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Edexcel C4 2006 June Q7
15 marks Standard +0.3
7.
\includegraphics[max width=\textwidth, alt={}]{c0c6303b-f527-4e68-91bc-5c9c6ffa8de8-11_209_212_214_863}
At time \(t\) seconds the length of the side of a cube is \(x \mathrm {~cm}\), the surface area of the cube is \(S \mathrm {~cm} ^ { 2 }\), and the volume of the cube is \(V \mathrm {~cm} ^ { 3 }\). The surface area of the cube is increasing at a constant rate of \(8 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }\).
Show that
  1. \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { k } { x }\), where \(k\) is a constant to be found,
  2. \(\frac { \mathrm { d } V } { \mathrm {~d} t } = 2 V ^ { \frac { 1 } { 3 } }\). Given that \(V = 8\) when \(t = 0\),
  3. solve the differential equation in part (b), and find the value of \(t\) when \(V = 16 \sqrt { } 2\).
Edexcel C4 2007 June Q1
5 marks Moderate -0.3
1. $$f ( x ) = ( 3 + 2 x ) ^ { - 3 } , \quad | x | < \frac { 3 } { 2 }$$ Find the binomial expansion of \(\mathrm { f } ( x )\), in ascending powers of \(x\), as far as the term in \(x ^ { 3 }\).
Give each coefficient as a simplified fraction.
(5)
Edexcel C4 2007 June Q2
6 marks Standard +0.3
2. Use the substitution \(u = 2 ^ { x }\) to find the exact value of $$\int _ { 0 } ^ { 1 } \frac { 2 ^ { x } } { \left( 2 ^ { x } + 1 \right) ^ { 2 } } d x$$
Edexcel C4 2007 June Q3
7 marks Standard +0.3
3. (a) Find \(\int x \cos 2 x d x\).
(b) Hence, using the identity \(\cos 2 x = 2 \cos ^ { 2 } x - 1\), deduce \(\int x \cos ^ { 2 } x \mathrm {~d} x\).
Edexcel C4 2007 June Q4
10 marks Standard +0.3
4. $$\frac { 2 \left( 4 x ^ { 2 } + 1 \right) } { ( 2 x + 1 ) ( 2 x - 1 ) } \equiv A + \frac { B } { ( 2 x + 1 ) } + \frac { C } { ( 2 x - 1 ) } .$$
  1. Find the values of the constants \(A , B\) and \(C\).
  2. Hence show that the exact value of \(\int _ { 1 } ^ { 2 } \frac { 2 \left( 4 x ^ { 2 } + 1 \right) } { ( 2 x + 1 ) ( 2 x - 1 ) } \mathrm { d } x\) is \(2 + \ln k\), giving the value of the constant \(k\).
Edexcel C4 2007 June Q5
10 marks Standard +0.3
5. The line \(l _ { 1 }\) has equation \(\mathbf { r } = \left( \begin{array} { r } 1 \\ 0 \\ - 1 \end{array} \right) + \lambda \left( \begin{array} { l } 1 \\ 1 \\ 0 \end{array} \right)\). The line \(l _ { 2 }\) has equation \(\mathbf { r } = \left( \begin{array} { l } 1 \\ 3 \\ 6 \end{array} \right) + \mu \left( \begin{array} { r } 2 \\ 1 \\ - 1 \end{array} \right)\).
  1. Show that \(l _ { 1 }\) and \(l _ { 2 }\) do not meet. The point \(A\) is on \(l _ { 1 }\) where \(\lambda = 1\), and the point \(B\) is on \(l _ { 2 }\) where \(\mu = 2\).
  2. Find the cosine of the acute angle between \(A B\) and \(l _ { 1 }\).
Edexcel C4 2007 June Q6
12 marks Standard +0.3
6. A curve has parametric equations $$x = \tan ^ { 2 } t , \quad y = \sin t , \quad 0 < t < \frac { \pi } { 2 }$$
  1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\). You need not simplify your answer.
  2. Find an equation of the tangent to the curve at the point where \(t = \frac { \pi } { 4 }\). Give your answer in the form \(y = a x + b\), where \(a\) and \(b\) are constants to be determined.
  3. Find a cartesian equation of the curve in the form \(y ^ { 2 } = \mathrm { f } ( x )\).
    \section*{LU}
Edexcel C4 2007 June Q7
11 marks Standard +0.3
7. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3b73fe78-cc47-4615-9cfb-0b8d9ec0ffda-09_627_606_244_667} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows part of the curve with equation \(y = \sqrt { } ( \tan x )\). The finite region \(R\), which is bounded by the curve, the \(x\)-axis and the line \(x = \frac { \pi } { 4 }\), is shown shaded in Figure 1.
  1. Given that \(y = \sqrt { } ( \tan x )\), complete the table with the values of \(y\) corresponding to \(x = \frac { \pi } { 16 } , \frac { \pi } { 8 }\) and \(\frac { 3 \pi } { 16 }\), giving your answers to 5 decimal places.
    \(x\)0\(\frac { \pi } { 16 }\)\(\frac { \pi } { 8 }\)\(\frac { 3 \pi } { 16 }\)\(\frac { \pi } { 4 }\)
    \(y\)01
  2. Use the trapezium rule with all the values of \(y\) in the completed table to obtain an estimate for the area of the shaded region \(R\), giving your answer to 4 decimal places. The region \(R\) is rotated through \(2 \pi\) radians around the \(x\)-axis to generate a solid of revolution.
  3. Use integration to find an exact value for the volume of the solid generated. \section*{LO}
Edexcel C4 2007 June Q8
14 marks Standard +0.3
8. A population growth is modelled by the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = k P ,$$ where \(P\) is the population, \(t\) is the time measured in days and \(k\) is a positive constant.
Given that the initial population is \(P _ { 0 }\),
  1. solve the differential equation, giving \(P\) in terms of \(P _ { 0 } , k\) and \(t\). Given also that \(k = 2.5\),
  2. find the time taken, to the nearest minute, for the population to reach \(2 P _ { 0 }\). In an improved model the differential equation is given as $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \lambda P \cos \lambda t$$ where \(P\) is the population, \(t\) is the time measured in days and \(\lambda\) is a positive constant.
    Given, again, that the initial population is \(P _ { 0 }\) and that time is measured in days,
  3. solve the second differential equation, giving \(P\) in terms of \(P _ { 0 } , \lambda\) and \(t\). Given also that \(\lambda = 2.5\),
  4. find the time taken, to the nearest minute, for the population to reach \(2 P _ { 0 }\) for the first time, using the improved model.
Edexcel C4 2008 June Q1
4 marks Moderate -0.8
1. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fb1924cc-9fa3-4fde-ba4d-6fb095f7f70b-02_519_451_210_749} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows part of the curve with equation \(y = \mathrm { e } ^ { 0.5 x ^ { 2 } }\). The finite region \(R\), shown shaded in Figure 1, is bounded by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 2\).
  1. Complete the table with the values of \(y\) corresponding to \(x = 0.8\) and \(x = 1.6\).
    \(x\)00.40.81.21.62
    \(y\)\(\mathrm { e } ^ { 0 }\)\(\mathrm { e } ^ { 0.08 }\)\(\mathrm { e } ^ { 0.72 }\)\(\mathrm { e } ^ { 2 }\)
  2. Use the trapezium rule with all the values in the table to find an approximate value for the area of \(R\), giving your answer to 4 significant figures.
Edexcel C4 2008 June Q2
6 marks Moderate -0.3
2. (a) Use integration by parts to find \(\int x \mathrm { e } ^ { x } \mathrm {~d} x\).
(b) Hence find \(\int x ^ { 2 } \mathrm { e } ^ { x } \mathrm {~d} x\).
Edexcel C4 2008 June Q3
8 marks Standard +0.3
3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fb1924cc-9fa3-4fde-ba4d-6fb095f7f70b-04_444_705_205_623} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Figure 2 shows a right circular cylindrical metal rod which is expanding as it is heated. After \(t\) seconds the radius of the rod is \(x \mathrm {~cm}\) and the length of the rod is \(5 x \mathrm {~cm}\). The cross-sectional area of the rod is increasing at the constant rate of \(0.032 \mathrm {~cm} ^ { 2 } \mathrm {~s} ^ { - 1 }\).
  1. Find \(\frac { \mathrm { d } x } { \mathrm {~d} t }\) when the radius of the rod is 2 cm , giving your answer to 3 significant figures.
  2. Find the rate of increase of the volume of the rod when \(x = 2\).
    \section*{LU}
Edexcel C4 2008 June Q4
9 marks Standard +0.3
4. A curve has equation \(3 x ^ { 2 } - y ^ { 2 } + x y = 4\). The points \(P\) and \(Q\) lie on the curve. The gradient of the tangent to the curve is \(\frac { 8 } { 3 }\) at \(P\) and at \(Q\).
  1. Use implicit differentiation to show that \(y - 2 x = 0\) at \(P\) and at \(Q\).
  2. Find the coordinates of \(P\) and \(Q\).
Edexcel C4 2008 June Q5
9 marks Standard +0.3
5. (a) Expand \(\frac { 1 } { \sqrt { } ( 4 - 3 x ) }\), where \(| x | < \frac { 4 } { 3 }\), in ascending powers of \(x\) up to and including the term in \(x ^ { 2 }\). Simplify each term.
(b) Hence, or otherwise, find the first 3 terms in the expansion of \(\frac { x + 8 } { \sqrt { } ( 4 - 3 x ) }\) as a series in ascending powers of \(x\).
Edexcel C4 2008 June Q6
12 marks Challenging +1.2
6. With respect to a fixed origin \(O\), the lines \(l _ { 1 }\) and \(l _ { 2 }\) are given by the equations $$\begin{array} { l l } l _ { 1 } : & \mathbf { r } = ( - 9 \mathbf { i } + 10 \mathbf { k } ) + \lambda ( 2 \mathbf { i } + \mathbf { j } - \mathbf { k } ) \\ l _ { 2 } : & \mathbf { r } = ( 3 \mathbf { i } + \mathbf { j } + 17 \mathbf { k } ) + \mu ( 3 \mathbf { i } - \mathbf { j } + 5 \mathbf { k } ) \end{array}$$ where \(\lambda\) and \(\mu\) are scalar parameters.
  1. Show that \(l _ { 1 }\) and \(l _ { 2 }\) meet and find the position vector of their point of intersection.
  2. Show that \(l _ { 1 }\) and \(l _ { 2 }\) are perpendicular to each other. The point \(A\) has position vector \(5 \mathbf { i } + 7 \mathbf { j } + 3 \mathbf { k }\).
  3. Show that \(A\) lies on \(l _ { 1 }\). The point \(B\) is the image of \(A\) after reflection in the line \(l _ { 2 }\).
  4. Find the position vector of \(B\).
Edexcel C4 2008 June Q7
11 marks Standard +0.8
7. (a) Express \(\frac { 2 } { 4 - y ^ { 2 } }\) in partial fractions.
(b) Hence obtain the solution of $$2 \cot x \frac { \mathrm {~d} y } { \mathrm {~d} x } = \left( 4 - y ^ { 2 } \right)$$ for which \(y = 0\) at \(x = \frac { \pi } { 3 }\), giving your answer in the form \(\sec ^ { 2 } x = \mathrm { g } ( y )\).
Edexcel C4 2008 June Q8
16 marks Standard +0.3
8. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fb1924cc-9fa3-4fde-ba4d-6fb095f7f70b-11_639_972_228_484} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows the curve \(C\) with parametric equations $$x = 8 \cos t , \quad y = 4 \sin 2 t , \quad 0 \leqslant t \leqslant \frac { \pi } { 2 } .$$ The point \(P\) lies on \(C\) and has coordinates \(( 4,2 \sqrt { } 3 )\).
  1. Find the value of \(t\) at the point \(P\). The line \(l\) is a normal to \(C\) at \(P\).
  2. Show that an equation for \(l\) is \(y = - x \sqrt { 3 } + 6 \sqrt { 3 }\). The finite region \(R\) is enclosed by the curve \(C\), the \(x\)-axis and the line \(x = 4\), as shown shaded in Figure 3.
  3. Show that the area of \(R\) is given by the integral \(\int _ { \frac { \pi } { 3 } } ^ { \frac { \pi } { 2 } } 64 \sin ^ { 2 } t \cos t \mathrm {~d} t\).
  4. Use this integral to find the area of \(R\), giving your answer in the form \(a + b \sqrt { } 3\), where \(a\) and \(b\) are constants to be determined.
Edexcel C4 2009 June Q1
6 marks Standard +0.3
1. $$f ( x ) = \frac { 1 } { \sqrt { ( 4 + x ) } } , \quad | x | < 4$$ Find the binomial expansion of \(\mathrm { f } ( x )\) in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\). Give each coefficient as a simplified fraction.
(6)
Edexcel C4 2009 June Q2
8 marks Moderate -0.3
2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c2622c33-9436-4254-a728-10ba4703a28c-03_655_1079_207_427} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows the finite region \(R\) bounded by the \(x\)-axis, the \(y\)-axis and the curve with equation \(y = 3 \cos \left( \frac { x } { 3 } \right) , 0 \leqslant x \leqslant \frac { 3 \pi } { 2 }\).
The table shows corresponding values of \(x\) and \(y\) for \(y = 3 \cos \left( \frac { x } { 3 } \right)\).
\(x\)0\(\frac { 3 \pi } { 8 }\)\(\frac { 3 \pi } { 4 }\)\(\frac { 9 \pi } { 8 }\)\(\frac { 3 \pi } { 2 }\)
\(y\)32.771642.121320
  1. Complete the table above giving the missing value of \(y\) to 5 decimal places.
  2. Using the trapezium rule, with all the values of \(y\) from the completed table, find an approximation for the area of \(R\), giving your answer to 3 decimal places.
  3. Use integration to find the exact area of \(R\).
Edexcel C4 2009 June Q3
10 marks Moderate -0.3
3. $$\mathrm { f } ( x ) = \frac { 4 - 2 x } { ( 2 x + 1 ) ( x + 1 ) ( x + 3 ) } = \frac { A } { 2 x + 1 } + \frac { B } { x + 1 } + \frac { C } { x + 3 }$$
  1. Find the values of the constants \(A , B\) and \(C\).
    1. Hence find \(\int f ( x ) \mathrm { d } x\).
    2. Find \(\int _ { 0 } ^ { 2 } \mathrm { f } ( x ) \mathrm { d } x\) in the form \(\ln k\), where \(k\) is a constant.
Edexcel C4 2009 June Q4
9 marks Standard +0.3
4. The curve \(C\) has the equation \(y \mathrm { e } ^ { - 2 x } = 2 x + y ^ { 2 }\).
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). The point \(P\) on \(C\) has coordinates \(( 0,1 )\).
  2. Find the equation of the normal to \(C\) at \(P\), giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.
Edexcel C4 2009 June Q5
10 marks Standard +0.3
5. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c2622c33-9436-4254-a728-10ba4703a28c-09_735_1222_205_358} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} Figure 2 shows a sketch of the curve with parametric equations $$x = 2 \cos 2 t , \quad y = 6 \sin t , \quad 0 \leqslant t \leqslant \frac { \pi } { 2 }$$
  1. Find the gradient of the curve at the point where \(t = \frac { \pi } { 3 }\).
  2. Find a cartesian equation of the curve in the form $$y = \mathrm { f } ( x ) , \quad - k \leqslant x \leqslant k$$ stating the value of the constant \(k\).
  3. Write down the range of \(\mathrm { f } ( x )\).
Edexcel C4 2009 June Q6
8 marks Standard +0.3
6. (a) Find \(\int \sqrt { } ( 5 - x ) \mathrm { d } x\).
(2) \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c2622c33-9436-4254-a728-10ba4703a28c-11_503_1270_370_335} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows a sketch of the curve with equation $$y = ( x - 1 ) \sqrt { } ( 5 - x ) , \quad 1 \leqslant x \leqslant 5$$ (b) (i) Using integration by parts, or otherwise, find $$\int ( x - 1 ) \sqrt { } ( 5 - x ) \mathrm { d } x$$ (ii) Hence find \(\int _ { 1 } ^ { 5 } ( x - 1 ) \sqrt { } ( 5 - x ) \mathrm { d } x\).
Edexcel C4 2009 June Q7
14 marks Standard +0.3
7. Relative to a fixed origin \(O\), the point \(A\) has position vector \(( 8 \mathbf { i } + 13 \mathbf { j } - 2 \mathbf { k } )\), the point \(B\) has position vector ( \(10 \mathbf { i } + 14 \mathbf { j } - 4 \mathbf { k }\) ), and the point \(C\) has position vector \(( 9 \mathbf { i } + 9 \mathbf { j } + 6 \mathbf { k } )\). The line \(l\) passes through the points \(A\) and \(B\).
  1. Find a vector equation for the line \(l\).
  2. Find \(| \overrightarrow { C B } |\).
  3. Find the size of the acute angle between the line segment \(C B\) and the line \(l\), giving your answer in degrees to 1 decimal place.
  4. Find the shortest distance from the point \(C\) to the line \(l\). The point \(X\) lies on \(l\). Given that the vector \(\overrightarrow { C X }\) is perpendicular to \(l\),
  5. find the area of the triangle \(C X B\), giving your answer to 3 significant figures.
Edexcel C4 2009 June Q8
10 marks Standard +0.8
8. (a) Using the identity \(\cos 2 \theta = 1 - 2 \sin ^ { 2 } \theta\), find \(\int \sin ^ { 2 } \theta \mathrm {~d} \theta\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c2622c33-9436-4254-a728-10ba4703a28c-15_516_580_383_680} \captionsetup{labelformat=empty} \caption{Figure 4}
\end{figure} Figure 4 shows part of the curve \(C\) with parametric equations $$x = \tan \theta , \quad y = 2 \sin 2 \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$ The finite shaded region \(S\) shown in Figure 4 is bounded by \(C\), the line \(x = \frac { 1 } { \sqrt { 3 } }\) and the \(x\)-axis. This shaded region is rotated through \(2 \pi\) radians about the \(x\)-axis to form a solid of revolution.
(b) Show that the volume of the solid of revolution formed is given by the integral $$k \int _ { 0 } ^ { \frac { \pi } { 6 } } \sin ^ { 2 } \theta \mathrm {~d} \theta$$ where \(k\) is a constant.
(c) Hence find the exact value for this volume, giving your answer in the form \(p \pi ^ { 2 } + q \pi \sqrt { } 3\), where \(p\) and \(q\) are constants.