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ZIMSEC A Level · N2012

Pure Mathematics Paper 1 November 2012

Questions
73
Total marks
120

Sit this paper online

Questions
73
Pass mark
44
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]trigonometry / cosine rule
In triangle ABCABC, AB=mAB = m cm, AC=12AC = 12 cm and angle BA^C=60°B\hat{A}C = 60°. Given that the area of the triangle is 675\sqrt{675} cm2^2, find the exact value of mm.
  1. Am=5m = 5
  2. Bm=109m = \sqrt{109}
  3. Cm=15m = 15
  4. Dm=53m = 5\sqrt{3}

Question 102

[1 marks]trigonometry / cosine rule
In triangle ABCABC, AB=5AB = 5 cm, AC=12AC = 12 cm and angle BA^C=60°B\hat{A}C = 60°. Find the exact length of BCBC.
  1. A109\sqrt{109} cm
  2. B1313 cm
  3. C229\sqrt{229} cm
  4. D77 cm

Question 103

[2 marks]trigonometry / cosine rule
Using the cosine rule with AB=5AB = 5 cm, AC=12AC = 12 cm and angle BA^C=60°B\hat{A}C = 60°, calculate n2=BC2n^2 = BC^2 (before taking the square root).

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Question 201

[1 marks]numerical methods / Newton-Raphson
Let f(x)=6sin⁡(π+x2)−3f(x) = 6\sin(\pi + x^2) - 3, with xx in radians. Which values show that the equation 6sin⁡(π+x2)=36\sin(\pi + x^2) = 3 has a root between 1.91.9 and 22?
  1. Af(1.9)=3f(1.9) = 3 and f(2)=6f(2) = 6
  2. Bf(1.9)=0.299f(1.9) = 0.299 and f(2)=1.54f(2) = 1.54
  3. Cf(1.9)=−0.299f(1.9) = -0.299 and f(2)=1.54f(2) = 1.54
  4. Df(1.9)=−0.299f(1.9) = -0.299 and f(2)=−1.54f(2) = -1.54

Question 202

[1 marks]numerical methods / Newton-Raphson
Starting with x1=1.9x_1 = 1.9, apply the Newton-Raphson method once to f(x)=6sin⁡(π+x2)−3f(x) = 6\sin(\pi + x^2) - 3 and give the second estimate to 3 significant figures.
  1. A1.881.88
  2. B1.911.91
  3. C1.951.95
  4. D2.012.01

Question 203

[2 marks]numerical methods / Newton-Raphson
Differentiate f(x)=6sin⁡(π+x2)−3f(x) = 6\sin(\pi + x^2) - 3 with respect to xx to find f′(x)f'(x).

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Question 204

[2 marks]numerical methods / Newton-Raphson
Using f′(x)=12xcos⁡(π+x2)f'(x) = 12x\cos(\pi+x^2), evaluate f′(1.9)f'(1.9) correct to 3 significant figures (the value used in the Newton-Raphson formula).

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Question 301

[1 marks]differentiation / Maclaurin series
Given y=esin⁡3xy = e^{\sin 3x}, find dydx\dfrac{dy}{dx}.
  1. Aecos⁡3xe^{\cos 3x}
  2. Bcos⁡3x esin⁡3x\cos 3x\, e^{\sin 3x}
  3. C3esin⁡3x3e^{\sin 3x}
  4. D3cos⁡3x esin⁡3x3\cos 3x\, e^{\sin 3x}

Question 302

[1 marks]differentiation / Maclaurin series
Find the Maclaurin series of f(x)=esin⁡3xf(x) = e^{\sin 3x} up to and including the term in x2x^2.
  1. A1+3x+9x21 + 3x + 9x^2
  2. B1+3x+92x21 + 3x + \tfrac{9}{2}x^2
  3. C1+x+32x21 + x + \tfrac{3}{2}x^2
  4. D3x+92x23x + \tfrac{9}{2}x^2

Question 303

[2 marks]differentiation / Maclaurin series
For y=esin⁡3xy = e^{\sin 3x}, evaluate y′(0)y'(0), the value of dydx\frac{dy}{dx} at x=0x=0 (needed to build the Maclaurin series).

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Question 304

[2 marks]differentiation / Maclaurin series
For y=esin⁡3xy = e^{\sin 3x}, evaluate y′′(0)y''(0), the value of d2ydx2\frac{d^2y}{dx^2} at x=0x=0.

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Question 401

[1 marks]parametric differentiation
A curve is given parametrically by y=et2y = e^{t^2} and x=et2−1x = e^{t^2} - 1 for 1≤t≤21 \le t \le 2. Find dydx\dfrac{dy}{dx}.
  1. A2tet22te^{t^2}
  2. Bet2e^{t^2}
  3. C12t\dfrac{1}{2t}
  4. D11

Question 402

[1 marks]parametric differentiation
The curve y=et2y = e^{t^2}, x=et2−1x = e^{t^2} - 1 for 1≤t≤21 \le t \le 2 is a straight line segment. What are the exact coordinates of its end points?
  1. A(e−1,e)(e - 1, e) and (e2−1,e2)(e^2 - 1, e^2)
  2. B(e,e−1)(e, e - 1) and (e4,e4−1)(e^4, e^4 - 1)
  3. C(0,1)(0, 1) and (e4−1,e4)(e^4 - 1, e^4)
  4. D(e−1,e)(e - 1, e) and (e4−1,e4)(e^4 - 1, e^4)

Question 403

[2 marks]parametric differentiation
For x=et2−1x = e^{t^2} - 1 and y=et2y = e^{t^2} with 1≤t≤21 \le t \le 2, find dydt\frac{dy}{dt} in terms of tt.

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Question 404

[2 marks]parametric differentiation
The graph of yy against xx for 1≤t≤21 \le t \le 2, where dydx=1\frac{dy}{dx} = 1 throughout, is:
  1. Aan exponential curve
  2. Ba straight line of gradient −1-1
  3. Ca parabola opening upward
  4. Da straight line of gradient 1

Question 501

[1 marks]geometric progressions
The 6th term of a geometric progression is 181\tfrac{1}{81} and the 3rd term is −13-\tfrac{1}{3}. Find the common ratio and the first term.
  1. Ar=−3r = -3, a=−127a = -\tfrac{1}{27}
  2. Br=−13r = -\tfrac{1}{3}, a=3a = 3
  3. Cr=−13r = -\tfrac{1}{3}, a=−3a = -3
  4. Dr=13r = \tfrac{1}{3}, a=−3a = -3

Question 502

[1 marks]geometric progressions
A geometric progression has first term −3-3 and common ratio −13-\tfrac{1}{3}. Find its sum to infinity.
  1. A94\tfrac{9}{4}
  2. B−49-\tfrac{4}{9}
  3. C−94-\tfrac{9}{4}
  4. D−92-\tfrac{9}{2}

Question 503

[1 marks]geometric progressions
Using (6th term) ÷\div (3rd term) =r3= r^3, evaluate r3r^3 for this progression.

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Question 504

[2 marks]geometric progressions
Using a=−3a = -3 and r=−13r = -\frac{1}{3}, find the exact value of the 4th term of the progression.

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Question 505

[2 marks]geometric progressions
Using a=−3a = -3 and r=−13r = -\frac{1}{3}, find S4S_4, the sum of the first 4 terms, as an exact fraction.

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Question 601

[1 marks]exponential functions / tangents
The curve y=p+eqxy = p + e^{qx} crosses the yy-axis at A(0,32)A\left(0, \tfrac{3}{2}\right) and has gradient −2-2 there. Find pp and qq.
  1. Ap=−2p = -2, q=12q = \tfrac{1}{2}
  2. Bp=32p = \tfrac{3}{2}, q=−2q = -2
  3. Cp=12p = \tfrac{1}{2}, q=−2q = -2
  4. Dp=12p = \tfrac{1}{2}, q=2q = 2

Question 602

[1 marks]exponential functions / tangents
State the equation of the asymptote of the curve y=12+e−2xy = \tfrac{1}{2} + e^{-2x}.
  1. Ax=12x = \tfrac{1}{2}
  2. By=32y = \tfrac{3}{2}
  3. Cy=12y = \tfrac{1}{2}
  4. Dy=0y = 0

Question 603

[1 marks]exponential functions / tangents
The curve y=p+eqxy = p + e^{qx} passes through (0,32)(0, \frac{3}{2}). Using y(0)=p+1=32y(0) = p + 1 = \frac{3}{2}, find pp.

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Question 604

[2 marks]exponential functions / tangents
Differentiate y=p+eqxy = p + e^{qx} with respect to xx to find dydx\frac{dy}{dx} in terms of qq and xx.

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Question 701

[1 marks]complex numbers / Argand diagram
A complex number z1z_1 has modulus 2 and lies 30°30° below the negative real axis on an Argand diagram. Express z1z_1 in the form a+iba + ib with exact real aa and bb.
  1. A−3−i-\sqrt{3} - i
  2. B−1−3i-1 - \sqrt{3}i
  3. C3−i\sqrt{3} - i
  4. D−3+i-\sqrt{3} + i

Question 702

[1 marks]complex numbers / Argand diagram
Given z1=−3−iz_1 = -\sqrt{3} - i, find the exact value of w=−83 iz1w = \dfrac{-8\sqrt{3}\,i}{z_1} in the form a+iba + ib.
  1. A23+6i2\sqrt{3} + 6i
  2. B−23−6i-2\sqrt{3} - 6i
  3. C23−6i2\sqrt{3} - 6i
  4. D6+23i6 + 2\sqrt{3}i

Question 703

[1 marks]complex numbers / Argand diagram
State the principal argument of z1z_1, in degrees, given z1z_1 has modulus 2 and lies 30°30° below the negative real axis.

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Question 704

[2 marks]complex numbers / Argand diagram
Find the modulus of w=23+6iw = 2\sqrt{3} + 6i.

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Question 705

[1 marks]complex numbers / Argand diagram
Find the argument of w=23+6iw = 2\sqrt{3} + 6i, in degrees.

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Question 801

[1 marks]polynomials / factor theorem
Given that 2x−12x - 1 is a factor of f(x)=2x3+bx2+x−2f(x) = 2x^3 + bx^2 + x - 2, find the value of bb.
  1. Ab=3b = 3
  2. Bb=−5b = -5
  3. Cb=2b = 2
  4. Db=5b = 5

Question 802

[1 marks]polynomials / factor theorem
Factorise f(x)=2x3+5x2+x−2f(x) = 2x^3 + 5x^2 + x - 2 completely.
  1. A(2x−1)(x−1)(x−2)(2x - 1)(x - 1)(x - 2)
  2. B(x−1)(2x2+7x+2)(x - 1)(2x^2 + 7x + 2)
  3. C(2x−1)(x+1)(x+2)(2x - 1)(x + 1)(x + 2)
  4. D(2x+1)(x−1)(x−2)(2x + 1)(x - 1)(x - 2)

Question 803

[1 marks]polynomials / factor theorem
Find the values of xx for which f(x)=(2x−1)(x+1)(x+2)<0f(x) = (2x - 1)(x + 1)(x + 2) < 0.
  1. Ax<−2x < -2 or −1<x<12-1 < x < \tfrac{1}{2}
  2. Bx<−2x < -2 only
  3. C−2<x<12-2 < x < \tfrac{1}{2}
  4. D−2<x<−1-2 < x < -1 or x>12x > \tfrac{1}{2}

Question 804

[2 marks]polynomials / factor theorem
Dividing f(x)=2x3+5x2+x−2f(x) = 2x^3 + 5x^2 + x - 2 by (2x−1)(2x-1) gives quotient x2+ax+bx^2 + ax + b. State the quotient.

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Question 805

[1 marks]polynomials / factor theorem
State the xx-intercepts of y=f(x)=(2x−1)(x+1)(x+2)y = f(x) = (2x-1)(x+1)(x+2), in ascending order.

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Question 806

[1 marks]polynomials / factor theorem
Evaluate f(0)f(0) for f(x)=2x3+5x2+x−2f(x) = 2x^3 + 5x^2 + x - 2.

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Question 901

[1 marks]series / arc length
A string of length 220 cm is divided into nn parts in the ratio 1:2:3:…:n1 : 2 : 3 : \ldots : n. The length of the longest segment is
  1. A220nn+1\dfrac{220n}{n+1} cm
  2. B220n\dfrac{220}{n} cm
  3. C440n(n+1)\dfrac{440}{n(n+1)} cm
  4. D440n+1\dfrac{440}{n+1} cm

Question 902

[1 marks]series / arc length
A 220 cm string is bent into a circle so that its ends just touch. Find the radius of the circle, correct to 3 significant figures.
  1. A17.517.5 cm
  2. B35.035.0 cm
  3. C70.070.0 cm
  4. D110110 cm

Question 903

[1 marks]series / arc length
The largest segment of a string bent into a circle subtends an angle of approximately 12.57n+1\dfrac{12.57}{n+1} radians at the centre. Find the smallest value of nn for which this angle is less than π12\dfrac{\pi}{12} radians.
  1. An=48n = 48
  2. Bn=47n = 47
  3. Cn=49n = 49
  4. Dn=12n = 12

Question 904

[2 marks]series / arc length
Using 2πr=2202\pi r = 220, state the exact radius rr of the circle as a fraction of π\pi.

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Question 905

[2 marks]series / arc length
Using the longest-segment formula 440n+1\frac{440}{n+1}, find its length in cm (2 d.p.) when n=48n = 48.

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Question 906

[2 marks]series / arc length
For n=48n = 48, use S=rθS = r\theta with the longest segment length ≈8.98\approx 8.98 cm and r≈35.0r \approx 35.0 cm to find θ\theta in radians, correct to 3 s.f.

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Question 1001

[1 marks]integration (area and volume)
The region RR is enclosed by the curve y=1+3x−2y = 1 + \sqrt{3x - 2}, the xx-axis and the lines x=23x = \tfrac{2}{3} and x=2x = 2. Calculate the area of RR.
  1. A169\tfrac{16}{9} units2^2
  2. B43\tfrac{4}{3} units2^2
  3. C689\tfrac{68}{9} units2^2
  4. D289\tfrac{28}{9} units2^2

Question 1002

[1 marks]integration (area and volume)
The region enclosed by y=1+3x−2y = 1 + \sqrt{3x - 2}, the xx-axis and the lines x=23x = \tfrac{2}{3} and x=2x = 2 is rotated completely about the xx-axis. Find the volume generated.
  1. A68π9\tfrac{68\pi}{9} units3^3
  2. B4π3\tfrac{4\pi}{3} units3^3
  3. C28π9\tfrac{28\pi}{9} units3^3
  4. D32π9\tfrac{32\pi}{9} units3^3

Question 1003

[2 marks]integration (area and volume)
Evaluate x+29(3x−2)3/2x + \frac{2}{9}(3x-2)^{3/2} at x=2x = 2 (the antiderivative used for the area, before subtracting the lower limit).

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Question 1004

[1 marks]integration (area and volume)
Evaluate x+29(3x−2)3/2x + \frac{2}{9}(3x-2)^{3/2} at x=23x = \frac{2}{3}.

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Question 1005

[1 marks]integration (area and volume)
Evaluate [1+3x−2]2[1 + \sqrt{3x-2}]^2 at x=2x = 2 (the integrand used for the volume).

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Question 1006

[2 marks]integration (area and volume)
Expanding [1+3x−2]2=1+23x−2+(3x−2)[1+\sqrt{3x-2}]^2 = 1 + 2\sqrt{3x-2} + (3x-2), evaluate the middle term 23x−22\sqrt{3x-2} at x=2x = 2.

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Question 1101

[1 marks]vectors in 3D
The points AA and CC have position vectors 8j−4k8\mathbf{j} - 4\mathbf{k} and −2i+3j+5k-2\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}. Find a unit vector parallel to AC→\overrightarrow{AC}.
  1. A1110(−2i−5j+9k)\tfrac{1}{\sqrt{110}}(-2\mathbf{i} - 5\mathbf{j} + 9\mathbf{k})
  2. B1110(−2i−5j+9k)\tfrac{1}{110}(-2\mathbf{i} - 5\mathbf{j} + 9\mathbf{k})
  3. C1110(2i+5j−9k)\tfrac{1}{\sqrt{110}}(2\mathbf{i} + 5\mathbf{j} - 9\mathbf{k})
  4. D178(−2i−5j+9k)\tfrac{1}{\sqrt{78}}(-2\mathbf{i} - 5\mathbf{j} + 9\mathbf{k})

Question 1102

[1 marks]vectors in 3D
Points AA, BB, CC have position vectors 8j−4k8\mathbf{j} - 4\mathbf{k}, pi+3j+4kp\mathbf{i} + 3\mathbf{j} + 4\mathbf{k} and −2i+3j+5k-2\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}. Find the positive value of pp for which BABA is perpendicular to BCBC.
  1. Ap=−4p = -4
  2. Bp=2p = 2
  3. Cp=8p = 8
  4. Dp=4p = 4

Question 1103

[1 marks]vectors in 3D
In triangle ABCABC, BABA is perpendicular to BCBC with ∣BA→∣=93|\overrightarrow{BA}| = \sqrt{93} and ∣BC→∣=17|\overrightarrow{BC}| = \sqrt{17}. Find the area of the triangle correct to 2 decimal places.
  1. A790.50790.50 units2^2
  2. B9.949.94 units2^2
  3. C19.8819.88 units2^2
  4. D39.7639.76 units2^2

Question 1104

[2 marks]vectors in 3D
Solving p2+2p−8=0p^2 + 2p - 8 = 0 (from BA→⋅BC→=0\overrightarrow{BA}\cdot\overrightarrow{BC} = 0) gives two roots. Find the negative root (not used in this question).

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Question 1105

[3 marks]vectors in 3D
With p=2p = 2, find ∣BA→∣|\overrightarrow{BA}|, correct to 3 significant figures, where BA→=−2i+5j−8k\overrightarrow{BA} = -2\mathbf{i} + 5\mathbf{j} - 8\mathbf{k}.

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Question 1106

[2 marks]vectors in 3D
With p=2p = 2, find ∣BC→∣|\overrightarrow{BC}|, correct to 3 significant figures, where BC→=−4i+k\overrightarrow{BC} = -4\mathbf{i} + \mathbf{k}.

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Question 1201

[1 marks]coordinate geometry / circles
A square of perimeter 12212\sqrt{2} cm has centre E(4,3)E(4, 3). Find the equation of the circle which passes through all four corners of the square, in the form x2+y2+ax+by+c=0x^2 + y^2 + ax + by + c = 0.
  1. Ax2+y2−8x−6y+25=0x^2 + y^2 - 8x - 6y + 25 = 0
  2. Bx2+y2−8x−6y+16=0x^2 + y^2 - 8x - 6y + 16 = 0
  3. Cx2+y2−8x−6y−9=0x^2 + y^2 - 8x - 6y - 9 = 0
  4. Dx2+y2+8x+6y+16=0x^2 + y^2 + 8x + 6y + 16 = 0

Question 1202

[1 marks]coordinate geometry / circles
A chord ADAD of a circle of radius 3 cm subtends a right angle at the centre. Find the area of the minor segment cut off, correct to 2 decimal places.
  1. A7.077.07 cm2^2
  2. B1.291.29 cm2^2
  3. C2.572.57 cm2^2
  4. D4.504.50 cm2^2

Question 1203

[2 marks]coordinate geometry / circles
A square has perimeter 12212\sqrt{2} cm. Find the length of one side.

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Question 1204

[1 marks]coordinate geometry / circles
Find the diagonal length dd of the square (used to find the circle's radius), given side 323\sqrt2 cm.

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Question 1205

[1 marks]coordinate geometry / circles
State the gradient of line BCBC.

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Question 1206

[1 marks]coordinate geometry / circles
State the gradient of line BABA.

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Question 1207

[1 marks]coordinate geometry / circles
State the equation of the tangent to the circle at point A=(4,6)A = (4, 6).

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Question 1301

[1 marks]differential equations
Ants bring food to a hive at a constant 500500 g per day and consume it at 12x\tfrac{1}{2}x g per day, where xx g is the amount present after tt days. Write down the differential equation satisfied by xx and tt.
  1. Adxdt=500+12x\dfrac{dx}{dt} = 500 + \tfrac{1}{2}x
  2. Bdxdt=12x−500\dfrac{dx}{dt} = \tfrac{1}{2}x - 500
  3. Cdxdt=500x−12\dfrac{dx}{dt} = 500x - \tfrac{1}{2}
  4. Ddxdt=500−12x\dfrac{dx}{dt} = 500 - \tfrac{1}{2}x

Question 1302

[1 marks]differential equations
Solve dxdt=500−12x\dfrac{dx}{dt} = 500 - \tfrac{1}{2}x given that x=400x = 400 when t=0t = 0, expressing xx in terms of tt.
  1. Ax=400e−t/2x = 400e^{-t/2}
  2. Bx=1000−600e−t/2x = 1000 - 600e^{-t/2}
  3. Cx=1000−400e−t/2x = 1000 - 400e^{-t/2}
  4. Dx=1000+600e−t/2x = 1000 + 600e^{-t/2}

Question 1303

[1 marks]differential equations
For a hive in which x=1000−600e−t/2x = 1000 - 600e^{-t/2} grammes of food is present after tt days, the maximum amount of food the hive can hold is
  1. A600600 g, because that is the initial deficit
  2. B10001000 g, because dxdt→0\tfrac{dx}{dt} \to 0 as x→1000x \to 1000
  3. C500500 g, because that is the daily delivery rate
  4. Dunbounded, because food arrives at a constant rate

Question 1304

[3 marks]differential equations
Find, correct to 3 significant figures, the time tt at which x=980x = 980 grammes, given x=1000−600e−t/2x = 1000 - 600e^{-t/2}.

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Question 1305

[2 marks]differential equations
Using x=400x = 400 when t=0t = 0, find the value of CC in x=1000−Ce−t/2x = 1000 - Ce^{-t/2}.

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Question 1306

[2 marks]differential equations
Find dxdt\frac{dx}{dt}, correct to the nearest whole number, when x=600x = 600 grammes.

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Question 1401

[1 marks]trigonometry / R-formula and transformations
Express 32(cos⁡x+sin⁡x)3\sqrt{2}(\cos x + \sin x) in the form Rcos⁡(x−α)R\cos(x - \alpha), where R>0R > 0 and 0°<α<90°0° < \alpha < 90°.
  1. A18cos⁡(x−45°)18\cos(x - 45°)
  2. B6cos⁡(x−45°)6\cos(x - 45°)
  3. C32cos⁡(x−45°)3\sqrt{2}\cos(x - 45°)
  4. D6cos⁡(x−30°)6\cos(x - 30°)

Question 1402

[1 marks]trigonometry / R-formula and transformations
Solve 2(cos⁡x+sin⁡x)=3\sqrt{2}(\cos x + \sin x) = \sqrt{3} for 0°≤x≤360°0° \le x \le 360°.
  1. Ax=75°x = 75° and x=345°x = 345°
  2. Bx=15°x = 15° and x=75°x = 75°
  3. Cx=30°x = 30° and x=60°x = 60°
  4. Dx=15°x = 15° and x=285°x = 285°

Question 1403

[1 marks]trigonometry / R-formula and transformations
State the maximum and minimum values of y=6+32(cos⁡x+sin⁡x)y = 6 + 3\sqrt{2}(\cos x + \sin x).
  1. Amaximum 6+326 + 3\sqrt{2}, minimum 6−326 - 3\sqrt{2}
  2. Bmaximum 1212, minimum 00
  3. Cmaximum 66, minimum −6-6
  4. Dmaximum 1818, minimum −6-6

Question 1404

[2 marks]trigonometry / R-formula and transformations
The graph of y=6+32(cos⁡x+sin⁡x)y = 6 + 3\sqrt2(\cos x + \sin x) is obtained from y=cos⁡xy = \cos x by which sequence of transformations?
  1. Atranslate 45°45° in the positive xx-direction, stretch parallel to the yy-axis by factor 66, then translate 66 units in the positive yy-direction
  2. Bstretch parallel to the yy-axis by factor 66, translate 45°45° in the negative xx-direction, then translate 66 units in the negative yy-direction
  3. Ctranslate 45°45° in the positive xx-direction, stretch parallel to the xx-axis by factor 66, then translate 66 units in the positive yy-direction
  4. Dtranslate 66 units in the positive xx-direction, stretch parallel to the yy-axis by factor 4545, then translate 4545 units in the positive yy-direction

Question 1405

[2 marks]trigonometry / R-formula and transformations
Find yy, correct to 1 decimal place, when x=0°x = 0° for y=6+32(cos⁡x+sin⁡x)y = 6 + 3\sqrt2(\cos x + \sin x).

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Question 1406

[2 marks]trigonometry / R-formula and transformations
Find the value of xx (0°≤x≤360°0° \le x \le 360°) at which y=6+32(cos⁡x+sin⁡x)=0y = 6 + 3\sqrt2(\cos x + \sin x) = 0.

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Question 1407

[2 marks]trigonometry / R-formula and transformations
Find the value of xx (0°≤x≤360°0° \le x \le 360°) at which y=6+32(cos⁡x+sin⁡x)y = 6 + 3\sqrt2(\cos x + \sin x) attains its maximum value.

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