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.
Paper 1 — Section A: Pure Mathematics
Answer all questions. Pure content; calculator and AQA formulae booklet permitted.
11 PastPaper.question · 53 PastPaper.marks
PastPaper.question 1 · Multiple choice
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The point \((2,k)\) lies on the curve \(y=3x^2-5x+1\). Find \(k\). Circle your answer.
A.1
B.3
C.7
D.-3
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PastPaper.workedSolution
\(k=3(4)-5(2)+1=12-10+1=3\).
PastPaper.markingScheme
B1 for 3.
PastPaper.question 2 · Multiple choice
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State the period of \(y=\tan 3x\). Circle your answer.
A.\(120^\circ\)
B.\(60^\circ\)
C.\(180^\circ\)
D.\(30^\circ\)
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PastPaper.workedSolution
\(\tan x\) has period \(180^\circ\); \(\tan 3x\) has period \(\tfrac{180^\circ}{3}=60^\circ\).
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B1 for 60°.
PastPaper.question 3 · Short
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Find the first three terms, in ascending powers of \(x\), of the binomial expansion of \((1+2x)^6\).
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 \(AB\).
\(f(x)=x^2-5x\). (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.
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.
Prove that the product of any two consecutive even integers is divisible by 8.
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PastPaper.workedSolution
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.
PastPaper.markingScheme
M1 represent as 2n, 2n+2; M1 expand to 4n(n+1); A1 n(n+1) even; A1 =8k; A1 conclude.
PastPaper.question 9 · Structured
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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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PastPaper.workedSolution
(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.
PastPaper.markingScheme
M1 form equation; A1 k=0.123; M1A1 (b); M1 set =5000; A1 ≈10 years.
PastPaper.question 10 · Structured
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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.
M1 y at x=5; M1 gradient; A1 tangent; M1 set y=0; A1 (17/4, 0).
PastPaper.question 11 · Short
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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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PastPaper.workedSolution
Tangency: \(x^2-3x+7=2x+c\Rightarrow x^2-5x+(7-c)=0\) has equal roots, so discriminant \(=0\): \(25-4(7-c)=0\Rightarrow 4c=3\Rightarrow c=\tfrac34\).
PastPaper.markingScheme
M1 equate; M1 discriminant =0; A1 form equation; A1 c=3/4.
Paper 1 — Section B: Mechanics
Answer all questions. Take g = 9.8 m s⁻² unless otherwise stated.
6 PastPaper.question · 27 PastPaper.marks
PastPaper.question 1 · Multiple choice
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A car decelerates uniformly from \(20\,\text{m s}^{-1}\) to rest in 8 seconds. Find the magnitude of its deceleration. Circle your answer.
A.\(2.5\,\text{m s}^{-2}\)
B.\(160\,\text{m s}^{-2}\)
C.\(0.4\,\text{m s}^{-2}\)
D.\(2\,\text{m s}^{-2}\)
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PastPaper.workedSolution
\(a=\tfrac{20}{8}=2.5\,\text{m s}^{-2}\).
PastPaper.markingScheme
B1 for 2.5 m s⁻².
PastPaper.question 2 · Structured
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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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PastPaper.workedSolution
(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 \(s=-10\): \(v^2=14^2-2(9.8)(-10)=196+196=392\Rightarrow v=19.8\,\text{m s}^{-1}\).
PastPaper.markingScheme
M1A1 max height; M1 use s=-10; A1 19.8.
PastPaper.question 3 · Structured
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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.
B1 resultant; M1 magnitude; M1 divide by mass; A1 1.28.
PastPaper.question 4 · Structured
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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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PastPaper.workedSolution
(a) \((5-3)g=8a\Rightarrow a=2.45\,\text{m s}^{-2}\). (b) For the 3 kg mass: \(T-3g=3a\Rightarrow T=3(9.8+2.45)=36.75\,\text{N}\).
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.
M1 v=0; A1 both times; M1A1 acceleration; M1 integrate; A1 -3 m.
PastPaper.question 6 · Structured
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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.
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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PastPaper.workedSolution
(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.
PastPaper.markingScheme
B1 A=56; M1 equation; A1 k=0.116; M1A1 (c); M1 set =30; A1 ≈17 min.
PastPaper.question 6 · Structured
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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.
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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PastPaper.workedSolution
(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 \(=\int_0^3\big((2x+1)-(x^2-x+1)\big)dx=\int_0^3(-x^2+3x)dx=[-\tfrac{x^3}{3}+\tfrac{3x^2}{2}]_0^3=-9+13.5=\tfrac92\).
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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PastPaper.workedSolution
(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).
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.
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) \(P(\text{exactly two }Y)\).
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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PastPaper.workedSolution
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.