1205

[2 marks]vectors
With AB→=−4i−3j+(b−3)k\overrightarrow{AB}=-4\mathbf i-3\mathbf j+(b-3)\mathbf k, write down an expression for ∣AB→∣2|\overrightarrow{AB}|^2, in terms of bb.

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Question 1206

[1 marks]vectors
Given ∣AB→∣2=25+(b−3)2|\overrightarrow{AB}|^2=25+(b-3)^2, state its minimum value (occurring at b=3b=3).

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Question 1207

[2 marks]vectors
Evaluate OA→⋅OB→\overrightarrow{OA}\cdot\overrightarrow{OB} in terms of bb (using OA→=3i+4j+(2−b)k\overrightarrow{OA}=3\mathbf i+4\mathbf j+(2-b)\mathbf k, OB→=−i+j−k\overrightarrow{OB}=-\mathbf i+\mathbf j-\mathbf k), before setting it to zero to find bb.

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Question 1301

[1 marks]differential equations / exponential decay
A wet substance loses moisture at a rate proportional to its moisture content MM. Solve the resulting differential equation given that M=m0M = m_0 when t=0t = 0.
  1. AM=m0ln⁡(kt)M = m_0\ln(kt)
  2. BM=m0ktM = \dfrac{m_0}{kt}
  3. CM=m0+ktM = m_0 + kt
  4. DM=m0ektM = m_0e^{kt}

Question 1302

[1 marks]differential equations / exponential decay
A substance loses half of its moisture in the first hour, so that M=m0e−tln⁡2M = m_0 e^{-t\ln 2}. Find the time it takes to lose 95%95\% of its moisture.
  1. A4.34.3 hours
  2. B5.05.0 hours
  3. C19.019.0 hours
  4. D3.03.0 hours

Question 1303

[2 marks]differential equations / exponential decay
Separating variables in dMdt=kM\dfrac{dM}{dt}=kM and integrating gives ln⁡M=kt+c\ln M=kt+c. Using M=m0M=m_0 when t=0t=0, state cc in terms of m0m_0.

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Question 1304

[2 marks]differential equations / exponential decay
Given M=m0ektM=m_0e^{kt} and that half the moisture is lost in the first hour (M=0.5m0M=0.5m_0 at t=1t=1), evaluate eke^k.

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Question 1305

[2 marks]differential equations / exponential decay
Using ek=0.5e^k=0.5, evaluate kk, correct to 4 decimal places.

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Question 1306

[2 marks]differential equations / exponential decay
To lose 95% of moisture, 5% remains, i.e. M=0.05m0M=0.05m_0. State the resulting equation in the form ekt=pe^{kt}=p. State pp.

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Question 1307

[2 marks]differential equations / exponential decay
Using k≈−0.6931k\approx-0.6931 and ekt=0.05e^{kt}=0.05, evaluate ktkt (i.e. ln⁡0.05\ln0.05), correct to 4 decimal places.

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Question 1401

[1 marks]Newton-Raphson method
Let f(x)=tan⁡x−4xf(x) = \tan x - 4x. Which calculation shows that the positive root of tan⁡x=4x\tan x = 4x lies between x=1x = 1 and x=1.5x = 1.5?
  1. Af(1)=1.557f(1) = 1.557 and f(1.5)=14.10f(1.5) = 14.10
  2. Bf(1)=2.443f(1) = 2.443 and f(1.5)=−8.10f(1.5) = -8.10
  3. Cf(1)=−2.443f(1) = -2.443 and f(1.5)=8.10f(1.5) = 8.10
  4. Df(1)=−2.443f(1) = -2.443 and f(1.5)=−8.10f(1.5) = -8.10

Question 1402

[1 marks]Newton-Raphson method
Applying the Newton-Raphson method to f(x)=tan⁡x−4xf(x) = \tan x - 4x gives which iterative formula?
  1. Axn+1=xntan⁡2xn−tan⁡xn+xntan⁡2xn+3x_{n+1} = \dfrac{x_n\tan^2 x_n - \tan x_n + x_n}{\tan^2 x_n + 3}
  2. Bxn+1=tan⁡xn−4xntan⁡2xn−3x_{n+1} = \dfrac{\tan x_n - 4x_n}{\tan^2 x_n - 3}
  3. Cxn+1=xntan⁡2xn−tan⁡xn+xntan⁡2xn−3x_{n+1} = \dfrac{x_n\tan^2 x_n - \tan x_n + x_n}{\tan^2 x_n - 3}
  4. Dxn+1=xntan⁡2xn+tan⁡xn−xntan⁡2xn−3x_{n+1} = \dfrac{x_n\tan^2 x_n + \tan x_n - x_n}{\tan^2 x_n - 3}

Question 1403

[1 marks]Newton-Raphson method
Starting with x1=1.4x_1 = 1.4, use the Newton-Raphson iteration for tan⁡x=4x\tan x = 4x to estimate the root correct to 4 decimal places.
  1. A1.39321.3932
  2. B1.39351.3935
  3. C1.40001.4000
  4. D1.50001.5000

Question 1404

[2 marks]Newton-Raphson method
At x=0.1x=0.1 radians, evaluate tan⁡x\tan x, correct to 4 decimal places (used to compare y=tan⁡xy=\tan x and y=4xy=4x near the origin).

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Question 1405

[1 marks]Newton-Raphson method
Comparing tan⁡(0.1)≈0.1003\tan(0.1)\approx0.1003 with 4(0.1)=0.44(0.1)=0.4, which is true just after x=0x=0?
  1. Atan⁡x=4x\tan x=4x
  2. Btan⁡x\tan x is undefined
  3. Ctan⁡x<4x\tan x<4x
  4. Dtan⁡x>4x\tan x>4x

Question 1406

[2 marks]Newton-Raphson method
As x→π2−x\to\dfrac{\pi}2^-, tan⁡x→∞\tan x\to\infty while 4x→4×π2=2π4x\to4\times\dfrac{\pi}2=2\pi (a finite value). State 2π2\pi, correct to 4 decimal places.

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Question 1407

[2 marks]Newton-Raphson method
For f(x)=tan⁡x−4xf(x)=\tan x-4x, evaluate f(1)f(1), correct to 3 decimal places.

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Question 1408

[2 marks]Newton-Raphson method
For f(x)=tan⁡x−4xf(x)=\tan x-4x, evaluate f(1.5)f(1.5), correct to 2 decimal places.

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Question 1501

[1 marks]functions / composite and inverse functions
Given f(x)=2x−3f(x) = 2x - 3 and g(x)=3x2+2x−2g(x) = 3x^2 + 2x - 2, express gf(x)gf(x) in its simplest form.
  1. A12x2−32x+2512x^2 - 32x + 25
  2. B6x2+4x−76x^2 + 4x - 7
  3. C12x2+32x+1912x^2 + 32x + 19
  4. D12x2−32x+1912x^2 - 32x + 19

Question 1502

[1 marks]functions / composite and inverse functions
Find the minimum value of gf(x)=12x2−32x+19gf(x) = 12x^2 - 32x + 19.
  1. A−43-\tfrac{4}{3}
  2. B−73-\tfrac{7}{3}
  3. C43\tfrac{4}{3}
  4. D1919

Question 1503

[1 marks]functions / composite and inverse functions
The function g(x)=3x2+2x−2g(x) = 3x^2 + 2x - 2 is defined for x≥kx \ge k. State the minimum value of kk for which gg has an inverse, and give g−1(x)g^{-1}(x).
  1. Ak=13k = \tfrac{1}{3}; g−1(x)=133x+7+13g^{-1}(x) = \tfrac{1}{3}\sqrt{3x+7} + \tfrac{1}{3}
  2. Bk=−13k = -\tfrac{1}{3}; g−1(x)=133x+7−13g^{-1}(x) = \tfrac{1}{3}\sqrt{3x+7} - \tfrac{1}{3}
  3. Ck=−73k = -\tfrac{7}{3}; g−1(x)=3x+7−13g^{-1}(x) = \sqrt{3x+7} - \tfrac{1}{3}
  4. Dk=−13k = -\tfrac{1}{3}; g−1(x)=133x−7−13g^{-1}(x) = \tfrac{1}{3}\sqrt{3x-7} - \tfrac{1}{3}

Question 1504

[2 marks]functions / composite and inverse functions
For f(x)=2x−3f(x)=2x-3, state f−1(x)f^{-1}(x).

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Question 1505

[2 marks]functions / composite and inverse functions
For f(x)=2x−3f(x)=2x-3 and f−1(x)=(x+3)/2f^{-1}(x)=(x+3)/2, state the point where the graphs of y=f(x)y=f(x) and y=f−1(x)y=f^{-1}(x) intersect (on the line y=xy=x), as 'x,y'.

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Question 1506

[2 marks]functions / composite and inverse functions
For gf(x)=12x2−32x+19gf(x)=12x^2-32x+19, state the value of xx at which the minimum occurs.

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Question 1507

[2 marks]functions / composite and inverse functions
Completing the square, g(x)=3x2+2x−2=3(x+13)2+dg(x)=3x^2+2x-2=3\left(x+\dfrac13\right)^2+d for some constant dd. State dd.

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Question 1508

[1 marks]functions / composite and inverse functions
Hence state the xx-coordinate of the vertex of g(x)=3x2+2x−2g(x)=3x^2+2x-2 (the minimum value of kk for which gg has an inverse on x≥kx\ge k).

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Question 1601

[1 marks]exponential functions / areas and volumes
The curve y=e2x−e3xy = e^{2x} - e^{3x} has a maximum point PP. Find the xx-coordinate of PP.
  1. A00
  2. Bln⁡16\ln\tfrac{1}{6}
  3. Cln⁡32\ln\tfrac{3}{2}
  4. Dln⁡23\ln\tfrac{2}{3}

Question 1602

[1 marks]exponential functions / areas and volumes
Calculate the area of the region bounded by y=e2x−e3xy = e^{2x} - e^{3x}, the xx-axis, from x=ln⁡23x = \ln\tfrac{2}{3} to x=0x = 0.
  1. A781\tfrac{7}{81} square units
  2. B16\tfrac{1}{6} square units
  3. C1081\tfrac{10}{81} square units
  4. D7162\tfrac{7}{162} square units

Question 1603

[1 marks]exponential functions / areas and volumes
The region bounded by y=e2x−e3xy = e^{2x} - e^{3x} and the xx-axis between x=ln⁡23x = \ln\tfrac{2}{3} and x=0x = 0 is rotated completely about the xx-axis. Find the volume, correct to 3 significant figures.
  1. A0.01670.0167 cubic units
  2. B0.04320.0432 cubic units
  3. C0.1670.167 cubic units
  4. D0.005330.00533 cubic units

Question 1604

[2 marks]exponential functions / areas and volumes
For y=e2x−e3xy=e^{2x}-e^{3x}, state dydx\dfrac{dy}{dx}.

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Question 1605

[2 marks]exponential functions / areas and volumes
Setting dydx=2e2x−3e3x=0\dfrac{dy}{dx}=2e^{2x}-3e^{3x}=0 and dividing throughout by e2xe^{2x}, state the resulting equation in the form 2=cex2=ce^x. State cc.

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Question 1606

[2 marks]exponential functions / areas and volumes
Evaluate [e2x2−e3x3]\left[\dfrac{e^{2x}}2-\dfrac{e^{3x}}3\right] at x=0x=0 (one part of the area calculation for region R).

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Question 1607

[2 marks]exponential functions / areas and volumes
Evaluate [e2x2−e3x3]\left[\dfrac{e^{2x}}2-\dfrac{e^{3x}}3\right] at x=ln⁡23x=\ln\dfrac23, as a single fraction.

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Question 1608

[1 marks]exponential functions / areas and volumes
Expanding (e2x−e3x)2(e^{2x}-e^{3x})^2 (needed for the volume integral) gives e4x−2e5x+e6xe^{4x}-2e^{5x}+e^{6x}. State the coefficient of e5xe^{5x} in this expansion.

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