Questions - Edexcel - A-Level Maths

Edexcel C4 2006 January Q8

12 marks Standard +0.3

8. \begin{figure}[h]

\captionsetup{labelformat=empty} \caption{Figure 2} \includegraphics[alt={},max width=\textwidth]{9bf05d7e-7bb9-40f6-b626-69a8a6eda5a5-10_545_979_285_552}

\end{figure} The curve shown in Figure 2 has parametric equations $$x = t - 2 \sin t , \quad y = 1 - 2 \cos t , \quad 0 \leqslant t \leqslant 2 \pi$$

Show that the curve crosses the $x$-axis where $t = \frac { \pi } { 3 }$ and $t = \frac { 5 \pi } { 3 }$. The finite region $R$ is enclosed by the curve and the $x$-axis, as shown shaded in Figure 2.
Show that the area of $R$ is given by the integral $$\int _ { \frac { \pi } { 3 } } ^ { \frac { 5 \pi } { 3 } } ( 1 - 2 \cos t ) ^ { 2 } \mathrm {~d} t$$
Use this integral to find the exact value of the shaded area.

Edexcel C4 2007 January Q1

5 marks Moderate -0.3

1. $$f ( x ) = ( 2 - 5 x ) ^ { - 2 } , \quad | x | < \frac { 2 } { 5 }$$ Find the binomial expansion of $\mathrm { f } ( x )$, in ascending powers of $x$, as far as the term in $x ^ { 3 }$, giving each coefficient as a simplified fraction.
(5)

Edexcel C4 2007 January Q2

7 marks Standard +0.3

2. \begin{figure}[h]

\captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{d366e541-15f6-4fb5-9afb-faf6120f1a1c-03_502_917_296_548}

\end{figure} The curve with equation $y = \frac { 1 } { 3 ( 1 + 2 x ) } , x > - \frac { 1 } { 2 }$, is shown in Figure 1.
The region bounded by the lines $x = - \frac { 1 } { 4 } , x = \frac { 1 } { 2 }$, the $x$-axis and the curve is shown shaded in Figure 1. This region is rotated through 360 degrees about the $x$-axis.

Use calculus to find the exact value of the volume of the solid generated. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 2} \includegraphics[alt={},max width=\textwidth]{d366e541-15f6-4fb5-9afb-faf6120f1a1c-03_383_447_1411_753}
\end{figure} Figure 2 shows a paperweight with axis of symmetry $A B$ where $A B = 3 \mathrm {~cm}$. $A$ is a point on the top surface of the paperweight, and $B$ is a point on the base of the paperweight. The paperweight is geometrically similar to the solid in part (a).
Find the volume of this paperweight.

Edexcel C4 2007 January Q3

9 marks Standard +0.3

A curve has parametric equations

$$x = 7 \cos t - \cos 7 t , y = 7 \sin t - \sin 7 t , \quad \frac { \pi } { 8 } < t < \frac { \pi } { 3 }$$

Find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$. You need not simplify your answer.
Find an equation of the normal to the curve at the point where $t = \frac { \pi } { 6 }$. Give your answer in its simplest exact form.

Edexcel C4 2007 January Q4

12 marks Standard +0.3

4. (a) Express $\frac { 2 x - 1 } { ( x - 1 ) ( 2 x - 3 ) }$ in partial fractions.
(b) Given that $x \geqslant 2$, find the general solution of the differential equation $$( 2 x - 3 ) ( x - 1 ) \frac { \mathrm { d } y } { \mathrm {~d} x } = ( 2 x - 1 ) y$$ (c) Hence find the particular solution of this differential equation that satisfies $y = 10$ at $x = 2$, giving your answer in the form $y = \mathrm { f } ( x )$.

Edexcel C4 2007 January Q5

7 marks Standard +0.3

5. A set of curves is given by the equation $\sin x + \cos y = 0.5$.

Use implicit differentiation to find an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$. For $- \pi < x < \pi$ and $- \pi < y < \pi$,
find the coordinates of the points where $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$.

Edexcel C4 2007 January Q6

6 marks Moderate -0.3

6. (a) Given that $y = 2 ^ { x }$, and using the result $2 ^ { x } = \mathrm { e } ^ { x \ln 2 }$, or otherwise, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 ^ { x } \ln 2$.
(b) Find the gradient of the curve with equation $y = 2 ^ { \left( x ^ { 2 } \right) }$ at the point with coordinates $( 2,16 )$.

Edexcel C4 2007 January Q7

14 marks Standard +0.3

7. The point $A$ has position vector $\mathbf { a } = 2 \mathbf { i } + 2 \mathbf { j } + \mathbf { k }$ and the point $B$ has position vector $\mathbf { b } = \mathbf { i } + \mathbf { j } - 4 \mathbf { k }$, relative to an origin $O$.

Find the position vector of the point $C$, with position vector $\mathbf { c }$, given by $$\mathbf { c } = \mathbf { a } + \mathbf { b } .$$
Show that $O A C B$ is a rectangle, and find its exact area. The diagonals of the rectangle, $A B$ and $O C$, meet at the point $D$.
Write down the position vector of the point $D$.
Find the size of the angle $A D C$.

Edexcel C4 2007 January Q8

15 marks Standard +0.3

8. $$I = \int _ { 0 } ^ { 5 } \mathrm { e } ^ { \sqrt { } ( 3 x + 1 ) } \mathrm { d } x$$

Given that $y = \mathrm { e } ^ { \sqrt { } ( 3 x + 1 ) }$, complete the table with the values of $y$ corresponding to $x = 2$, 3 and 4.
$x$ 0 1 2 3 4 5
$y$ $\mathrm { e } ^ { 1 }$ $\mathrm { e } ^ { 2 }$ $\mathrm { e } ^ { 4 }$
Use the trapezium rule, with all the values of $y$ in the completed table, to obtain an estimate for the original integral $I$, giving your answer to 4 significant figures.
Use the substitution $t = \sqrt { } ( 3 x + 1 )$ to show that $I$ may be expressed as $\int _ { a } ^ { b } k t e ^ { t } \mathrm {~d} t$, giving the values of $a , b$ and $k$.
Use integration by parts to evaluate this integral, and hence find the value of $I$ correct to 4 significant figures, showing all the steps in your working.

Edexcel C4 2008 January Q1

6 marks Moderate -0.8

1. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{ac7d862f-d10d-45ed-9077-ae4c7413cbf6-02_390_675_246_630} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} The curve shown in Figure 1 has equation $y = \mathrm { e } ^ { x } \sqrt { } ( \sin x ) , 0 \leqslant x \leqslant \pi$. The finite region $R$ bounded by the curve and the $x$-axis is shown shaded in Figure 1.

Complete the table below with the values of $y$ corresponding to $x = \frac { \pi } { 4 }$ and $\frac { \pi } { 2 }$, giving your answers to 5 decimal places.
$x$ 0 $\frac { \pi } { 4 }$ $\frac { \pi } { 2 }$ $\frac { 3 \pi } { 4 }$ $\pi$
$y$ 0 8.87207 0
Use the trapezium rule, with all the values in the completed table, to obtain an estimate for the area of the region $R$. Give your answer to 4 decimal places.

Edexcel C4 2008 January Q2

7 marks Standard +0.3

2. (a) Use the binomial theorem to expand $$( 8 - 3 x ) ^ { \frac { 1 } { 3 } } , \quad | x | < \frac { 8 } { 3 }$$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$, giving each term as a simplified fraction.
(b) Use your expansion, with a suitable value of $x$, to obtain an approximation to $\sqrt [ 3 ] { } ( 7.7 )$. Give your answer to 7 decimal places.

Edexcel C4 2008 January Q3

5 marks Standard +0.2

3. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{ac7d862f-d10d-45ed-9077-ae4c7413cbf6-04_493_490_278_712} \captionsetup{labelformat=empty} \caption{Figure 2}

\end{figure} The curve shown in Figure 2 has equation $y = \frac { 1 } { ( 2 x + 1 ) }$. The finite region bounded by the curve, the $x$-axis and the lines $x = a$ and $x = b$ is shown shaded in Figure 2. This region is rotated through $360 ^ { \circ }$ about the $x$-axis to generate a solid of revolution. Find the volume of the solid generated. Express your answer as a single simplified fraction, in terms of $a$ and $b$.

Edexcel C4 2008 January Q4

9 marks Moderate -0.3

4. (i) Find $\int \ln \left( \frac { x } { 2 } \right) \mathrm { d } x$.
(ii) Find the exact value of $\int _ { \frac { \pi } { 4 } } ^ { \frac { \pi } { 2 } } \sin ^ { 2 } x \mathrm {~d} x$.

Edexcel C4 2008 January Q5

9 marks Standard +0.3

5. A curve is described by the equation $$x ^ { 3 } - 4 y ^ { 2 } = 12 x y$$

Find the coordinates of the two points on the curve where $x = - 8$.
Find the gradient of the curve at each of these points.

Edexcel C4 2008 January Q6

11 marks Standard +0.3

6. The points $A$ and $B$ have position vectors $2 \mathbf { i } + 6 \mathbf { j } - \mathbf { k }$ and $3 \mathbf { i } + 4 \mathbf { j } + \mathbf { k }$ respectively. The line $l _ { 1 }$ passes through the points $A$ and $B$.

Find the vector $\overrightarrow { A B }$.
Find a vector equation for the line $l _ { 1 }$. A second line $l _ { 2 }$ passes through the origin and is parallel to the vector $\mathbf { i } + \mathbf { k }$. The line $l _ { 1 }$ meets the line $l _ { 2 }$ at the point $C$.
Find the acute angle between $l _ { 1 }$ and $l _ { 2 }$.
Find the position vector of the point $C$.

Edexcel C4 2008 January Q7

15 marks Standard +0.3

7. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{ac7d862f-d10d-45ed-9077-ae4c7413cbf6-09_559_864_255_530} \captionsetup{labelformat=empty} \caption{Figure 3}

\end{figure} The curve $C$ has parametric equations $$x = \ln ( t + 2 ) , \quad y = \frac { 1 } { ( t + 1 ) } , \quad t > - 1$$ The finite region $R$ between the curve $C$ and the $x$-axis, bounded by the lines with equations $x = \ln 2$ and $x = \ln 4$, is shown shaded in Figure 3.

Show that the area of $R$ is given by the integral $$\int _ { 0 } ^ { 2 } \frac { 1 } { ( t + 1 ) ( t + 2 ) } \mathrm { d } t$$
Hence find an exact value for this area.
Find a cartesian equation of the curve $C$, in the form $y = \mathrm { f } ( x )$.
State the domain of values for $x$ for this curve. $\_\_\_\_$}

Edexcel C4 2008 January Q8

13 marks Standard +0.3

8. Liquid is pouring into a large vertical circular cylinder at a constant rate of $1600 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$ and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is $4000 \mathrm {~cm} ^ { 2 }$.

Show that at time $t$ seconds, the height $h \mathrm {~cm}$ of liquid in the cylinder satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - k \sqrt { } h \text {, where } k \text { is a positive constant. }$$ When $h = 25$, water is leaking out of the hole at $400 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$.
Show that $k = 0.02$
Separate the variables of the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.4 - 0.02 \sqrt { } h$$ to show that the time taken to fill the cylinder from empty to a height of 100 cm is given by $$\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { } h } \mathrm {~d} h$$ Using the substitution $h = ( 20 - x ) ^ { 2 }$, or otherwise,
find the exact value of $\int _ { 0 } ^ { 100 } \frac { 50 } { 20 - \sqrt { h } } \mathrm {~d} h$.
Hence find the time taken to fill the cylinder from empty to a height of 100 cm , giving your answer in minutes and seconds to the nearest second.

Edexcel C4 2009 January Q1

7 marks Moderate -0.3

A curve $C$ has the equation $y ^ { 2 } - 3 y = x ^ { 3 } + 8$.

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
Hence find the gradient of $C$ at the point where $y = 3$.

Edexcel C4 2009 January Q3

14 marks Standard +0.3

3. $$f ( x ) = \frac { 27 x ^ { 2 } + 32 x + 16 } { ( 3 x + 2 ) ^ { 2 } ( 1 - x ) } , \quad | x | < \frac { 2 } { 3 }$$ Given that $\mathrm { f } ( x )$ can be expressed in the form $$f ( x ) = \frac { A } { ( 3 x + 2 ) } + \frac { B } { ( 3 x + 2 ) ^ { 2 } } + \frac { C } { ( 1 - x ) }$$

find the values of $B$ and $C$ and show that $A = 0$.
Hence, or otherwise, find the series expansion of $\mathrm { f } ( x )$, in ascending powers of $x$, up to and including the term in $x ^ { 2 }$. Simplify each term.
Find the percentage error made in using the series expansion in part (b) to estimate the value of $\mathrm { f } ( 0.2 )$. Give your answer to 2 significant figures. \section*{LU}

Edexcel C4 2009 January Q4

13 marks Standard +0.3

4. With respect to a fixed origin $O$ the lines $l _ { 1 }$ and $l _ { 2 }$ are given by the equations $$l _ { 1 } : \quad \mathbf { r } = \left( \begin{array} { c }

Edexcel C4 2009 January Q5

7 marks Standard +0.3

5. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{a5579938-e202-4543-8513-6483ede49850-09_696_686_196_626} \captionsetup{labelformat=empty} \caption{Figure 2}

\end{figure} A container is made in the shape of a hollow inverted right circular cone. The height of the container is 24 cm and the radius is 16 cm , as shown in Figure 2. Water is flowing into the container. When the height of water is $h \mathrm {~cm}$, the surface of the water has radius $r \mathrm {~cm}$ and the volume of water is $V \mathrm {~cm} ^ { 3 }$.

Show that $V = \frac { 4 \pi h ^ { 3 } } { 27 }$.
[0pt] [The volume $V$ of a right circular cone with vertical height $h$ and base radius $r$ is given by the formula $V = \frac { 1 } { 3 } \pi r ^ { 2 } h$.] Water flows into the container at a rate of $8 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$.
Find, in terms of $\pi$, the rate of change of $h$ when $h = 12$.

Edexcel C4 2009 January Q6

13 marks Standard +0.3

6. (a) Find $\int \tan ^ { 2 } x \mathrm {~d} x$.
(b) Use integration by parts to find $\int \frac { 1 } { x ^ { 3 } } \ln x \mathrm {~d} x$.
(c) Use the substitution $u = 1 + e ^ { x }$ to show that $$\int \frac { \mathrm { e } ^ { 3 x } } { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } - \mathrm { e } ^ { x } + \ln \left( 1 + \mathrm { e } ^ { x } \right) + k$$ where $k$ is a constant.

Edexcel C4 2009 January Q17

Standard +0.3

17 \end{array} \right) + \lambda \left( \begin{array} { c } - 2
1
- 4 \end{array} \right) \quad l _ { 2 } : \quad \mathbf { r } = \left( \begin{array} { c } - 5
11
p \end{array} \right) + \mu \left( \begin{array} { l } q
2
2 \end{array} \right)$$ where $\lambda$ and $\mu$ are parameters and $p$ and $q$ are constants. Given that $l _ { 1 }$ and $l _ { 2 }$ are perpendicular,

show that $q = - 3$. Given further that $l _ { 1 }$ and $l _ { 2 }$ intersect, find
the value of $p$,
the coordinates of the point of intersection. The point $A$ lies on $l _ { 1 }$ and has position vector $\left( \begin{array} { c } 9 \\ 3 \\ 13 \end{array} \right)$. The point $C$ lies on $l _ { 2 }$.\\ Given that a circle, with centre $C$, cuts the line $l _ { 1 }$ at the points $A$ and $B$,
find the position vector of $B$.\\ 5. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a5579938-e202-4543-8513-6483ede49850-09_696_686_196_626} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} A container is made in the shape of a hollow inverted right circular cone. The height of the container is 24 cm and the radius is 16 cm , as shown in Figure 2. Water is flowing into the container. When the height of water is $h \mathrm {~cm}$, the surface of the water has radius $r \mathrm {~cm}$ and the volume of water is $V \mathrm {~cm} ^ { 3 }$.
Show that $V = \frac { 4 \pi h ^ { 3 } } { 27 }$.\\[0pt] [The volume $V$ of a right circular cone with vertical height $h$ and base radius $r$ is given by the formula $V = \frac { 1 } { 3 } \pi r ^ { 2 } h$.] Water flows into the container at a rate of $8 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$.
Find, in terms of $\pi$, the rate of change of $h$ when $h = 12$. 6. (a) Find $\int \tan ^ { 2 } x \mathrm {~d} x$.
Use integration by parts to find $\int \frac { 1 } { x ^ { 3 } } \ln x \mathrm {~d} x$.
Use the substitution $u = 1 + e ^ { x }$ to show that $$\int \frac { \mathrm { e } ^ { 3 x } } { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } - \mathrm { e } ^ { x } + \ln \left( 1 + \mathrm { e } ^ { x } \right) + k$$ where $k$ is a constant.\\ 7. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a5579938-e202-4543-8513-6483ede49850-13_511_714_237_612} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} The curve $C$ shown in Figure 3 has parametric equations $$x = t ^ { 3 } - 8 t , \quad y = t ^ { 2 }$$ where $t$ is a parameter. Given that the point $A$ has parameter $t = - 1$,
find the coordinates of $A$. The line $l$ is the tangent to $C$ at $A$.
Show that an equation for $l$ is $2 x - 5 y - 9 = 0$. The line $l$ also intersects the curve at the point $B$.
Find the coordinates of $B$.

Edexcel C4 2010 January Q1

9 marks Moderate -0.3

(a) Find the binomial expansion of

$$\sqrt { } ( 1 - 8 x ) , \quad | x | < \frac { 1 } { 8 }$$ in ascending powers of $x$ up to and including the term in $x ^ { 3 }$, simplifying each term.
(b) Show that, when $x = \frac { 1 } { 100 }$, the exact value of $\sqrt { } ( 1 - 8 x )$ is $\frac { \sqrt { } 23 } { 5 }$.
(c) Substitute $x = \frac { 1 } { 100 }$ into the binomial expansion in part (a) and hence obtain an approximation to $\sqrt { } 23$. Give your answer to 5 decimal places.

Edexcel C4 2010 January Q2

13 marks Moderate -0.3

2. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{5ef3ae4a-a06d-48c1-8b79-7d7c3f95d120-03_623_1176_196_374} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} Figure 1 shows a sketch of the curve with equation $y = x \ln x , x \geqslant 1$. The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the $x$-axis and the line $x = 4$. The table shows corresponding values of $x$ and $y$ for $y = x \ln x$.

$x$	1	1.5	2	2.5	3	3.5	4
$y$	0	0.608			3.296	4.385	5.545

Complete the table with the values of $y$ corresponding to $x = 2$ and $x = 2.5$, giving your answers to 3 decimal places.
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. Use integration by parts to find $\int x \ln x \mathrm {~d} x$.
2. Hence find the exact area of $R$, giving your answer in the form $\frac { 1 } { 4 } ( a \ln 2 + b )$, where $a$ and $b$ are integers.

\(x\)	0	1	2	3	4	5
\(y\)	\(\mathrm { e } ^ { 1 }\)	\(\mathrm { e } ^ { 2 }\)				\(\mathrm { e } ^ { 4 }\)

\(x\)	0	\(\frac { \pi } { 4 }\)	\(\frac { \pi } { 2 }\)	\(\frac { 3 \pi } { 4 }\)	\(\pi\)
\(y\)	0			8.87207	0

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