AQA Paper 1 (Paper 1) 2020 June

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

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. $$\mathrm { f } ( x ) = \frac { 1 } { x } \quad \mathrm { f } ( x ) = x \quad \mathrm { f } ( x ) = x ^ { 3 } \quad \mathrm { f } ( x ) = \frac { 2 x + 1 } { x + 2 }$$

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

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 $$4 - | 2 x - 6 | > 2$$

5 Prove that, for integer values of \(n\) such that \(0 \leq n < 4\) $$2 ^ { n + 2 } > 3 ^ { n }$$

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. 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.

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
2. 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).

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] \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 $$\frac { 2 x ^ { 2 } + x } { ( x + 1 ) ( x + 2 ) ^ { 2 } } \equiv \frac { 1 } { x + 1 } - \frac { 6 } { ( x + 2 ) ^ { 2 } }$$

9

1. 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.

10

1. An arithmetic series is given by $$\sum _ { r = 5 } ^ { 20 } ( 4 r + 1 )$$ 10
2. 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

   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
5. (ii) The 40th term of the series is 4 times the 2nd term. Find the values of \(b\) and \(c\).
   [0pt] [4 marks]

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] \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
4. 1. an underestimate.
      11
5. (ii) an overestimate.

12 A curve \(C\) has equation $$x ^ { 3 } \sin y + \cos y = A x$$ where \(A\) is a constant.
\(C\) passes through the point \(P \left( \sqrt { 3 } , \frac { \pi } { 6 } \right)\)
12

1. Show that \(A = 2\)
   12
2. 1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 - 3 x ^ { 2 } \sin y } { x ^ { 3 } \cos y - \sin y }\)
      12
3. (ii) Hence, find the gradient of the curve at \(P\).
   12
4. (iii) The tangent to \(C\) at \(P\) intersects the \(x\)-axis at \(Q\).
   Find the exact \(x\)-coordinate of \(Q\).

13 The function f is defined by $$\mathrm { f } ( x ) = \frac { 2 x + 3 } { x - 2 } \quad x \in \mathbb { R } , x \neq 2$$ 13

1. 1. Find f-1
      13
2. (ii) Write down an expression for \(\mathrm { ff } ( x )\).
   13
3. The function g is defined by $$g ( x ) = \frac { 2 x ^ { 2 } - 5 x } { 2 } \quad x \in \mathbb { R } , 0 \leq x \leq 4$$ 13
4. 1. Find the range of g .
      13
5. (ii) Determine whether g has an inverse.
   Fully justify your answer.
6. Show that $$g f ( x ) = \frac { 48 + 29 x - 2 x ^ { 2 } } { 2 x ^ { 2 } - 8 x + 8 }$$ 13
7. It can be shown that fg is given by $$f g ( x ) = \frac { 4 x ^ { 2 } - 10 x + 6 } { 2 x ^ { 2 } - 5 x - 4 }$$ with domain \(\{ x \in \mathbb { R } : 0 \leq x \leq 4 , x \neq a \}\)
   Find the value of \(a\).
   Fully justify your answer.

14 The function f is defined by $$f ( x ) = 3 ^ { x } \sqrt { x } - 1 \quad \text { where } x \geq 0$$ 14

1. \(\quad \mathrm { f } ( x ) = 0\) has a single solution at the point \(x = \alpha\)
   By considering a suitable change of sign, show that \(\alpha\) lies between 0 and 1
   14
2. 1. Show that $$\mathrm { f } ^ { \prime } ( x ) = \frac { 3 ^ { x } ( 1 + x \ln 9 ) } { 2 \sqrt { x } }$$

      14 (ii) Use the Newton-Raphson method with \(x _ { 1 } = 1\) to find \(x _ { 3 }\), an approximation for \(\alpha\). Give your answer to five decimal places. [2 marks] \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) 14 (iii) Explain why the Newton-Raphson method fails to find \(\alpha\) with \(x _ { 1 } = 0\) [2 marks] \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)

15 The region enclosed between the curves \(y = \mathrm { e } ^ { x } , y = 6 - \mathrm { e } ^ { \overline { 2 } }\) and the line \(x = 0\) is shown shaded in the diagram below.
\includegraphics[max width=\textwidth, alt={}, center]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-28_1155_1009_424_513} Show that the exact area of the shaded region is $$6 \ln 4 - 5$$ Fully justify your answer.
\includegraphics[max width=\textwidth, alt={}, center]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-30_2491_1736_219_139}