AQA AS-Level · Thinka 原創模擬試題

2023 AQA AS-Level Mathematics 7356 模擬試題連答案詳解

Thinka Jun 2023 AQA AS Level-Style Mock — Mathematics 7356

160 180 分鐘2023
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 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.
11 題目 · 53
題目 1 · 選擇題
1
The point \((2,k)\) lies on the curve \(y=3x^2-5x+1\). Find \(k\). Circle your answer.
  1. A.1
  2. B.3
  3. C.7
  4. D.-3
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解題

\(k=3(4)-5(2)+1=12-10+1=3\).

評分準則

B1 for 3.
題目 2 · 選擇題
1
State the period of \(y=\tan 3x\). Circle your answer.
  1. A.\(120^\circ\)
  2. B.\(60^\circ\)
  3. C.\(180^\circ\)
  4. D.\(30^\circ\)
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解題

\(\tan x\) has period \(180^\circ\); \(\tan 3x\) has period \(\tfrac{180^\circ}{3}=60^\circ\).

評分準則

B1 for 60°.
題目 3 · Short
3
Find the first three terms, in ascending powers of \(x\), of the binomial expansion of \((1+2x)^6\).
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解題

\((1+2x)^6=1+\binom{6}{1}(2x)+\binom{6}{2}(2x)^2+\dots=1+12x+15(4x^2)+\dots=1+12x+60x^2+\dots\)

評分準則

M1 binomial structure; A1 \(12x\); A1 \(60x^2\).
題目 4 · Short
5
Solve \(2\cos^2 x+3\sin x=3\) for \(0^\circ\le x\le360^\circ\).
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解題

\(2(1-\sin^2 x)+3\sin x=3\Rightarrow 2\sin^2 x-3\sin x+1=0\Rightarrow(2\sin x-1)(\sin x-1)=0\). So \(\sin x=\tfrac12\Rightarrow x=30^\circ,150^\circ\); \(\sin x=1\Rightarrow x=90^\circ\).

評分準則

M1 identity; M1 quadratic in sin; A1 factorise; A1 30° & 150°; A1 90°.
題目 5 · Structured
6
The points are \(A(1,3)\) and \(B(7,11)\). (a) Find the midpoint of \(AB\). (b) Find the gradient of \(AB\). (c) Find the equation of the perpendicular bisector of \(AB\).
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解題

(a) Midpoint \((4,7)\). (b) Gradient \(=\tfrac{11-3}{7-1}=\tfrac{8}{6}=\tfrac43\). (c) Perpendicular gradient \(-\tfrac34\): \(y-7=-\tfrac34(x-4)\Rightarrow y=-\tfrac34x+10\).

評分準則

B1 midpoint; M1A1 gradient; M1 perpendicular gradient; A1 equation; B1 simplified.
題目 6 · Structured
7
\(f(x)=x^2-5x\). (a) Use differentiation from first principles to show that \(f'(x)=2x-5\). (b) Find the gradient at \(x=4\). (c) Find the value of \(x\) at which the gradient is 7.
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解題

(a) \(f'(x)=\lim_{h\to0}\tfrac{(x+h)^2-5(x+h)-(x^2-5x)}{h}=\lim_{h\to0}\tfrac{2xh+h^2-5h}{h}=\lim_{h\to0}(2x-5+h)=2x-5\). (b) \(2(4)-5=3\). (c) \(2x-5=7\Rightarrow x=6\).

評分準則

M1 form difference quotient; M1 expand & simplify; A1 limit =2x-5; B1 (b); M1A1 (c).
題目 7 · Structured
7
A curve has equation \(y=3x^2-4x+1\). (a) Find \(\displaystyle\int y\,dx\). (b) Find the area under the curve between \(x=1\) and \(x=3\). (c) Find the x-coordinate of the point where the gradient of the curve is zero.
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解題

(a) \(\int(3x^2-4x+1)dx=x^3-2x^2+x+c\). (b) \([x^3-2x^2+x]_1^3=(27-18+3)-(1-2+1)=12\). (c) \(\tfrac{dy}{dx}=6x-4=0\Rightarrow x=\tfrac23\).

評分準則

M1A1 integral; M1 limits; A1 12; M1A1 (c).
題目 8 · Proof
5
Prove that the product of any two consecutive even integers is divisible by 8.
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解題

Two consecutive even integers can be written \(2n\) and \(2n+2\). Their product is \(2n(2n+2)=4n(n+1)\). Since \(n(n+1)\) is a product of consecutive integers it is even, say \(2k\). Hence the product \(=4(2k)=8k\), divisible by 8.

評分準則

M1 represent as 2n, 2n+2; M1 expand to 4n(n+1); A1 n(n+1) even; A1 =8k; A1 conclude.
題目 9 · Structured
8
The value of a car is modelled by \(V=18000e^{-kt}\), where \(V\) is in pounds and \(t\) is the age in years. When the car is 4 years old it is worth £11000. (a) Find \(k\) to 3 significant figures. (b) Find the value when the car is 7 years old. (c) Find, to the nearest year, when the value falls to £5000.
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解題

(a) \(11000=18000e^{-4k}\Rightarrow e^{-4k}=\tfrac{11}{18}\Rightarrow k=\tfrac14\ln\tfrac{18}{11}=0.123\). (b) \(V=18000e^{-0.1231\times7}=18000(0.4223)\approx£7600\). (c) \(5000=18000e^{-kt}\Rightarrow t=\tfrac{\ln(18000/5000)}{0.1231}=10.4\), about 10 years.

評分準則

M1 form equation; A1 k=0.123; M1A1 (b); M1 set =5000; A1 ≈10 years.
題目 10 · Structured
6
A curve has equation \(y=x^2-6x+8\). (a) Find the equation of the tangent to the curve at \(x=5\). (b) Find the coordinates of the point where this tangent crosses the x-axis.
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解題

At \(x=5\), \(y=3\). \(\tfrac{dy}{dx}=2x-6=4\). (a) Tangent: \(y-3=4(x-5)\Rightarrow y=4x-17\). (b) \(y=0\Rightarrow x=\tfrac{17}{4}=4.25\).

評分準則

M1 y at x=5; M1 gradient; A1 tangent; M1 set y=0; A1 (17/4, 0).
題目 11 · Short
4
The line \(y=2x+c\) is a tangent to the curve \(y=x^2-3x+7\). Find the value of \(c\).
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解題

Tangency: \(x^2-3x+7=2x+c\Rightarrow x^2-5x+(7-c)=0\) has equal roots, so discriminant \(=0\): \(25-4(7-c)=0\Rightarrow 4c=3\Rightarrow c=\tfrac34\).

評分準則

M1 equate; M1 discriminant =0; A1 form equation; A1 c=3/4.

卷一 — 乙部: Mechanics

Answer all questions. Take g = 9.8 m s⁻² unless otherwise stated.
6 題目 · 27
題目 1 · 選擇題
1
A car decelerates uniformly from \(20\,\text{m s}^{-1}\) to rest in 8 seconds. Find the magnitude of its deceleration. Circle your answer.
  1. A.\(2.5\,\text{m s}^{-2}\)
  2. B.\(160\,\text{m s}^{-2}\)
  3. C.\(0.4\,\text{m s}^{-2}\)
  4. D.\(2\,\text{m s}^{-2}\)
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解題

\(a=\tfrac{20}{8}=2.5\,\text{m s}^{-2}\).

評分準則

B1 for 2.5 m s⁻².
題目 2 · Structured
4
A stone is projected vertically upwards at \(14\,\text{m s}^{-1}\) from the top of a cliff \(10\,\text{m}\) above the sea (take \(g=9.8\)). (a) Find the greatest height above the cliff top. (b) Find the speed with which it hits the sea.
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解題

(a) \(v^2=u^2-2gs\): \(0=14^2-2(9.8)s\Rightarrow s=10\,\text{m}\). (b) Taking up positive, at the sea \(s=-10\): \(v^2=14^2-2(9.8)(-10)=196+196=392\Rightarrow v=19.8\,\text{m s}^{-1}\).

評分準則

M1A1 max height; M1 use s=-10; A1 19.8.
題目 3 · Structured
4
Two forces \((4\mathbf{i}-3\mathbf{j})\,\text{N}\) and \((\mathbf{i}+7\mathbf{j})\,\text{N}\) act on a particle of mass \(5\,\text{kg}\). (a) Find the resultant force. (b) Find the magnitude of the acceleration.
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解題

(a) Resultant \(=(5\mathbf{i}+4\mathbf{j})\,\text{N}\). (b) \(|\mathbf{F}|=\sqrt{25+16}=\sqrt{41}\); \(a=\tfrac{\sqrt{41}}{5}=1.28\,\text{m s}^{-2}\).

評分準則

B1 resultant; M1 magnitude; M1 divide by mass; A1 1.28.
題目 4 · Structured
6
Two particles of masses \(5\,\text{kg}\) and \(3\,\text{kg}\) are connected by a light inextensible string over a smooth pulley and released from rest. (a) Find the acceleration. (b) Find the tension.
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解題

(a) \((5-3)g=8a\Rightarrow a=2.45\,\text{m s}^{-2}\). (b) For the 3 kg mass: \(T-3g=3a\Rightarrow T=3(9.8+2.45)=36.75\,\text{N}\).

評分準則

M1 system equation; A1 a; M1 single-mass equation; A1 T; B1 method.
題目 5 · Structured
6
A particle moves in a straight line with velocity \(v=2t^2-10t+8\;\text{m s}^{-1}\). (a) Find the times when it is at rest. (b) Find the acceleration when \(t=4\). (c) Find the displacement during the first 3 seconds.
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解題

(a) \(2t^2-10t+8=0\Rightarrow t^2-5t+4=0\Rightarrow(t-1)(t-4)=0\), \(t=1,4\). (b) \(a=4t-10\); at \(t=4\), \(a=6\). (c) \(s=\int_0^3(2t^2-10t+8)dt=[\tfrac{2t^3}{3}-5t^2+8t]_0^3=18-45+24=-3\,\text{m}\).

評分準則

M1 v=0; A1 both times; M1A1 acceleration; M1 integrate; A1 -3 m.
題目 6 · Structured
6
A lorry of mass \(2000\,\text{kg}\) tows a trailer of mass \(800\,\text{kg}\) by a light rigid tow-bar. The engine produces a forward force of \(6000\,\text{N}\); resistances are \(500\,\text{N}\) on the lorry and \(300\,\text{N}\) on the trailer. (a) Find the acceleration. (b) Find the tension in the tow-bar.
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解題

(a) Whole system: \(6000-800=2800a\Rightarrow a=\tfrac{5200}{2800}=1.857\,\text{m s}^{-2}\). (b) Trailer: \(T-300=800a\Rightarrow T=300+800(1.857)=1786\approx1790\,\text{N}\).

評分準則

M1 system equation; A1 a; M1 trailer equation; A1 T≈1790; B1 method.

卷二 — 甲部: Pure Mathematics

Answer all questions. Pure content; calculator and AQA formulae booklet permitted.
10 題目 · 53
題目 1 · 選擇題
1
Find \(\displaystyle\int(6x^2-4)\,dx\). Circle your answer.
  1. A.\(2x^3-4x+c\)
  2. B.\(12x+c\)
  3. C.\(2x^3-4+c\)
  4. D.\(6x^3-4x+c\)
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解題

\(\int(6x^2-4)dx=2x^3-4x+c\).

評分準則

B1 for \(2x^3-4x+c\).
題目 2 · 選擇題
1
Express \(\sqrt{50}+\sqrt{8}\) in the form \(k\sqrt2\). Circle your answer.
  1. A.\(7\sqrt2\)
  2. B.\(\sqrt{58}\)
  3. C.\(10\sqrt2\)
  4. D.\(58\)
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解題

\(\sqrt{50}+\sqrt8=5\sqrt2+2\sqrt2=7\sqrt2\).

評分準則

B1 for \(7\sqrt2\).
題目 3 · Structured
5
\(f(x)=2x^3-3x^2-11x+6\). (a) Show that \((x-3)\) is a factor. (b) Factorise \(f(x)\) completely.
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解題

(a) \(f(3)=54-27-33+6=0\), so \((x-3)\) is a factor. (b) \(f(x)=(x-3)(2x^2+3x-2)=(x-3)(2x-1)(x+2)\).

評分準則

M1 evaluate f(3); A1 conclude factor; M1 divide; A1 quadratic; A1 full factorisation.
題目 4 · Structured
4
The quadratic equation \(x^2+kx+(k+3)=0\) has no real roots. Find the set of values of \(k\).
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解題

No real roots \(\Rightarrow\) discriminant \(<0\): \(k^2-4(k+3)<0\Rightarrow k^2-4k-12<0\Rightarrow(k-6)(k+2)<0\Rightarrow -2

評分準則

M1 discriminant <0; A1 \(k^2-4k-12<0\); M1 factorise; A1 \(-2
題目 5 · Structured
8
A cup of coffee cools according to \(T=22+Ae^{-kt}\), where \(T\,^\circ\text{C}\) is the temperature \(t\) minutes after it is made. Initially \(T=78\), and after 6 minutes \(T=50\). (a) Find \(A\). (b) Find \(k\) to 3 significant figures. (c) Find the temperature after 15 minutes. (d) Find, to the nearest minute, the time to reach \(30^\circ\text{C}\).
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解題

(a) \(78=22+A\Rightarrow A=56\). (b) \(50=22+56e^{-6k}\Rightarrow e^{-6k}=\tfrac12\Rightarrow k=\tfrac{\ln2}{6}=0.116\). (c) \(T=22+56e^{-0.1155\times15}=22+56(0.1768)=31.9^\circ\text{C}\). (d) \(30=22+56e^{-kt}\Rightarrow e^{-kt}=\tfrac{1}{7}\Rightarrow t=\tfrac{\ln7}{0.1155}=16.8\), about 17 min.

評分準則

B1 A=56; M1 equation; A1 k=0.116; M1A1 (c); M1 set =30; A1 ≈17 min.
題目 6 · Structured
7
In triangle \(PQR\), \(PQ=9\,\text{cm}\), \(QR=6\,\text{cm}\) and angle \(PQR=110^\circ\). (a) Find \(PR\). (b) Find the area of the triangle. (c) Find angle \(QPR\) to 1 decimal place.
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解題

(a) \(PR^2=9^2+6^2-2(9)(6)\cos110^\circ=117+36.94=153.9\Rightarrow PR=12.4\,\text{cm}\). (b) Area \(=\tfrac12(9)(6)\sin110^\circ=25.4\,\text{cm}^2\). (c) \(\tfrac{\sin QPR}{6}=\tfrac{\sin110^\circ}{12.4}\Rightarrow\sin QPR=0.4543\Rightarrow QPR=27.0^\circ\).

評分準則

M1A1 cosine rule; A1 PR; M1A1 area; M1 sine rule; A1 27.0°.
題目 7 · Structured
10
The line \(y=2x+1\) and the curve \(y=x^2-x+1\) intersect at two points. (a) Find the coordinates of the points of intersection. (b) Write down the inequalities that define the enclosed region. (c) Find the exact area enclosed.
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解題

(a) \(x^2-x+1=2x+1\Rightarrow x^2-3x=0\Rightarrow x(x-3)=0\); points \((0,1),(3,7)\). (b) \(0\le x\le3\) and \(x^2-x+1\le y\le 2x+1\). (c) Area \(=\int_0^3\big((2x+1)-(x^2-x+1)\big)dx=\int_0^3(-x^2+3x)dx=[-\tfrac{x^3}{3}+\tfrac{3x^2}{2}]_0^3=-9+13.5=\tfrac92\).

評分準則

M1 equate; A1 quadratic; A1 both points; B1 inequalities; M1 integrate difference; M1 limits; A1 9/2.
題目 8 · Structured
5
A curve has equation \(y=x^3-6x^2+9x+2\). (a) Find \(\dfrac{dy}{dx}\). (b) Find the coordinates of the stationary points. (c) Use the second derivative to determine the nature of each.
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解題

(a) \(\tfrac{dy}{dx}=3x^2-12x+9=3(x-1)(x-3)\). (b) \(x=1\Rightarrow y=6\); \(x=3\Rightarrow y=2\). (c) \(\tfrac{d^2y}{dx^2}=6x-12\); at \(x=1\) it is \(-6\) (max), at \(x=3\) it is \(6\) (min).

評分準則

M1A1 derivative; M1 points; A1 both; M1 second derivative; A1 natures.
題目 9 · Structured
8
In an arithmetic series the 3rd term is 13 and the 8th term is 38. (a) Find the first term and common difference. (b) Find the sum of the first 20 terms. (c) Find which term of the series is equal to 98.
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解題

(a) \(a+2d=13\), \(a+7d=38\Rightarrow 5d=25\Rightarrow d=5\), \(a=3\). (b) \(S_{20}=\tfrac{20}{2}(2(3)+19(5))=10(101)=1010\). (c) \(3+(n-1)5=98\Rightarrow n-1=19\Rightarrow n=20\).

評分準則

M1 two equations; A1 d=5; A1 a=3; M1A1 sum; M1A1 nth term.
題目 10 · Short
4
Express \(\dfrac{7}{3-\sqrt2}\) in the form \(a+b\sqrt2\), where \(a\) and \(b\) are integers.
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解題

Multiply by the conjugate: \(\dfrac{7(3+\sqrt2)}{(3-\sqrt2)(3+\sqrt2)}=\dfrac{7(3+\sqrt2)}{9-2}=3+\sqrt2\).

評分準則

M1 multiply by conjugate; A1 denominator =7; A1 a=3; A1 b=1.

卷二 — 乙部: Statistics

Answer all questions. A calculator with statistical functions may be used.
6 題目 · 27
題目 1 · 選擇題
1
Selecting every 10th person from an ordered list of customers is an example of which sampling method? Circle your answer.
  1. A.Systematic
  2. B.Cluster
  3. C.Stratified
  4. D.Quota
查看答案詳解

解題

Choosing members at a fixed interval through an ordered list is systematic sampling.

評分準則

B1 systematic.
題目 2 · 選擇題
1
Which measure of average is most affected by a single extreme outlier? Circle your answer.
  1. A.Median
  2. B.Mode
  3. C.Mean
  4. D.Modal class
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解題

The mean uses every value, so a single extreme value shifts it the most.

評分準則

B1 mean.
題目 3 · Structured
5
Six data values are \(4, 7, 9, 10, 13, 17\). (a) Find the mean. (b) Find the standard deviation to 2 decimal places.
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解題

(a) Mean \(=\tfrac{60}{6}=10\). (b) Squared deviations: \(36,9,1,0,9,49\), sum \(=104\); variance \(=\tfrac{104}{6}=17.33\); s.d. \(=4.16\).

評分準則

M1A1 mean; M1 squared deviations; M1 variance; A1 4.16.
題目 4 · Structured
7
On each trial one of three outcomes occurs independently with \(P(X)=0.25\), \(P(Y)=0.4\), \(P(Z)=0.35\). For four trials find: (a) \(P(\text{no }Z)\); (b) \(P(\text{at least one }X)\); (c) \(P(\text{exactly two }Y)\).
查看答案詳解

解題

(a) \((0.65)^4=0.1785\). (b) \(1-(0.75)^4=1-0.3164=0.6836\). (c) \(\binom{4}{2}(0.4)^2(0.6)^2=6(0.16)(0.36)=0.3456\).

評分準則

M1A1 (a); M1A1 (b); M1 binomial term; A1 (c).
題目 5 · Structured
6
The discrete random variable \(X\) has \(P(X=x)=k(5-x)\) for \(x=1,2,3,4\). (a) Find \(k\). (b) Find \(E(X)\). (c) Find \(P(X\ge3)\).
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解題

(a) \(k(4+3+2+1)=1\Rightarrow10k=1\Rightarrow k=0.1\). (b) \(E(X)=0.1(1\cdot4+2\cdot3+3\cdot2+4\cdot1)=0.1(20)=2\). (c) \(P(X\ge3)=k(2+1)=0.3\).

評分準則

M1 sum to 1; A1 k=0.1; M1A1 E(X); A1 0.3.
題目 6 · Structured
7
A drug company claims its treatment has a success rate greater than 0.6. In a trial of 25 patients, 20 are treated successfully. Test the claim at the 5% significance level.
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解題

Let \(p\) be the success probability. \(H_0:p=0.6\), \(H_1:p>0.6\). Under \(H_0\), \(X\sim B(25,0.6)\). \(P(X\ge20)=1-P(X\le19)=0.0173\). Since \(0.0173<0.05\), reject \(H_0\): there is evidence at the 5% level that the success rate exceeds 0.6.

評分準則

B1 hypotheses; B1 distribution; M1 \(P(X\ge20)\); A1 0.0173; M1 compare; A1 conclusion in context.

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