Questions C4 (1162 questions)

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Edexcel C4 2005 June Q7
  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
  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
  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
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
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
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
  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
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 \(\frac { 3 } { 2 } \ln 3\).
Edexcel C4 2006 June Q7
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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.