An original Thinka practice paper modelled on the structure and difficulty of the Jun 2022 AQA AS Level Mathematics 7356 paper. Not affiliated with or reproduced from AQA.
卷一 — 甲部: Pure Mathematics
Answer all questions. Pure content; calculator and AQA formulae booklet permitted.
10 题目 · 53 分
题目 1 · 選擇題
1 分
Express \(\log_3 5 + \log_3 2\) as a single logarithm. Circle your answer.
Solve \(2\sin^2 x = 3\cos x\) for \(0^\circ \le x \le 360^\circ\).
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解题
Using \(\sin^2 x = 1-\cos^2 x\): \(2-2\cos^2 x = 3\cos x\Rightarrow 2\cos^2 x+3\cos x-2=0\Rightarrow (2\cos x-1)(\cos x+2)=0\). So \(\cos x=\tfrac12\) (reject \(\cos x=-2\)), giving \(x=60^\circ,300^\circ\).
评分标准
M1 use of identity; M1 form quadratic in \(\cos x\); A1 \(\cos x=\tfrac12\); A1 60°; A1 300°.
题目 5 · Structured
6 分
A circle has the points \(A(-1,2)\) and \(B(5,10)\) as the ends of a diameter. (a) Write down the centre. (b) Show that the radius is 5. (c) Find the equation of the circle. (d) Determine whether the point \((6,6)\) lies inside, on, or outside the circle.
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解题
(a) Centre = midpoint = \((2,6)\). (b) \(AB=\sqrt{6^2+8^2}=10\), so radius \(=5\). (c) \((x-2)^2+(y-6)^2=25\). (d) \((6-2)^2+(6-6)^2=16<25\), so the point is inside.
The curve \(y=9-x^2\) meets the x-axis at two points. (a) Find the x-coordinates of these points. (b) Find the exact area enclosed between the curve and the x-axis.
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解题
(a) \(9-x^2=0\Rightarrow x=\pm3\). (b) Area \(=\int_{-3}^{3}(9-x^2)\,dx = \left[9x-\tfrac{x^3}{3}\right]_{-3}^{3} = (27-9)-(-27+9)=18-(-18)=36\).
评分标准
M1A1 intercepts; M1 set up definite integral; A1 integrate; M1 limits; A1 36.
题目 7 · Structured
8 分
A curve has equation \(y=2x^3-9x^2+12x\). (a) Find \(\dfrac{dy}{dx}\). (b) Find the coordinates of the stationary points. (c) Determine the nature of each stationary point.
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解题
(a) \(\tfrac{dy}{dx}=6x^2-18x+12=6(x-1)(x-2)\). (b) Stationary where \(x=1\) (\(y=5\)) and \(x=2\) (\(y=4\)). (c) \(\tfrac{d^2y}{dx^2}=12x-18\); at \(x=1\) it is \(-6<0\) (maximum), at \(x=2\) it is \(6>0\) (minimum).
评分标准
M1A1 derivative; M1 solve =0; A1 both points; M1 second derivative; A1 both natures.
题目 8 · Proof
5 分
Prove that, for every integer \(n\), \(n^3-n\) is divisible by 6.
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解题
\(n^3-n=n(n^2-1)=(n-1)n(n+1)\), the product of three consecutive integers. Among any three consecutive integers at least one is divisible by 2 and exactly one is divisible by 3. Hence the product is divisible by \(2\times3=6\).
评分标准
M1 factorise; A1 three consecutive integers; A1 divisible by 2; A1 divisible by 3; A1 conclude divisible by 6.
题目 9 · Structured
9 分
A curve has equation \(y=x^2-4x+5\). (a) Find the equation of the tangent to the curve at \(x=3\). (b) Find the equation of the normal at \(x=3\). (c) The normal meets the curve again; find the coordinates of that point.
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解题
At \(x=3\), \(y=2\). \(\tfrac{dy}{dx}=2x-4=2\). (a) Tangent: \(y-2=2(x-3)\Rightarrow y=2x-4\). (b) Normal gradient \(-\tfrac12\): \(y=-\tfrac12 x+\tfrac72\). (c) Set \(x^2-4x+5=-\tfrac12 x+\tfrac72\Rightarrow 2x^2-7x+3=0\Rightarrow(2x-1)(x-3)=0\), other root \(x=\tfrac12\), \(y=\tfrac{13}{4}\).
评分标准
M1A1 tangent; M1A1 normal; M1 equate to curve; M1 solve quadratic; A1 other point.
题目 10 · Structured
8 分
A population is modelled by \(P=500e^{kt}\), where \(t\) is in years. After 3 years the population is 800. (a) Find \(k\) to 3 significant figures. (b) Estimate the population after 10 years. (c) Find, to the nearest year, when the population reaches 2000.
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解题
(a) \(800=500e^{3k}\Rightarrow k=\tfrac{1}{3}\ln 1.6=0.157\). (b) \(P=500e^{0.157\times10}=500e^{1.566}\approx2395\). (c) \(2000=500e^{kt}\Rightarrow e^{kt}=4\Rightarrow t=\tfrac{\ln4}{0.1566}\approx8.85\), so about 9 years.
评分标准
M1 form equation; A1 k=0.157; M1A1 (b); M1 set =2000; A1 ≈9 years.
卷一 — 乙部: Mechanics
Answer all questions. Take g = 9.8 m s⁻² unless otherwise stated.
6 题目 · 27 分
题目 1 · 選擇題
1 分
A particle starts from rest and moves in a straight line with constant acceleration. Which graph best shows its velocity \(v\) against time \(t\)? Circle your answer.
A.A straight line through the origin with positive gradient
B.A horizontal line
C.A curve of increasing gradient
D.A straight line with positive intercept
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解题
Constant acceleration from rest gives \(v=at\): a straight line through the origin with positive gradient.
评分标准
B1 straight line through the origin.
题目 2 · Structured
4 分
A ball is projected vertically upwards from ground level at \(21\,\text{m s}^{-1}\) (take \(g=9.8\)). (a) Find the maximum height reached. (b) Find the total time before it returns to the ground.
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解题
(a) \(v^2=u^2-2gs\): \(0=21^2-2(9.8)s\Rightarrow s=\tfrac{441}{19.6}=22.5\,\text{m}\). (b) Time up: \(0=21-9.8t\Rightarrow t=2.143\,\text{s}\); total \(=2\times2.143=4.29\,\text{s}\).
评分标准
M1A1 max height; M1 time up; A1 total time.
题目 3 · Structured
4 分
A particle of mass \(2\,\text{kg}\) is acted on by forces \(\mathbf{F_1}=(3\mathbf{i}+4\mathbf{j})\,\text{N}\) and \(\mathbf{F_2}=(5\mathbf{i}-2\mathbf{j})\,\text{N}\). (a) Find the resultant force. (b) Find the magnitude of the acceleration.
B1 resultant; M1 magnitude; M1 divide by mass; A1 4.12.
题目 4 · Structured
6 分
Two particles of masses \(3\,\text{kg}\) and \(5\,\text{kg}\) are connected by a light inextensible string passing over a smooth fixed pulley, and released from rest. (a) Find the acceleration of the system. (b) Find the tension in the string.
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解题
(a) For the system, \((5-3)g=(5+3)a\Rightarrow a=\tfrac{2\times9.8}{8}=2.45\,\text{m s}^{-2}\). (b) For the 3 kg mass: \(T-3g=3a\Rightarrow T=3(9.8)+3(2.45)=36.75\,\text{N}\).
评分标准
M1 equation of motion for system; A1 a=2.45; M1 equation for one mass; A1 T=36.75; A1 consistent check.
题目 5 · Structured
6 分
A particle moves in a straight line with velocity \(v=3t^2-12t+9\;\text{m s}^{-1}\). (a) Find the times when the particle is instantaneously at rest. (b) Find its acceleration when \(t=3\). (c) Find its displacement during the first 4 seconds.
M1 set v=0; A1 both times; M1A1 acceleration; M1 integrate; A1 4 m.
题目 6 · Structured
6 分
A car of mass \(1200\,\text{kg}\) tows a trailer of mass \(400\,\text{kg}\) by a light rigid tow-bar along a straight horizontal road. The driving force is \(4000\,\text{N}\); the resistances are \(450\,\text{N}\) on the car and \(150\,\text{N}\) on the trailer. (a) Find the acceleration. (b) Find the tension in the tow-bar.
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解题
(a) For the whole system: \(4000-(450+150)=1600a\Rightarrow a=\tfrac{3400}{1600}=2.125\,\text{m s}^{-2}\). (b) For the trailer: \(T-150=400a\Rightarrow T=150+400(2.125)=1000\,\text{N}\).
A bottle of water cools according to \(T=20+Ae^{-kt}\), where \(T\,^\circ\text{C}\) is the temperature \(t\) minutes after it leaves a fridge. Initially \(T=80\), and after 5 minutes \(T=50\). (a) Find \(A\). (b) Find \(k\) to 3 significant figures. (c) Find the temperature after 12 minutes. (d) Find, to the nearest minute, the time taken to reach \(30^\circ\text{C}\).
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解题
(a) At \(t=0\): \(80=20+A\Rightarrow A=60\). (b) \(50=20+60e^{-5k}\Rightarrow e^{-5k}=\tfrac12\Rightarrow k=\tfrac{\ln2}{5}=0.139\). (c) \(T=20+60e^{-0.1386\times12}=20+60(0.1895)=31.4^\circ\text{C}\). (d) \(30=20+60e^{-kt}\Rightarrow e^{-kt}=\tfrac16\Rightarrow t=\tfrac{\ln6}{0.1386}=12.9\), about 13 min.
评分标准
B1 A=60; M1 form equation; A1 k=0.139; M1A1 (c); M1 set =30; A1 ≈13 min.
题目 6 · Structured
7 分
In triangle \(ABC\), \(AB=7\,\text{cm}\), \(BC=8\,\text{cm}\) and angle \(ABC=60^\circ\). (a) Find the length \(AC\). (b) Find the area of the triangle. (c) Find angle \(BAC\) to 1 decimal place.
The line \(y=x+1\) and the curve \(y=x^2-2x-3\) intersect at two points. (a) Find the coordinates of the points of intersection. (b) Write down the inequalities that define the region enclosed between the line and the curve. (c) Find the exact area of the enclosed region.
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解题
(a) \(x^2-2x-3=x+1\Rightarrow x^2-3x-4=0\Rightarrow(x-4)(x+1)=0\); points \((-1,0),(4,5)\). (b) \(-1\le x\le4\) and \(x^2-2x-3\le y\le x+1\). (c) Area \(=\int_{-1}^{4}\big((x+1)-(x^2-2x-3)\big)dx=\int_{-1}^{4}(-x^2+3x+4)dx=\tfrac{125}{6}\).
A curve has equation \(y=2x^3-3x^2-12x\). (a) Find \(\dfrac{dy}{dx}\). (b) Find the coordinates of the stationary points. (c) Determine the nature of each.
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解题
(a) \(\tfrac{dy}{dx}=6x^2-6x-12=6(x-2)(x+1)\). (b) \(x=2\Rightarrow y=-20\); \(x=-1\Rightarrow y=7\). (c) \(\tfrac{d^2y}{dx^2}=12x-6\); at \(x=2\) positive (min), at \(x=-1\) negative (max).
A geometric series has second term 6 and sum to infinity 27. (a) Show that the common ratio satisfies \(9r^2-9r+2=0\). (b) Find the two possible values of \(r\). (c) For \(r=\tfrac13\), find the first term and the sum of the first 5 terms.
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解题
Let first term \(a\). \(ar=6\) and \(\tfrac{a}{1-r}=27\). So \(a=\tfrac6r\) and \(\tfrac{6/r}{1-r}=27\Rightarrow6=27r(1-r)\Rightarrow27r^2-27r+6=0\Rightarrow9r^2-9r+2=0\). (b) \((3r-1)(3r-2)=0\Rightarrow r=\tfrac13,\tfrac23\). (c) \(r=\tfrac13\Rightarrow a=18\); \(S_5=18\cdot\tfrac{1-(1/3)^5}{1-1/3}=27\cdot\tfrac{242}{243}\approx26.9\).
评分标准
M1 two equations; M1 eliminate a; A1 given quadratic; M1 factorise; A1 both r; M1 a=18; A1 S₅.
题目 10 · Structured
4 分
Express \(\dfrac{5}{2-\sqrt3}\) in the form \(a+b\sqrt3\), where \(a\) and \(b\) are integers.
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解题
Multiply numerator and denominator by \(2+\sqrt3\): \(\dfrac{5(2+\sqrt3)}{(2-\sqrt3)(2+\sqrt3)}=\dfrac{5(2+\sqrt3)}{4-3}=5(2+\sqrt3)=10+5\sqrt3\).
Answer all questions. A calculator with statistical functions may be used.
6 题目 · 27 分
题目 1 · 選擇題
1 分
A school has 40 classes of 30 students. A researcher selects 3 whole classes at random and surveys every student in them. This sampling method is: circle your answer.
A.Cluster
B.Stratified
C.Systematic
D.Quota
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解题
Selecting whole intact groups at random is cluster sampling.
评分标准
B1 cluster.
题目 2 · 選擇題
1 分
On a box plot the median is much closer to the lower quartile than to the upper quartile. The distribution is best described as: circle your answer.
A.Symmetric
B.Positively skewed
C.Negatively skewed
D.Uniform
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解题
A longer tail to the right (median near \(Q_1\)) indicates positive skew.
评分标准
B1 positively skewed.
题目 3 · Structured
5 分
The times (in minutes) for six visits are: \(12, 15, 18, 11, 14, 20\). (a) Find the mean. (b) Find the standard deviation to 2 decimal places.
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解题
(a) Mean \(=\tfrac{90}{6}=15\). (b) Squared deviations: \(9,0,9,16,1,25\), sum \(=60\); variance \(=\tfrac{60}{6}=10\); s.d. \(=\sqrt{10}=3.16\).
On each gym visit a member independently chooses an apple, banana or cake, with \(P(A)=0.3\), \(P(B)=0.45\), \(P(C)=0.25\). For four randomly chosen visits, find the probability that the member chose: (a) at least one apple; (b) the same item on all four visits; (c) apple exactly twice and cake exactly twice.
A coin is suspected of being biased towards heads. It is tossed 20 times and 15 heads are obtained. Test, at the 5% significance level, whether there is evidence that the probability of a head exceeds 0.5.
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解题
Let \(p=P(\text{head})\). \(H_0:p=0.5\), \(H_1:p>0.5\). Under \(H_0\), \(X\sim B(20,0.5)\). \(P(X\ge15)=1-P(X\le14)=0.0207\). Since \(0.0207<0.05\), reject \(H_0\): there is evidence at the 5% level that the coin is biased towards heads.