Danho
ZIMSEC A Level · J2017

Pure Mathematics Paper 1 June 2017

Questions
79
Total marks
120

Sit this paper online

Questions
79
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]Modulus / inequalities
Solve the inequality ∣x−1∣<x|x-1| < x.
  1. A0<x<10 < x < 1
  2. Bx>1x > 1
  3. Cx>12x > \tfrac{1}{2}
  4. Dx<12x < \tfrac{1}{2}

Question 102

[2 marks]Modulus / inequalities
For x≥1x \ge 1, ∣x−1∣=x−1|x-1| = x-1. Which statement correctly explains why every x≥1x \ge 1 satisfies ∣x−1∣<x|x-1| < x?
  1. AIt simplifies to −1<0-1<0, true for all real xx
  2. BIt only holds when xx is a whole number, not all reals
  3. CIt requires x≠1x \neq 1 in addition to x≥1x \ge 1 to hold
  4. DIt simplifies to x<−1x<-1, which never happens for x≥1x \ge 1

Question 201

[1 marks]Polynomials / factor theorem
Factorise completely 3x3+x2−8x+43x^3 + x^2 - 8x + 4.
  1. A(x−1)(3x−2)(x+2)(x-1)(3x-2)(x+2)
  2. B(x−1)(3x−4)(x+1)(x-1)(3x-4)(x+1)
  3. C(x+1)(3x−2)(x−2)(x+1)(3x-2)(x-2)
  4. D(x−1)(3x+2)(x−2)(x-1)(3x+2)(x-2)

Question 202

[1 marks]Polynomials / factor theorem
Confirm (x−1)(x-1) is a factor of 3x3+x2−8x+43x^3+x^2-8x+4 by evaluating the polynomial at x=1x=1.

Answer this when you sit the paper.

Question 203

[2 marks]Polynomials / factor theorem
Divide 3x3+x2−8x+43x^3+x^2-8x+4 by (x−1)(x-1). State the resulting quadratic quotient in the form 3x2+bx+c3x^2+bx+c.

Answer this when you sit the paper.

Question 301

[1 marks]Indices / quadratic in disguise
Solve the equation y2/3−5y1/3+6=0y^{2/3} - 5y^{1/3} + 6 = 0.
  1. Ay=−8y = -8 or y=−27y = -27
  2. By=2y = 2 or y=3y = 3
  3. Cy=8y = 8 or y=27y = 27
  4. Dy=4y = 4 or y=9y = 9

Question 302

[2 marks]Indices / quadratic in disguise
Expand and simplify (x1/3−2)(x2/3+2x1/3+1)(x^{1/3}-2)(x^{2/3}+2x^{1/3}+1).
  1. Ax2/3−8x^{2/3} - 8
  2. Bx−3x1/3−2x - 3x^{1/3} - 2
  3. Cx−2x - 2
  4. Dx−x1/3−2x - x^{1/3} - 2

Question 303

[1 marks]Indices / quadratic in disguise
Solving u2−5u+6=0u^2-5u+6=0 where u=y1/3u=y^{1/3} gives two values of uu. Which pair is correct?
  1. Au=6u=6 and u=−1u=-1
  2. Bu=−2u=-2 and u=−3u=-3
  3. Cu=2u=2 and u=3u=3
  4. Du=1u=1 and u=6u=6

Question 304

[2 marks]Indices / quadratic in disguise
Using the expansion x−3x1/3−2x-3x^{1/3}-2 from part (a), find its value when x=8x=8.

Answer this when you sit the paper.

Question 401

[1 marks]Small changes / rates
The radius of a sphere increases from 11 cm to 11.02 cm. Find the approximate percentage increase in its volume.
  1. A0.18%0.18\%
  2. B0.36%0.36\%
  3. C1.82%1.82\%
  4. D0.55%0.55\%

Question 402

[1 marks]Small changes / rates
For V=43πr3V=\tfrac43\pi r^3, find dVdr\dfrac{dV}{dr}.
  1. A2πr22\pi r^2
  2. B4πr34\pi r^3
  3. C43πr2\tfrac43\pi r^2
  4. D4πr24\pi r^2

Question 403

[2 marks]Small changes / rates
Using δV≈4πr2 δr\delta V \approx 4\pi r^2\,\delta r with r=11r=11 and δr=0.02\delta r=0.02, find δV\delta V as a multiple of π\pi (e.g. 9.68π9.68\pi).

Answer this when you sit the paper.

Question 404

[2 marks]Small changes / rates
Find δV\delta V to 3 significant figures as a decimal (take π≈3.142\pi \approx 3.142), using δV≈4πr2 δr\delta V \approx 4\pi r^2\,\delta r with r=11r=11, δr=0.02\delta r=0.02.

Answer this when you sit the paper.

Question 501

[1 marks]Binomial series / approximation
Expand 11−2x\dfrac{1}{\sqrt{1-2x}} in ascending powers of xx up to and including the term in x3x^3.
  1. A1+x+12x2+12x31 + x + \tfrac{1}{2}x^2 + \tfrac{1}{2}x^3
  2. B1+2x+3x2+4x31 + 2x + 3x^2 + 4x^3
  3. C1−x+32x2−52x31 - x + \tfrac{3}{2}x^2 - \tfrac{5}{2}x^3
  4. D1+x+32x2+52x31 + x + \tfrac{3}{2}x^2 + \tfrac{5}{2}x^3

Question 502

[1 marks]Binomial series / approximation
State the range of values of xx for which the expansion of 11−2x\dfrac{1}{\sqrt{1-2x}} is valid.
  1. A∣x∣<1|x| < 1
  2. B∣x∣<12|x| < \tfrac{1}{2}
  3. Cx>0x > 0
  4. D∣x∣<2|x| < 2

Question 503

[1 marks]Binomial series / approximation
By substituting x=−16x = -\tfrac{1}{6} into the expansion of 11−2x\dfrac{1}{\sqrt{1-2x}}, find an approximation for 3\sqrt{3} correct to 5 decimal places.
  1. A0.863420.86342
  2. B1.725001.72500
  3. C1.726851.72685
  4. D1.732051.73205

Question 504

[2 marks]Binomial series / approximation
To approximate 3\sqrt3 using the expansion of (1−2x)−1/2(1-2x)^{-1/2}, which value of xx is substituted, and what does 1−2x1-2x equal there?
  1. Ax=16x=\tfrac16, giving 1−2x=231-2x=\tfrac23
  2. Bx=−16x=-\tfrac16, giving 1−2x=431-2x=\tfrac43
  3. Cx=−13x=-\tfrac13, giving 1−2x=531-2x=\tfrac53
  4. Dx=12x=\tfrac12, giving 1−2x=01-2x=0

Question 505

[1 marks]Binomial series / approximation
In the expansion (1−2x)−1/2=1+x+32x2+52x3+⋯(1-2x)^{-1/2}=1+x+\tfrac32x^2+\tfrac52x^3+\cdots, state the coefficient of x2x^2.

Answer this when you sit the paper.

Question 601

[1 marks]Vectors in 3D
P(1,2,1)P(1,2,1), Q(4,7,8)Q(4,7,8) and R(6,4,12)R(6,4,12) are three vertices of a parallelogram PQRSPQRS. Find the position vector of SS.
  1. A−3i+j−5k-3\mathbf{i} + \mathbf{j} - 5\mathbf{k}
  2. B5i+2j+11k5\mathbf{i} + 2\mathbf{j} + 11\mathbf{k}
  3. C3i−j+5k3\mathbf{i} - \mathbf{j} + 5\mathbf{k}
  4. D9i+9j+19k9\mathbf{i} + 9\mathbf{j} + 19\mathbf{k}

Question 602

[1 marks]Vectors in 3D
MM and NN are the midpoints of PQPQ and QRQR in parallelogram PQRSPQRS, where P(1,2,1)P(1,2,1) and R(6,4,12)R(6,4,12). Find the unit vector in the direction of MN→\overrightarrow{MN}.
  1. A12150(5i+2j+11k)\tfrac{1}{2\sqrt{150}}(5\mathbf{i} + 2\mathbf{j} + 11\mathbf{k})
  2. B1150(5i+2j+11k)\tfrac{1}{\sqrt{150}}(5\mathbf{i} + 2\mathbf{j} + 11\mathbf{k})
  3. C1150(5i+2j+11k)\tfrac{1}{150}(5\mathbf{i} + 2\mathbf{j} + 11\mathbf{k})
  4. D1150(−5i−2j−11k)\tfrac{1}{\sqrt{150}}(-5\mathbf{i} - 2\mathbf{j} - 11\mathbf{k})

Question 603

[2 marks]Vectors in 3D
For P(1,2,1)P(1,2,1), Q(4,7,8)Q(4,7,8), R(6,4,12)R(6,4,12), which pair correctly gives RQ⃗\vec{RQ} and PR⃗\vec{PR}?
  1. ARQ⃗=(−2,3,−4)\vec{RQ}=(-2,3,-4), PR⃗=(5,2,11)\vec{PR}=(5,2,11)
  2. BRQ⃗=(−2,3,−4)\vec{RQ}=(-2,3,-4), PR⃗=(−5,−2,−11)\vec{PR}=(-5,-2,-11)
  3. CRQ⃗=(3,−2,−4)\vec{RQ}=(3,-2,-4), PR⃗=(5,11,2)\vec{PR}=(5,11,2)
  4. DRQ⃗=(2,−3,4)\vec{RQ}=(2,-3,4), PR⃗=(5,2,11)\vec{PR}=(5,2,11)

Question 604

[2 marks]Vectors in 3D
Find ∣PR⃗∣|\vec{PR}|, the magnitude of PR⃗=5i+2j+11k\vec{PR}=5\mathbf{i}+2\mathbf{j}+11\mathbf{k}, in simplified surd form.

Answer this when you sit the paper.

Question 701

[1 marks]Completing the square / transformations
Express f(x)=x2−8x+10f(x) = x^2 - 8x + 10 in the form (x+a)2+b(x+a)^2 + b.
  1. A(x+4)2−6(x+4)^2 - 6
  2. B(x−4)2+6(x-4)^2 + 6
  3. C(x−4)2−6(x-4)^2 - 6
  4. D(x−8)2−54(x-8)^2 - 54

Question 702

[1 marks]Completing the square / transformations
Describe the transformations that map y=x2y = x^2 onto y=x2−8x+10y = x^2 - 8x + 10.
  1. ATranslation 8 units in the +x+x direction, then 10 units in the +y+y direction
  2. BReflection in the yy-axis, then translation 6 units down
  3. CTranslation 4 units in the +x+x direction, then 6 units in the −y-y direction
  4. DTranslation 4 units in the −x-x direction, then 6 units in the +y+y direction

Question 703

[2 marks]Completing the square / transformations
For f(x)=x2−8x+10=(x−4)2−6f(x)=x^2-8x+10=(x-4)^2-6, which option correctly states the least value and the line of symmetry?
  1. Aleast value −6-6, line of symmetry x=4x=4
  2. Bleast value 44, line of symmetry x=−6x=-6
  3. Cleast value 1010, line of symmetry x=8x=8
  4. Dleast value −6-6, line of symmetry x=−4x=-4

Question 704

[1 marks]Completing the square / transformations
Which single transformation maps y=x2y=x^2 onto y=(x−4)2y=(x-4)^2 (ignoring the vertical shift)?
  1. Atranslation of 4 units in the positive yy-direction
  2. Btranslation of 4 units in the positive xx-direction
  3. Cstretch of scale factor 4 parallel to the xx-axis
  4. Dtranslation of 4 units in the negative xx-direction

Question 705

[2 marks]Completing the square / transformations
Which single transformation maps y=(x−4)2y=(x-4)^2 onto y=(x−4)2−6y=(x-4)^2-6?
  1. Astretch of scale factor 6 parallel to the yy-axis
  2. Btranslation of 6 units in the positive yy-direction
  3. Ctranslation of 6 units in the negative xx-direction
  4. Dtranslation of 6 units in the negative yy-direction

Question 801

[1 marks]Trigonometry / circle geometry
Solve 2cos⁡2x+3sin⁡x=32\cos^2 x + 3\sin x = 3 for 0<x<π0 < x < \pi.
  1. Ax=π6, π2, 5π6x = \tfrac{\pi}{6},\ \tfrac{\pi}{2},\ \tfrac{5\pi}{6}
  2. Bx=π6, 5π6x = \tfrac{\pi}{6},\ \tfrac{5\pi}{6}
  3. Cx=π3, 2π3x = \tfrac{\pi}{3},\ \tfrac{2\pi}{3}
  4. Dx=π2x = \tfrac{\pi}{2} only

Question 802

[1 marks]Trigonometry / circle geometry
A circle of radius rr has diameter BOCBOC, and AA lies on the circle with angle AO^B=θA\hat{O}B = \theta. The area of the region bounded by the diameter and the arc, less triangle AOBAOB, is
  1. A12r2(π+sin⁡θ)\tfrac{1}{2}r^2(\pi + \sin\theta)
  2. Br2(π−sin⁡θ)r^2(\pi - \sin\theta)
  3. C12r2(θ−sin⁡θ)\tfrac{1}{2}r^2(\theta - \sin\theta)
  4. D12r2(π−sin⁡θ)\tfrac{1}{2}r^2(\pi - \sin\theta)

Question 803

[2 marks]Trigonometry / circle geometry
Solving 2sin⁡2x−3sin⁡x+1=02\sin^2x-3\sin x+1=0 gives two values of sin⁡x\sin x. Which pair is correct?
  1. Asin⁡x=1\sin x = 1 and sin⁡x=0.5\sin x = 0.5
  2. Bsin⁡x=0.5\sin x = 0.5 and sin⁡x=0.25\sin x = 0.25
  3. Csin⁡x=1\sin x = 1 and sin⁡x=−0.5\sin x = -0.5
  4. Dsin⁡x=−1\sin x = -1 and sin⁡x=0.5\sin x = 0.5

Question 804

[2 marks]Trigonometry / circle geometry
In the shaded-region diagram, a semicircle has diameter BOCBOC (radius rr). State its area in terms of rr and π\pi.

Answer this when you sit the paper.

Question 805

[2 marks]Trigonometry / circle geometry
In the shaded-region diagram, triangle AOBAOB has OA=OB=rOA=OB=r and angle AO^B=θA\hat{O}B=\theta. State its area in terms of rr and θ\theta.

Answer this when you sit the paper.

Question 901

[1 marks]Partial fractions / integration
Express f(x)=x3+2x2+x+2x2+xf(x) = \dfrac{x^3 + 2x^2 + x + 2}{x^2 + x} in the form x+1+Ax+Bx+1x + 1 + \dfrac{A}{x} + \dfrac{B}{x+1}.
  1. AA=1A = 1, B=−1B = -1
  2. BA=2A = 2, B=−2B = -2
  3. CA=−2A = -2, B=2B = 2
  4. DA=2A = 2, B=2B = 2

Question 902

[1 marks]Partial fractions / integration
Evaluate ∫12(x+1+2x−2x+1)dx\displaystyle\int_1^2 \left(x + 1 + \dfrac{2}{x} - \dfrac{2}{x+1}\right)dx.
  1. A52+2ln⁡34\tfrac{5}{2} + 2\ln\tfrac{3}{4}
  2. B52+2ln⁡43\tfrac{5}{2} + 2\ln\tfrac{4}{3}
  3. C32+2ln⁡43\tfrac{3}{2} + 2\ln\tfrac{4}{3}
  4. D52+4ln⁡43\tfrac{5}{2} + 4\ln\tfrac{4}{3}

Question 903

[2 marks]Partial fractions / integration
In f(x)=x+1+Ax+Bx+1f(x)=x+1+\dfrac{A}{x}+\dfrac{B}{x+1}, which pair (A,B)(A,B) is correct for f(x)=x3+2x2+x+2x2+xf(x)=\dfrac{x^3+2x^2+x+2}{x^2+x}?
  1. AA=1A=1, B=−1B=-1
  2. BA=−2A=-2, B=2B=2
  3. CA=2A=2, B=2B=2
  4. DA=2A=2, B=−2B=-2

Question 904

[2 marks]Partial fractions / integration
Find the indefinite integral ∫(x+1+2x−2x+1)dx\displaystyle\int\left(x+1+\dfrac{2}{x}-\dfrac{2}{x+1}\right)dx (omit the constant of integration).

Answer this when you sit the paper.

Question 905

[2 marks]Partial fractions / integration
Evaluate ∫122x dx\displaystyle\int_1^2 \dfrac{2}{x}\,dx (just this term of the integrand) to 3 significant figures.

Answer this when you sit the paper.

Question 1001

[1 marks]Complex numbers
Given z1=2−3iz_1 = 2 - 3i and z2=1+3iz_2 = 1 + 3i, express z1z2\dfrac{z_1}{z_2} in the form x+iyx + iy.
  1. A−710+910i-\tfrac{7}{10} + \tfrac{9}{10}i
  2. B710−910i\tfrac{7}{10} - \tfrac{9}{10}i
  3. C−710−910i-\tfrac{7}{10} - \tfrac{9}{10}i
  4. D−75−95i-\tfrac{7}{5} - \tfrac{9}{5}i

Question 1002

[1 marks]Complex numbers
Given z1z2=−710−910i\dfrac{z_1}{z_2} = -\tfrac{7}{10} - \tfrac{9}{10}i, find ∣z1z2∣\left|\dfrac{z_1}{z_2}\right|.
  1. A13010\tfrac{130}{10}
  2. B11032\tfrac{1}{10}\sqrt{32}
  3. C110130\tfrac{1}{10}\sqrt{130}
  4. D130\sqrt{130}

Question 1003

[1 marks]Complex numbers
Find arg⁡(z1z2)\arg\left(\dfrac{z_1}{z_2}\right) where z1z2=−710−910i\dfrac{z_1}{z_2} = -\tfrac{7}{10} - \tfrac{9}{10}i, giving your answer in radians in the range 00 to 2π2\pi.
  1. A0.910.91
  2. B2.232.23
  3. C4.054.05
  4. D5.385.38

Question 1004

[1 marks]Complex numbers
For z1=2−3iz_1=2-3i and z2=1+3iz_2=1+3i, find z1z2z_1 z_2 (the product, not the quotient) in the form x+iyx+iy.

Answer this when you sit the paper.

Question 1005

[2 marks]Complex numbers
Which pair correctly gives ∣z1∣|z_1| and ∣z2∣|z_2| for z1=2−3iz_1=2-3i, z2=1+3iz_2=1+3i?
  1. A13\sqrt{13} and 10\sqrt{10}
  2. B55 and 1010
  3. C5\sqrt5 and 10\sqrt{10}
  4. D13\sqrt{13} and 8\sqrt8

Question 1006

[1 marks]Complex numbers
The point z1z2=−710−910i\dfrac{z_1}{z_2}=-\tfrac{7}{10}-\tfrac{9}{10}i lies in which quadrant of the Argand diagram?
  1. Athird quadrant
  2. Bfirst quadrant
  3. Csecond quadrant
  4. Dfourth quadrant

Question 1007

[2 marks]Complex numbers
Express arg⁡(z1z2)\arg\left(\dfrac{z_1}{z_2}\right) in degrees, to 1 decimal place, in the range −180°-180° to 180°180°.

Answer this when you sit the paper.

Question 1101

[1 marks]Numerical methods / Newton-Raphson
Let f(x)=ln⁡x−xx−1f(x) = \ln x - \dfrac{x}{x-1}. Which values show that a root of ln⁡x=xx−1\ln x = \dfrac{x}{x-1} lies between x=0.3x = 0.3 and x=0.45x = 0.45?
  1. Af(0.3)=0.775f(0.3) = 0.775 and f(0.45)=−0.0196f(0.45) = -0.0196
  2. Bf(0.3)=−0.775f(0.3) = -0.775 and f(0.45)=−0.0196f(0.45) = -0.0196
  3. Cf(0.3)=−1.204f(0.3) = -1.204 and f(0.45)=−0.798f(0.45) = -0.798
  4. Df(0.3)=−0.775f(0.3) = -0.775 and f(0.45)=0.0196f(0.45) = 0.0196

Question 1102

[1 marks]Numerical methods / Newton-Raphson
Taking x0=0.45x_0 = 0.45, use the Newton-Raphson method twice on ln⁡x=xx−1\ln x = \dfrac{x}{x-1} and give the root correct to 4 decimal places.
  1. A0.30000.3000
  2. B0.44640.4464
  3. C0.44650.4465
  4. D0.45000.4500

Question 1103

[2 marks]Numerical methods / Newton-Raphson
Find f′(x)f'(x) for f(x)=ln⁡x−xx−1f(x)=\ln x-\dfrac{x}{x-1}.

Answer this when you sit the paper.

Question 1104

[2 marks]Numerical methods / Newton-Raphson
Taking x0=0.45x_0=0.45, perform one Newton-Raphson iteration on f(x)=ln⁡x−xx−1f(x)=\ln x-\dfrac{x}{x-1} to find x1x_1, correct to 6 decimal places.

Answer this when you sit the paper.

Question 1105

[2 marks]Numerical methods / Newton-Raphson
Find f(0.45)f(0.45) for f(x)=ln⁡x−xx−1f(x)=\ln x-\dfrac{x}{x-1}, correct to 4 decimal places.

Answer this when you sit the paper.

Question 1106

[2 marks]Numerical methods / Newton-Raphson
The values f(0.3)<0f(0.3)<0 and f(0.45)>0f(0.45)>0 (with ff continuous on [0.3,0.45][0.3,0.45]) guarantee a root in that interval by which result?
  1. Athe Fundamental Theorem of Calculus
  2. Bthe Mean Value Theorem
  3. Cthe Intermediate Value Theorem
  4. DRolle's theorem

Question 1201

[1 marks]Coordinate geometry / circles
Find the centre and radius of the circle x2+y2−4x+6y=12x^2 + y^2 - 4x + 6y = 12.
  1. Acentre (−2,3)(-2, 3), radius 55
  2. Bcentre (2,−3)(2, -3), radius 2525
  3. Ccentre (4,−6)(4, -6), radius 55
  4. Dcentre (2,−3)(2, -3), radius 55

Question 1202

[1 marks]Coordinate geometry / circles
The circle x2+y2−4x+6y=12x^2 + y^2 - 4x + 6y = 12 meets the line 3y=x+43y = x + 4 at AA and BB. Find the length of the chord ABAB.
  1. A10\sqrt{10}
  2. B1010
  3. C2102\sqrt{10}
  4. D5\sqrt{5}

Question 1203

[1 marks]Coordinate geometry / circles
A chord ABAB of length 10\sqrt{10} is drawn in a circle of radius 5 with centre CC. Find the area of triangle ABCABC.
  1. A1515 units2^2
  2. B152\tfrac{15}{2} units2^2
  3. C452\tfrac{45}{2} units2^2
  4. D154\tfrac{15}{4} units2^2

Question 1204

[2 marks]Coordinate geometry / circles
Solving y2−3y+2=0y^2-3y+2=0 (from substituting the line into the circle) gives two values of yy. Which pair is correct?
  1. Ay=1y=1 and y=−2y=-2
  2. By=2y=2 and y=4y=4
  3. Cy=1y=1 and y=2y=2
  4. Dy=−1y=-1 and y=−2y=-2

Question 1205

[2 marks]Coordinate geometry / circles
Using x=3y−4x=3y-4, which pair of coordinates correctly gives AA and BB?
  1. AA(2,−2)A(2,-2) and B(1,−1)B(1,-1)
  2. BA(−2,2)A(-2,2) and B(1,1)B(1,1)
  3. CA(2,2)A(2,2) and B(1,−1)B(1,-1)
  4. DA(2,2)A(2,2) and B(−1,1)B(-1,1)

Question 1206

[2 marks]Coordinate geometry / circles
Find the perpendicular distance hh from centre C(2,−3)C(2,-3) to chord ABAB, using h2+(102)2=52h^2+\left(\tfrac{\sqrt{10}}{2}\right)^2=5^2, to 3 significant figures.

Answer this when you sit the paper.

Question 1207

[1 marks]Coordinate geometry / circles
The chord ABAB has length 10\sqrt{10}. Find (AB)2(AB)^2.

Answer this when you sit the paper.

Question 1301

[1 marks]Series (GP and AP)
A series has rr-th term Ur=2(13)rU_r = 2\left(\tfrac{1}{3}\right)^r. Find nn such that ∑r=1nUr=8081\displaystyle\sum_{r=1}^{n}U_r = \tfrac{80}{81}.
  1. An=5n = 5
  2. Bn=4n = 4
  3. Cn=81n = 81
  4. Dn=3n = 3

Question 1302

[1 marks]Series (GP and AP)
For an arithmetic progression the sum of the first nn terms is Sn=32n2−2nS_n = \tfrac{3}{2}n^2 - 2n. Find the nn-th term.
  1. A6n+72\dfrac{6n+7}{2}
  2. B3n−23n - 2
  3. C3n−42\dfrac{3n-4}{2}
  4. D6n−72\dfrac{6n-7}{2}

Question 1303

[2 marks]Series (GP and AP)
Which option gives the first three terms U1,U2,U3U_1,U_2,U_3 of Ur=2(13)rU_r=2\left(\tfrac13\right)^r?
  1. A13,19,127\tfrac13,\tfrac19,\tfrac{1}{27}
  2. B2,23,292,\tfrac23,\tfrac29
  3. C23,29,227\tfrac23,\tfrac29,\tfrac{2}{27}
  4. D23,19,127\tfrac23,\tfrac19,\tfrac{1}{27}

Question 1304

[2 marks]Series (GP and AP)
For the arithmetic progression with Sn=32n2−2nS_n=\tfrac32n^2-2n, which option gives U1,U2,U3U_1,U_2,U_3?
  1. A−0.5, 2, 4.5-0.5,\ 2,\ 4.5
  2. B0.5, 2.5, 5.50.5,\ 2.5,\ 5.5
  3. C1.5, −2, 7.51.5,\ -2,\ 7.5
  4. D−0.5, 2.5, 5.5-0.5,\ 2.5,\ 5.5

Question 1305

[2 marks]Series (GP and AP)
For Sn=32n2−2nS_n=\tfrac32n^2-2n, find S4S_4.

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Question 1306

[2 marks]Series (GP and AP)
Hence find U4=S4−S3U_4=S_4-S_3 for Sn=32n2−2nS_n=\tfrac32n^2-2n.

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Question 1401

[1 marks]Curve sketching / areas and volumes
Solve the inequality f(x)<0f(x) < 0 where f(x)=x(x+1)(2−x)f(x) = x(x+1)(2-x).
  1. A−1<x<2-1 < x < 2
  2. Bx>2x > 2 only
  3. Cx<−1x < -1 or 0<x<20 < x < 2
  4. D−1<x<0-1 < x < 0 or x>2x > 2

Question 1402

[1 marks]Curve sketching / areas and volumes
Find the area of the region bounded by y=x(x+1)(2−x)y = x(x+1)(2-x) and the xx-axis between x=−1x = -1 and x=0x = 0.
  1. A512\tfrac{5}{12} units2^2
  2. B−512-\tfrac{5}{12} units2^2
  3. C712\tfrac{7}{12} units2^2
  4. D56\tfrac{5}{6} units2^2

Question 1403

[1 marks]Curve sketching / areas and volumes
The region bounded by y=x(x+1)(2−x)y = x(x+1)(2-x) and the xx-axis between x=−1x = -1 and x=0x = 0 is rotated completely about the xx-axis. Find the volume generated.
  1. A512π\tfrac{5}{12}\pi
  2. B22105π\tfrac{22}{105}\pi
  3. C22105\tfrac{22}{105}
  4. D44105π\tfrac{44}{105}\pi

Question 1404

[2 marks]Curve sketching / areas and volumes
For y=x(x+1)(2−x)y=x(x+1)(2-x), which option correctly gives its roots and its behaviour as x→∞x\to\infty?
  1. Aroots 1,0,−21,0,-2; y→−∞y\to-\infty
  2. Broots −1,0,2-1,0,2; y→0y\to0
  3. Croots −1,0,2-1,0,2; y→−∞y\to-\infty
  4. Droots −1,0,2-1,0,2; y→+∞y\to+\infty

Question 1405

[1 marks]Curve sketching / areas and volumes
Expand x(x+1)(2−x)x(x+1)(2-x) as a cubic ax3+bx2+cxax^3+bx^2+cx. State the coefficient of x2x^2.

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Question 1406

[2 marks]Curve sketching / areas and volumes
Let F(x)=−x44+x33+x2F(x)=-\tfrac{x^4}{4}+\tfrac{x^3}{3}+x^2 be an antiderivative of −x3+x2+2x-x^3+x^2+2x. Find F(−1)F(-1).

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Question 1407

[2 marks]Curve sketching / areas and volumes
Find [f(x)]2[f(x)]^2 where f(x)=−x3+x2+2xf(x)=-x^3+x^2+2x, at x=−0.5x=-0.5, to 3 decimal places.

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Question 1408

[3 marks]Curve sketching / areas and volumes
Which definite integral correctly represents the volume generated when the region between x=−1x=-1 and x=0x=0 under y=x(x+1)(2−x)y=x(x+1)(2-x) is rotated about the xx-axis?
  1. A∫−10[f(x)]2 dx\displaystyle\int_{-1}^{0}[f(x)]^2\,dx
  2. Bπ∫−10[f(x)]2 dx\pi\displaystyle\int_{-1}^{0}[f(x)]^2\,dx
  3. Cπ∫−10f(x) dx\pi\displaystyle\int_{-1}^{0}f(x)\,dx
  4. D2π∫−10xf(x) dx2\pi\displaystyle\int_{-1}^{0}xf(x)\,dx

Question 1501

[1 marks]Differential equations
In a population of fixed size pp, xx people have disease A and y=p−xy = p - x have disease B, and the rate of spread satisfies dxdt∝xy\dfrac{dx}{dt} \propto xy. Which differential equation follows?
  1. Adxdt=cx(p−x)\dfrac{dx}{dt} = \dfrac{c}{x(p-x)}
  2. Bdxdt=cx(p−x)\dfrac{dx}{dt} = cx(p - x)
  3. Cdxdt=c(p−x)\dfrac{dx}{dt} = c(p - x)
  4. Ddxdt=cxp\dfrac{dx}{dt} = cxp

Question 1502

[1 marks]Differential equations
To separate dxdt=cx(p−x)\dfrac{dx}{dt} = cx(p - x), the fraction 1x(p−x)\dfrac{1}{x(p-x)} is written in partial fractions as
  1. A1p(1x+1p−x)\dfrac{1}{p}\left(\dfrac{1}{x} + \dfrac{1}{p-x}\right)
  2. B1p(1x−1p−x)\dfrac{1}{p}\left(\dfrac{1}{x} - \dfrac{1}{p-x}\right)
  3. C1p2(1x+1p−x)\dfrac{1}{p^2}\left(\dfrac{1}{x} + \dfrac{1}{p-x}\right)
  4. D1x+1p−x\dfrac{1}{x} + \dfrac{1}{p-x}

Question 1503

[1 marks]Differential equations
Given x=p10x=\tfrac{p}{10} at t=0t=0, so y=p−x=9p10y=p-x=\tfrac{9p}{10}, find xp−x\dfrac{x}{p-x} at t=0t=0.

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Question 1504

[2 marks]Differential equations
Integrating 1p(1x+1p−x)dx=c dt\dfrac1p\left(\dfrac1x+\dfrac{1}{p-x}\right)dx=c\,dt gives 1pln⁡(xp−x)=ct+C\dfrac1p\ln\left(\dfrac{x}{p-x}\right)=ct+C. Using xp−x=19\dfrac{x}{p-x}=\tfrac19 at t=0t=0, find CC in terms of pp.

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Question 1505

[1 marks]Differential equations
Rearranging xp−x=19epct\dfrac{x}{p-x}=\tfrac19e^{pct} and solving for xx gives x=pepct9+epctx=\dfrac{pe^{pct}}{9+e^{pct}}. State the constant added to epcte^{pct} in the denominator.

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Question 1506

[2 marks]Differential equations
Separating variables in dxdt=cx(p−x)\dfrac{dx}{dt}=cx(p-x) gives which equation?
  1. A∫dxx(p−x)=∫c dt\displaystyle\int\dfrac{dx}{x(p-x)}=\displaystyle\int c\,dt
  2. B∫x(p−x) dx=∫c dt\displaystyle\int x(p-x)\,dx=\displaystyle\int c\,dt
  3. C∫dxx=∫c(p−x) dt\displaystyle\int\dfrac{dx}{x}=\displaystyle\int c(p-x)\,dt
  4. D∫dpx(p−x)=∫c dt\displaystyle\int\dfrac{dp}{x(p-x)}=\displaystyle\int c\,dt

Question 1507

[2 marks]Differential equations
Since y=p−xy=p-x, and dxdt=cx(p−x)=cxy\dfrac{dx}{dt}=cx(p-x)=cxy, find dydt\dfrac{dy}{dt} in terms of cc, xx and yy.

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Question 1508

[3 marks]Differential equations
Using x=pepct9+epctx=\dfrac{pe^{pct}}{9+e^{pct}}, find the time tt at which y=p10y=\tfrac{p}{10} (i.e. x=9p10x=\tfrac{9p}{10}), giving the answer in the form t=kln⁡3pct=\dfrac{k\ln3}{pc}. State the value of kk.

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