AQA AS-Level · Thinka-original Practice Paper

2023 AQA AS-Level Mathematics 7356 Practice Paper with Answers

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

160 marks180 mins2023

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 Question · 53 marks

Question 1 · Multiple choice

1 marks

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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Worked solution

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

Marking scheme

B1 for 3.

Question 2 · Multiple choice

1 marks

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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Worked solution

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

Marking scheme

B1 for 60°.

Question 3 · Short

3 marks

Find the first three terms, in ascending powers of \(x\), of the binomial expansion of \((1+2x)^6\).

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Worked solution

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

Marking scheme

M1 binomial structure; A1 \(12x\); A1 \(60x^2\).

Question 4 · Short

5 marks

Solve \(2\cos^2 x+3\sin x=3\) for \(0^\circ\le x\le360^\circ\).

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Worked solution

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

Marking scheme

M1 identity; M1 quadratic in sin; A1 factorise; A1 30° & 150°; A1 90°.

Question 5 · Structured

6 marks

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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Worked solution

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

Marking scheme

B1 midpoint; M1A1 gradient; M1 perpendicular gradient; A1 equation; B1 simplified.

Question 6 · Structured

7 marks

\(f(x)=x^2-5x\). (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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Worked solution

(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) \(2x-5=7\Rightarrow x=6\).

Marking scheme

M1 form difference quotient; M1 expand & simplify; A1 limit =2x-5; B1 (b); M1A1 (c).

Question 7 · Structured

7 marks

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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Worked solution

(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) \(\tfrac{dy}{dx}=6x-4=0\Rightarrow x=\tfrac23\).

Marking scheme

M1A1 integral; M1 limits; A1 12; M1A1 (c).

Question 8 · Proof

5 marks

Prove that the product of any two consecutive even integers is divisible by 8.

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Worked solution

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.

Marking scheme

M1 represent as 2n, 2n+2; M1 expand to 4n(n+1); A1 n(n+1) even; A1 =8k; A1 conclude.

Question 9 · Structured

8 marks

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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Worked solution

(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.

Marking scheme

M1 form equation; A1 k=0.123; M1A1 (b); M1 set =5000; A1 ≈10 years.

Question 10 · Structured

6 marks

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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Worked solution

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

Marking scheme

M1 y at x=5; M1 gradient; A1 tangent; M1 set y=0; A1 (17/4, 0).

Question 11 · Short

4 marks

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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Worked solution

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

Marking scheme

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 Question · 27 marks

Question 1 · Multiple choice

1 marks

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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Worked solution

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

Marking scheme

B1 for 2.5 m s⁻².

Question 2 · Structured

4 marks

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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Worked solution

(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 \(s=-10\): \(v^2=14^2-2(9.8)(-10)=196+196=392\Rightarrow v=19.8\,\text{m s}^{-1}\).

Marking scheme

M1A1 max height; M1 use s=-10; A1 19.8.

Question 3 · Structured

4 marks

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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Worked solution

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

Marking scheme

B1 resultant; M1 magnitude; M1 divide by mass; A1 1.28.

Question 4 · Structured

6 marks

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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Worked solution

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

Marking scheme

M1 system equation; A1 a; M1 single-mass equation; A1 T; B1 method.

Question 5 · Structured

6 marks

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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Worked solution

(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) \(s=\int_0^3(2t^2-10t+8)dt=[\tfrac{2t^3}{3}-5t^2+8t]_0^3=18-45+24=-3\,\text{m}\).

Marking scheme

M1 v=0; A1 both times; M1A1 acceleration; M1 integrate; A1 -3 m.

Question 6 · Structured

6 marks

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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Worked solution

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

Marking scheme

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

Paper 2 — Section A: Pure Mathematics

Answer all questions. Pure content; calculator and AQA formulae booklet permitted.

10 Question · 53 marks

Question 1 · Multiple choice

1 marks

Find \(\displaystyle\int(6x^2-4)\,dx\). Circle your answer.

A.\(2x^3-4x+c\)
B.\(12x+c\)
C.\(2x^3-4+c\)
D.\(6x^3-4x+c\)

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Worked solution

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

Marking scheme

B1 for \(2x^3-4x+c\).

Question 2 · Multiple choice

1 marks

Express \(\sqrt{50}+\sqrt{8}\) in the form \(k\sqrt2\). Circle your answer.

A.\(7\sqrt2\)
B.\(\sqrt{58}\)
C.\(10\sqrt2\)
D.\(58\)

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Worked solution

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

Marking scheme

B1 for \(7\sqrt2\).

Question 3 · Structured

5 marks

\(f(x)=2x^3-3x^2-11x+6\). (a) Show that \((x-3)\) is a factor. (b) Factorise \(f(x)\) completely.

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Worked solution

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

Marking scheme

M1 evaluate f(3); A1 conclude factor; M1 divide; A1 quadratic; A1 full factorisation.

Question 4 · Structured

4 marks

The quadratic equation \(x^2+kx+(k+3)=0\) has no real roots. Find the set of values of \(k\).

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Worked solution

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

Marking scheme

M1 discriminant <0; A1 \(k^2-4k-12<0\); M1 factorise; A1 \(-2

Question 5 · Structured

8 marks

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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Worked solution

(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.

Marking scheme

B1 A=56; M1 equation; A1 k=0.116; M1A1 (c); M1 set =30; A1 ≈17 min.

Question 6 · Structured

7 marks

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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Worked solution

(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) \(\tfrac{\sin QPR}{6}=\tfrac{\sin110^\circ}{12.4}\Rightarrow\sin QPR=0.4543\Rightarrow QPR=27.0^\circ\).

Marking scheme

M1A1 cosine rule; A1 PR; M1A1 area; M1 sine rule; A1 27.0°.

Question 7 · Structured

10 marks

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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Worked solution

(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 \(=\int_0^3\big((2x+1)-(x^2-x+1)\big)dx=\int_0^3(-x^2+3x)dx=[-\tfrac{x^3}{3}+\tfrac{3x^2}{2}]_0^3=-9+13.5=\tfrac92\).

Marking scheme

M1 equate; A1 quadratic; A1 both points; B1 inequalities; M1 integrate difference; M1 limits; A1 9/2.

Question 8 · Structured

5 marks

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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Worked solution

(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).

Marking scheme

M1A1 derivative; M1 points; A1 both; M1 second derivative; A1 natures.

Question 9 · Structured

8 marks

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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Worked solution

(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) \(3+(n-1)5=98\Rightarrow n-1=19\Rightarrow n=20\).

Marking scheme

M1 two equations; A1 d=5; A1 a=3; M1A1 sum; M1A1 nth term.

Question 10 · Short

4 marks

Express \(\dfrac{7}{3-\sqrt2}\) in the form \(a+b\sqrt2\), where \(a\) and \(b\) are integers.

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Worked solution

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

Marking scheme

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

Paper 2 — Section B: Statistics

Answer all questions. A calculator with statistical functions may be used.

6 Question · 27 marks

Question 1 · Multiple choice

1 marks

Selecting every 10th person from an ordered list of customers is an example of which sampling method? Circle your answer.

A.Systematic
B.Cluster
C.Stratified
D.Quota

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Worked solution

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

Marking scheme

B1 systematic.

Question 2 · Multiple choice

1 marks

Which measure of average is most affected by a single extreme outlier? Circle your answer.

A.Median
B.Mode
C.Mean
D.Modal class

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Worked solution

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

Marking scheme

B1 mean.

Question 3 · Structured

5 marks

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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Worked solution

(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. \(=4.16\).

Marking scheme

M1A1 mean; M1 squared deviations; M1 variance; A1 4.16.

Question 4 · Structured

7 marks

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) \(P(\text{exactly two }Y)\).

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Worked solution

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

Marking scheme

M1A1 (a); M1A1 (b); M1 binomial term; A1 (c).

Question 5 · Structured

6 marks

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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Worked solution

(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) \(P(X\ge3)=k(2+1)=0.3\).

Marking scheme

M1 sum to 1; A1 k=0.1; M1A1 E(X); A1 0.3.

Question 6 · Structured

7 marks

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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Worked solution

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.

Marking scheme

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

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