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AQA Paper 1 2018 June Q15
6 marks Moderate -0.5
15 A curve has equation \(y = x ^ { 3 } - 48 x\) The point \(A\) on the curve has \(x\) coordinate - 4
The point \(B\) on the curve has \(x\) coordinate \(- 4 + h\) 15
  1. Show that the gradient of the line \(A B\) is \(h ^ { 2 } - 12 h\) 15
  2. Explain how the result of part (a) can be used to show that \(A\) is a stationary point on the curve. \includegraphics[max width=\textwidth, alt={}, center]{85b10472-8149-4387-999f-2ef153f1a105-25_2488_1719_219_150} \includegraphics[max width=\textwidth, alt={}, center]{85b10472-8149-4387-999f-2ef153f1a105-26_2488_1719_219_150} \includegraphics[max width=\textwidth, alt={}, center]{85b10472-8149-4387-999f-2ef153f1a105-27_2488_1719_219_150} \includegraphics[max width=\textwidth, alt={}, center]{85b10472-8149-4387-999f-2ef153f1a105-28_2498_1721_213_150}
AQA Paper 1 2019 June Q1
1 marks Easy -1.8
1 Given that \(a > 0\), determine which of these expressions is not equivalent to the others. Circle your answer.
[0pt] [1 mark] $$- 2 \log _ { 10 } \left( \frac { 1 } { a } \right) \quad 2 \log _ { 10 } ( a ) \quad \log _ { 10 } \left( a ^ { 2 } \right) \quad - 4 \log _ { 10 } ( \sqrt { a } )$$
AQA Paper 1 2019 June Q2
1 marks Easy -1.8
2 Given \(y = \mathrm { e } ^ { k x }\), where \(k\) is a constant, find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) Circle your answer. $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { k x } \quad \frac { \mathrm {~d} y } { \mathrm {~d} x } = k \mathrm { e } ^ { k x } \quad \frac { \mathrm {~d} y } { \mathrm {~d} x } = k x \mathrm { e } ^ { k x - 1 } \quad \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { \mathrm { e } ^ { k x } } { k }$$
AQA Paper 1 2019 June Q3
1 marks Easy -1.8
3 The diagram below shows a sector of a circle. \includegraphics[max width=\textwidth, alt={}, center]{6b1312f4-9a5c-4465-8129-7d37e99efefe-02_375_406_1647_817} The radius of the circle is 4 cm and \(\theta = 0.8\) radians. Find the area of the sector. Circle your answer.
[0pt] [1 mark] \(1.28 \mathrm {~cm} ^ { 2 }\) \(3.2 \mathrm {~cm} ^ { 2 }\) \(6.4 \mathrm {~cm} ^ { 2 }\) \(12.8 \mathrm {~cm} ^ { 2 }\)
AQA Paper 1 2019 June Q4
4 marks Moderate -0.3
4 The point \(A\) has coordinates \(( - 1 , a )\) and the point \(B\) has coordinates \(( 3 , b )\) The line \(A B\) has equation \(5 x + 4 y = 17\) Find the equation of the perpendicular bisector of the points \(A\) and \(B\).
AQA Paper 1 2019 June Q5
7 marks Standard +0.8
5 An arithmetic sequence has first term \(a\) and common difference \(d\). The sum of the first 16 terms of the sequence is 260 5
  1. Show that \(4 a + 30 d = 65\) 5
  2. Given that the sum of the first 60 terms is 315 , find the sum of the first 41 terms.
    5
  3. \(\quad S _ { n }\) is the sum of the first \(n\) terms of the sequence. Explain why the value you found in part (b) is the maximum value of \(S _ { n }\)
AQA Paper 1 2019 June Q6
8 marks Moderate -0.3
6 The function f is defined by $$\mathrm { f } ( x ) = \frac { 1 } { 2 } \left( x ^ { 2 } + 1 \right) , x \geq 0$$ 6
  1. Find the range of f . 6
    1. Find \(\mathrm { f } ^ { - 1 } ( x )\) 6
  3. (ii) State the range of \(\mathrm { f } ^ { - 1 } ( x )\) 6
  4. State the transformation which maps the graph of \(y = \mathrm { f } ( x )\) onto the graph of \(y = \mathrm { f } ^ { - 1 } ( x )\) 6
  5. Find the coordinates of the point of intersection of the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\)
AQA Paper 1 2019 June Q7
11 marks Standard +0.3
7
  1. By sketching the graphs of \(y = \frac { 1 } { x }\) and \(y = \sec 2 x\) on the axes below, show that the equation $$\frac { 1 } { x } = \sec 2 x$$ has exactly one solution for \(x > 0\) \includegraphics[max width=\textwidth, alt={}, center]{6b1312f4-9a5c-4465-8129-7d37e99efefe-08_675_771_689_639} 7
  2. By considering a suitable change of sign, show that the solution to the equation lies between 0.4 and 0.6
    7
  3. Show that the equation can be rearranged to give $$x = \frac { 1 } { 2 } \cos ^ { - 1 } x$$ 7
    1. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \cos ^ { - 1 } x _ { n }$$ with \(x _ { 1 } = 0.4\), to find \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to four decimal places.
      7
  5. (ii) On the graph below, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\).
    [0pt] [2 marks] \includegraphics[max width=\textwidth, alt={}, center]{6b1312f4-9a5c-4465-8129-7d37e99efefe-09_954_1600_1717_223}
AQA Paper 1 2019 June Q8
4 marks Standard +0.3
8 \(\quad \mathrm { P } ( n ) = \sum _ { k = 0 } ^ { n } k ^ { 3 } - \sum _ { k = 0 } ^ { n - 1 } k ^ { 3 }\) where \(n\) is a positive integer.
8
  1. Find \(\mathrm { P } ( 3 )\) and \(\mathrm { P } ( 10 )\) 8
  2. Solve the equation \(\mathrm { P } ( n ) = 1.25 \times 10 ^ { 8 }\)
AQA Paper 1 2019 June Q9
5 marks Standard +0.3
9 Prove that the sum of a rational number and an irrational number is always irrational.
10
The volume of a spherical bubble is increasing at a constant rate.
Show that the rate of increase of the radius, \(r\), of the bubble is inversely proportional to \(r ^ { 2 }\) \(\text { Volume of a sphere } = \frac { 4 } { 3 } \pi r ^ { 3 }\)
\includegraphics[max width=\textwidth, alt={}]{6b1312f4-9a5c-4465-8129-7d37e99efefe-13_2488_1716_219_153}
Jodie is attempting to use differentiation from first principles to prove that the gradient of \(y = \sin x\) is zero when \(x = \frac { \pi } { 2 }\) Jodie's teacher tells her that she has made mistakes starting in Step 4 of her working. Her working is shown below. \includegraphics[max width=\textwidth, alt={}, center]{6b1312f4-9a5c-4465-8129-7d37e99efefe-14_645_1298_584_370} Step 1 Gradient of chord \(A B = \frac { \sin \left( \frac { \pi } { 2 } + h \right) - \sin \left( \frac { \pi } { 2 } \right) } { h }\) Step 2 \(= \frac { \sin \left( \frac { \pi } { 2 } \right) \cos ( h ) + \cos \left( \frac { \pi } { 2 } \right) \sin ( h ) - \sin \left( \frac { \pi } { 2 } \right) } { h }\) Step 3 $$= \sin \left( \frac { \pi } { 2 } \right) \left( \frac { \cos ( h ) - 1 } { h } \right) + \cos \left( \frac { \pi } { 2 } \right) \frac { \sin ( h ) } { h }$$ Step 4
For gradient of curve at \(A\),
let \(h = 0\) then \(\frac { \cos ( h ) - 1 } { h } = 0\) and \(\frac { \sin ( h ) } { h } = 0\) Step 5
Hence the gradient of the curve at \(A\) is given by \(\sin \left( \frac { \pi } { 2 } \right) \times 0 + \cos \left( \frac { \pi } { 2 } \right) \times 0 = 0\) Complete Steps 4 and 5 of Jodie's working below, to correct her proof. Step 4 For gradient of curve at \(A\), Step 5 Hence the gradient of the curve at \(A\) is given by
AQA Paper 1 2019 June Q12
7 marks Standard +0.3
12
  1. Show that the equation $$2 \cot ^ { 2 } x + 2 \operatorname { cosec } ^ { 2 } x = 1 + 4 \operatorname { cosec } x$$ can be written in the form $$a \operatorname { cosec } ^ { 2 } x + b \operatorname { cosec } x + c = 0$$ 12
  2. Hence, given \(x\) is obtuse and $$2 \cot ^ { 2 } x + 2 \operatorname { cosec } ^ { 2 } x = 1 + 4 \operatorname { cosec } x$$ find the exact value of \(\tan x\) Fully justify your answer.
    13A curve, \(C\), has equation
    \(y = \frac { \mathrm { e } ^ { 3 x - 5 } } { x ^ { 2 } }\)
    Show that \(C\) has exactly one stationary point.
    Fully justify your answer.
    \includegraphics[max width=\textwidth, alt={}]{6b1312f4-9a5c-4465-8129-7d37e99efefe-19_2488_1716_219_153}
AQA Paper 1 2019 June Q14
10 marks Standard +0.3
14 The graph of \(y = \frac { 2 x ^ { 3 } } { x ^ { 2 } + 1 }\) is shown for \(0 \leq x \leq 4\) \includegraphics[max width=\textwidth, alt={}, center]{6b1312f4-9a5c-4465-8129-7d37e99efefe-20_1022_640_411_701} Caroline is attempting to approximate the shaded area, \(A\), under the curve using the trapezium rule by splitting the area into \(n\) trapezia. 14
  1. When \(n = 4\) 14
    1. State the number of ordinates that Caroline uses. 14
  3. (ii) Calculate the area that Caroline should obtain using this method.
    Give your answer correct to two decimal places.
    14
  4. Show that the exact area of \(A\) is $$16 - \ln 17$$ Fully justify your answer.
    14
  		5. Explain what would happen to Caroline's answer to part (a)(ii) as \(n \rightarrow \infty\)[1 mark]Do not write outside the box
    \includegraphics[max width=\textwidth, alt={}]{6b1312f4-9a5c-4465-8129-7d37e99efefe-23_2488_1716_219_153}
AQA Paper 1 2019 June Q15
10 marks Moderate -0.3
15
  1. At time \(t\) hours after a high tide, the height, \(h\) metres, of the tide and the velocity, \(v\) knots, of the tidal flow can be modelled using the parametric equations $$\begin{aligned} & v = 4 - \left( \frac { 2 t } { 3 } - 2 \right) ^ { 2 } \\ & h = 3 - 2 \sqrt [ 3 ] { t - 3 } \end{aligned}$$ High tides and low tides occur alternately when the velocity of the tidal flow is zero.
    A high tide occurs at 2 am.
    15
    1. Use the model to find the height of this high tide.
      15
  3. (ii) Find the time of the first low tide after 2 am.
    15
  4. (iii) Find the height of this low tide.
    15
  5. Use the model to find the height of the tide when it is flowing with maximum velocity.
    15
  6. Comment on the validity of the model.
AQA Paper 1 2019 June Q16
16 marks Challenging +1.2
16
  1. \(\quad y = \mathrm { e } ^ { - x } ( \sin x + \cos x )\) Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) Simplify your answer.
    16
  2. Hence, show that $$\int \mathrm { e } ^ { - x } \sin x \mathrm {~d} x = a \mathrm { e } ^ { - x } ( \sin x + \cos x ) + c$$ where \(a\) is a rational number.
    16
  3. A sketch of the graph of \(y = \mathrm { e } ^ { - x } \sin x\) for \(x \geq 0\) is shown below. \(A _ { 1 } , A _ { 2 } , \ldots , A _ { n } , \ldots\) The areas of the finite regions bounded by the curve and the \(x\)-axis are denoted by \includegraphics[max width=\textwidth, alt={}, center]{6b1312f4-9a5c-4465-8129-7d37e99efefe-27_974_1507_502_262} 16
    1. Find the exact value of the area \(A _ { 1 }\) 16
  5. (ii) Show that $$\frac { A _ { 2 } } { A _ { 1 } } = \mathrm { e } ^ { - \pi }$$ 16
  6. (iii) Given that $$\frac { A _ { n + 1 } } { A _ { n } } = \mathrm { e } ^ { - \pi }$$ show that the exact value of the total area enclosed between the curve and the \(x\)-axis is $$\frac { 1 + \mathrm { e } ^ { \pi } } { 2 \left( \mathrm { e } ^ { \pi } - 1 \right) }$$ \includegraphics[max width=\textwidth, alt={}, center]{6b1312f4-9a5c-4465-8129-7d37e99efefe-30_2488_1719_219_150} \includegraphics[max width=\textwidth, alt={}, center]{6b1312f4-9a5c-4465-8129-7d37e99efefe-31_2488_1719_219_150} \includegraphics[max width=\textwidth, alt={}, center]{6b1312f4-9a5c-4465-8129-7d37e99efefe-32_2496_1721_214_148}
AQA Paper 1 2020 June Q1
2 marks Easy -1.2
1 The first three terms, in ascending powers of \(x\), of the binomial expansion of \(( 9 + 2 x ) ^ { \frac { 1 } { 2 } }\) are given by $$( 9 + 2 x ) ^ { \frac { 1 } { 2 } } \approx a + \frac { x } { 3 } - \frac { x ^ { 2 } } { 54 }$$ where \(a\) is a constant. 1
  1. State the range of values of \(x\) for which this expansion is valid.
    Circle your answer. \(| x | < \frac { 2 } { 9 }\) \(| x | < \frac { 2 } { 3 }\) \(| x | < 1\) \(| x | < \frac { 9 } { 2 }\) 1
  2. Find the value of \(a\).
    Circle your answer.
    [0pt] [1 mark]
    1239
AQA Paper 1 2020 June Q2
1 marks Easy -1.8
2 A student is searching for a solution to the equation \(\mathrm { f } ( x ) = 0\) He correctly evaluates $$f ( - 1 ) = - 1 \text { and } f ( 1 ) = 1$$ and concludes that there must be a root between - 1 and 1 due to the change of sign.
Select the function \(\mathrm { f } ( x )\) for which the conclusion is incorrect.
Circle your answer. $$\mathrm { f } ( x ) = \frac { 1 } { x } \quad \mathrm { f } ( x ) = x \quad \mathrm { f } ( x ) = x ^ { 3 } \quad \mathrm { f } ( x ) = \frac { 2 x + 1 } { x + 2 }$$
AQA Paper 1 2020 June Q3
1 marks Easy -1.2
3 The diagram shows a sector \(O A B\) of a circle with centre \(O\) and radius 2 \includegraphics[max width=\textwidth, alt={}, center]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-03_374_455_1187_790} The angle \(A O B\) is \(\theta\) radians and the perimeter of the sector is 6
Find the value of \(\theta\) Circle your answer.
[0pt] [1 mark]
1 \(\sqrt { 3 }\) 2
3
AQA Paper 1 2020 June Q4
5 marks Easy -1.2
4
  1. Sketch the graph of \includegraphics[max width=\textwidth, alt={}, center]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-04_933_1093_349_475} 4
  2. Solve the inequality $$4 - | 2 x - 6 | > 2$$
AQA Paper 1 2020 June Q5
2 marks Easy -1.8
5 Prove that, for integer values of \(n\) such that \(0 \leq n < 4\) $$2 ^ { n + 2 } > 3 ^ { n }$$
AQA Paper 1 2020 June Q6
4 marks Easy -1.2
6 Four students, Tom, Josh, Floella and Georgia are attempting to complete the indefinite integral $$\int \frac { 1 } { x } \mathrm {~d} x \quad \text { for } x > 0$$ Each of the students' solutions is shown below: $$\begin{array} { l l } \text { Tom } & \int \frac { 1 } { x } \mathrm {~d} x = \ln x \\ \text { Josh } & \int \frac { 1 } { x } \mathrm {~d} x = k \ln x \\ \text { Floella } & \int \frac { 1 } { x } \mathrm {~d} x = \ln A x \\ \text { Georgia } & \int \frac { 1 } { x } \mathrm {~d} x = \ln x + c \end{array}$$ 6
    1. Explain what is wrong with Tom's answer. 6
  2. (ii) Explain what is wrong with Josh's answer.
    6
  3. Explain why Floella and Georgia's answers are equivalent.
AQA Paper 1 2020 June Q7
4 marks Moderate -0.8
7 Consecutive terms of a sequence are related by $$u _ { n + 1 } = 3 - \left( u _ { n } \right) ^ { 2 }$$ 7
  1. In the case that \(u _ { 1 } = 2\) 7
    1. Find \(u _ { 3 }\) 7
  3. (ii) Find \(u _ { 50 }\) 7
  4. State a different value for \(u _ { 1 }\) which gives the same value for \(u _ { 50 }\) as found in part (a)(ii).
AQA Paper 1 2020 June Q8
7 marks Standard +0.3
8 Mike, an amateur astronomer who lives in the South of England, wants to know how the number of hours of darkness changes through the year. On various days between February and September he records the length of time, \(H\) hours, of darkness along with \(t\), the number of days after 1 January. His results are shown in Figure 1 below. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-08_940_1541_696_246}
\end{figure} Mike models this data using the equation $$H = 3.87 \sin \left( \frac { 2 \pi ( t + 101.75 ) } { 365 } \right) + 11.7$$ 8
  1. Find the minimum number of hours of darkness predicted by Mike's model. Give your answer to the nearest minute.
    [0pt] [2 marks] 8
  2. Find the maximum number of consecutive days where the number of hours of darkness predicted by Mike's model exceeds 14
    8
  3. Mike's friend Sofia, who lives in Spain, also records the number of hours of darkness on various days throughout the year. Her results are shown in Figure 2 below. \begin{figure}[h]
    \captionsetup{labelformat=empty} \caption{Figure 2} \includegraphics[alt={},max width=\textwidth]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-10_933_1537_561_248}
    \end{figure} Sofia attempts to model her data by refining Mike's model.
    She decides to increase the 3.87 value, leaving everything else unchanged.
    Explain whether Sofia's refinement is appropriate. \includegraphics[max width=\textwidth, alt={}, center]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-11_2488_1730_219_141} \(9 \quad\) Chloe is attempting to write \(\frac { 2 x ^ { 2 } + x } { ( x + 1 ) ( x + 2 ) ^ { 2 } }\) as partial fractions, with constant numerators. Her incorrect attempt is shown below. Step 1 $$\frac { 2 x ^ { 2 } + x } { ( x + 1 ) ( x + 2 ) ^ { 2 } } \equiv \frac { A } { x + 1 } + \frac { B } { ( x + 2 ) ^ { 2 } }$$ Step 2 $$2 x ^ { 2 } + x \equiv A ( x + 2 ) ^ { 2 } + B ( x + 1 )$$ Step 3 $$\begin{aligned} & \text { Let } x = - 1 \Rightarrow A = 1 \\ & \text { Let } x = - 2 \Rightarrow B = - 6 \end{aligned}$$ Answer $$\frac { 2 x ^ { 2 } + x } { ( x + 1 ) ( x + 2 ) ^ { 2 } } \equiv \frac { 1 } { x + 1 } - \frac { 6 } { ( x + 2 ) ^ { 2 } }$$
AQA Paper 1 2020 June Q9
7 marks Moderate -0.3
9
    1. By using a counter example, show that the answer obtained by Chloe cannot be correct.
      9
  2. (ii) Explain her mistake in Step 1.
    9
  3. Write \(\frac { 2 x ^ { 2 } + x } { ( x + 1 ) ( x + 2 ) ^ { 2 } }\) as partial fractions, with constant numerators.
AQA Paper 1 2020 June Q10
12 marks Moderate -0.8
10
  1. An arithmetic series is given by $$\sum _ { r = 5 } ^ { 20 } ( 4 r + 1 )$$ 10
    1. Write down the first term of the series.
      10
  3. (ii) Write down the common difference of the series.
    10
  4. (iii) Find the number of terms of the series.
    10
  5. A different arithmetic series is given by \(\sum _ { r = 10 } ^ { 100 } ( b r + c )\)
    where \(b\) and \(c\) are constants.
    The sum of this series is 7735
    10
  6. (ii) The 40th term of the series is 4 times the 2nd term. Find the values of \(b\) and \(c\).
    [0pt] [4 marks]
AQA Paper 1 2020 June Q11
9 marks Moderate -0.3
11 The region \(R\) enclosed by the lines \(x = 1 , x = 6 , y = 0\) and the curve $$y = \ln ( 8 - x )$$ is shown shaded in Figure 3 below. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 3} \includegraphics[alt={},max width=\textwidth]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-17_419_869_598_587}
\end{figure} All distances are measured in centimetres.
11
  1. Use a single trapezium to find an approximate value of the area of the shaded region, giving your answer in \(\mathrm { cm } ^ { 2 }\) to two decimal places.
    [0pt] [2 marks]
    \section*{Question 11 continues on the next page} 11
  2. Shape \(B\) is made from four copies of region \(R\) as shown in Figure 4 below. \begin{figure}[h]
    \captionsetup{labelformat=empty} \caption{Figure 4} \includegraphics[alt={},max width=\textwidth]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-18_707_711_438_667}
    \end{figure} Shape \(B\) is cut from metal of thickness 2 mm
    The metal has a density of \(10.5 \mathrm {~g} / \mathrm { cm } ^ { 3 }\) Use the trapezium rule with six ordinates to calculate an approximate value of the mass of Shape B. Give your answer to the nearest gram.
    11
  3. Without further calculation, give one reason why the mass found in part (b) may be:
    11
    1. an underestimate.
      11
  5. (ii) an overestimate.