Questions - Edexcel C4 - A-Level Maths

$x$	0	$\frac { 3 \pi } { 8 }$	$\frac { 3 \pi } { 4 }$	$\frac { 9 \pi } { 8 }$	$\frac { 3 \pi } { 2 }$
$y$	3	2.77164	2.12132		0

Complete the table above giving the missing value of $y$ to 5 decimal places.
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.
Use integration to find the exact area of $R$.

Edexcel C4 2009 June Q3

3. $$\mathrm { f } ( x ) = \frac { 4 - 2 x } { ( 2 x + 1 ) ( x + 1 ) ( x + 3 ) } = \frac { A } { 2 x + 1 } + \frac { B } { x + 1 } + \frac { C } { x + 3 }$$

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

4. The curve $C$ has the equation $y \mathrm { e } ^ { - 2 x } = 2 x + y ^ { 2 }$.

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$. The point $P$ on $C$ has coordinates $( 0,1 )$.
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

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 $$x = 2 \cos 2 t , \quad y = 6 \sin t , \quad 0 \leqslant t \leqslant \frac { \pi } { 2 }$$

Find the gradient of the curve at the point where $t = \frac { \pi } { 3 }$.
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 $k$.
Write down the range of $\mathrm { f } ( x )$.

Edexcel C4 2009 June Q6

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

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 $B$.

Find a vector equation for the line $l$.
Find $| \overrightarrow { C B } |$.
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.
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 $l$,
find the area of the triangle $C X B$, giving your answer to 3 significant figures.

Edexcel C4 2009 June Q8

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.

Edexcel C4 2010 June Q1

1. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{280ae2a5-7344-4ba3-907f-235fba3fd5b3-02_684_767_274_589} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} Figure 1 shows part of the curve with equation $y = \sqrt { } \left( 0.75 + \cos ^ { 2 } x \right)$. The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the $y$-axis, the $x$-axis and the line with equation $x = \frac { \pi } { 3 }$.

Complete the table with values of $y$ corresponding to $x = \frac { \pi } { 6 }$ and $x = \frac { \pi } { 4 }$.
$x$ 0 $\frac { \pi } { 12 }$ $\frac { \pi } { 6 }$ $\frac { \pi } { 4 }$ $\frac { \pi } { 3 }$
$y$ 1.3229 1.2973 1
Use the trapezium rule
1. with the values of $y$ at $x = 0 , x = \frac { \pi } { 6 }$ and $x = \frac { \pi } { 3 }$ to find an estimate of the area of $R$. Give your answer to 3 decimal places.
2. with the values of $y$ at $x = 0 , x = \frac { \pi } { 12 } , x = \frac { \pi } { 6 } , x = \frac { \pi } { 4 }$ and $x = \frac { \pi } { 3 }$ to find a further estimate of the area of $R$. Give your answer to 3 decimal places.
  (6) \section*{LU}

Edexcel C4 2010 June Q2

2. Using the substitution $u = \cos x + 1$, or otherwise, show that $$\int _ { 0 } ^ { \frac { \pi } { 2 } } \mathrm { e } ^ { \cos x + 1 } \sin x \mathrm {~d} x = \mathrm { e } ( \mathrm { e } - 1 )$$ (6)

Edexcel C4 2010 June Q3

3. A curve $C$ has equation $$2 ^ { x } + y ^ { 2 } = 2 x y$$ Find the exact value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at the point on $C$ with coordinates $( 3,2 )$.

Edexcel C4 2010 June Q4

4. A curve $C$ has parametric equations $$x = \sin ^ { 2 } t , \quad y = 2 \tan t , \quad 0 \leqslant t < \frac { \pi } { 2 }$$

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$. The tangent to $C$ at the point where $t = \frac { \pi } { 3 }$ cuts the $x$-axis at the point $P$.
Find the $x$-coordinate of $P$.
\section*{LU}

Edexcel C4 2010 June Q5

5. $$\frac { 2 x ^ { 2 } + 5 x - 10 } { ( x - 1 ) ( x + 2 ) } \equiv A + \frac { B } { x - 1 } + \frac { C } { x + 2 }$$

Find the values of the constants $A , B$ and $C$.
Hence, or otherwise, expand $\frac { 2 x ^ { 2 } + 5 x - 10 } { ( x - 1 ) ( x + 2 ) }$ in ascending powers of $x$, as far as the term in $x ^ { 2 }$. Give each coefficient as a simplified fraction.

Edexcel C4 2010 June Q6

6. $$f ( \theta ) = 4 \cos ^ { 2 } \theta - 3 \sin ^ { 2 } \theta$$

Show that $f ( \theta ) = \frac { 1 } { 2 } + \frac { 7 } { 2 } \cos 2 \theta$.
Hence, using calculus, find the exact value of $\int _ { 0 } ^ { \frac { \pi } { 2 } } \theta \mathrm { f } ( \theta ) \mathrm { d } \theta$.

Edexcel C4 2010 June Q7

7. The line $l _ { 1 }$ has equation $\mathbf { r } = \left( \begin{array} { r } 2
3
- 4 \end{array} \right) + \lambda \left( \begin{array} { l } 1
2
1 \end{array} \right)$, where $\lambda$ is a scalar parameter. The line $l _ { 2 }$ has equation $\mathbf { r } = \left( \begin{array} { r } 0
9
- 3 \end{array} \right) + \mu \left( \begin{array} { l } 5
0
2 \end{array} \right)$, where $\mu$ is a scalar parameter.
Given that $l _ { 1 }$ and $l _ { 2 }$ meet at the point $C$, find

the coordinates of $C$. The point $A$ is the point on $l _ { 1 }$ where $\lambda = 0$ and the point $B$ is the point on $l _ { 2 }$ where $\mu = - 1$.
Find the size of the angle $A C B$. Give your answer in degrees to 2 decimal places.
Hence, or otherwise, find the area of the triangle $A B C$.

Edexcel C4 2010 June Q8

8. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{280ae2a5-7344-4ba3-907f-235fba3fd5b3-12_474_837_283_610} \captionsetup{labelformat=empty} \caption{Figure 2}

\end{figure} Figure 2 shows a cylindrical water tank. The diameter of a circular cross-section of the tank is 6 m . Water is flowing into the tank at a constant rate of $0.48 \pi \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }$. At time $t$ minutes, the depth of the water in the tank is $h$ metres. There is a tap at a point $T$ at the bottom of the tank. When the tap is open, water leaves the tank at a rate of $0.6 \pi h \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }$.

Show that $t$ minutes after the tap has been opened $$75 \frac { \mathrm {~d} h } { \mathrm {~d} t } = ( 4 - 5 h )$$ When $t = 0 , h = 0.2$
Find the value of $t$ when $h = 0.5$

Edexcel C4 2011 June Q1

1. $$\frac { 9 x ^ { 2 } } { ( x - 1 ) ^ { 2 } ( 2 x + 1 ) } = \frac { A } { ( x - 1 ) } + \frac { B } { ( x - 1 ) ^ { 2 } } + \frac { C } { ( 2 x + 1 ) }$$ Find the values of the constants $A , B$ and $C$.

Edexcel C4 2011 June Q2

2. $$f ( x ) = \frac { 1 } { \sqrt { } \left( 9 + 4 x ^ { 2 } \right) } , \quad | x | < \frac { 3 } { 2 }$$ Find the first three non-zero terms of the binomial expansion of $\mathrm { f } ( x )$ in ascending powers of $x$. Give each coefficient as a simplified fraction.

Edexcel C4 2011 June Q3

3. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{9d513d77-b8f9-4223-832f-f566c5f50457-04_391_741_274_605} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} A hollow hemispherical bowl is shown in Figure 1. Water is flowing into the bowl. When the depth of the water is $h \mathrm {~m}$, the volume $V \mathrm {~m} ^ { 3 }$ is given by $$V = \frac { 1 } { 12 } \pi \cdot h ^ { 2 } ( 3 - 4 h ) , \quad 0 \leqslant h \leqslant 0.25$$

Find, in terms of $\pi , \frac { \mathrm { d } V } { \mathrm {~d} h }$ when $h = 0.1$ Water flows into the bowl at a rate of $\frac { \pi } { 800 } \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 1 }$.
Find the rate of change of $h$, in $\mathrm { m } \mathrm { s } ^ { - 1 }$, when $h = 0.1$

Edexcel C4 2011 June Q4

4. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{9d513d77-b8f9-4223-832f-f566c5f50457-05_673_1058_264_443} \captionsetup{labelformat=empty} \caption{Figure 2}

\end{figure} Figure 2 shows a sketch of the curve with equation $y = x ^ { 3 } \ln \left( x ^ { 2 } + 2 \right) , x \geqslant 0$.
The finite region $R$, shown shaded in Figure 2, is bounded by the curve, the $x$-axis and the line $x = \sqrt { } 2$. The table below shows corresponding values of $x$ and $y$ for $y = x ^ { 3 } \ln \left( x ^ { 2 } + 2 \right)$.

$x$	0	$\frac { \sqrt { } 2 } { 4 }$	$\frac { \sqrt { } 2 } { 2 }$	$\frac { 3 \sqrt { } 2 } { 4 }$	$\sqrt { } 2$
$y$	0		0.3240		3.9210

Complete the table above giving the missing values 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 2 decimal places.
Use the substitution $u = x ^ { 2 } + 2$ to show that the area of $R$ is $$\frac { 1 } { 2 } \int _ { 2 } ^ { 4 } ( u - 2 ) \ln u \mathrm {~d} u$$
Hence, or otherwise, find the exact area of $R$.

Edexcel C4 2011 June Q5

Find the gradient of the curve with equation

$$\ln y = 2 x \ln x , \quad x > 0 , y > 0$$ at the point on the curve where $x = 2$. Give your answer as an exact value.

Edexcel C4 2011 June Q6

6. 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} { r } 6
- 3
- 2 \end{array} \right) + \lambda \left( \begin{array} { r } - 1
2
3 \end{array} \right) , \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { r } - 5

Edexcel C4 2011 June Q7

7. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{9d513d77-b8f9-4223-832f-f566c5f50457-10_643_999_276_475} \captionsetup{labelformat=empty} \caption{Figure 3}

\end{figure} Figure 3 shows part of the curve $C$ with parametric equations $$x = \tan \theta , \quad y = \sin \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$ The point $P$ lies on $C$ and has coordinates $\left( \sqrt { } 3 , \frac { 1 } { 2 } \sqrt { } 3 \right)$.

Find the value of $\theta$ at the point $P$. The line $l$ is a normal to $C$ at $P$. The normal cuts the $x$-axis at the point $Q$.
Show that $Q$ has coordinates $( k \sqrt { } 3,0 )$, giving the value of the constant $k$. The finite shaded region $S$ shown in Figure 3 is bounded by the curve $C$, the line $x = \sqrt { } 3$ and the $x$-axis. This shaded region is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution.
Find the volume of the solid of revolution, giving your answer in the form $p \pi \sqrt { } 3 + q \pi ^ { 2 }$, where $p$ and $q$ are constants.

Edexcel C4 2011 June Q15

15
3 \end{array} \right) + \mu \left( \begin{array} { r } 2
- 3
1 \end{array} \right)$$ where $\lambda$ and $\mu$ are scalar parameters.

Show that $l _ { 1 }$ and $l _ { 2 }$ meet and find the position vector of their point of intersection $A$.
Find, to the nearest $0.1 ^ { \circ }$, the acute angle between $l _ { 1 }$ and $l _ { 2 }$. The point $B$ has position vector $\left( \begin{array} { r } 5
- 1
1 \end{array} \right)$.
Show that $B$ lies on $l _ { 1 }$.
Find the shortest distance from $B$ to the line $l _ { 2 }$, giving your answer to 3 significant figures.
7. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9d513d77-b8f9-4223-832f-f566c5f50457-10_643_999_276_475} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows part of the curve $C$ with parametric equations $$x = \tan \theta , \quad y = \sin \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$ The point $P$ lies on $C$ and has coordinates $\left( \sqrt { } 3 , \frac { 1 } { 2 } \sqrt { } 3 \right)$.
Find the value of $\theta$ at the point $P$. The line $l$ is a normal to $C$ at $P$. The normal cuts the $x$-axis at the point $Q$.
Show that $Q$ has coordinates $( k \sqrt { } 3,0 )$, giving the value of the constant $k$. The finite shaded region $S$ shown in Figure 3 is bounded by the curve $C$, the line $x = \sqrt { } 3$ and the $x$-axis. This shaded region is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution.
Find the volume of the solid of revolution, giving your answer in the form $p \pi \sqrt { } 3 + q \pi ^ { 2 }$, where $p$ and $q$ are constants. 8. (a) Find $\int ( 4 y + 3 ) ^ { - \frac { 1 } { 2 } } \mathrm {~d} y$
Given that $y = 1.5$ at $x = - 2$, solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \sqrt { } ( 4 y + 3 ) } { x ^ { 2 } }$$ giving your answer in the form $y = \mathrm { f } ( x )$.

Edexcel C4 2012 June Q1

1. $$\mathrm { f } ( x ) = \frac { 1 } { x ( 3 x - 1 ) ^ { 2 } } = \frac { A } { x } + \frac { B } { ( 3 x - 1 ) } + \frac { C } { ( 3 x - 1 ) ^ { 2 } }$$

Find the values of the constants $A , B$ and $C$.
1. Hence find $\int \mathrm { f } ( x ) \mathrm { d } x$.
2. Find $\int _ { 1 } ^ { 2 } \mathrm { f } ( x ) \mathrm { d } x$, leaving your answer in the form $a + \ln b$, where $a$ and $b$ are constants. 1
  $f ( x ) = \frac { 1 } { x ( 3 x - 1 ) ^ { 2 } } = \frac { A } { x } + \frac { } { ( 3 x }$
Find the values of the constants $A , B$ and $C$.

\(x\)	0	\(\frac { 3 \pi } { 8 }\)	\(\frac { 3 \pi } { 4 }\)	\(\frac { 9 \pi } { 8 }\)	\(\frac { 3 \pi } { 2 }\)
\(y\)	3	2.77164	2.12132		0

\(x\)	0	\(\frac { \pi } { 12 }\)	\(\frac { \pi } { 6 }\)	\(\frac { \pi } { 4 }\)	\(\frac { \pi } { 3 }\)
\(y\)	1.3229	1.2973			1

\(x\)	0	\(\frac { \sqrt { } 2 } { 4 }\)	\(\frac { \sqrt { } 2 } { 2 }\)	\(\frac { 3 \sqrt { } 2 } { 4 }\)	\(\sqrt { } 2\)
\(y\)	0		0.3240		3.9210

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