AQA Paper 1 (Paper 1) 2024 June

1 Find the coefficient of \(x\) in the expansion of $$\left( 4 x ^ { 3 } - 5 x ^ { 2 } + 3 x - 2 \right) \left( x ^ { 5 } + 4 x + 1 \right)$$ Circle your answer.
[0pt] [1 mark]
\(\begin{array} { l l l l } - 5 & - 2 & 7 & 11 \end{array}\)

2 The function f is defined by \(\mathrm { f } ( x ) = \mathrm { e } ^ { x } + 1\) for \(x \in \mathbb { R }\) Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) Tick ( ✓ ) one box.
\(\mathrm { f } ^ { - 1 } ( x ) = \ln ( x - 1 )\)
\includegraphics[max width=\textwidth, alt={}, center]{0320e0a6-adc0-440a-b1da-d1a49fe06179-03_113_113_534_804}
\(\mathrm { f } ^ { - 1 } ( x ) = \ln ( x ) - 1\)
\includegraphics[max width=\textwidth, alt={}, center]{0320e0a6-adc0-440a-b1da-d1a49fe06179-03_109_113_689_804}
\(\mathrm { f } ^ { - 1 } ( x ) = \frac { 1 } { \mathrm { e } ^ { x } + 1 }\)
\includegraphics[max width=\textwidth, alt={}, center]{0320e0a6-adc0-440a-b1da-d1a49fe06179-03_113_108_840_804}
\(\mathrm { f } ^ { - 1 } ( x ) = \frac { x - 1 } { \mathrm { e } }\)
\includegraphics[max width=\textwidth, alt={}, center]{0320e0a6-adc0-440a-b1da-d1a49fe06179-03_108_109_991_808}

3 The expression $$\frac { 12 x ^ { 2 } + 3 x + 7 } { 3 x - 5 }$$ can be written as $$A x + B + \frac { C } { 3 x - 5 }$$ State the value of \(A\) Circle your answer.
[0pt] [1 mark] $$\begin{array} { l l l l } 3 & 4 & 7 & 9 \end{array}$$

4 One of the diagrams below shows the graph of \(y = \arccos x\) Identify the graph of \(y = \arccos x\) Tick ( ✓ ) one box.
[0pt] [1 mark]
\includegraphics[max width=\textwidth, alt={}, center]{0320e0a6-adc0-440a-b1da-d1a49fe06179-05_1339_1545_555_315}

5 Solve the equation $$\sin ^ { 2 } x = 1$$ for \(0 ^ { \circ } < x < 360 ^ { \circ }\)
[0pt] [3 marks]

6 Use the chain rule to find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(y = \left( x ^ { 3 } + 5 x \right) ^ { 7 }\)

7 Show that $$\frac { 3 + \sqrt { 8 n } } { 1 + \sqrt { 2 n } }$$ can be written as $$\frac { 4 n - 3 + \sqrt { 2 n } } { 2 n - 1 }$$ where \(n\) is a positive integer.
[0pt] [4 marks]

8

1. Find the first three terms, in ascending powers of \(x\), in the expansion of $$( 2 + k x ) ^ { 5 }$$ where \(k\) is a positive constant.
   [0pt] [3 marks]
   8
2. Hence, given that the coefficient of \(x\) is four times the coefficient of \(x ^ { 2 }\), find the value of \(k\)

9

1. Show that, for small values of \(\theta\) measured in radians $$\cos 4 \theta + 2 \sin 3 \theta - \tan 2 \theta \approx A + B \theta + C \theta ^ { 2 }$$ where \(A , B\) and \(C\) are constants to be found.
   [0pt] [3 marks]
   9
2. Use your answer to part (a) to find an approximation for $$\cos 0.28 + 2 \sin 0.21 - \tan 0.14$$ Give your answer to three decimal places.
   [0pt] [2 marks]

10

1. An arithmetic sequence has 300 terms. The first term of the sequence is - 7 and the last term is 32 Find the sum of the 300 terms.
   [0pt] [2 marks]
   10
2. A school holds a raffle at its summer fair. There are nine prizes.
   The total value of the prizes is \(\pounds 1260\)
   The values of the prizes form an arithmetic sequence.
   The top prize has the highest value, and the bottom prize has the least value.
   The value of the top prize is six times the value of the bottom prize.
   Find the value of the top prize.

11 It is given that $$f ( x ) = x ( x - a ) ( x - 6 )$$ where \(0 < a < 6\) 11

1. Sketch the graph of \(y = \mathrm { f } ( x )\) on the axes below.
   \includegraphics[max width=\textwidth, alt={}, center]{0320e0a6-adc0-440a-b1da-d1a49fe06179-14_1207_1105_733_447} 11
2. Sketch the graph of \(y = \mathrm { f } ( - 2 x )\) on the axes below.
   \includegraphics[max width=\textwidth, alt={}, center]{0320e0a6-adc0-440a-b1da-d1a49fe06179-15_1207_1107_413_445}

12 The terms, \(u _ { n }\), of a periodic sequence are defined by $$u _ { 1 } = 3 \text { and } u _ { n + 1 } = \frac { - 6 } { u _ { n } }$$ 12

1. \(\quad\) Find \(u _ { 2 } , u _ { 3 }\) and \(u _ { 4 }\)
   [0pt] [2 marks]
   12
2. State the period of the sequence.
   12
3. Find the value of \(\sum _ { n = 1 } ^ { 101 } u _ { n }\)
   [0pt] [2 marks]

13

1. It is given that $$P ( x ) = 4 x ^ { 3 } + 8 x ^ { 2 } + 11 x + 4$$ Use the factor theorem to show that \(( 2 x + 1 )\) is a factor of \(\mathrm { P } ( x )\)
   13
2. Express \(\mathrm { P } ( x )\) in the form $$\mathrm { P } ( x ) = ( 2 x + 1 ) \left( a x ^ { 2 } + b x + c \right)$$ where \(a\), \(b\) and \(c\) are constants to be found.
   13
3. Given that \(n\) is a positive integer, use your answer to part (b) to explain why \(4 n ^ { 3 } + 8 n ^ { 2 } + 11 n + 4\) is never prime.
   [0pt] [2 marks]

14

1. The equation $$x ^ { 3 } = \mathrm { e } ^ { 6 - 2 x }$$ has a single solution, \(x = \alpha\)
   By considering a suitable change of sign, show that \(\alpha\) lies between 0 and 4
   14
2. Show that the equation \(x ^ { 3 } = \mathrm { e } ^ { 6 - 2 x }\) can be rearranged to give $$x = 3 - \frac { 3 } { 2 } \ln x$$ 14
3. 1. Use the iterative formula $$x _ { n + 1 } = 3 - \frac { 3 } { 2 } \ln x _ { n }$$ with \(x _ { 1 } = 4\), to find \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\)
      Give your answers to three decimal places.
      14
4. (ii) Figure 1 below shows a sketch of parts of the graphs of $$y = 3 - \frac { 3 } { 2 } \ln x \text { and } y = x$$ On Figure 1, draw a staircase or cobweb diagram to show how convergence takes place.
   Label, on the \(x\)-axis, the positions of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\)
   [0pt] [2 marks] \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{0320e0a6-adc0-440a-b1da-d1a49fe06179-22_1328_1390_744_395}
   \end{figure} 14
5. (iii) Explain why the iterative formula $$x _ { n + 1 } = 3 - \frac { 3 } { 2 } \ln x _ { n }$$ fails to converge to \(\alpha\) when the starting value is \(x _ { 1 } = 0\)

15

1. Show that the expression $$\sin 2 \theta \operatorname { cosec } \theta + \cos 2 \theta \sec \theta$$ can be written as $$4 \cos \theta - \sec \theta$$ where \(\sin \theta \neq 0\) and \(\cos \theta \neq 0\)
   [0pt] [4 marks]
   15
2. A student is attempting to solve the equation $$\sin 2 \theta \operatorname { cosec } \theta + \cos 2 \theta \sec \theta = 3 \text { for } 0 ^ { \circ } \leq \theta \leq 360 ^ { \circ }$$ They use the result from part (a), and write the following incorrect solution: $$\sin 2 \theta \operatorname { cosec } \theta + \cos 2 \theta \sec \theta = 3$$ Step \(1 \quad 4 \cos \theta - \sec \theta = 3\)
   Step \(24 \cos \theta - \frac { 1 } { \cos \theta } - 3 = 0\)
   Step \(34 \cos ^ { 2 } \theta - 3 \cos \theta - 1 = 0\) Step \(4 \cos \theta = 1\) or \(\cos \theta = - 0.25\) Step \(5 \theta = 0 ^ { \circ } , 104.5 ^ { \circ } , 255.5 ^ { \circ } , 360 ^ { \circ }\) 15
3. 1. Explain why the student should reject one of their values for \(\cos \theta\) in Step 4. 15
4. (ii) State the correct solutions to the equation $$\sin 2 \theta \operatorname { cosec } \theta + \cos 2 \theta \sec \theta = 3 \text { for } 0 ^ { \circ } \leq \theta \leq 360 ^ { \circ }$$ Figure 2 below shows a 1.5 metre length of pipe. \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 2} \includegraphics[alt={},max width=\textwidth]{0320e0a6-adc0-440a-b1da-d1a49fe06179-26_335_693_502_740}
   \end{figure} The symmetrical cross-section of the pipe is shown below, in Figure 3, where \(x\) and \(y\) are measured in centimetres. \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 3} \includegraphics[alt={},max width=\textwidth]{0320e0a6-adc0-440a-b1da-d1a49fe06179-26_652_734_1247_717}
   \end{figure} Use the trapezium rule, with the values shown in the table below, to find the best estimate for the volume of the pipe.

   \(\boldsymbol { x }\)	0	0.4	0.8	1.2	1.6	2
   \(\boldsymbol { y }\)	- 3	- 2.943	- 2.752	- 2.353	- 1.572	0

17 The function f is defined by $$\mathrm { f } ( x ) = | x | + 1 \text { for } x \in \mathbb { R }$$ The function g is defined by $$g ( x ) = \ln x$$ where g has its greatest possible domain. 17

1. Using set notation, state the range of f 17
2. State the domain of g
   17
3. The composite function h is given by $$\mathrm { h } ( x ) = \operatorname { gf } ( x ) \text { for } x \in \mathbb { R }$$ 17
4. 1. Write down an expression for \(\mathrm { h } ( x )\) in terms of \(x\)
      17
5. (ii) Determine if h has an inverse. Fully justify your answer.
   [0pt] [2 marks]

18

1. Use a suitable substitution to show that $$\int _ { 0 } ^ { 4 } ( 4 x + 1 ) ( 2 x + 1 ) ^ { \frac { 1 } { 2 } } \mathrm {~d} x$$ can be written as $$\frac { 1 } { 2 } \int _ { a } ^ { 9 } \left( 2 u ^ { \frac { 3 } { 2 } } - u ^ { \frac { 1 } { 2 } } \right) \mathrm { d } u$$ where \(a\) is a constant to be found.
   18
2. Hence, or otherwise, show that $$\int _ { 0 } ^ { 4 } ( 4 x + 1 ) ( 2 x + 1 ) ^ { \frac { 1 } { 2 } } \mathrm {~d} x = \frac { 1322 } { 15 }$$ 18
3. A graph has the equation $$y = ( 4 x + 1 ) \sqrt { 2 x + 1 }$$ A student uses four rectangles to approximate the area under the graph between the lines \(x = 0\) and \(x = 4\) The rectangles are all the same width.
   All the rectangles are drawn under the curve as shown in the diagram below.
   \includegraphics[max width=\textwidth, alt={}, center]{0320e0a6-adc0-440a-b1da-d1a49fe06179-32_1031_698_744_735} The total area of the four rectangles is \(A\) The student decides to improve their approximation by increasing the number of rectangles used. Explain why the value of the student's improved approximation will be greater than \(A\), but less than \(\frac { 1322 } { 15 }\)

19 A curve has equation $$y ^ { 3 } \mathrm { e } ^ { 2 x } + 2 y - 16 x = k$$ where \(k\) is a constant. The curve has a stationary point on the \(y\)-axis.
Determine the value of \(k\)
2 A gardener stores rainwater in a cylindrical container. The container has a height of 130 centimetres.
The gardener empties the water from the container through a hose.
The hose is attached 5 centimetres from the bottom of the container.
At time \(t\) minutes after the hose is switched on, the depth of water, \(h\) centimetres, in the container decreases at a rate which is proportional to \(h - 5\) Initially the container of water is full, and the depth of water is decreasing at a rate of 1.5 centimetres per minute.

20

1. Show that $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - 0.012 ( h - 5 )$$ 20
2. Solve the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - 0.012 ( h - 5 )$$ to find an expression for \(h\) in terms of \(t\)
   [0pt] [5 marks]
   20
3. Find the time taken for the container to be half empty. Give your answer to the nearest minute.
   \includegraphics[max width=\textwidth, alt={}, center]{0320e0a6-adc0-440a-b1da-d1a49fe06179-38_2495_1915_169_123} Question number Additional page, if required. Write the question numbers in the left-hand margin.