AQA A-Level · Thinka 原創模擬試題

2023 AQA A-Level Mathematics 7357 模擬試題連答案詳解

Thinka Jun 2023 AQA A Level-Style Mock — Mathematics 7357

300 360 分鐘2023
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 AQA A Level Mathematics 7357 paper. Not affiliated with or reproduced from AQA.

卷一: Pure Mathematics

Answer all questions. Entirely pure content; calculator and AQA formulae booklet permitted.
11 題目 · 100
題目 1 · 選擇題
1
Differentiate \(y=\ln(5x)\). Circle your answer.
  1. A.\(\tfrac5x\)
  2. B.\(\tfrac1x\)
  3. C.\(\tfrac{1}{5x}\)
  4. D.\(\ln5\)
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解題

\(\tfrac{d}{dx}\ln(5x)=\tfrac{5}{5x}=\tfrac1x\).

評分準則

B1 for \(\tfrac1x\).
題目 2 · 選擇題
1
Find \(\displaystyle\int e^{3x}\,dx\). Circle your answer.
  1. A.\(3e^{3x}+c\)
  2. B.\(\tfrac13 e^{3x}+c\)
  3. C.\(e^{3x}+c\)
  4. D.\(3e^{x}+c\)
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解題

\(\int e^{3x}dx=\tfrac13 e^{3x}+c\).

評分準則

B1 for \(\tfrac13 e^{3x}+c\).
題目 3 · Short
5
Find the first three terms, in ascending powers of \(x\), of the binomial expansion of \((3+2x)^6\).
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解題

\((3+2x)^6=3^6+\binom61 3^5(2x)+\binom62 3^4(2x)^2+\dots=729+6(243)(2)x+15(81)(4)x^2=729+2916x+4860x^2\).

評分準則

M1 binomial structure; A1 729; A1 2916x; M1 third-term values; A1 4860x².
題目 4 · Structured
12
A circle \(C\) has equation \(x^2+y^2-10x+2y+1=0\). (a) Find the centre and radius. (b) Show that the point \((5,4)\) lies on \(C\). (c) Find the equation of the tangent to \(C\) at \((5,4)\). (d) Find the length of the tangent to \(C\) from the external point \((12,3)\).
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解題

(a) \((x-5)^2+(y+1)^2=25\): centre \((5,-1)\), radius \(5\). (b) \(25+16-50+8+1=0\) ✓. (c) The radius from \((5,-1)\) to \((5,4)\) is vertical, so the tangent is horizontal: \(y=4\). (d) Tangent length \(=\sqrt{(12-5)^2+(3+1)^2-25}=\sqrt{49+16-25}=\sqrt{40}=6.32\).

評分準則

M1A1 centre & radius; M1A1 (b); M1 radius direction; A1 tangent y=4; M1A1 tangent length.
題目 5 · Structured
10
(a) An arithmetic series has first term 4 and common difference 6. Find (i) the 15th term and (ii) the sum of the first 15 terms. (b) A geometric series has first term 4 and common ratio 1.05. Find (i) the 10th term and (ii) the sum of the first 10 terms. (c) State, with a reason, whether the geometric series converges.
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解題

(a)(i) \(4+14(6)=88\); (ii) \(S_{15}=\tfrac{15}{2}(8+84)=690\). (b)(i) \(4(1.05)^9=6.20\); (ii) \(S_{10}=4\cdot\tfrac{1.05^{10}-1}{0.05}=50.3\). (c) No: \(|r|=1.05>1\), so it diverges.

評分準則

M1A1 (a); M1A1 (b)(i); M1A1 (b)(ii); B1 (c) with reason.
題目 6 · Structured
12
\(f(x)=x^3-7x+2\). (a) Show that a root of \(f(x)=0\) lies between 2 and 3. (b) Apply the Newton–Raphson method with \(x_0=2.5\) to find \(x_1\) and \(x_2\). (c) The iteration \(x_{n+1}=\sqrt[3]{7x_n-2}\) is used with \(x_0=2.5\). Find the root to 3 decimal places.
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解題

(a) \(f(2)=-4<0\), \(f(3)=8>0\): sign change implies a root in \((2,3)\). (b) \(f'(x)=3x^2-7\); \(x_1=2.5-\tfrac{0.125}{11.75}=2.4894\); \(x_2=2.4894-\tfrac{f(2.4894)}{f'(2.4894)}=2.4893\). (c) \(x_1=\sqrt[3]{15.5}=2.494,\,x_2=2.490,\,x_3=2.489,\dots\to 2.489\).

評分準則

M1 evaluate ends; A1 sign change; M1 NR formula; A1 x₁; A1 x₂; M1 iterate; A1 2.489.
題目 7 · Structured
10
\(p(x)=2x^3+ax^2+bx-6\). Given that \((x-2)\) and \((x+1)\) are factors: (a) find \(a\) and \(b\); (b) factorise \(p(x)\) completely; (c) solve \(p(x)=0\).
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解題

(a) \(p(2)=16+4a+2b-6=0\Rightarrow2a+b=-5\); \(p(-1)=-2+a-b-6=0\Rightarrow a-b=8\). Adding, \(3a=3\Rightarrow a=1,\,b=-7\). (b) \(p(x)=2x^3+x^2-7x-6=(x-2)(x+1)(2x+3)\). (c) \(x=2,-1,-\tfrac32\).

評分準則

M1 p(2)=0; M1 p(-1)=0; A1 a,b; M1 divide; A1 full factorisation; A1 three roots.
題目 8 · Structured
8
(a) A geometric series has first term 6 and common ratio \(\tfrac23\). Find its sum to infinity. (b) Solve \(\tan 2x=1\) for \(0\le x\le\pi\). (c) Solve \(4\sin x=3\cos x\) for \(0\le x\le2\pi\), to 2 decimal places.
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解題

(a) \(\tfrac{6}{1-2/3}=18\). (b) \(2x=\tfrac{\pi}{4},\tfrac{5\pi}{4}\Rightarrow x=\tfrac{\pi}{8},\tfrac{5\pi}{8}\). (c) \(\tan x=\tfrac34\Rightarrow x=0.6435\) or \(0.6435+\pi=3.785\).

評分準則

B1 (a); M1 2x values; A1 both x; M1 tan x; A1 0.64; A1 3.79.
題目 9 · Structured
10
(a) Prove the identity \(\dfrac{1+\sin x}{\cos x}+\dfrac{\cos x}{1+\sin x}\equiv\dfrac{2}{\cos x}\). (b) Hence solve \(\dfrac{1+\sin x}{\cos x}+\dfrac{\cos x}{1+\sin x}=4\) for \(0
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解題

(a) Common denominator: \(\tfrac{(1+\sin x)^2+\cos^2x}{\cos x(1+\sin x)}=\tfrac{1+2\sin x+\sin^2x+\cos^2x}{\cos x(1+\sin x)}=\tfrac{2+2\sin x}{\cos x(1+\sin x)}=\tfrac{2}{\cos x}\). (b) \(\tfrac{2}{\cos x}=4\Rightarrow\cos x=\tfrac12\Rightarrow x=\tfrac{\pi}{3},\tfrac{5\pi}{3}\). (c) \(\tfrac{1+\cos2x}{\sin2x}=\tfrac{2\cos^2x}{2\sin x\cos x}=\cot x\).

評分準則

M1 combine; M1 use identity; A1 result; M1A1 (b); M1A1 (c).
題目 10 · Structured
15
(a) Find \(\displaystyle\int\left(6x^2-\tfrac{4}{x^2}+e^{3x}\right)dx\). (b) Use the substitution \(u=x^2+4\) to find \(\displaystyle\int\tfrac{2x}{\sqrt{x^2+4}}\,dx\). (c) Find the area enclosed between the curve \(y=x^2+3\) and the line \(y=4x\).
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解題

(a) \(2x^3+\tfrac{4}{x}+\tfrac13 e^{3x}+c\). (b) \(u=x^2+4,\,du=2x\,dx\Rightarrow\int u^{-1/2}du=2\sqrt u=2\sqrt{x^2+4}+c\). (c) Intersect: \(x^2+3=4x\Rightarrow x=1,3\); area \(=\int_1^3(4x-x^2-3)dx=[2x^2-\tfrac{x^3}{3}-3x]_1^3=0-(-\tfrac43)=\tfrac43\).

評分準則

M1A1A1 (a); M1 substitution; A1 (b); M1 limits 1,3; M1 integrate difference; A1 4/3.
題目 11 · Structured
16
A curve has equation \(y=2x^3-9x^2+12x-3\). (a) Find \(\dfrac{dy}{dx}\). (b) Find the stationary points and determine their nature. (c) Find the coordinates of the point of inflection. (d) Find the equation of the tangent at \(x=0\). (e) State the set of values of \(x\) for which the curve is increasing.
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解題

(a) \(\tfrac{dy}{dx}=6x^2-18x+12=6(x-1)(x-2)\). (b) \(x=1\Rightarrow(1,2)\); \(x=2\Rightarrow(2,1)\). \(\tfrac{d^2y}{dx^2}=12x-18\): at \(x=1\) negative (max), at \(x=2\) positive (min). (c) \(12x-18=0\Rightarrow x=1.5,\,y=1.5\). (d) At \(x=0\), \(y=-3\), gradient \(12\): \(y=12x-3\). (e) Increasing where \(\tfrac{dy}{dx}>0\): \(x<1\) or \(x>2\).

評分準則

M1A1 derivative; M1 stationary; A1 both; M1A1 natures; M1A1 inflection; M1A1 tangent; B1 increasing set.

卷二: Pure & Mechanics

Answer all questions. Section A is pure; Section B is mechanics (take g = 9.8 m s⁻²).
14 題目 · 100
題目 1 · 選擇題
1
Find \(\displaystyle\int\tfrac1x\,dx\). Circle your answer.
  1. A.\(\ln|x|+c\)
  2. B.\(-\tfrac{1}{x^2}+c\)
  3. C.\(\tfrac{1}{2}x^2+c\)
  4. D.\(x\ln x+c\)
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解題

\(\int\tfrac1x dx=\ln|x|+c\).

評分準則

B1 for \(\ln|x|+c\).
題目 2 · Short
3
Express \(\dfrac{4x+5}{(x+1)(x-2)}\) in partial fractions.
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解題

Let \(\tfrac{4x+5}{(x+1)(x-2)}=\tfrac{A}{x+1}+\tfrac{B}{x-2}\). Then \(4x+5=A(x-2)+B(x+1)\). \(x=-1:1=-3A\Rightarrow A=-\tfrac13\); \(x=2:13=3B\Rightarrow B=\tfrac{13}{3}\).

評分準則

M1 set up identity; M1 substitute; A1 A=-1/3, B=13/3.
題目 3 · Proof
6
Prove by contradiction that \(\sqrt5\) is irrational.
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解題

Assume \(\sqrt5=\tfrac pq\) in lowest terms. Then \(p^2=5q^2\), so \(5\mid p^2\Rightarrow5\mid p\). Write \(p=5k\): \(25k^2=5q^2\Rightarrow q^2=5k^2\), so \(5\mid q\). Then 5 divides both \(p\) and \(q\), contradicting lowest terms. Hence \(\sqrt5\) is irrational.

評分準則

B1 assume rational; M1 \(p^2=5q^2\); A1 5|p; M1 substitute; A1 5|q; A1 contradiction.
題目 4 · Structured
10
A closed box has a square base of side \(x\,\text{cm}\) and volume \(1000\,\text{cm}^3\). (a) Show that the total surface area is \(S=2x^2+\dfrac{4000}{x}\). (b) Find the value of \(x\) that minimises \(S\). (c) Find the minimum surface area.
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解題

(a) Height \(h=\tfrac{1000}{x^2}\); \(S=2x^2+4xh=2x^2+\tfrac{4000}{x}\). (b) \(\tfrac{dS}{dx}=4x-\tfrac{4000}{x^2}=0\Rightarrow x^3=1000\Rightarrow x=10\). (c) \(S=200+400=600\,\text{cm}^2\) (\(\tfrac{d^2S}{dx^2}>0\), minimum).

評分準則

M1 height; A1 show S; M1 differentiate; M1 solve =0; A1 x=10; A1 S=600.
題目 5 · Structured
10
(a) A population grows so that \(\dfrac{dP}{dt}=0.4P\). Show that \(P=Ae^{0.4t}\). (b) Given \(P=150\) when \(t=0\), find \(P\) when \(t=5\). (c) Solve the differential equation \(\dfrac{dy}{dx}=4x^3y\), giving \(y\) in terms of \(x\).
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解題

(a) \(\int\tfrac{dP}{P}=\int0.4\,dt\Rightarrow\ln P=0.4t+c\Rightarrow P=Ae^{0.4t}\). (b) \(A=150\Rightarrow P=150e^{2}=1108\). (c) \(\int\tfrac{dy}{y}=\int4x^3dx\Rightarrow\ln y=x^4+c\Rightarrow y=Be^{x^4}\).

評分準則

M1 separate; A1 show; M1 A=150; A1 1108; M1 separate; A1 \(y=Be^{x^4}\).
題目 6 · Structured
10
(a) Solve \(3^{2x}-12\cdot3^{x}+27=0\). (b) Solve \(\log_3 x+\log_3(x+6)=3\). (c) Evaluate \(\log_2 32-\log_2 4\).
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解題

(a) Let \(y=3^x\): \(y^2-12y+27=0\Rightarrow(y-3)(y-9)=0\Rightarrow3^x=3\,(x=1)\) or \(3^x=9\,(x=2)\). (b) \(\log_3 x(x+6)=3\Rightarrow x^2+6x=27\Rightarrow(x+9)(x-3)=0\Rightarrow x=3\) (reject \(-9\)). (c) \(5-2=3\).

評分準則

M1 substitution; A1 both x; M1 combine logs; M1 solve; A1 x=3; A1 (c).
題目 7 · Structured
11
(a) Express \(5\sin x-12\cos x\) in the form \(R\sin(x-\alpha)\), \(R>0\), \(0<\alpha<90^\circ\). (b) State the maximum value and the value of \(x\) in \(0\le x\le360^\circ\) at which it occurs. (c) Solve \(5\sin x-12\cos x=6.5\) for \(0\le x\le360^\circ\).
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解題

(a) \(R=\sqrt{5^2+12^2}=13\), \(\tan\alpha=\tfrac{12}{5}\Rightarrow\alpha=67.38^\circ\): \(13\sin(x-67.38^\circ)\). (b) Max \(13\) when \(x-67.38^\circ=90^\circ\Rightarrow x=157.4^\circ\). (c) \(\sin(x-67.38^\circ)=0.5\Rightarrow x-67.38^\circ=30^\circ,150^\circ\Rightarrow x=97.4^\circ,217.4^\circ\).

評分準則

M1 R=13; M1 α; A1 form; M1A1 (b); M1 solve; A1 both x.
題目 8 · 選擇題
1
Find the weight of a particle of mass \(8\,\text{kg}\) (take \(g=9.8\)). Circle your answer.
  1. A.\(8\,\text{N}\)
  2. B.\(0.82\,\text{N}\)
  3. C.\(78.4\,\text{N}\)
  4. D.\(9.8\,\text{N}\)
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解題

Weight \(=mg=8\times9.8=78.4\,\text{N}\).

評分準則

B1 for 78.4 N.
題目 9 · Structured
6
A particle moves in a straight line with acceleration \(a=(12t-6)\,\text{m s}^{-2}\) and velocity \(4\,\text{m s}^{-1}\) at \(t=0\). (a) Find \(v\) in terms of \(t\). (b) Find \(v\) when \(t=2\). (c) Find the displacement during the first 2 seconds.
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解題

(a) \(v=\int(12t-6)dt=6t^2-6t+C\); \(v(0)=4\Rightarrow C=4\), so \(v=6t^2-6t+4\). (b) \(v(2)=24-12+4=16\). (c) \(s=\int_0^2(6t^2-6t+4)dt=[2t^3-3t^2+4t]_0^2=16-12+8=12\,\text{m}\).

評分準則

M1 integrate a; A1 v; B1 (b); M1 integrate v; A1 12 m.
題目 10 · Structured
9
A particle of mass \(3\,\text{kg}\) lies on a smooth plane inclined at \(30^\circ\). It is connected by a light inextensible string over a smooth pulley at the top of the plane to a mass of \(5\,\text{kg}\) hanging freely. The system is released from rest. (a) Find the acceleration. (b) Find the tension. (c) State one assumption about the string.
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解題

For the hanging mass: \(5g-T=5a\). On the plane: \(T-3g\sin30^\circ=3a\). Adding: \(5g-1.5g=8a\Rightarrow a=\tfrac{34.3}{8}=4.29\,\text{m s}^{-2}\). Then \(T=5g-5a=49-21.44=27.6\,\text{N}\). (c) The string is light and inextensible.

評分準則

M1 hanging-mass equation; M1 on-plane equation; A1 add; A1 a=4.29; M1 T; A1 27.6; B1 assumption.
題目 11 · Structured
10
A uniform beam \(AB\) of length \(10\,\text{m}\) and weight \(500\,\text{N}\) rests horizontally on supports at \(A\) and \(B\). A load of \(300\,\text{N}\) is placed \(3\,\text{m}\) from \(A\). (a) Find the reaction at each support. (b) Find how far from \(A\) the load must be placed for the reactions to be equal.
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解題

(a) Moments about \(A\): \(R_B(10)=500(5)+300(3)=3400\Rightarrow R_B=340\,\text{N}\); \(R_A=500+300-340=460\,\text{N}\). (b) Equal reactions \(=400\) each: \(400(10)=500(5)+300d\Rightarrow300d=1500\Rightarrow d=5\,\text{m}\).

評分準則

M1 moments about A; A1 R_B; M1 resolve; A1 R_A; M1 set equal; A1 d=5.
題目 12 · Structured
8
A particle has position vector \(\mathbf{r}=(t^2-4t)\mathbf{i}+(t^2-2t)\mathbf{j}\) (\(t\) in seconds). (a) Find its velocity when \(t=1\). (b) Find its speed when \(t=1\). (c) Find the time at which it is moving parallel to \(\mathbf{j}\).
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解題

\(\mathbf{v}=(2t-4)\mathbf{i}+(2t-2)\mathbf{j}\). (a) At \(t=1\): \(\mathbf{v}=-2\mathbf{i}+0\mathbf{j}\). (b) Speed \(=\sqrt{(-2)^2+0^2}=2\,\text{m s}^{-1}\). (c) Parallel to \(\mathbf{j}\) when the \(\mathbf{i}\)-component is zero: \(2t-4=0\Rightarrow t=2\).

評分準則

M1 differentiate; A1 v at t=1; M1A1 speed; M1 set i-comp =0; A1 t=2.
題目 13 · Structured
6
A ball is thrown horizontally with speed \(20\,\text{m s}^{-1}\) from a point \(25\,\text{m}\) above horizontal ground (take \(g=9.8\)). (a) Find the time to reach the ground. (b) Find the horizontal distance travelled.
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解題

(a) \(25=\tfrac12(9.8)t^2\Rightarrow t^2=5.102\Rightarrow t=2.26\,\text{s}\). (b) \(x=20(2.26)=45.2\,\text{m}\).

評分準則

M1 vertical equation; A1 t=2.26; M1 horizontal motion; A1 45.2 m; B1 method.
題目 14 · Structured
9
A particle \(P\) of mass \(4\,\text{kg}\) lies on a smooth horizontal table. It is connected by a light inextensible string passing over a smooth pulley at the edge of the table to a particle \(Q\) of mass \(6\,\text{kg}\) hanging freely. The system is released from rest. (a) Find the acceleration. (b) Find the tension in the string. (c) State one assumption you have made about the pulley.
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解題

For \(Q\): \(6g-T=6a\). For \(P\): \(T=4a\). Adding: \(6g=10a\Rightarrow a=\tfrac{58.8}{10}=5.88\,\text{m s}^{-2}\). Then \(T=4(5.88)=23.52\,\text{N}\). (c) The pulley is smooth, so the tension is the same on both sides of the string.

評分準則

M1 equation for Q; M1 equation for P; A1 add; A1 a=5.88; M1 T; A1 23.52; B1 assumption.

Paper 3: Pure & Statistics

Answer all questions. Section A is pure; Section B is statistics (Large Data Set context).
13 題目 · 100
題目 1 · 選擇題
1
State the period of \(y=\sin(2\pi x)\). Circle your answer.
  1. A.\(2\pi\)
  2. B.\(\tfrac12\)
  3. C.\(1\)
  4. D.\(\pi\)
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解題

Period \(=\tfrac{2\pi}{2\pi}=1\).

評分準則

B1 for 1.
題目 2 · Structured
10
Differentiate each of the following. (a) \(y=(3x-2)^4\). (b) \(y=x^2\ln x\). (c) \(y=\dfrac{x}{x^2+4}\). (d) Hence state the gradient of the curve in part (a) at \(x=1\).
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解題

(a) Chain rule: \(12(3x-2)^3\). (b) Product rule: \(2x\ln x+x\). (c) Quotient rule: \(\tfrac{(x^2+4)-x(2x)}{(x^2+4)^2}=\tfrac{4-x^2}{(x^2+4)^2}\). (d) At \(x=1\): \(12(1)^3=12\).

評分準則

M1A1 (a); M1A1 (b); M1A1 (c); M1A1 (d).
題目 3 · Structured
9
Find each integral. (a) \(\displaystyle\int(2x+5)^3\,dx\). (b) \(\displaystyle\int\tfrac{9}{3x+1}\,dx\). (c) \(\displaystyle\int_1^2\tfrac{6}{x^2}\,dx\).
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解題

(a) \(\tfrac{(2x+5)^4}{8}+c\). (b) \(3\ln|3x+1|+c\). (c) \([-\tfrac{6}{x}]_1^2=-3-(-6)=3\).

評分準則

M1A1 (a); M1A1 (b); M1 antiderivative; A1 (c)=3.
題目 4 · Structured
10
(a) Solve \(2\cos 2x=\sqrt3\) for \(0\le x\le2\pi\). (b) Prove that \(\tan^2 x+1\equiv\sec^2 x\). (c) Solve \(2\sin x+1=0\) for \(0\le x\le2\pi\).
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解題

(a) \(\cos2x=\tfrac{\sqrt3}{2}\Rightarrow2x=\tfrac{\pi}{6},\tfrac{11\pi}{6},\tfrac{13\pi}{6},\tfrac{23\pi}{6}\Rightarrow x=\tfrac{\pi}{12},\tfrac{11\pi}{12},\tfrac{13\pi}{12},\tfrac{23\pi}{12}\). (b) \(\tan^2x+1=\tfrac{\sin^2x}{\cos^2x}+1=\tfrac{\sin^2x+\cos^2x}{\cos^2x}=\tfrac{1}{\cos^2x}=\sec^2x\). (c) \(\sin x=-\tfrac12\Rightarrow x=\tfrac{7\pi}{6},\tfrac{11\pi}{6}\).

評分準則

M1 cos2x value; A1 all four x; M1A1 (b); M1A1 (c).
題目 5 · Structured
10
A radioactive sample decays as \(m=m_0e^{-kt}\), with a half-life of 12 years. (a) Find \(k\) to 3 significant figures. (b) Given \(m_0=80\,\text{g}\), find the mass after 20 years. (c) Find, to the nearest year, the time to fall to \(20\,\text{g}\). (d) Find the rate of decay at \(t=0\).
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解題

(a) \(e^{-12k}=\tfrac12\Rightarrow k=\tfrac{\ln2}{12}=0.0578\). (b) \(m=80e^{-0.0578\times20}=80e^{-1.155}=25.2\,\text{g}\). (c) \(20=80e^{-kt}\Rightarrow t=\tfrac{\ln4}{0.0578}=24\,\text{years}\). (d) \(\tfrac{dm}{dt}=-km_0\) at \(t=0=-0.0578(80)=-4.62\,\text{g/year}\).

評分準則

M1A1 k; M1A1 (b); M1A1 (c); M1A1 (d).
題目 6 · Structured
10
(a) An arithmetic series is \(5,9,13,\dots\). Find the sum of the first 25 terms. (b) A geometric series is \(200,150,112.5,\dots\). Find its sum to infinity. (c) Use the arithmetic-series formula to prove that \(1+3+5+\dots+(2n-1)=n^2\). (d) Find the first term of the series in (a) that exceeds 100.
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解題

(a) \(a=5,d=4\): \(S_{25}=\tfrac{25}{2}(10+24(4))=\tfrac{25}{2}(106)=1325\). (b) \(r=0.75\): \(\tfrac{200}{1-0.75}=800\). (c) AP with \(a=1,d=2\): \(S_n=\tfrac n2(2+(n-1)2)=n^2\). (d) \(5+(n-1)4>100\Rightarrow n>24.75\Rightarrow n=25\) (term \(=101\)).

評分準則

M1A1 (a); M1A1 (b); M1A1 (c); M1A1 (d).
題目 7 · 選擇題
1
For \(X\sim B(25,0.2)\), state the mean \(E(X)\). Circle your answer.
  1. A.\(0.2\)
  2. B.\(5\)
  3. C.\(4\)
  4. D.\(20\)
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解題

\(E(X)=np=25\times0.2=5\).

評分準則

B1 for 5.
題目 8 · Structured
6
(a) Define a simple random sample. (b) A college has 600 students, each with an ID number. Describe how to take a simple random sample of 60. (c) State one advantage of stratified sampling here.
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解題

(a) A simple random sample is one in which every possible sample of the required size is equally likely (every member has an equal chance of selection). (b) Number students 1–600; use random numbers to choose 60 distinct IDs, ignoring repeats and numbers above 600. (c) Stratified sampling represents each subgroup (e.g. year group) in proportion, reducing bias.

評分準則

B1 definition; M1 numbering; A1 random selection; B1 advantage; B1 reason; B1 clarity.
題目 9 · Structured
10
Eight values are \(10,12,14,15,18,20,24,28\). (a) Find the median. (b) Find \(Q_1\), \(Q_3\) and the IQR. (c) Find the mean. (d) Find the standard deviation to 2 decimal places. (e) Using the \(1.5\times\text{IQR}\) rule, determine whether 28 is an outlier.
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解題

(a) Median \(=\tfrac{15+18}{2}=16.5\). (b) \(Q_1=12,\,Q_3=22\) (mean of 20 and 24), \(\text{IQR}=10\). (c) Mean \(=\tfrac{141}{8}=17.625\). (d) \(\sum x^2=2749\); variance \(=\tfrac{2749}{8}-17.625^2=32.98\); s.d. \(=5.74\). (e) \(Q_3+1.5(10)=37>28\), so 28 is not an outlier.

評分準則

B1 median; B1 quartiles; B1 IQR; M1A1 mean; M1A1 s.d.; B1 outlier.
題目 10 · Structured
10
The masses of items are modelled by \(X\sim N(50,6^2)\) (in grams). (a) Find \(P(X>56)\). (b) Find \(P(44
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解題

(a) \(z=\tfrac{56-50}{6}=1\Rightarrow P(Z>1)=0.1587\). (b) \(z=-1\) and \(z=1.5\Rightarrow\Phi(1.5)-\Phi(-1)=0.9332-0.1587=0.7745\). (c) \(P(X>x)=0.05\Rightarrow z=1.645\Rightarrow x=50+1.645(6)=59.9\,\text{g}\).

評分準則

M1 standardise; A1 (a); M1A1 (b); M1 z=1.645; A1 (c).
題目 11 · Structured
10
(a) For \(X\sim B(12,0.35)\) find (i) \(P(X=4)\) and (ii) \(P(X\le2)\). (b) A coin-like process is claimed to succeed with probability greater than 0.35. In 12 trials there are 9 successes. Test at the 5% level whether the success probability exceeds 0.35.
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解題

(a)(i) \(\binom{12}{4}0.35^4 0.65^8=0.2367\). (ii) \(P(X\le2)=0.1513\). (b) \(H_0:p=0.35\), \(H_1:p>0.35\), \(X\sim B(12,0.35)\). \(P(X\ge9)=1-P(X\le8)=0.0028<0.05\), so reject \(H_0\): evidence the success probability exceeds 0.35.

評分準則

M1A1 (a)(i); A1 (a)(ii); B1 hypotheses; M1 P(X≥9); A1 0.0028; A1 conclusion.
題目 12 · Structured
7
Events \(A\) and \(B\) satisfy \(P(A)=0.6\), \(P(B)=0.3\) and \(P(A\cap B)=0.18\). (a) Find \(P(A\cup B)\). (b) Find \(P(B\mid A)\). (c) Determine, with justification, whether \(A\) and \(B\) are independent.
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解題

(a) \(P(A\cup B)=0.6+0.3-0.18=0.72\). (b) \(P(B\mid A)=\tfrac{0.18}{0.6}=0.3\). (c) \(P(A)P(B)=0.6(0.3)=0.18=P(A\cap B)\), so \(A\) and \(B\) are independent.

評分準則

M1A1 (a); M1A1 (b); M1 compare; A1 conclusion.
題目 13 · Structured
6
The discrete random variable \(X\) has \(P(X=0)=0.1,\,P(X=1)=0.3,\,P(X=2)=0.4,\,P(X=3)=0.2\). (a) Find \(E(X)\). (b) Find \(\text{Var}(X)\). (c) Find \(E(3X-2)\).
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解題

(a) \(E(X)=0(0.1)+1(0.3)+2(0.4)+3(0.2)=1.7\). (b) \(E(X^2)=0+0.3+1.6+1.8=3.7\); \(\text{Var}(X)=3.7-1.7^2=0.81\). (c) \(E(3X-2)=3(1.7)-2=3.1\).

評分準則

M1A1 E(X); M1 E(X²); A1 Var; A1 (c).

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