Questions - Edexcel - A-Level Maths

Edexcel C4 2014 June Q4

10 marks Standard +0.3

4. (a) Express $\frac { 25 } { x ^ { 2 } ( 2 x + 1 ) }$ in partial fractions. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{e14881c1-5ba5-4868-92ee-8bc58d4884dc-06_623_849_408_561} \captionsetup{labelformat=empty} \caption{Figure 2}

\end{figure} Figure 2 shows a sketch of part of the curve $C$ with equation $y = \frac { 5 } { x \sqrt { } ( 2 x + 1 ) } , x > 0$ The finite region $R$ is bounded by the curve $C$, the $x$-axis, the line with equation $x = 1$ and the line with equation $x = 4$ This region is shown shaded in Figure 2 The region $R$ is rotated through $360 ^ { \circ }$ about the $x$-axis.
(b) Use calculus to find the exact volume of the solid of revolution generated, giving your answer in the form $a + b \ln c$, where $a , b$ and $c$ are constants.

Edexcel C4 2014 June Q5

6 marks Moderate -0.3

5. At time $t$ seconds the radius of a sphere is $r \mathrm {~cm}$, its volume is $V \mathrm {~cm} ^ { 3 }$ and its surface area is $S \mathrm {~cm} ^ { 2 }$. [You are given that $V = \frac { 4 } { 3 } \pi r ^ { 3 }$ and that $S = 4 \pi r ^ { 2 }$ ] The volume of the sphere is increasing uniformly at a constant rate of $3 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$.

Find $\frac { \mathrm { d } r } { \mathrm {~d} t }$ when the radius of the sphere is 4 cm , giving your answer to 3 significant figures.
Find the rate at which the surface area of the sphere is increasing when the radius is 4 cm .

Edexcel C4 2014 June Q6

10 marks Standard +0.2

6. With respect to a fixed origin, the point $A$ with position vector $\mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k }$ lies on the line $l _ { 1 }$ with equation $$\mathbf { r } = \left( \begin{array} { l } 1 \\ 2 \\ 3 \end{array} \right) + \lambda \left( \begin{array} { r } 0 \\ 2 \\ - 1 \end{array} \right) , \quad \text { where } \lambda \text { is a scalar parameter, }$$ and the point $B$ with position vector $4 \mathbf { i } + p \mathbf { j } + 3 \mathbf { k }$, where $p$ is a constant, lies on the line $l _ { 2 }$ with equation $$\mathbf { r } = \left( \begin{array} { l }

Edexcel C4 2014 June Q7

10 marks Challenging +1.2

7
0
7 \end{array} \right) + \mu \left( \begin{array} { r } 3
- 5
4 \end{array} \right) , \quad \text { where } \mu \text { is a scalar parameter. }$$

Find the value of the constant $p$.
Show that $l _ { 1 }$ and $l _ { 2 }$ intersect and find the position vector of their point of intersection, $C$.
Find the size of the angle $A C B$, giving your answer in degrees to 3 significant figures.
Find the area of the triangle $A B C$, giving your answer to 3 significant figures.\\ 7. The rate of increase of the number, $N$, of fish in a lake is modelled by the differential equation $$\frac { \mathrm { d } N } { \mathrm {~d} t } = \frac { ( k t - 1 ) ( 5000 - N ) } { t } \quad t > 0 , \quad 0 < N < 5000$$ In the given equation, the time $t$ is measured in years from the start of January 2000 and $k$ is a positive constant.
By solving the differential equation, show that $$N = 5000 - A t \mathrm { e } ^ { - k t }$$ where $A$ is a positive constant. After one year, at the start of January 2001, there are 1200 fish in the lake. After two years, at the start of January 2002, there are 1800 fish in the lake.
Find the exact value of the constant $A$ and the exact value of the constant $k$.
Hence find the number of fish in the lake after five years. Give your answer to the nearest hundred fish.

Edexcel C4 2014 June Q8

12 marks Standard +0.3

8. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{e14881c1-5ba5-4868-92ee-8bc58d4884dc-13_808_965_248_502} \captionsetup{labelformat=empty} \caption{Figure 3}

\end{figure} The curve shown in Figure 3 has parametric equations $$x = t - 4 \sin t , y = 1 - 2 \cos t , \quad - \frac { 2 \pi } { 3 } \leqslant t \leqslant \frac { 2 \pi } { 3 }$$ The point $A$, with coordinates ( $k , 1$ ), lies on the curve. Given that $k > 0$

find the exact value of $k$,
find the gradient of the curve at the point $A$. There is one point on the curve where the gradient is equal to $- \frac { 1 } { 2 }$
Find the value of $t$ at this point, showing each step in your working and giving your answer to 4 decimal places.
[0pt] [Solutions based entirely on graphical or numerical methods are not acceptable.]

Edexcel C4 2014 June Q1

7 marks Standard +0.3

A curve $C$ has the equation

$$x ^ { 3 } + 2 x y - x - y ^ { 3 } - 20 = 0$$

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
Find an equation of the tangent to $C$ at the point $( 3 , - 2 )$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.

Edexcel C4 2014 June Q3

11 marks Moderate -0.3

3
3
\end{tabular}}
\hline \multicolumn{2}{|l|}{\begin{tabular}{l}

Edexcel C4 2014 June Q4

5 marks Moderate -0.3

4
4
\end{tabular}}
\hline 5 &
\hline \multicolumn{2}{|l|}{\begin{tabular}{l}

Edexcel C4 2014 June Q6

12 marks Moderate -0.3

6
6
\end{tabular}}
\hline 7 &
\hline 8 &
\hline &
\hline \multicolumn{2}{|c|}{}
\hline \multicolumn{2}{|c|}{}
\hline \multicolumn{2}{|c|}{}
\hline \multicolumn{2}{|c|}{}
\hline \multicolumn{2}{|c|}{}
\hline \multicolumn{2}{|c|}{}
\hline \multicolumn{2}{|c|}{}
\hline \multicolumn{2}{|c|}{}
\hline \multicolumn{2}{|c|}{}
\hline \multicolumn{2}{|c|}{}
\hline \multicolumn{2}{|c|}{}
\hline \multicolumn{2}{|c|}{}
\hline Total &
\hline \end{tabular} \end{center} Turn over

A curve $C$ has the equation

$$x ^ { 3 } + 2 x y - x - y ^ { 3 } - 20 = 0$$

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
Find an equation of the tangent to $C$ at the point $( 3 , - 2 )$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
2. Given that the binomial expansion of $( 1 + k x ) ^ { - 4 } , | k x | < 1$, is $$1 - 6 x + A x ^ { 2 } + \ldots$$
find the value of the constant $k$,
find the value of the constant $A$, giving your answer in its simplest form.
3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a9963b13-7db4-4a1d-8c75-829f4aade994-05_659_865_269_550} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a sketch of part of the curve with equation $y = \frac { 10 } { 2 x + 5 \sqrt { } x } , x > 0$ The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the $x$-axis, and the lines with equations $x = 1$ and $x = 4$ The table below shows corresponding values of $x$ and $y$ for $y = \frac { 10 } { 2 x + 5 \sqrt { } x }$
$x$ 1 2 3 4
$y$ 1.42857 0.90326 0.55556
Complete the table above by giving the missing value of $y$ to 5 decimal places.
Use the trapezium rule, with all the values of $y$ in the completed table, to find an estimate for the area of $R$, giving your answer to 4 decimal places.
By reference to the curve in Figure 1, state, giving a reason, whether your estimate in part (b) is an overestimate or an underestimate for the area of $R$.
Use the substitution $u = \sqrt { } x$, or otherwise, to find the exact value of $$\int _ { 1 } ^ { 4 } \frac { 10 } { 2 x + 5 \sqrt { x } } d x$$ 4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a9963b13-7db4-4a1d-8c75-829f4aade994-07_618_703_246_625} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} A vase with a circular cross-section is shown in Figure 2. Water is flowing into the vase. When the depth of the water is $h \mathrm {~cm}$, the volume of water $V \mathrm {~cm} ^ { 3 }$ is given by $$V = 4 \pi h ( h + 4 ) , \quad 0 \leqslant h \leqslant 25$$ Water flows into the vase at a constant rate of $80 \pi \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$ Find the rate of change of the depth of the water, in $\mathrm { cms } ^ { - 1 }$, when $h = 6$ 5. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a9963b13-7db4-4a1d-8c75-829f4aade994-08_675_1262_267_340} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows a sketch of the curve $C$ with parametric equations $$x = 4 \cos \left( t + \frac { \pi } { 6 } \right) , \quad y = 2 \sin t , \quad 0 \leqslant t < 2 \pi$$
Show that $$x + y = 2 \sqrt { 3 } \cos t$$
Show that a cartesian equation of $C$ is $$( x + y ) ^ { 2 } + a y ^ { 2 } = b$$ where $a$ and $b$ are integers to be determined. \includegraphics[max width=\textwidth, alt={}, center]{a9963b13-7db4-4a1d-8c75-829f4aade994-09_104_51_2617_1900}
6. (i) Find $$\int x \mathrm { e } ^ { 4 x } \mathrm {~d} x$$ (ii) Find $$\int \frac { 8 } { ( 2 x - 1 ) ^ { 3 } } \mathrm {~d} x , \quad x > \frac { 1 } { 2 }$$ (iii) Given that $y = \frac { \pi } { 6 }$ at $x = 0$, solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { e } ^ { x } \operatorname { cosec } 2 y \operatorname { cosec } y$$

Edexcel C4 2014 June Q7

15 marks Challenging +1.2

7. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{a9963b13-7db4-4a1d-8c75-829f4aade994-12_681_1203_258_376} \captionsetup{labelformat=empty} \caption{Figure 4}

\end{figure} Figure 4 shows a sketch of part of the curve $C$ with parametric equations $$x = 3 \tan \theta , \quad y = 4 \cos ^ { 2 } \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$ The point $P$ lies on $C$ and has coordinates ( 3,2 ). The line $l$ is the normal to $C$ at $P$. The normal cuts the $x$-axis at the point $Q$.

Find the $x$ coordinate of the point $Q$. The finite region $S$, shown shaded in Figure 4, is bounded by the curve $C$, the $x$-axis, the $y$-axis and the line $l$. This shaded region is rotated $2 \pi$ radians about the $x$-axis to form a solid of revolution.
Find the exact value of the volume of the solid of revolution, giving your answer in the form $p \pi + q \pi ^ { 2 }$, where $p$ and $q$ are rational numbers to be determined.
[0pt] [You may use the formula $V = \frac { 1 } { 3 } \pi r ^ { 2 } h$ for the volume of a cone.]

Edexcel C4 2014 June Q8

15 marks Standard +0.3

8. Relative to a fixed origin $O$, the point $A$ has position vector $\left( \begin{array} { r } - 2 \\ 4 \\ 7 \end{array} \right)$ and the point $B$ has position vector $\left( \begin{array} { r } - 1 \\ 3 \\ 8 \end{array} \right)$ The line $l _ { 1 }$ passes through the points $A$ and $B$.

Find the vector $\overrightarrow { A B }$.
Hence find a vector equation for the line $l _ { 1 }$ The point $P$ has position vector $\left( \begin{array} { l } 0 \\ 2 \\ 3 \end{array} \right)$ Given that angle $P B A$ is $\theta$,
show that $\cos \theta = \frac { 1 } { 3 }$ The line $l _ { 2 }$ passes through the point $P$ and is parallel to the line $l _ { 1 }$
Find a vector equation for the line $l _ { 2 }$ The points $C$ and $D$ both lie on the line $l _ { 2 }$ Given that $A B = P C = D P$ and the $x$ coordinate of $C$ is positive,
find the coordinates of $C$ and the coordinates of $D$.
find the exact area of the trapezium $A B C D$, giving your answer as a simplified surd.

Edexcel C4 2015 June Q1

8 marks Moderate -0.8

(a) Find the binomial expansion of

$$( 4 + 5 x ) ^ { \frac { 1 } { 2 } } , \quad | x | < \frac { 4 } { 5 }$$ in ascending powers of $x$, up to and including the term in $x ^ { 2 }$.
Give each coefficient in its simplest form.
(b) Find the exact value of $( 4 + 5 x ) ^ { \frac { 1 } { 2 } }$ when $x = \frac { 1 } { 10 }$ Give your answer in the form $k \sqrt { 2 }$, where $k$ is a constant to be determined.
(c) Substitute $x = \frac { 1 } { 10 }$ into your binomial expansion from part (a) and hence find an approximate value for $\sqrt { 2 }$ Give your answer in the form $\frac { p } { q }$ where $p$ and $q$ are integers.

Edexcel C4 2015 June Q2

11 marks Standard +0.3

2. The curve $C$ has equation $$x ^ { 2 } - 3 x y - 4 y ^ { 2 } + 64 = 0$$

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
Find the coordinates of the points on $C$ where $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ (Solutions based entirely on graphical or numerical methods are not acceptable.)

Edexcel C4 2015 June Q3

8 marks Standard +0.3

3. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{89d4a7a5-3f4f-4d16-b14e-a27243cedd78-05_620_867_301_536} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} Figure 1 shows a sketch of part of the curve with equation $y = 4 x - x \mathrm { e } ^ { \frac { 1 } { 2 } x } , x \geqslant 0$ The curve meets the $x$-axis at the origin $O$ and cuts the $x$-axis at the point $A$.

Find, in terms of $\ln 2$, the $x$ coordinate of the point $A$.
Find $$\int x \mathrm { e } ^ { \frac { 1 } { 2 } x } \mathrm {~d} x$$ The finite region $R$, shown shaded in Figure 1, is bounded by the $x$-axis and the curve with equation $$y = 4 x - x \mathrm { e } ^ { \frac { 1 } { 2 } x } , x \geqslant 0$$
Find, by integration, the exact value for the area of $R$. Give your answer in terms of $\ln 2$

Edexcel C4 2015 June Q4

11 marks Standard +0.3

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 }

Edexcel C4 2015 June Q5

6 marks Standard +0.3

5
- 3
p \end{array} \right) + \lambda \left( \begin{array} { r } 0
1
- 3 \end{array} \right) , \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { r }

Edexcel C4 2015 June Q6

8 marks Standard +0.8

6. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{89d4a7a5-3f4f-4d16-b14e-a27243cedd78-11_666_993_244_392} \captionsetup{labelformat=empty} \caption{Diagram not to scale}

\end{figure} Figure 2 Figure 2 shows a sketch of the curve with equation $y = \sqrt { ( 3 - x ) ( x + 1 ) } , 0 \leqslant x \leqslant 3$ The finite region $R$, shown shaded in Figure 2, is bounded by the curve, the $x$-axis, and the $y$-axis.

Use the substitution $x = 1 + 2 \sin \theta$ to show that $$\int _ { 0 } ^ { 3 } \sqrt { ( 3 - x ) ( x + 1 ) } d x = k \int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 2 } } \cos ^ { 2 } \theta d \theta$$ where $k$ is a constant to be determined.
Hence find, by integration, the exact area of $R$.

Edexcel C4 2015 June Q8

10 marks Standard +0.3

8
5
- 2 \end{array} \right) + \mu \left( \begin{array} { r } 3
4
- 5 \end{array} \right)$$ where $\lambda$ and $\mu$ are scalar parameters and $p$ is a constant. The lines $l _ { 1 }$ and $l _ { 2 }$ intersect at the point $A$.

Find the coordinates of $A$.
Find the value of the constant $p$.
Find the acute angle between $l _ { 1 }$ and $l _ { 2 }$, giving your answer in degrees to 2 decimal places. The point $B$ lies on $l _ { 2 }$ where $\mu = 1$
Find the shortest distance from the point $B$ to the line $l _ { 1 }$, giving your answer to 3 significant figures.
1. A curve $C$ has parametric equations
$$x = 4 t + 3 , \quad y = 4 t + 8 + \frac { 5 } { 2 t } , \quad t \neq 0$$
Find the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at the point on $C$ where $t = 2$, giving your answer as a fraction in its simplest form.
Show that the cartesian equation of the curve $C$ can be written in the form $$y = \frac { x ^ { 2 } + a x + b } { x - 3 } , \quad x \neq 3$$ where $a$ and $b$ are integers to be determined.\\ 6. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{89d4a7a5-3f4f-4d16-b14e-a27243cedd78-11_666_993_244_392} \captionsetup{labelformat=empty} \caption{Diagram not to scale}
\end{figure} Figure 2 Figure 2 shows a sketch of the curve with equation $y = \sqrt { ( 3 - x ) ( x + 1 ) } , 0 \leqslant x \leqslant 3$\\ The finite region $R$, shown shaded in Figure 2, is bounded by the curve, the $x$-axis, and the $y$-axis.
Use the substitution $x = 1 + 2 \sin \theta$ to show that $$\int _ { 0 } ^ { 3 } \sqrt { ( 3 - x ) ( x + 1 ) } d x = k \int _ { - \frac { \pi } { 6 } } ^ { \frac { \pi } { 2 } } \cos ^ { 2 } \theta d \theta$$ where $k$ is a constant to be determined.
Hence find, by integration, the exact area of $R$. 7. (a) Express $\frac { 2 } { P ( P - 2 ) }$ in partial fractions. A team of biologists is studying a population of a particular species of animal. The population is modelled by the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 2 } P ( P - 2 ) \cos 2 t , t \geqslant 0$$ where $P$ is the population in thousands, and $t$ is the time measured in years since the start of the study. Given that $P = 3$ when $t = 0$,
solve this differential equation to show that $$P = \frac { 6 } { 3 - \mathrm { e } ^ { \frac { 1 } { 2 } \sin 2 t } }$$
find the time taken for the population to reach 4000 for the first time. Give your answer in years to 3 significant figures.\\ 8. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{89d4a7a5-3f4f-4d16-b14e-a27243cedd78-15_696_1418_287_262} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows a sketch of part of the curve $C$ with equation $$y = 3 ^ { x }$$ The point $P$ lies on $C$ and has coordinates $( 2,9 )$.
The line $l$ is a tangent to $C$ at $P$. The line $l$ cuts the $x$-axis at the point $Q$.
Find the exact value of the $x$ coordinate of $Q$. The finite region $R$, shown shaded in Figure 3, is bounded by the curve $C$, the $x$-axis, the $y$-axis and the line $l$. This region $R$ is rotated through $360 ^ { \circ }$ about the $x$-axis.
Use integration to find the exact value of the volume of the solid generated. Give your answer in the form $\frac { p } { q }$ where $p$ and $q$ are exact constants.
[0pt] [You may assume the formula $V = \frac { 1 } { 3 } \pi r ^ { 2 } h$ for the volume of a cone.]

Edexcel C4 2016 June Q1

6 marks Moderate -0.3

Use the binomial series to find the expansion of

$$\frac { 1 } { ( 2 + 5 x ) ^ { 3 } } , \quad | x | < \frac { 2 } { 5 }$$ in ascending powers of $x$, up to and including the term in $x ^ { 3 }$.
Give each coefficient as a fraction in its simplest form.
(6)

Edexcel C4 2016 June Q2

9 marks Moderate -0.3

2. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{cbfbb690-bc85-46e5-a97f-35df4b6f1c84-03_712_1091_248_470} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} Figure 1 shows a sketch of part of the curve with equation $y = x ^ { 2 } \ln x , x \geqslant 1$ The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the $x$-axis and the line $x = 2$ The table below shows corresponding values of $x$ and $y$ for $y = x ^ { 2 } \ln x$

$x$	1	1.2	1.4	1.6	1.8	2
$y$	0	0.2625		1.2032	1.9044	2.7726

Complete the table above, giving the missing value of $y$ to 4 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 3 decimal places.
Use integration to find the exact value for the area of $R$.

Edexcel C4 2016 June Q3

9 marks Standard +0.3

The curve $C$ has equation

$$2 x ^ { 2 } y + 2 x + 4 y - \cos ( \pi y ) = 17$$

Use implicit differentiation to find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$. The point $P$ with coordinates $\left( 3 , \frac { 1 } { 2 } \right)$ lies on $C$.
The normal to $C$ at $P$ meets the $x$-axis at the point $A$.
Find the $x$ coordinate of $A$, giving your answer in the form $\frac { a \pi + b } { c \pi + d }$, where $a , b , c$ and $d$ are integers to be determined.

Edexcel C4 2016 June Q4

7 marks Moderate -0.8

4. The rate of decay of the mass of a particular substance is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - \frac { 5 } { 2 } x , \quad t \geqslant 0$$ where $x$ is the mass of the substance measured in grams and $t$ is the time measured in days.
Given that $x = 60$ when $t = 0$,

solve the differential equation, giving $x$ in terms of $t$. You should show all steps in your working and give your answer in its simplest form.
Find the time taken for the mass of the substance to decay from 60 grams to 20 grams. Give your answer to the nearest minute.

Edexcel C4 2016 June Q5

6 marks Standard +0.3

5. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{cbfbb690-bc85-46e5-a97f-35df4b6f1c84-09_605_1131_248_466} \captionsetup{labelformat=empty} \caption{Figure 2}

\end{figure} Figure 2 shows a sketch of the curve $C$ with parametric equations $$x = 4 \tan t , \quad y = 5 \sqrt { 3 } \sin 2 t , \quad 0 \leqslant t < \frac { \pi } { 2 }$$ The point $P$ lies on $C$ and has coordinates $\left( 4 \sqrt { 3 } , \frac { 15 } { 2 } \right)$.

Find the exact value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at the point $P$. Give your answer as a simplified surd. The point $Q$ lies on the curve $C$, where $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$
Find the exact coordinates of the point $Q$.

Edexcel C4 2016 June Q6

15 marks Standard +0.8

6. (i) Given that $y > 0$, find $$\int \frac { 3 y - 4 } { y ( 3 y + 2 ) } d y$$ (ii) (a) Use the substitution $x = 4 \sin ^ { 2 } \theta$ to show that $$\int _ { 0 } ^ { 3 } \sqrt { \left( \frac { x } { 4 - x } \right) } \mathrm { d } x = \lambda \int _ { 0 } ^ { \frac { \pi } { 3 } } \sin ^ { 2 } \theta \mathrm {~d} \theta$$ where $\lambda$ is a constant to be determined.
(b) Hence use integration to find $$\int _ { 0 } ^ { 3 } \sqrt { \left( \frac { x } { 4 - x } \right) } d x$$ giving your answer in the form $a \pi + b$, where $a$ and $b$ are exact constants.

Edexcel C4 2016 June Q7

8 marks Standard +0.3

7. (a) Find $$\int ( 2 x - 1 ) ^ { \frac { 3 } { 2 } } d x$$ giving your answer in its simplest form. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{cbfbb690-bc85-46e5-a97f-35df4b6f1c84-13_727_1177_596_370} \captionsetup{labelformat=empty} \caption{Figure 3}

\end{figure} Figure 3 shows a sketch of part of the curve $C$ with equation $$y = ( 2 x - 1 ) ^ { \frac { 3 } { 4 } } , \quad x \geqslant \frac { 1 } { 2 }$$ The curve $C$ cuts the line $y = 8$ at the point $P$ with coordinates $( k , 8 )$, where $k$ is a constant.
(b) Find the value of $k$. The finite region $S$, shown shaded in Figure 3, is bounded by the curve $C$, the $x$-axis, the $y$-axis and the line $y = 8$. This region is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution.
(c) Find the exact value of the volume of the solid generated.

Questions — Edexcel (9685 questions)

Browse by module

Browse by board

Edexcel C4 2014 June Q4

Edexcel C4 2014 June Q5

Edexcel C4 2014 June Q6

Edexcel C4 2014 June Q7

Edexcel C4 2014 June Q8

Edexcel C4 2014 June Q1

Edexcel C4 2014 June Q3

Edexcel C4 2014 June Q4

Edexcel C4 2014 June Q6

Edexcel C4 2014 June Q7

Edexcel C4 2014 June Q8

Edexcel C4 2015 June Q1

Edexcel C4 2015 June Q2

Edexcel C4 2015 June Q3

Edexcel C4 2015 June Q4

Edexcel C4 2015 June Q5

Edexcel C4 2015 June Q6

Edexcel C4 2015 June Q8

Edexcel C4 2016 June Q1

Edexcel C4 2016 June Q2

Edexcel C4 2016 June Q3

Edexcel C4 2016 June Q4

Edexcel C4 2016 June Q5

Edexcel C4 2016 June Q6

Edexcel C4 2016 June Q7

\(x\)	1	2	3	4
\(y\)	1.42857	0.90326		0.55556