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Edexcel C4 2013 January Q6
9 marks Standard +0.3
6. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a98d4a7f-1e6d-4294-9b5c-c945e8fbe83e-09_862_1534_219_205} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows a sketch of part of the curve with equation \(y = 1 - 2 \cos x\), where \(x\) is measured in radians. The curve crosses the \(x\)-axis at the point \(A\) and at the point \(B\).
  1. Find, in terms of \(\pi\), the \(x\) coordinate of the point \(A\) and the \(x\) coordinate of the point \(B\). The finite region \(S\) enclosed by the curve and the \(x\)-axis is shown shaded in Figure 3. The region \(S\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
  2. Find, by integration, the exact value of the volume of the solid generated.
Edexcel C4 2013 January Q7
14 marks Standard +0.3
7. With respect to a fixed origin \(O\), the lines \(l _ { 1 }\) and \(l _ { 2 }\) are given by the equations $$\begin{aligned} & l _ { 1 } : \mathbf { r } = ( 9 \mathbf { i } + 13 \mathbf { j } - 3 \mathbf { k } ) + \lambda ( \mathbf { i } + 4 \mathbf { j } - 2 \mathbf { k } ) \\ & l _ { 2 } : \mathbf { r } = ( 2 \mathbf { i } - \mathbf { j } + \mathbf { k } ) + \mu ( 2 \mathbf { i } + \mathbf { j } + \mathbf { k } ) \end{aligned}$$ where \(\lambda\) and \(\mu\) are scalar parameters.
  1. Given that \(l _ { 1 }\) and \(l _ { 2 }\) meet, find the position vector of their point of intersection.
  2. Find the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\), giving your answer in degrees to 1 decimal place. Given that the point \(A\) has position vector \(4 \mathbf { i } + 16 \mathbf { j } - 3 \mathbf { k }\) and that the point \(P\) lies on \(l _ { 1 }\) such that \(A P\) is perpendicular to \(l _ { 1 }\),
  3. find the exact coordinates of \(P\).
Edexcel C4 2013 January Q8
9 marks Standard +0.3
8. A bottle of water is put into a refrigerator. The temperature inside the refrigerator remains constant at \(3 ^ { \circ } \mathrm { C }\) and \(t\) minutes after the bottle is placed in the refrigerator the temperature of the water in the bottle is \(\theta ^ { \circ } \mathrm { C }\). The rate of change of the temperature of the water in the bottle is modelled by the differential equation, $$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = \frac { ( 3 - \theta ) } { 125 }$$
  1. By solving the differential equation, show that, $$\theta = A \mathrm { e } ^ { - 0.008 t } + 3$$ where \(A\) is a constant. Given that the temperature of the water in the bottle when it was put in the refrigerator was \(16 ^ { \circ } \mathrm { C }\),
  2. find the time taken for the temperature of the water in the bottle to fall to \(10 ^ { \circ } \mathrm { C }\), giving your answer to the nearest minute.
Edexcel C4 2014 January Q1
8 marks Standard +0.3
  1. (a) Find the binomial expansion of
$$\frac { 1 } { ( 4 + 3 x ) ^ { 3 } } , \quad | x | < \frac { 4 } { 3 }$$ in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\).
Give each coefficient as a simplified fraction. In the binomial expansion of $$\frac { 1 } { ( 4 - 9 x ) ^ { 3 } } , \quad | x | < \frac { 4 } { 9 }$$ the coefficient of \(x ^ { 2 }\) is \(A\).
(b) Using your answer to part (a), or otherwise, find the value of \(A\). Give your answer as a simplified fraction.
Edexcel C4 2014 January Q2
10 marks Standard +0.3
2. (i) Find $$\int x \cos \left( \frac { x } { 2 } \right) \mathrm { d } x$$ (ii) (a) Express \(\frac { 1 } { x ^ { 2 } ( 1 - 3 x ) }\) in partial fractions.
(b) Hence find, for \(0 < x < \frac { 1 } { 3 }\) $$\int \frac { 1 } { x ^ { 2 } ( 1 - 3 x ) } \mathrm { d } x$$
Edexcel C4 2014 January Q3
7 marks Moderate -0.3
  1. The number of bacteria, \(N\), present in a liquid culture at time \(t\) hours after the start of a scientific study is modelled by the equation
$$N = 5000 ( 1.04 ) ^ { t } , \quad t \geqslant 0$$ where \(N\) is a continuous function of \(t\).
  1. Find the number of bacteria present at the start of the scientific study.
  2. Find the percentage increase in the number of bacteria present from \(t = 0\) to \(t = 2\) Given that \(N = 15000\) when \(t = T\),
  3. find the value of \(\frac { \mathrm { d } N } { \mathrm {~d} t }\) when \(t = T\), giving your answer to 3 significant figures.
Edexcel C4 2014 January Q4
11 marks Standard +0.2
4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{245bbe52-3a14-4494-af17-7711caf79b22-10_752_1182_226_395} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \frac { 4 \mathrm { e } ^ { - x } } { 3 \sqrt { } \left( 1 + 3 \mathrm { e } ^ { - x } \right) }\) The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis, the line \(x = - 3 \ln 2\) and the \(y\)-axis. The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 4 \mathrm { e } ^ { - x } } { 3 \sqrt { } \left( 1 + 3 \mathrm { e } ^ { - x } \right) }\)
\(x\)\(- 3 \ln 2\)\(- 2 \ln 2\)\(- \ln 2\)0
\(y\)2.13331.00790.6667
  1. Complete the table above by giving the missing value of \(y\) to 4 decimal places.
  2. Use the trapezium rule, with all the values of \(y\) in the completed table, to obtain an estimate for the area of \(R\), giving your answer to 2 decimal places.
    1. Using the substitution \(u = 1 + 3 \mathrm { e } ^ { - x }\), or otherwise, find $$\int \frac { 4 \mathrm { e } ^ { - x } } { 3 \sqrt { } \left( 1 + 3 \mathrm { e } ^ { - x } \right) } \mathrm { d } x$$
    2. Hence find the value of the area of \(R\).
Edexcel C4 2014 January Q5
6 marks Moderate -0.3
  1. Given that \(y = 2\) at \(x = \frac { \pi } { 8 }\), solve the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 3 y ^ { 2 } } { 2 \sin ^ { 2 } 2 x }$$ giving your answer in the form \(y = \mathrm { f } ( x )\). \includegraphics[max width=\textwidth, alt={}, center]{245bbe52-3a14-4494-af17-7711caf79b22-17_81_102_2649_1779}
Edexcel C4 2014 January Q6
5 marks Moderate -0.3
6. Oil is leaking from a storage container onto a flat section of concrete at a rate of \(0.48 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). The leaking oil spreads to form a pool with an increasing circular cross-section. The pool has a constant uniform thickness of 3 mm . Find the rate at which the radius \(r\) of the pool of oil is increasing at the instant when \(r = 5 \mathrm {~cm}\). Give your answer, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), to 3 significant figures. \includegraphics[max width=\textwidth, alt={}, center]{245bbe52-3a14-4494-af17-7711caf79b22-19_104_95_2617_1786}
Edexcel C4 2014 January Q7
13 marks Challenging +1.2
7. The curve \(C\) has parametric equations $$x = 2 \cos t , \quad y = \sqrt { 3 } \cos 2 t , \quad 0 \leqslant t \leqslant \pi$$ where \(t\) is a parameter.
  1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\). The point \(P\) lies on \(C\) where \(t = \frac { 2 \pi } { 3 }\) The line \(l\) is a normal to \(C\) at \(P\).
  2. Show that an equation for \(l\) is $$2 x - 2 \sqrt { 3 } y - 1 = 0$$ The line \(l\) intersects the curve \(C\) again at the point \(Q\).
  3. Find the exact coordinates of \(Q\). You must show clearly how you obtained your answers. \includegraphics[max width=\textwidth, alt={}, center]{245bbe52-3a14-4494-af17-7711caf79b22-23_106_63_2595_1882}
Edexcel C4 2014 January Q8
15 marks Standard +0.3
8. With respect to a fixed origin \(O\), the lines \(l _ { 1 }\) and \(l _ { 2 }\) are given by the equations $$l _ { 1 } : \mathbf { r } = \left( \begin{array} { r } 2 \\ - 3 \\ 4 \end{array} \right) + \lambda \left( \begin{array} { r } - 1 \\ 2 \\ 1 \end{array} \right) , \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { r } 2 \\ - 3 \\ 4 \end{array} \right) + \mu \left( \begin{array} { r } 5 \\ - 2 \\ 5 \end{array} \right)$$ where \(\lambda\) and \(\mu\) are scalar parameters.
  1. Find, to the nearest \(0.1 ^ { \circ }\), the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\) The point \(A\) has position vector \(\left( \begin{array} { l } 0 \\ 1 \\ 6 \end{array} \right)\).
  2. Show that \(A\) lies on \(l _ { 1 }\) The lines \(l _ { 1 }\) and \(l _ { 2 }\) intersect at the point \(X\).
  3. Write down the coordinates of \(X\).
  4. Find the exact value of the distance \(A X\). The distinct points \(B _ { 1 }\) and \(B _ { 2 }\) both lie on the line \(l _ { 2 }\) Given that \(A X = X B _ { 1 } = X B _ { 2 }\)
  5. find the area of the triangle \(A B _ { 1 } B _ { 2 }\) giving your answer to 3 significant figures. Given that the \(x\) coordinate of \(B _ { 1 }\) is positive,
  6. find the exact coordinates of \(B _ { 1 }\) and the exact coordinates of \(B _ { 2 }\) \includegraphics[max width=\textwidth, alt={}, center]{245bbe52-3a14-4494-af17-7711caf79b22-28_96_59_2478_1834}
Edexcel C4 2005 June Q1
5 marks Moderate -0.3
  1. Use the binomial theorem to expand
$$\sqrt { } ( 4 - 9 x ) , \quad | x | < \frac { 4 } { 9 }$$ in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\), simplifying each term.
Edexcel C4 2005 June Q2
7 marks Standard +0.3
2. A curve has equation $$x ^ { 2 } + 2 x y - 3 y ^ { 2 } + 16 = 0 .$$ Find the coordinates of the points on the curve where \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\).
Edexcel C4 2005 June Q3
8 marks Moderate -0.3
3. (a) Express \(\frac { 5 x + 3 } { ( 2 x - 3 ) ( x + 2 ) }\) in partial fractions.
(b) Hence find the exact value of \(\int _ { 2 } ^ { 6 } \frac { 5 x + 3 } { ( 2 x - 3 ) ( x + 2 ) } \mathrm { d } x\), giving your answer as a single logarithm.
Edexcel C4 2005 June Q4
7 marks Standard +0.3
4. Use the substitution \(x = \sin \theta\) to find the exact value of $$\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x$$
Edexcel C4 2005 June Q5
10 marks Standard +0.3
5. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{7fa2c564-d1e5-4fd0-a690-e3189daea332-06_586_1079_260_427}
\end{figure} Figure 1 shows the graph of the curve with equation $$y = x \mathrm { e } ^ { 2 x } , \quad x \geqslant 0$$ The finite region \(R\) bounded by the lines \(x = 1\), the \(x\)-axis and the curve is shown shaded in Figure 1.
  1. Use integration to find the exact value for the area of \(R\).
  2. Complete the table with the values of \(y\) corresponding to \(x = 0.4\) and 0.8 .
    \(x\)00.20.40.60.81
    \(y = x \mathrm { e } ^ { 2 x }\)00.298361.992077.38906
  3. Use the trapezium rule with all the values in the table to find an approximate value for this area, giving your answer to 4 significant figures.
Edexcel C4 2005 June Q6
12 marks Standard +0.3
  1. A curve has parametric equations
$$x = 2 \cot t , \quad y = 2 \sin ^ { 2 } t , \quad 0 < t \leqslant \frac { \pi } { 2 }$$
  1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of the parameter \(t\).
  2. Find an equation of the tangent to the curve at the point where \(t = \frac { \pi } { 4 }\).
  3. Find a cartesian equation of the curve in the form \(y = \mathrm { f } ( x )\). State the domain on which the curve is defined.
Edexcel C4 2005 June Q7
13 marks Standard +0.3
  1. The line \(l _ { 1 }\) has vector equation
$$\mathbf { r } = \left( \begin{array} { l } 3 \\ 1 \\ 2 \end{array} \right) + \lambda \left( \begin{array} { r } 1 \\ - 1 \\ 4 \end{array} \right)$$ and the line \(l _ { 2 }\) has vector equation $$\mathbf { r } = \left( \begin{array} { r } 0 \\ 4 \\ - 2 \end{array} \right) + \mu \left( \begin{array} { r } 1 \\ - 1 \\ 0 \end{array} \right) ,$$ where \(\lambda\) and \(\mu\) are parameters.
The lines \(l _ { 1 }\) and \(l _ { 2 }\) intersect at the point \(B\) and the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\) is \(\theta\).
  1. Find the coordinates of \(B\).
  2. Find the value of \(\cos \theta\), giving your answer as a simplified fraction. The point \(A\), which lies on \(l _ { 1 }\), has position vector \(\mathbf { a } = 3 \mathbf { i } + \mathbf { j } + 2 \mathbf { k }\).
    The point \(C\), which lies on \(l _ { 2 }\), has position vector \(\mathbf { c } = 5 \mathbf { i } - \mathbf { j } - 2 \mathbf { k }\).
    The point \(D\) is such that \(A B C D\) is a parallelogram.
  3. Show that \(| \overrightarrow { A B } | = | \overrightarrow { B C } |\).
  4. Find the position vector of the point \(D\).
Edexcel C4 2005 June Q8
13 marks Moderate -0.3
  1. Liquid is pouring into a container at a constant rate of \(20 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) and is leaking out at a rate proportional to the volume of liquid already in the container.
    1. Explain why, at time \(t\) seconds, the volume, \(V \mathrm {~cm} ^ { 3 }\), of liquid in the container satisfies the differential equation
    $$\frac { \mathrm { d } V } { \mathrm {~d} t } = 20 - k V$$ where \(k\) is a positive constant. The container is initially empty.
  2. By solving the differential equation, show that $$V = A + B \mathrm { e } ^ { - k t }$$ giving the values of \(A\) and \(B\) in terms of \(k\). Given also that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = 10\) when \(t = 5\),
  3. find the volume of liquid in the container at 10 s after the start.
Edexcel C4 2006 June Q1
7 marks Moderate -0.3
  1. A curve \(C\) is described by the equation
$$3 x ^ { 2 } - 2 y ^ { 2 } + 2 x - 3 y + 5 = 0$$ Find an equation of the normal to \(C\) at the point ( 0,1 ), giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.
Edexcel C4 2006 June Q2
9 marks Standard +0.3
2. $$f ( x ) = \frac { 3 x - 1 } { ( 1 - 2 x ) ^ { 2 } } , \quad | x | < \frac { 1 } { 2 }$$ Given that, for \(x \neq \frac { 1 } { 2 } , \quad \frac { 3 x - 1 } { ( 1 - 2 x ) ^ { 2 } } = \frac { A } { ( 1 - 2 x ) } + \frac { B } { ( 1 - 2 x ) ^ { 2 } } , \quad\) where \(A\) and \(B\) are constants,
  1. find the values of \(A\) and \(B\).
  2. Hence, or otherwise, find the series expansion of \(\mathrm { f } ( x )\), in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\), simplifying each term.
    (6)
Edexcel C4 2006 June Q3
9 marks Moderate -0.3
3. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{c0c6303b-f527-4e68-91bc-5c9c6ffa8de8-04_423_777_306_569}
\end{figure} The curve with equation \(y = 3 \sin \frac { x } { 2 } , 0 \leqslant x \leqslant 2 \pi\), is shown in Figure 1. The finite region enclosed by the curve and the \(x\)-axis is shaded.
  1. Find, by integration, the area of the shaded region. This region is rotated through \(2 \pi\) radians about the \(x\)-axis.
  2. Find the volume of the solid generated.
Edexcel C4 2006 June Q4
9 marks Moderate -0.3
4. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 2} \includegraphics[alt={},max width=\textwidth]{c0c6303b-f527-4e68-91bc-5c9c6ffa8de8-05_480_1059_313_438}
\end{figure} The curve shown in Figure 2 has parametric equations $$x = \sin t , y = \sin \left( t + \frac { \pi } { 6 } \right) , \quad - \frac { \pi } { 2 } < t < \frac { \pi } { 2 }$$
  1. Find an equation of the tangent to the curve at the point where \(t = \frac { \pi } { 6 }\).
  2. Show that a cartesian equation of the curve is $$y = \frac { \sqrt { } 3 } { 2 } x + \frac { 1 } { 2 } \sqrt { } \left( 1 - x ^ { 2 } \right) , \quad - 1 < x < 1$$
Edexcel C4 2006 June Q5
13 marks Standard +0.3
  1. The point \(A\), with coordinates \(( 0 , a , b )\) lies on the line \(l _ { 1 }\), which has equation
$$\mathbf { r } = 6 \mathbf { i } + 19 \mathbf { j } - \mathbf { k } + \lambda ( \mathbf { i } + 4 \mathbf { j } - 2 \mathbf { k } )$$
  1. Find the values of \(a\) and \(b\). The point \(P\) lies on \(l _ { 1 }\) and is such that \(O P\) is perpendicular to \(l _ { 1 }\), where \(O\) is the origin.
  2. Find the position vector of point \(P\). Given that \(B\) has coordinates \(( 5,15,1 )\),
  3. show that the points \(A , P\) and \(B\) are collinear and find the ratio \(A P : P B\).
Edexcel C4 2006 June Q6
13 marks Moderate -0.3
6. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 3} \includegraphics[alt={},max width=\textwidth]{c0c6303b-f527-4e68-91bc-5c9c6ffa8de8-09_442_805_283_555}
\end{figure} Figure 3 shows a sketch of the curve with equation \(y = ( x - 1 ) \ln x , \quad x > 0\).
  1. Complete the table with the values of \(y\) corresponding to \(x = 1.5\) and \(x = 2.5\).
    \(x\)11.522.53
    \(y\)0\(\ln 2\)\(2 \ln 3\)
    Given that \(I = \int _ { 1 } ^ { 3 } ( x - 1 ) \ln x \mathrm {~d} x\),
  2. use the trapezium rule
    1. with values of \(y\) at \(x = 1,2\) and 3 to find an approximate value for \(I\) to 4 significant figures,
    2. with values of \(y\) at \(x = 1,1.5,2,2.5\) and 3 to find another approximate value for \(I\) to 4 significant figures.
  3. Explain, with reference to Figure 3, why an increase in the number of values improves the accuracy of the approximation.
  4. Show, by integration, that the exact value of \(\int _ { 1 } ^ { 3 } ( x - 1 ) \ln x \mathrm {~d} x\) is \(\frac { 3 } { 2 } \ln 3\).