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ZIMSEC A Level · J2007

Pure Mathematics Paper 1 June 2007

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]inequalities
Solve the inequality ∣2x+3∣>7|2x + 3| > 7.
  1. A−5<x<2-5 < x < 2
  2. Bx>2x > 2 or x<−2x < -2
  3. Cx>2x > 2 or x<−5x < -5
  4. Dx>4x > 4 or x<−10x < -10

Question 102

[2 marks]inequalities
Solve 2x+3=72x+3=7 and 2x+3=−72x+3=-7 (the boundary equations from ∣2x+3∣=7|2x+3|=7). State both solutions as 'larger, smaller'.

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Question 201

[1 marks]binomial expansion
Find the coefficient of x6x^6 in the expansion of (9+x2)12(9+x^2)^{\frac12} in ascending powers of xx.
  1. A−13888-\dfrac{1}{3888}
  2. B13888\dfrac{1}{3888}
  3. C111664\dfrac{1}{11664}
  4. D11944\dfrac{1}{1944}

Question 202

[2 marks]binomial expansion
In the binomial expansion of (1+u)1/2(1+u)^{1/2} up to the u3u^3 term, state the coefficient of the u3u^3 term as a fraction (used later to find the x6x^6 coefficient in (9+x2)1/2(9+x^2)^{1/2}).

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Question 301

[1 marks]exponential equations
Find the exact value of the solution to the equation 2e2x−7ex−4=02e^{2x} - 7e^x - 4 = 0.
  1. Ax=ln⁡14x = \ln\dfrac14
  2. Bx=4x = 4
  3. Cx=ln⁡4x = \ln 4
  4. Dx=−ln⁡2x = -\ln 2

Question 302

[2 marks]exponential equations
The equation 2e2x−7ex−4=02e^{2x}-7e^x-4=0 becomes 2u2−7u−4=02u^2-7u-4=0 with u=exu=e^x, which factorises as (2u+1)(u−4)=0(2u+1)(u-4)=0. State the root of the quadratic that must be rejected, since u=ex>0u=e^x>0.

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Question 401

[1 marks]trigonometric equations
Express cos⁡θ−2sin⁡θ\cos\theta - 2\sin\theta in the form Rcos⁡(θ+α)R\cos(\theta+\alpha), where R>0R>0 and 0∘<α<90∘0^\circ<\alpha<90^\circ.
  1. AR=5R=\sqrt5, α≈26.57∘\alpha \approx 26.57^\circ
  2. BR=5R=5, α≈63.43∘\alpha \approx 63.43^\circ
  3. CR=3R=\sqrt3, α≈63.43∘\alpha \approx 63.43^\circ
  4. DR=5R=\sqrt5, α≈63.43∘\alpha \approx 63.43^\circ

Question 402

[1 marks]trigonometric equations
Given that cos⁡θ−2sin⁡θ=5cos⁡(θ+63.43∘)\cos\theta - 2\sin\theta = \sqrt5\cos(\theta+63.43^\circ), solve cos⁡θ−2sin⁡θ=0.2\cos\theta - 2\sin\theta = 0.2 for 0∘≤θ≤360∘0^\circ \le \theta \le 360^\circ.
  1. Aθ≈41.4∘\theta \approx 41.4^\circ or θ≈231.7∘\theta \approx 231.7^\circ
  2. Bθ≈21.4∘\theta \approx 21.4^\circ or θ≈211.7∘\theta \approx 211.7^\circ
  3. Cθ≈84.9∘\theta \approx 84.9^\circ or θ≈275.1∘\theta \approx 275.1^\circ
  4. Dθ≈31.7∘\theta \approx 31.7^\circ or θ≈211.7∘\theta \approx 211.7^\circ

Question 403

[2 marks]trigonometric equations
Given 5cos⁡(θ+α)=0.2\sqrt5\cos(\theta+\alpha)=0.2 with α≈63.43∘\alpha\approx63.43^\circ, evaluate cos⁡−1(0.2/5)\cos^{-1}(0.2/\sqrt5) in degrees, correct to 1 decimal place (the principal value of θ+α\theta+\alpha).

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Question 404

[2 marks]trigonometric equations
Continuing from cos⁡(θ+α)=0.2/5\cos(\theta+\alpha)=0.2/\sqrt5 with α≈63.43∘\alpha\approx63.43^\circ and principal value θ+α≈84.9∘\theta+\alpha\approx84.9^\circ, state the second value of θ+α\theta+\alpha in [0∘,360∘][0^\circ,360^\circ] (i.e. 360∘360^\circ minus the principal value), to 1 decimal place.

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Question 501

[1 marks]implicit differentiation / logarithms
A gas obeys the law ln⁡p+1.4ln⁡v=ln⁡c\ln p + 1.4\ln v = \ln c, where cc is a constant. Find dvdp\dfrac{dv}{dp} in terms of vv and pp.
  1. Advdp=−1.4vp\dfrac{dv}{dp} = -\dfrac{1.4v}{p}
  2. Bdvdp=−vp\dfrac{dv}{dp} = -\dfrac{v}{p}
  3. Cdvdp=v1.4p\dfrac{dv}{dp} = \dfrac{v}{1.4p}
  4. Ddvdp=−v1.4p\dfrac{dv}{dp} = -\dfrac{v}{1.4p}

Question 502

[1 marks]implicit differentiation / logarithms
Using dvdp=−v1.4p\dfrac{dv}{dp}=-\dfrac{v}{1.4p}, find the approximate percentage change in vv given that pp increases by 1%, stating whether it is an increase or a decrease.
  1. Aa decrease of approximately 7%7\%
  2. Ban increase of approximately 0.714%0.714\%
  3. Ca decrease of approximately 1.4%1.4\%
  4. Da decrease of approximately 0.714%0.714\%

Question 503

[2 marks]implicit differentiation / logarithms
A gas obeys ln⁡p+1.4ln⁡v=ln⁡c\ln p + 1.4\ln v = \ln c. Differentiating implicitly with respect to pp gives which equation?
  1. A1.4p+1vdvdp=0\dfrac{1.4}{p} + \dfrac{1}{v}\dfrac{dv}{dp} = 0
  2. B1p+1.4vdvdp=0\dfrac{1}{p} + \dfrac{1.4}{v}\dfrac{dv}{dp} = 0
  3. C1p−1.4vdvdp=0\dfrac{1}{p} - \dfrac{1.4}{v}\dfrac{dv}{dp} = 0
  4. D1p+1.4vdvdp=0\dfrac{1}{p} + 1.4v\dfrac{dv}{dp} = 0

Question 504

[2 marks]implicit differentiation / logarithms
Using dvdp=−v1.4p\dfrac{dv}{dp}=-\dfrac{v}{1.4p} and δpp=0.01\dfrac{\delta p}{p}=0.01 (a 1% increase in pp), state the exact fraction for dvdp⋅δpv\dfrac{dv}{dp}\cdot\dfrac{\delta p}{v} (the fractional change in vv), before converting to a percentage.

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Question 601

[1 marks]circle mensuration
A chord of a circle of radius rr subtends an angle of θ\theta radians at the centre. Which expression gives the area of the minor segment cut off by the chord?
  1. A12r2(θ−sin⁡θ)\dfrac12 r^2(\theta-\sin\theta)
  2. B12r2θ\dfrac12 r^2\theta
  3. C12r2sin⁡θ\dfrac12 r^2\sin\theta
  4. Dr2(θ−sin⁡θ)r^2(\theta-\sin\theta)

Question 602

[1 marks]circle mensuration
Two circles of radii 3 m and 4 m have centres A and B respectively, 5 m apart. The circles intersect at C and D. Why is each radius (AC and BC) a tangent to the other circle?
  1. ABecause C lies on both circles, every line drawn through C touches each circle at just that point, which is what makes a line tangent.
  2. BBecause triangle ACB is isosceles, its base angles are equal, which forces each radius to touch the other circle at a single point.
  3. CBecause 32+42=523^2+4^2=5^2, triangle ACB is right-angled at C, so AC and BC are perpendicular; a radius perpendicular to another circle's radius at the point of contact lies along that circle's tangent there.
  4. DBecause AC and BC are both shorter than AB, they lie outside both circles, which is the defining property of a tangent line.

Question 603

[1 marks]circle mensuration
Two circles of radii 3 m and 4 m have centres 5 m apart and intersect at C and D. Calculate the area common to the two circles.
  1. A≈4.03 m2\approx 4.03\text{ m}^2
  2. B≈6.64 m2\approx 6.64\text{ m}^2
  3. C≈18.64 m2\approx 18.64\text{ m}^2
  4. D≈2.62 m2\approx 2.62\text{ m}^2

Question 604

[2 marks]circle mensuration
Two circles, radii 3 m and 4 m, have centres AA and BB 5 m apart, intersecting at CC. Triangle ACBACB has sides AC=3AC=3, CB=4CB=4, AB=5AB=5. Calculate angle CABCAB in degrees, correct to 2 decimal places.

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Question 605

[2 marks]circle mensuration
For circle AA (radius 3 m) in the two-circle problem (centres 5 m apart, radii 3 m and 4 m), the segment cut off by chord CDCD subtends angle θ=2×53.13∘=106.26∘=1.8546\theta = 2\times53.13^\circ = 106.26^\circ = 1.8546 rad at the centre. Using segment area =12r2(θ−sin⁡θ)=\frac12 r^2(\theta-\sin\theta), calculate this segment's area in circle AA, correct to 2 decimal places (in m2^2).

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Question 701

[1 marks]sequences and series
Two sequences are defined for n=1,2,3,…n=1,2,3,\ldots by Un=10n−3U_n=10n-3 and Vn=4[1−(13)n]V_n=4\left[1-\left(\dfrac13\right)^n\right]. Which statement correctly describes their behaviour as n→∞n\to\infty?
  1. ABoth UnU_n and VnV_n diverge to +∞+\infty.
  2. BUnU_n converges to 10 while VnV_n diverges to +∞+\infty.
  3. CUnU_n diverges to +∞+\infty while VnV_n converges to 4.
  4. DUnU_n converges to −3-3 while VnV_n converges to 4.

Question 702

[1 marks]sequences and series
Given that Un=10n−3U_n=10n-3 and ∑n=1NUn=259\displaystyle\sum_{n=1}^{N}U_n=259, find NN.

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Question 703

[2 marks]sequences and series
For Un=10n−3U_n=10n-3, the sum ∑n=1NUn\sum_{n=1}^N U_n simplifies to 5N2+2N5N^2+2N. Evaluate this expression at N=5N=5.

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Question 704

[2 marks]sequences and series
Given 5N2+2N−259=05N^2+2N-259=0 (from ∑n=1NUn=259\sum_{n=1}^N U_n=259 with Un=10n−3U_n=10n-3), calculate the discriminant b2−4acb^2-4ac using a=5a=5, b=2b=2, c=−259c=-259.

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Question 705

[1 marks]sequences and series
State 5184\sqrt{5184} (the square root of the discriminant of 5N2+2N−259=05N^2+2N-259=0).

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Question 801

[1 marks]polynomial functions and curve sketching
Given f(x)=x3−4x2−3x+18f(x)=x^3-4x^2-3x+18, and that (x−3)(x-3) is a factor, f(x)f(x) factorises completely as (x−3)2(x+2)(x-3)^2(x+2). Which statement correctly describes where y=f(x)y=f(x) meets the axes?
  1. AIt touches the xx-axis at x=−2x=-2 (repeated root) and crosses at x=3x=3, and meets the yy-axis at (0,18)(0,18).
  2. BIt touches the xx-axis at x=3x=3 (repeated root) and crosses at x=−2x=-2, and meets the yy-axis at (0,−18)(0,-18).
  3. CIt touches the xx-axis at x=3x=3 (repeated root) and crosses at x=−2x=-2, and meets the yy-axis at (0,18)(0,18).
  4. DIt crosses the xx-axis at x=3x=3 and x=−2x=-2, both simple crossings, and meets the yy-axis at (0,−18)(0,-18).

Question 802

[1 marks]polynomial functions and curve sketching
Given f(x)=(x−3)2(x+2)f(x)=(x-3)^2(x+2), the curve y=f(x+3)y=f(x+3) is a horizontal translation of y=f(x)y=f(x). Where does y=f(x+3)y=f(x+3) meet the xx-axis?
  1. Ax=0x=0 (repeated root) and x=5x=5
  2. Bx=−3x=-3 (repeated root) and x=−8x=-8
  3. Cx=6x=6 (repeated root) and x=1x=1
  4. Dx=0x=0 (repeated root) and x=−5x=-5

Question 803

[1 marks]polynomial functions and curve sketching
Given f(x)=x3−4x2−3x+18f(x)=x^3-4x^2-3x+18 with stationary points at x=−13x=-\tfrac13 (local max, f(−13)≈18.52f(-\tfrac13)\approx18.52) and x=3x=3 (local min, f(3)=0f(3)=0), how many real roots does f(x)−4=0f(x)-4=0 have?
  1. A0
  2. B1
  3. C2
  4. D3

Question 804

[1 marks]polynomial functions and curve sketching
For f(x)=x3−4x2−3x+18f(x)=x^3-4x^2-3x+18, evaluate f(3)f(3).

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Question 805

[2 marks]polynomial functions and curve sketching
Given (x−3)(x-3) is a factor of f(x)=x3−4x2−3x+18f(x)=x^3-4x^2-3x+18, state the quadratic quotient obtained by dividing f(x)f(x) by (x−3)(x-3), in the form x2+bx+cx^2+bx+c.

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Question 806

[2 marks]polynomial functions and curve sketching
For f(x)=x3−4x2−3x+18f(x)=x^3-4x^2-3x+18, f′(x)=3x2−8x−3f'(x)=3x^2-8x-3. What are the two x-values where f′(x)=0f'(x)=0 (the turning points of ff)?
  1. Ax=3x=3 or x=−13x=-\dfrac13
  2. Bx=3x=3 or x=13x=\dfrac13
  3. Cx=83x=\dfrac83 or x=0x=0
  4. Dx=−3x=-3 or x=13x=\dfrac13

Question 901

[1 marks]numerical methods / differentiation
Taking x1=1.2x_1=1.2 as a first approximation to the root of x10=ln⁡x\dfrac{x}{10}=\ln x, use the Newton-Raphson method to find the root correct to 3 decimal places.

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Question 902

[1 marks]numerical methods / differentiation
Given V=πD36V=\dfrac{\pi D^3}{6} so that dVdD=πD22\dfrac{dV}{dD}=\dfrac{\pi D^2}{2}, estimate the maximum error in computing the volume of a sphere whose diameter is measured as 16 cm with a possible error of 0.01 cm.
  1. A≈0.16π cm3\approx 0.16\pi\ \text{cm}^3
  2. B≈0.64π cm3\approx 0.64\pi\ \text{cm}^3
  3. C≈2.56π cm3\approx 2.56\pi\ \text{cm}^3
  4. D≈1.28π cm3\approx 1.28\pi\ \text{cm}^3

Question 903

[2 marks]numerical methods / differentiation
Using x1=1.2x_1=1.2 as a first approximation to a root of x/10=ln⁡xx/10=\ln x, apply one Newton-Raphson step with g(x)=x/10−ln⁡xg(x)=x/10-\ln x to find x2x_2, correct to 3 decimal places.

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Question 904

[2 marks]numerical methods / differentiation
For V=16πD3V=\frac16\pi D^3 (the volume of a sphere in terms of diameter DD), state dVdD\dfrac{dV}{dD} in terms of DD.

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Question 905

[2 marks]numerical methods / differentiation
For g(x)=x/10−ln⁡xg(x)=x/10-\ln x (used in the Newton-Raphson method for x/10=ln⁡xx/10=\ln x), evaluate g′(1.2)g'(1.2) correct to 4 decimal places.

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Question 1001

[1 marks]vectors
The position vectors of AA, BB, CC are (314)\begin{pmatrix}3\\1\\4\end{pmatrix}, (730)\begin{pmatrix}7\\3\\0\end{pmatrix}, (157−8)\begin{pmatrix}15\\7\\-8\end{pmatrix} respectively. Find the angle between OB→\overrightarrow{OB} and AC→\overrightarrow{AC}.
  1. A≈41.9∘\approx 41.9^\circ
  2. B≈84.4∘\approx 84.4^\circ
  3. C≈48.1∘\approx 48.1^\circ
  4. D≈138.1∘\approx 138.1^\circ

Question 1002

[1 marks]vectors
Given AB→=(42−4)\overrightarrow{AB}=\begin{pmatrix}4\\2\\-4\end{pmatrix} and AC→=(126−12)\overrightarrow{AC}=\begin{pmatrix}12\\6\\-12\end{pmatrix}, which statement correctly shows that AA, BB, CC are collinear?
  1. ABoth AB→\overrightarrow{AB} and AC→\overrightarrow{AC} have a negative zz-component, so they point into the same region of space, showing the three points lie on one line.
  2. BAC→=3AB→\overrightarrow{AC}=3\overrightarrow{AB}, so AB→\overrightarrow{AB} and AC→\overrightarrow{AC} are parallel and share the point AA, meaning AA, BB, CC lie on one straight line.
  3. CAB→⋅AC→=48+12+48=108≠0\overrightarrow{AB}\cdot\overrightarrow{AC}=48+12+48=108\neq0, so the vectors point in the same direction, confirming the three points are collinear.
  4. D∣AC→∣=18|\overrightarrow{AC}|=18 is exactly three times ∣AB→∣=6|\overrightarrow{AB}|=6, so BB is the midpoint of ACAC, which proves collinearity.

Question 1003

[1 marks]vectors
Given that the vector (5−3p)\begin{pmatrix}5\\-3\\p\end{pmatrix} is perpendicular to OA→=(314)\overrightarrow{OA}=\begin{pmatrix}3\\1\\4\end{pmatrix}, find the value of pp.

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Question 1004

[2 marks]vectors
Given A=(3,1,4)A=(3,1,4) and C=(15,7,−8)C=(15,7,-8), find the column vector AC→=C−A\overrightarrow{AC}=C-A. State it as 'x, y, z'.

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Question 1005

[2 marks]vectors
Given OB→=(7,3,0)\overrightarrow{OB}=(7,3,0) and AC→=(12,6,−12)\overrightarrow{AC}=(12,6,-12), calculate the dot product OB→⋅AC→\overrightarrow{OB}\cdot\overrightarrow{AC}.

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Question 1006

[2 marks]vectors
Given A=(3,1,4)A=(3,1,4), B=(7,3,0)B=(7,3,0), C=(15,7,−8)C=(15,7,-8), so AB→=(4,2,−4)\overrightarrow{AB}=(4,2,-4) and AC→=(12,6,−12)\overrightarrow{AC}=(12,6,-12). State the scalar kk such that AC→=k⋅AB→\overrightarrow{AC}=k\cdot\overrightarrow{AB}.

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Question 1101

[1 marks]parametric equations / differentiation
A curve has parametric equations x=sin⁡ϕx=\sin\phi, y=1−cos⁡2ϕy=1-\cos^2\phi. Find dydx\dfrac{dy}{dx}.
  1. A2sin⁡ϕ2\sin\phi
  2. Bsin⁡2ϕ\sin2\phi
  3. C2cos⁡ϕ2\cos\phi
  4. Dcos⁡ϕ\cos\phi

Question 1102

[1 marks]parametric equations / differentiation
For the curve x=sin⁡ϕx=\sin\phi, y=1−cos⁡2ϕy=1-\cos^2\phi, find the equation of the tangent at the point QQ where ϕ=π/6\phi=\pi/6 (so x=12x=\tfrac12, y=14y=\tfrac14, and dydx=2sin⁡ϕ=1\dfrac{dy}{dx}=2\sin\phi=1 at QQ).
  1. Ay=−x+34y=-x+\dfrac34
  2. By=x−14y=x-\dfrac14
  3. Cy=2x−34y=2x-\dfrac34
  4. Dy=x+14y=x+\dfrac14

Question 1103

[1 marks]parametric equations / differentiation
The tangent (y=x−14y=x-\tfrac14) and normal (y=−x+34y=-x+\tfrac34) to the curve at QQ meet the yy-axis at points AA and BB respectively. State the coordinates of AA and BB.
  1. AA=(0,14)A=(0,\tfrac14), B=(0,−34)B=(0,-\tfrac34)
  2. BA=(−14,0)A=(-\tfrac14,0), B=(34,0)B=(\tfrac34,0)
  3. CA=(0,34)A=(0,\tfrac34), B=(0,−14)B=(0,-\tfrac14)
  4. DA=(0,−14)A=(0,-\tfrac14), B=(0,34)B=(0,\tfrac34)

Question 1104

[2 marks]parametric equations / differentiation
For x=sin⁡ϕx=\sin\phi, y=1−cos⁡2ϕy=1-\cos^2\phi, find dydϕ\dfrac{dy}{d\phi} in terms of ϕ\phi.

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Question 1105

[2 marks]parametric equations / differentiation
For x=sin⁡ϕx=\sin\phi, y=1−cos⁡2ϕy=1-\cos^2\phi at ϕ=π/6\phi=\pi/6, state the coordinates (x,y)(x,y) as 'x, y'.

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Question 1106

[2 marks]parametric equations / differentiation
At ϕ=π/6\phi=\pi/6 on the curve x=sin⁡ϕx=\sin\phi, y=1−cos⁡2ϕy=1-\cos^2\phi, given dydx=2sin⁡ϕ\dfrac{dy}{dx}=2\sin\phi, state the gradient of the tangent and the gradient of the normal, as 'tangent, normal'.

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Question 1201

[1 marks]coordinate geometry / circles
Line ll has equation 3y+2x=83y+2x=8 and point A=(1,15)A=(1,15). The line through AA perpendicular to ll is 2y−3x=272y-3x=27, meeting ll at (−5,6)(-5,6). Find the perpendicular distance of AA from ll.
  1. A9139\sqrt{13}
  2. B3133\sqrt{13}
  3. C13\sqrt{13}
  4. D6136\sqrt{13}

Question 1202

[1 marks]coordinate geometry / circles
Find the equation of the circle with centre A(1,15)A(1,15) that touches the line l: 3y+2x=8l:\ 3y+2x=8 (i.e. has radius equal to the perpendicular distance 3133\sqrt{13}).
  1. A(x−1)2+(y−15)2=117(x-1)^2+(y-15)^2=117
  2. B(x+1)2+(y+15)2=117(x+1)^2+(y+15)^2=117
  3. C(x−1)2+(y−15)2=117(x-1)^2+(y-15)^2=\sqrt{117}
  4. D(x−1)2+(y−15)2=13(x-1)^2+(y-15)^2=13

Question 1203

[2 marks]coordinate geometry / circles
State the gradient of the line ll: 3y+2x=83y+2x=8.

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Question 1204

[3 marks]coordinate geometry / circles
Solve simultaneously 3y+2x=83y+2x=8 and 2y−3x=272y-3x=27 (the line through A(1,15)A(1,15) perpendicular to ll). State the intersection point as 'x, y'.

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Question 1205

[2 marks]coordinate geometry / circles
Using A(1,15)A(1,15) and the intersection point (−5,6)(-5,6) of the two lines, compute (1−(−5))2+(15−6)2(1-(-5))^2+(15-6)^2.

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Question 1206

[2 marks]coordinate geometry / circles
Given 117=k13\sqrt{117}=k\sqrt{13}, state the integer kk.

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Question 1301

[1 marks]quadratics / completing the square
Given 2x2−6x+3=2(x−32)2−322x^2-6x+3=2(x-\tfrac32)^2-\tfrac32, state the coordinates of the turning point of y=2x2−6x+3y=2x^2-6x+3.
  1. A(−32,−32)(-\tfrac32,-\tfrac32)
  2. B(32,32)(\tfrac32,\tfrac32)
  3. C(32,−32)(\tfrac32,-\tfrac32)
  4. D(3,−32)(3,-\tfrac32)

Question 1302

[1 marks]quadratics / completing the square
Which sequence of three transformations correctly turns the graph of y=x2y=x^2 into the graph of y=2x2−6x+3=2(x−32)2−32y=2x^2-6x+3=2(x-\tfrac32)^2-\tfrac32?
  1. ATranslate 32\tfrac32 left; stretch vertically by factor 2; translate down 32\tfrac32.
  2. BStretch vertically by factor 2; translate 32\tfrac32 right; translate up 32\tfrac32.
  3. CTranslate 32\tfrac32 right; translate down 32\tfrac32; stretch vertically by factor 2.
  4. DTranslate 32\tfrac32 right; stretch vertically by factor 2; translate down 32\tfrac32.

Question 1303

[1 marks]quadratics / completing the square
Find the values of kk for which the equation 2x2−6x+3=kx+12x^2-6x+3=kx+1 has distinct real roots.
  1. Ak>10k>10 or k<2k<2
  2. Bk>−2k>-2 or k<−10k<-10
  3. C−10<k<−2-10<k<-2
  4. Dk>2k>2 or k<10k<10

Question 1304

[1 marks]quadratics / completing the square
To complete the square, 2x2−6x+32x^2-6x+3 is first written as 2(x2−3x)+32(x^2-3x)+3. State the number that must be added and subtracted inside the bracket to make x2−3xx^2-3x a perfect square.

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Question 1305

[2 marks]quadratics / completing the square
Rearranging 2x2−6x+3=kx+12x^2-6x+3=kx+1 into the form 2x2−(6+k)x+d=02x^2-(6+k)x+d=0, state the constant dd.

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Question 1306

[2 marks]quadratics / completing the square
The discriminant of 2x2−(6+k)x+2=02x^2-(6+k)x+2=0 has the form (6+k)2−m(6+k)^2-m. State mm.

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Question 1307

[2 marks]quadratics / completing the square
Solve (6+k)2=16(6+k)^2=16 for both boundary values of kk (where the discriminant of 2x2−(6+k)x+2=02x^2-(6+k)x+2=0 is zero). State them as 'smaller, larger'.

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Question 1308

[2 marks]quadratics / completing the square
For y=2x2−6x+3y=2x^2-6x+3 written as 2(x−32)2−322(x-\frac32)^2-\frac32, state the vertical stretch factor applied to y=x2y=x^2 and the x-coordinate of the axis of symmetry, as 'stretch, axis x'.

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Question 1401

[1 marks]integration / trapezium rule
Using the substitution u=2x+3u=2x+3, find ∫4x(2x+3)5 dx\displaystyle\int\dfrac{4x}{(2x+3)^5}\,dx.
  1. A−13(2x+3)3+34(2x+3)4+C-\dfrac{1}{3(2x+3)^3}+\dfrac{3}{4(2x+3)^4}+C
  2. B13(2x+3)3−34(2x+3)4+C\dfrac{1}{3(2x+3)^3}-\dfrac{3}{4(2x+3)^4}+C
  3. C−14(2x+3)3+35(2x+3)4+C-\dfrac{1}{4(2x+3)^3}+\dfrac{3}{5(2x+3)^4}+C
  4. D−13(2x+3)3−34(2x+3)4+C-\dfrac{1}{3(2x+3)^3}-\dfrac{3}{4(2x+3)^4}+C

Question 1402

[1 marks]integration / trapezium rule
For y=2xx−1y=\dfrac{2x}{x-1}, using the trapezium rule with 4 intervals of width 1 over [2,6][2,6] (ordinates at x=2,3,4,5,6x=2,3,4,5,6 giving y=4,3,83,52,125y=4,3,\tfrac83,\tfrac52,\tfrac{12}5), estimate the shaded area to 2 decimal places.
  1. A9.609.60
  2. B10.5710.57
  3. C11.3711.37
  4. D12.8012.80

Question 1403

[1 marks]integration / trapezium rule
Using 2xx−1=2+2x−1\dfrac{2x}{x-1}=2+\dfrac{2}{x-1}, find the exact area under y=2xx−1y=\dfrac{2x}{x-1} between x=2x=2 and x=6x=6.
  1. A8+ln⁡58+\ln5
  2. B4+2ln⁡54+2\ln5
  3. C8+2ln⁡58+2\ln5
  4. D8+2ln⁡38+2\ln 3

Question 1404

[2 marks]integration / trapezium rule
Using u=2x+3u=2x+3, integrate ∫u−4 du\int u^{-4}\,du.

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Question 1405

[2 marks]integration / trapezium rule
For y=2xx−1y=\dfrac{2x}{x-1}, state the ordinates y(2)y(2) and y(6)y(6), as 'y(2), y(6)'.

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Question 1406

[2 marks]integration / trapezium rule
For y=2xx−1y=\dfrac{2x}{x-1}, calculate y(3)+y(4)+y(5)y(3)+y(4)+y(5), correct to 2 decimal places.

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Question 1407

[2 marks]integration / trapezium rule
On y=2xx−1y=\dfrac{2x}{x-1} over [2,6][2,6], why is the trapezium rule estimate for the shaded area greater than the sum 62+83+104+125\dfrac{6}{2}+\dfrac{8}{3}+\dfrac{10}{4}+\dfrac{12}{5}?
  1. AThe curve is convex on [2,6][2,6], so each trapezium lies above the curve, giving an overestimate; the given sum instead uses the smaller left-hand ordinate on each interval, an underestimate by comparison.
  2. BThe given sum uses the same ordinates as the trapezium rule but divides by a different width, which happens to make it smaller for this particular curve.
  3. CThe curve is concave on [2,6][2,6], so each trapezium lies below the curve, making the trapezium estimate an underestimate rather than an overestimate.
  4. DThe trapezium rule overestimates any decreasing function purely because the function is decreasing, independent of how the curve bends.

Question 1408

[2 marks]integration / trapezium rule
For ∫26(2+2x−1)dx=[2x+2ln⁡(x−1)]26\displaystyle\int_2^6\left(2+\dfrac{2}{x-1}\right)dx=\big[2x+2\ln(x-1)\big]_2^6, evaluate 2x+2ln⁡(x−1)2x+2\ln(x-1) at x=6x=6, giving an exact answer.

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Question 1409

[1 marks]integration / trapezium rule
For [2x+2ln⁡(x−1)]\big[2x+2\ln(x-1)\big], evaluate this expression at x=2x=2.

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Question 1501

[1 marks]Maclaurin series / integration
Given f(x)=xe3xf(x)=xe^{3x} with f′(x)=3xe3x+e3xf'(x)=3xe^{3x}+e^{3x} and f′′(x)=3e3x(3x+2)f''(x)=3e^{3x}(3x+2), find f′′′(x)f'''(x) and hence f′′′(0)f'''(0).
  1. Af′′′(x)=27e3x(x+2)f'''(x)=27e^{3x}(x+2), so f′′′(0)=54f'''(0)=54
  2. Bf′′′(x)=9e3x(x+1)f'''(x)=9e^{3x}(x+1), so f′′′(0)=9f'''(0)=9
  3. Cf′′′(x)=27e3x(x+1)f'''(x)=27e^{3x}(x+1), so f′′′(0)=27f'''(0)=27
  4. Df′′′(x)=9e3x(3x+2)f'''(x)=9e^{3x}(3x+2), so f′′′(0)=18f'''(0)=18

Question 1502

[1 marks]Maclaurin series / integration
Find by integration the exact area in the first quadrant bounded by y=xe3xy=xe^{3x}, the xx-axis, and the line x=1x=1, correct to 3 decimal places.
  1. Ae3−13≈6.362\dfrac{e^3-1}{3}\approx6.362
  2. B2e3−19≈4.353\dfrac{2e^3-1}{9}\approx4.353
  3. Ce3+19≈2.454\dfrac{e^3+1}{9}\approx2.454
  4. D2e3+19≈4.575\dfrac{2e^3+1}{9}\approx4.575

Question 1503

[1 marks]Maclaurin series / integration
Integrating the Maclaurin series f(x)≈x+3x2+92x3+92x4f(x)\approx x+3x^2+\tfrac92x^3+\tfrac92x^4 from 00 to 11 gives an approximate area of 3.5253.525. Given the exact area is 4.5754.575 (to 3 d.p.), state the absolute error in the series approximation.
  1. A0.7750.775
  2. B1.0501.050
  3. C1.2901.290
  4. D8.1008.100

Question 1504

[2 marks]Maclaurin series / integration
Given f(x)=xe3xf(x)=xe^{3x} with f′′′(x)=27e3x(x+1)f'''(x)=27e^{3x}(x+1), find fiv(x)f^{iv}(x) and evaluate fiv(0)f^{iv}(0).

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Question 1505

[2 marks]Maclaurin series / integration
Given f(0)=0f(0)=0, f′(0)=1f'(0)=1, f′′(0)=6f''(0)=6, f′′′(0)=27f'''(0)=27, state the coefficients of x2x^2 and x3x^3 in the Maclaurin series of f(x)=xe3xf(x)=xe^{3x}, as 'c2, c3'.

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Question 1506

[2 marks]Maclaurin series / integration
Integrating xe3xxe^{3x} by parts, which expression is the antiderivative (before evaluating any limits)?
  1. Ax3e3x−19e3x+C\dfrac{x}{3}e^{3x} - \dfrac{1}{9}e^{3x} + C
  2. Bx3e3x+19e3x+C\dfrac{x}{3}e^{3x} + \dfrac{1}{9}e^{3x} + C
  3. Cx9e3x−13e3x+C\dfrac{x}{9}e^{3x} - \dfrac{1}{3}e^{3x} + C
  4. D3xe3x−9e3x+C3xe^{3x} - 9e^{3x} + C

Question 1507

[2 marks]Maclaurin series / integration
Evaluate e3e^3 correct to 3 decimal places.

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Question 1508

[2 marks]Maclaurin series / integration
Using the Maclaurin polynomial x+3x2+92x3+92x4x+3x^2+\frac92x^3+\frac92x^4 for f(x)=xe3xf(x)=xe^{3x}, evaluate ∫01(3x2+92x3)dx\displaystyle\int_0^1\left(3x^2+\frac92x^3\right)dx.

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Question 1509

[1 marks]Maclaurin series / integration
Using the same Maclaurin polynomial, evaluate ∫01(x+92x4)dx\displaystyle\int_0^1\left(x+\frac92x^4\right)dx.

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