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Edexcel C34 2017 June Q4
8 marks Standard +0.3
4. $$f ( x ) = \frac { 27 } { ( 3 - 5 x ) ^ { 2 } } \quad | x | < \frac { 3 } { 5 }$$
  1. Find the series expansion of \(\mathrm { f } ( x )\), in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\). Give each coefficient in its simplest form.
    (5) Use your answer to part (a) to find the series expansion in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\), of
  2. \(g ( x ) = \frac { 27 } { ( 3 + 5 x ) ^ { 2 } } \quad | x | < \frac { 3 } { 5 }\)
  3. \(\mathrm { h } ( x ) = \frac { 27 } { ( 3 - x ) ^ { 2 } } \quad | x | < 3\)
Edexcel C34 2017 June Q5
8 marks Standard +0.3
5. $$\frac { 6 - 5 x - 4 x ^ { 2 } } { ( 2 - x ) ( 1 + 2 x ) } \equiv A + \frac { B } { ( 2 - x ) } + \frac { C } { ( 1 + 2 x ) }$$
  1. Find the values of the constants \(A , B\) and \(C\). $$f ( x ) = \frac { 6 - 5 x - 4 x ^ { 2 } } { ( 2 - x ) ( 1 + 2 x ) } \quad x > 2$$
  2. Using part (a), find \(\mathrm { f } ^ { \prime } ( x )\).
  3. Prove that \(\mathrm { f } ( x )\) is a decreasing function.
Edexcel C34 2017 June Q6
6 marks Standard +0.3
  1. The line \(l _ { 1 }\) has vector equation \(\mathbf { r } = \left( \begin{array} { r } 5 \\ - 2 \\ 4 \end{array} \right) + \lambda \left( \begin{array} { r } 6 \\ 3 \\ - 1 \end{array} \right)\), where \(\lambda\) is a scalar parameter. The line \(l _ { 2 }\) has vector equation \(\mathbf { r } = \left( \begin{array} { r } 10 \\ 5 \\ - 3 \end{array} \right) + \mu \left( \begin{array} { l } 3 \\ 1 \\ 2 \end{array} \right)\), where \(\mu\) is a scalar parameter.
Justify, giving reasons in each case, whether the lines \(l _ { 1 }\) and \(l _ { 2 }\) are parallel, intersecting or skew.
(6)
Edexcel C34 2017 June Q7
9 marks Standard +0.3
  1. (a) Prove that
$$\frac { 1 - \cos 2 x } { 1 + \cos 2 x } \equiv \tan ^ { 2 } x , \quad x \neq ( 2 n + 1 ) 90 ^ { \circ } , n \in \mathbb { Z }$$ (b) Hence, or otherwise, solve, for \(- 90 ^ { \circ } < \theta < 90 ^ { \circ }\), $$\frac { 2 - 2 \cos 2 \theta } { 1 + \cos 2 \theta } - 2 = 7 \sec \theta$$ Give your answers in degrees to one decimal place.
(Solutions based entirely on graphical or numerical methods are not acceptable.)
Edexcel C34 2017 June Q8
7 marks Moderate -0.3
8. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{29b56d51-120a-4275-a761-8b8aed7bca54-24_560_1029_219_463} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \sqrt { \frac { x } { x ^ { 2 } + 1 } } , \quad x \geqslant 0\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the line with equation \(x = 2\), the \(x\)-axis and the line with equation \(x = 7\) The table below shows corresponding values of \(x\) and \(y\) for \(y = \sqrt { \frac { x } { x ^ { 2 } + 1 } }\)
\(x\)234567
\(y\)0.63250.54770.48510.43850.40270.3742
  1. Use the trapezium rule, with all the values of \(y\) in the table, to find an estimate for the area of \(R\), giving your answer to 3 decimal places. The region \(R\) is rotated \(360 ^ { \circ }\) about the \(x\)-axis to form a solid of revolution.
  2. Use calculus to find the exact volume of the solid of revolution formed. Write your answer in its simplest form. \includegraphics[max width=\textwidth, alt={}, center]{29b56d51-120a-4275-a761-8b8aed7bca54-24_2255_47_314_1979}
Edexcel C34 2017 June Q9
8 marks Standard +0.3
9. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{29b56d51-120a-4275-a761-8b8aed7bca54-28_615_328_210_808} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Figure 2 shows a sketch of a triangle \(A B C\).
Given \(\overrightarrow { A B } = 2 \mathbf { i } + 3 \mathbf { j } - 2 \mathbf { k }\) and \(\overrightarrow { A C } = 5 \mathbf { i } - 6 \mathbf { j } + \mathbf { k }\),
  1. find the size of angle \(C A B\), giving your answer in degrees to 2 decimal places,
  2. find the area of triangle \(A B C\), giving your answer to 2 decimal places. Using your answer to part (b), or otherwise,
  3. find the shortest distance from \(A\) to \(B C\), giving your answer to 2 decimal places.
Edexcel C34 2017 June Q10
9 marks Standard +0.3
  1. (a) Write \(2 \sin \theta - \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R\) and \(\alpha\) are constants, \(R > 0\) and \(0 < \alpha \leqslant 90 ^ { \circ }\). Give the exact value of \(R\) and give the value of \(\alpha\) to one decimal place.
    (3)
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{29b56d51-120a-4275-a761-8b8aed7bca54-32_513_1194_404_374} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows a sketch of the graph with equation \(y = 2 \sin \theta - \cos \theta , \quad 0 \leqslant \theta < 360 ^ { \circ }\) (b) Sketch the graph with equation $$y = | 2 \sin \theta - \cos \theta | , \quad 0 \leqslant \theta < 360 ^ { \circ }$$ stating the coordinates of all points at which the graph meets or cuts the coordinate axes. The temperature of a warehouse is modelled by the equation $$f ( t ) = 5 + \left| 2 \sin ( 15 t ) ^ { \circ } - \cos ( 15 t ) ^ { \circ } \right| , \quad 0 \leqslant t < 24$$ where \(\mathrm { f } ( t )\) is the temperature of the warehouse in degrees Celsius and \(t\) is the time measured in hours from midnight. State
(c) (i) the maximum value of \(f ( t )\),
(ii) the largest value of \(t\), for \(0 \leqslant t < 24\), at which this maximum value occurs. Give your answer to one decimal place.
Edexcel C34 2017 June Q11
11 marks Standard +0.3
11. $$y = \left( 2 x ^ { 2 } - 3 \right) \tan \left( \frac { 1 } { 2 } x \right) , \quad 0 < x < \pi$$
  1. Find the exact value of \(x\) when \(y = 0\) Given that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\) when \(x = \alpha\),
  2. show that $$2 \alpha ^ { 2 } - 3 + 4 \alpha \sin \alpha = 0$$ The iterative formula $$x _ { n + 1 } = \frac { 3 } { \left( 2 x _ { n } + 4 \sin x _ { n } \right) }$$ can be used to find an approximation for \(\alpha\).
  3. Taking \(x _ { 1 } = 0.7\), find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving each answer to 4 decimal places.
  4. By choosing a suitable interval, show that \(\alpha = 0.7283\) to 4 decimal places.
    \includegraphics[max width=\textwidth, alt={}]{29b56d51-120a-4275-a761-8b8aed7bca54-38_2253_50_314_1977}
Edexcel C34 2017 June Q12
14 marks Standard +0.3
12. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{29b56d51-120a-4275-a761-8b8aed7bca54-40_471_949_219_493} \captionsetup{labelformat=empty} \caption{Figure 4}
\end{figure} Figure 4 shows a right cylindrical water tank. The diameter of the circular cross section of the tank is 4 m and the height is 2.25 m . Water is flowing into the tank at a constant rate of \(0.4 \pi \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }\). There is a tap at a point \(T\) at the base of the tank. When the tap is open, water leaves the tank at a rate of \(0.2 \pi \sqrt { h } \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }\), where \(h\) is the height of the water in metres.
  1. Show that at time \(t\) minutes after the tap has been opened, the height \(h \mathrm {~m}\) of the water in the tank satisfies the differential equation $$20 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 2 - \sqrt { h }$$ At the instant when the tap is opened, \(t = 0\) and \(h = 0.16\)
  2. Use the differential equation to show that the time taken to fill the tank to a height of 2.25 m is given by $$\int _ { 0.16 } ^ { 2.25 } \frac { 20 } { 2 - \sqrt { h } } \mathrm {~d} h$$ Using the substitution \(h = ( 2 - x ) ^ { 2 }\), or otherwise,
  3. find the time taken to fill the tank to a height of 2.25 m . Give your answer in minutes to the nearest minute.
    (Solutions based entirely on graphical or numerical methods are not acceptable.)
Edexcel C34 2017 June Q13
9 marks Standard +0.8
13. Figure 5 A colony of ants is being studied. The number of ants in the colony is modelled by the equation $$P = 200 - \frac { 160 \mathrm { e } ^ { 0.6 t } } { 15 + \mathrm { e } ^ { 0.8 t } } \quad t \in \mathbb { R } , t \geqslant 0$$ where \(P\) is the number of ants, measured in thousands, \(t\) years after the study started. A sketch of the graph of \(P\) against \(t\) is shown in Figure 5
  1. Calculate the number of ants in the colony at the start of the study.
  2. Find \(\frac { \mathrm { d } P } { \mathrm {~d} t }\) The population of ants initially decreases, reaching a minimum value after \(T\) years, as shown in Figure 5
  3. Using your answer to part (b), calculate the value of \(T\) to 2 decimal places.
    (Solutions based entirely on graphical or numerical methods are not acceptable.)
Edexcel C34 2017 June Q14
16 marks Standard +0.8
14. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{29b56d51-120a-4275-a761-8b8aed7bca54-48_506_812_219_571} \captionsetup{labelformat=empty} \caption{Figure 6}
\end{figure} Figure 6 shows a sketch of the curve \(C\) with parametric equations $$x = 8 \cos ^ { 3 } \theta , \quad y = 6 \sin ^ { 2 } \theta , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$$ Given that the point \(P\) lies on \(C\) and has parameter \(\theta = \frac { \pi } { 3 }\)
  1. find the coordinates of \(P\). The line \(l\) is the normal to \(C\) at \(P\).
  2. Show that an equation of \(l\) is \(y = x + 3.5\) The finite region \(S\), shown shaded in Figure 6, is bounded by the curve \(C\), the line \(l\), the \(y\)-axis and the \(x\)-axis.
  3. Show that the area of \(S\) is given by $$4 + 144 \int _ { 0 } ^ { \frac { \pi } { 3 } } \left( \sin \theta \cos ^ { 2 } \theta - \sin \theta \cos ^ { 4 } \theta \right) d \theta$$
  4. Hence, by integration, find the exact area of \(S\).
    (Solutions based entirely on graphical or numerical methods are not acceptable.)
    END
Edexcel C34 2018 June Q1
6 marks Moderate -0.8
  1. (i) Find
$$\int \frac { 2 x ^ { 2 } + 5 x + 1 } { x ^ { 2 } } \mathrm {~d} x , \quad x > 0$$ (ii) Find $$\int x \cos 2 x \mathrm {~d} x$$
Edexcel C34 2018 June Q2
7 marks Moderate -0.3
2. A curve \(C\) has parametric equations $$x = \frac { 3 } { 2 } t - 5 , \quad y = 4 - \frac { 6 } { t } \quad t \neq 0$$
  1. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at \(t = 3\), giving your answer as a fraction in its simplest form.
  2. Show that a cartesian equation of \(C\) can be expressed in the form $$y = \frac { a x + b } { x + 5 } \quad x \neq k$$ where \(a , b\) and \(k\) are integers to be found.
Edexcel C34 2018 June Q3
7 marks Standard +0.3
3. $$f ( x ) = 2 ^ { x - 1 } - 4 + 1.5 x \quad x \in \mathbb { R }$$
  1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as $$x = \frac { 1 } { 3 } \left( 8 - 2 ^ { x } \right)$$ The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\), where \(\alpha = 1.6\) to one decimal place.
  2. Starting with \(x _ { 0 } = 1.6\), use the iteration formula $$x _ { n + 1 } = \frac { 1 } { 3 } \left( 8 - 2 ^ { x _ { n } } \right)$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.
  3. By choosing a suitable interval, prove that \(\alpha = 1.633\) to 3 decimal places.
Edexcel C34 2018 June Q4
10 marks Standard +0.3
4. (a) Find the binomial expansion of $$( 1 + p x ) ^ { - 4 } , \quad | p x | < 1$$ in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\), giving each coefficient as simply as possible in terms of the constant \(p\). $$f ( x ) = \frac { 3 + 4 x } { ( 1 + p x ) ^ { 4 } } \quad | p x | < 1$$ where \(p\) is a positive constant. In the series expansion of \(\mathrm { f } ( x )\), the coefficient of \(x ^ { 2 }\) is twice the coefficient of \(x\).
(b) Find the value of \(p\).
(c) Hence find the coefficient of \(x ^ { 3 }\) in the series expansion of \(\mathrm { f } ( x )\), giving your answer as a simplified fraction.
Edexcel C34 2018 June Q5
12 marks Standard +0.2
    1. The functions \(f\) and \(g\) are defined by
$$\begin{array} { l l } \mathrm { f } : x \rightarrow \mathrm { e } ^ { 2 x } - 5 , & x \in \mathbb { R } \\ \mathrm {~g} : x \rightarrow \ln ( 3 x - 1 ) , & x \in \mathbb { R } , x > \frac { 1 } { 3 } \end{array}$$
  1. Find \(\mathrm { f } ^ { - 1 }\) and state its domain.
  2. Find \(\mathrm { fg } ( 3 )\), giving your answer in its simplest form.
    (ii) (a) Sketch the graph with equation $$y = | 4 x - a |$$ where \(a\) is a positive constant. State the coordinates of each point where the graph cuts or meets the coordinate axes. Given that $$| 4 x - a | = 9 a$$ where \(a\) is a positive constant,
  3. find the possible values of $$| x - 6 a | + 3 | x |$$ giving your answers, in terms of \(a\), in their simplest form.
Edexcel C34 2018 June Q6
11 marks Standard +0.3
6. (a) Express \(\sqrt { 5 } \cos \theta - 2 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\) State the value of \(R\) and give the value of \(\alpha\) to 4 significant figures.
(b) Solve, for \(- \pi < \theta < \pi\), $$\sqrt { 5 } \cos \theta - 2 \sin \theta = 0.5$$ giving your answers to 3 significant figures. [Solutions based entirely on graphical or numerical methods are not acceptable.] $$\mathrm { f } ( x ) = A ( \sqrt { 5 } \cos \theta - 2 \sin \theta ) + B \quad \theta \in \mathbb { R }$$ where \(A\) and \(B\) are constants. Given that the range of f is $$- 15 \leqslant f ( x ) \leqslant 33$$ (c) find the value of \(B\) and the possible values of \(A\).
Edexcel C34 2018 June Q7
5 marks Standard +0.3
7. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a377da06-a968-438c-bec2-ae55283dae47-22_362_766_237_589} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a hemispherical bowl.
Water is flowing into the bowl at a constant rate of \(180 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
When the height of the water is \(h \mathrm {~cm}\), the volume of water \(V \mathrm {~cm} ^ { 3 }\) is given by $$V = \frac { 1 } { 3 } \pi h ^ { 2 } ( 90 - h ) , \quad 0 \leqslant h \leqslant 30$$ Find the rate of change of the height of the water, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), when \(h = 15\) Give your answer to 2 significant figures.
Edexcel C34 2018 June Q8
11 marks Standard +0.3
8. With respect to a fixed origin \(O\), the lines \(l _ { 1 }\) and \(l _ { 2 }\) are given by the equations $$l _ { 1 } : \mathbf { r } = \left( \begin{array} { r } 1 \\ - 3 \\ 2 \end{array} \right) + \lambda \left( \begin{array} { l } 1 \\ 2 \\ 3 \end{array} \right) , \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { l } 6 \\ 4 \\ 1 \end{array} \right) + \mu \left( \begin{array} { r } 1 \\ 1 \\ - 1 \end{array} \right)$$ where \(\lambda\) and \(\mu\) are scalar parameters.
  1. Show that \(l _ { 1 }\) and \(l _ { 2 }\) do not meet. The point \(P\) is on \(l _ { 1 }\) where \(\lambda = 0\), and the point \(Q\) is on \(l _ { 2 }\) where \(\mu = - 1\)
  2. Find the acute angle between the line segment \(P Q\) and \(l _ { 1 }\), giving your answer in degrees to 2 decimal places.
  3. Find the shortest distance from the point \(Q\) to the line \(l _ { 1 }\), giving your answer to 3 significant figures.
Edexcel C34 2018 June Q9
8 marks Standard +0.8
9. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a377da06-a968-438c-bec2-ae55283dae47-28_533_1095_258_365} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Diagram not drawn to scale
  1. Find $$\int \frac { 1 } { ( 2 x - 1 ) ^ { 2 } } d x$$ Figure 2 shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = \frac { 12 } { ( 2 x - 1 ) } \quad 1 \leqslant x \leqslant 5$$ The finite region \(R\), shown shaded in Figure 2, is bounded by the line with equation \(x = 1\), the curve with equation \(y = \mathrm { f } ( x )\) and the line with equation \(y = \frac { 4 } { 3 }\). The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.
  2. Find the exact value of the volume of the solid generated, giving your answer in its simplest form.
    \section*{Leave
    k}
Edexcel C34 2018 June Q10
7 marks Standard +0.8
10. The curve \(C\) satisfies the equation $$x \mathrm { e } ^ { 5 - 2 y } - y = 0 \quad x > 0 , \quad y > 0$$ The point \(P\) with coordinates ( \(2 \mathrm { e } ^ { - 1 } , 2\) ) lies on \(C\).
The tangent to \(C\) at \(P\) cuts the \(x\)-axis at the point \(A\) and cuts the \(y\)-axis at the point \(B\).
Given that \(O\) is the origin, find the exact area of triangle \(O A B\), giving your answer in its simplest form. \includegraphics[max width=\textwidth, alt={}]{a377da06-a968-438c-bec2-ae55283dae47-35_4_21_127_2042} L
Edexcel C34 2018 June Q11
9 marks Standard +0.8
11. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a377da06-a968-438c-bec2-ae55283dae47-36_601_1140_242_402} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure}
  1. By writing \(\sec \theta\) as \(\frac { 1 } { \cos \theta }\), show that when \(x = 3 \sec \theta\), $$\frac { \mathrm { d } x } { \mathrm {~d} \theta } = 3 \sec \theta \tan \theta$$ Figure 3 shows a sketch of part of the curve \(C\) with equation $$y = \frac { \sqrt { x ^ { 2 } - 9 } } { x } \quad x \geqslant 3$$ The finite region \(R\), shown shaded in Figure 3, is bounded by the curve \(C\), the \(x\)-axis and the line with equation \(x = 6\)
  2. Use the substitution \(x = 3 \sec \theta\) to find the exact value of the area of \(R\). [Solutions based entirely on graphical or numerical methods are not acceptable.]
Edexcel C34 2018 June Q12
9 marks Standard +0.8
12. (a) Show that $$\cot x - \tan x \equiv 2 \cot 2 x , \quad x \neq 90 n ^ { \circ } , n \in \mathbb { Z }$$ (b) Hence, or otherwise, solve, for \(0 \leqslant \theta < 180 ^ { \circ }\) $$5 + \cot \left( \theta - 15 ^ { \circ } \right) - \tan \left( \theta - 15 ^ { \circ } \right) = 0$$ giving your answers to one decimal place.
[0pt] [Solutions based entirely on graphical or numerical methods are not acceptable.]
Edexcel C34 2018 June Q13
11 marks Standard +0.3
13. (a) Express \(\frac { 1 } { ( 4 - x ) ( 2 - x ) }\) in partial fractions. The mass, \(x\) grams, of a substance at time \(t\) seconds after a chemical reaction starts is modelled by the differential equation
where \(k\) is a constant.
(b) solve the differential equation and show that the solution can be written as $$x = \frac { 4 - 4 \mathrm { e } ^ { 2 k t } } { 1 - 2 \mathrm { e } ^ { 2 k t } }$$ Given that \(k = 0.1\) (c) find the value of \(t\) when \(x = 1\), giving your answer, in seconds, to 3 significant figures. The mass, \(x\) grams, of a substance at time \(t\) seconds after a chemical reaction starts is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = k ( 4 - x ) ( 2 - x ) , \quad t \geqslant 0,0 \leqslant x < 2$$ where \(k\) is a constant. $$\text { Given that when } t = 0 , x = 0$$ (b) solve the differential equation and show that the solution can be written as
Edexcel C34 2018 June Q14
12 marks Standard +0.3
14. Given that $$y = \frac { \left( x ^ { 2 } - 4 \right) ^ { \frac { 1 } { 2 } } } { x ^ { 3 } } \quad x > 2$$
  1. show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { A x ^ { 2 } + 12 } { x ^ { 4 } \left( x ^ { 2 } - 4 \right) ^ { \frac { 1 } { 2 } } } \quad x > 2$$ where \(A\) is a constant to be found. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{a377da06-a968-438c-bec2-ae55283dae47-48_593_1134_865_395} \captionsetup{labelformat=empty} \caption{Figure 4}
    \end{figure} Figure 4 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\) where $$\mathrm { f } ( x ) = \frac { 24 \left( x ^ { 2 } - 4 \right) ^ { \frac { 1 } { 2 } } } { x ^ { 3 } } \quad x > 2$$
  2. Use your answer to part (a) to find the range of f.
  3. State a reason why f-1 does not exist.