Questions - Edexcel C34 - A-Level Maths

Show that $l _ { 1 }$ and $l _ { 2 }$ do not meet. The point $P$ is on $l _ { 1 }$ where $\lambda = 0$, and the point $Q$ is on $l _ { 2 }$ where $\mu = - 1$
Find the acute angle between the line segment $P Q$ and $l _ { 1 }$, giving your answer in degrees to 2 decimal places.
Find the shortest distance from the point $Q$ to the line $l _ { 1 }$, giving your answer to 3 significant figures.

Edexcel C34 2018 June Q9

9. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{a377da06-a968-438c-bec2-ae55283dae47-28_533_1095_258_365} \captionsetup{labelformat=empty} \caption{Figure 2}

\end{figure} Diagram not drawn to scale

Find $$\int \frac { 1 } { ( 2 x - 1 ) ^ { 2 } } d x$$ Figure 2 shows a sketch of the curve with equation $y = \mathrm { f } ( x )$ where $$f ( x ) = \frac { 12 } { ( 2 x - 1 ) } \quad 1 \leqslant x \leqslant 5$$ The finite region $R$, shown shaded in Figure 2, is bounded by the line with equation $x = 1$, the curve with equation $y = \mathrm { f } ( x )$ and the line with equation $y = \frac { 4 } { 3 }$. The region $R$ is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution.
Find the exact value of the volume of the solid generated, giving your answer in its simplest form.
\section*{Leave
k}

Edexcel C34 2018 June Q10

10. The curve $C$ satisfies the equation $$x \mathrm { e } ^ { 5 - 2 y } - y = 0 \quad x > 0 , \quad y > 0$$ The point $P$ with coordinates ( $2 \mathrm { e } ^ { - 1 } , 2$ ) lies on $C$.
The tangent to $C$ at $P$ cuts the $x$-axis at the point $A$ and cuts the $y$-axis at the point $B$.
Given that $O$ is the origin, find the exact area of triangle $O A B$, giving your answer in its simplest form.
\includegraphics[max width=\textwidth, alt={}]{a377da06-a968-438c-bec2-ae55283dae47-35_4_21_127_2042} L

Edexcel C34 2018 June Q11

11. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{a377da06-a968-438c-bec2-ae55283dae47-36_601_1140_242_402} \captionsetup{labelformat=empty} \caption{Figure 3}

\end{figure}

By writing $\sec \theta$ as $\frac { 1 } { \cos \theta }$, show that when $x = 3 \sec \theta$, $$\frac { \mathrm { d } x } { \mathrm {~d} \theta } = 3 \sec \theta \tan \theta$$ Figure 3 shows a sketch of part of the curve $C$ with equation $$y = \frac { \sqrt { x ^ { 2 } - 9 } } { x } \quad x \geqslant 3$$ The finite region $R$, shown shaded in Figure 3, is bounded by the curve $C$, the $x$-axis and the line with equation $x = 6$
Use the substitution $x = 3 \sec \theta$ to find the exact value of the area of $R$. [Solutions based entirely on graphical or numerical methods are not acceptable.]

Edexcel C34 2018 June Q12

12. (a) Show that $$\cot x - \tan x \equiv 2 \cot 2 x , \quad x \neq 90 n ^ { \circ } , n \in \mathbb { Z }$$ (b) Hence, or otherwise, solve, for $0 \leqslant \theta < 180 ^ { \circ }$ $$5 + \cot \left( \theta - 15 ^ { \circ } \right) - \tan \left( \theta - 15 ^ { \circ } \right) = 0$$ giving your answers to one decimal place.
[0pt] [Solutions based entirely on graphical or numerical methods are not acceptable.]

Edexcel C34 2018 June Q13

13. (a) Express $\frac { 1 } { ( 4 - x ) ( 2 - x ) }$ in partial fractions. The mass, $x$ grams, of a substance at time $t$ seconds after a chemical reaction starts is modelled by the differential equation
where $k$ is a constant.
(b) solve the differential equation and show that the solution can be written as $$x = \frac { 4 - 4 \mathrm { e } ^ { 2 k t } } { 1 - 2 \mathrm { e } ^ { 2 k t } }$$ Given that $k = 0.1$
(c) find the value of $t$ when $x = 1$, giving your answer, in seconds, to 3 significant figures. The mass, $x$ grams, of a substance at time $t$ seconds after a chemical reaction starts is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = k ( 4 - x ) ( 2 - x ) , \quad t \geqslant 0,0 \leqslant x < 2$$ where $k$ is a constant. $$\text { Given that when } t = 0 , x = 0$$ (b) solve the differential equation and show that the solution can be written as

Edexcel C34 2018 June Q14

14. Given that $$y = \frac { \left( x ^ { 2 } - 4 \right) ^ { \frac { 1 } { 2 } } } { x ^ { 3 } } \quad x > 2$$

show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { A x ^ { 2 } + 12 } { x ^ { 4 } \left( x ^ { 2 } - 4 \right) ^ { \frac { 1 } { 2 } } } \quad x > 2$$ where $A$ is a constant to be found. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a377da06-a968-438c-bec2-ae55283dae47-48_593_1134_865_395} \captionsetup{labelformat=empty} \caption{Figure 4}
\end{figure} Figure 4 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x )$ where $$\mathrm { f } ( x ) = \frac { 24 \left( x ^ { 2 } - 4 \right) ^ { \frac { 1 } { 2 } } } { x ^ { 3 } } \quad x > 2$$
Use your answer to part (a) to find the range of f.
State a reason why f-1 does not exist.

Edexcel C34 2019 June Q1

1. $$f ( x ) = 2 x ^ { 3 } + x - 20$$

Show that the equation $\mathrm { f } ( x ) = 0$ can be rewritten as $$x = \sqrt [ 3 ] { a - b x }$$ where $a$ and $b$ are positive constants to be determined.
Starting with $x _ { 1 } = 2.1$ use the iteration formula $x _ { n + 1 } = \sqrt [ 3 ] { a - b x _ { n } }$, with the numerical values of $a$ and $b$, to calculate the values of $x _ { 2 }$ and $x _ { 3 }$ giving your answers to 3 decimal places.
Using a suitable interval, show that 2.077 is a root of the equation $\mathrm { f } ( x ) = 0$ correct to 3 decimal places.
Hence state a root, to 3 decimal places, of the equation $$2 ( x + 2 ) ^ { 3 } + x - 18 = 0$$
VIIIV SIHI NI JIIYM ION OC VIIV SIHI NI JIIIM ION OC VIIV SIHI NI JIIYM ION OC

Edexcel C34 2019 June Q2

2. (a) Find $\int \frac { 4 x + 3 } { x } \mathrm {~d} x , \quad x > 0$
(b) Given that $y = 25$ at $x = 1$, solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { ( 4 x + 3 ) y ^ { \frac { 1 } { 2 } } } { x } \quad x > 0 , y > 0$$ giving your answer in the form $y = [ \mathrm { g } ( x ) ] ^ { 2 }$.
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\includegraphics[max width=\textwidth, alt={}, center]{a9870c94-0910-46ec-a54a-44a431cb324e-05_52_49_2777_1886}

Edexcel C34 2019 June Q3

3. A curve $C$ has parametric equations $$x = \sqrt { 3 } \tan \theta , \quad y = \sec ^ { 2 } \theta , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 3 }$$ The cartesian equation of $C$ is $$y = \mathrm { f } ( x ) , \quad 0 \leqslant x \leqslant k , \quad \text { where } k \text { is a constant }$$

State the value of $k$.
Find $\mathrm { f } ( x )$ in its simplest form.
Hence, or otherwise, find the gradient of the curve at the point where $\theta = \frac { \pi } { 6 }$

Edexcel C34 2019 June Q4

4. The curve $C$ has equation $$3 y \mathrm { e } ^ { - 2 x } = 4 x ^ { 2 } + y ^ { 2 } + 2$$

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$. The point $P$ on $C$ has coordinates $( 0,2 )$.
Find the equation of the normal to $C$ at $P$ giving your answer in the form $y = m x + c$, where $m$ and $c$ are constants to be found.
(3)
VIIIV SIHI NI JIIYM ION OC VIUV SIHI NI JIIIM ION OO VI4V SIHI NI JIIYM ION OO

Edexcel C34 2019 June Q5

5. A bath is filled with hot water. The temperature, $\theta ^ { \circ } \mathrm { C }$, of the water in the bath, $t$ minutes after the bath has been filled, is given by $$\theta = 20 + A \mathrm { e } ^ { - k t }$$ where $A$ and $k$ are positive constants. Given that the temperature of the water in the bath is initially $38 ^ { \circ } \mathrm { C }$,

find the value of $A$. The temperature of the water in the bath 16 minutes after the bath has been filled is $24.5 ^ { \circ } \mathrm { C }$.
Show that $k = \frac { 1 } { 8 } \ln 2$ Using the values for $k$ and $A$,
find the temperature of the water 40 minutes after the bath has been filled, giving your answer to 3 significant figures.
Explain why the temperature of the water in the bath cannot fall to $19 ^ { \circ } \mathrm { C }$.

Edexcel C34 2019 June Q6

6. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{a9870c94-0910-46ec-a54a-44a431cb324e-14_988_1120_123_395} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} Figure 1 shows a sketch of the graph with equation $y = | 4 x + 10 a |$, where $a$ is a positive constant. The graph cuts the $y$-axis at the point $P$ and meets the $x$-axis at the point $Q$ as shown.

1. State the coordinates of $P$.
2. State the coordinates of $Q$.
A copy of Figure 1 is shown on page 15. On this copy, sketch the graph with equation $$y = | x | - a$$ Show on the sketch the coordinates of each point where your graph cuts or meets the coordinate axes.
Hence, or otherwise, solve the equation $$| 4 x + 10 a | = | x | - a$$ giving your answers in terms of $a$. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a9870c94-0910-46ec-a54a-44a431cb324e-15_860_1128_447_392} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} $\_\_\_\_$ 7

Edexcel C34 2019 June Q7

7. (a) Express $5 \cos \theta - 3 \sin \theta$ in the form $R \cos ( \theta + \alpha )$, where $R > 0$ and $0 < \alpha < \frac { \pi } { 2 }$ Give the exact value of $R$ and give the value of $\alpha$, in radians, to 4 decimal places. The height of sea water, $H$ metres, on a harbour wall is modelled by the equation $$H = 6 + 2.5 \cos \left( \frac { 4 \pi t } { 25 } \right) - 1.5 \sin \left( \frac { 4 \pi t } { 25 } \right) , \quad 0 \leqslant t < 12$$ where $t$ is the number of hours after midday.
(b) Calculate the times at which the model predicts that the height of sea water on the harbour wall will be 4.6 metres. Give your answers to the nearest minute.
\includegraphics[max width=\textwidth, alt={}, center]{a9870c94-0910-46ec-a54a-44a431cb324e-18_2257_54_314_1977}

Edexcel C34 2019 June Q8

8. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{a9870c94-0910-46ec-a54a-44a431cb324e-22_524_1443_260_246} \captionsetup{labelformat=empty} \caption{Figure 2}

\end{figure} Figure 2 shows a sketch of part of the curve $y = \mathrm { f } ( x )$, where $$\mathrm { f } ( x ) = \frac { 6 x + 2 } { 3 x ^ { 2 } + 5 } , \quad x \in \mathbb { R }$$

Find $\mathrm { f } ^ { \prime } ( x )$, writing your answer as a single fraction in its simplest form. The curve has two turning points, a maximum at point $A$ and a minimum at point $B$, as shown in Figure 2.
Using part (a), find the coordinates of point $A$ and the coordinates of point $B$.
State the coordinates of the maximum turning point of the function with equation $$y = \mathrm { f } ( 2 x ) + 4 \quad x \in \mathbb { R }$$
Find the range of the function $$\operatorname { g } ( x ) = \frac { 6 x + 2 } { 3 x ^ { 2 } + 5 } , \quad x \leqslant 0$$

Edexcel C34 2019 June Q9

9. (a) Using the formula for $\sin ( A + B )$ and the relevant double angle formulae, find an
identity for $\sin 3 x$, giving your answer in the form $$\sin ( 3 x ) \equiv P \sin x + Q \sin ^ { 3 } x$$ where $P$ and $Q$ are constants to be determined.
(b) Hence, showing each step of your working, evaluate $$\int _ { \frac { \pi } { 6 } } ^ { \frac { \pi } { 2 } } \sin 3 x \cos x d x$$ (Solutions based entirely on graphical or numerical methods are not acceptable.)

VIIIV SIHI NI III M LON OC

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Edexcel C34 2019 June Q10

(a) Use the binomial series to find the expansion of

$$\frac { 1 } { ( 2 + 3 x ) ^ { 3 } } \quad | x | < \frac { 2 } { 3 }$$ in ascending powers of $x$, up to and including the term in $x ^ { 2 }$, giving each term as a simplified fraction.
(b) Hence or otherwise, find the coefficient of $x ^ { 2 }$ in the series expansion of

$\frac { 1 } { ( 2 + 6 x ) ^ { 3 } } \quad | x | < \frac { 1 } { 3 }$
$\frac { 4 - x } { ( 2 + 3 x ) ^ { 3 } } \quad | x | < \frac { 2 } { 3 }$

Edexcel C34 2019 June Q11

11. (a) Given $$\frac { 9 } { t ^ { 2 } ( t - 3 ) } \equiv \frac { A } { t } + \frac { B } { t ^ { 2 } } + \frac { C } { ( t - 3 ) }$$ find the value of the constants $A , B$ and $C$.
(b) $$I = \int _ { 4 } ^ { 12 } \frac { 9 } { t ^ { 2 } ( t - 3 ) } \mathrm { d } t$$ Find the exact value of $I$, giving your answer in the form $\ln ( a ) - b$, where $a$ and $b$ are positive constants. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{a9870c94-0910-46ec-a54a-44a431cb324e-34_535_880_959_525} \captionsetup{labelformat=empty} \caption{Figure 3}

\end{figure} Figure 3 shows a sketch of part of the curve $C$ with parametric equations $$x = 2 \ln ( t - 3 ) , \quad y = \frac { 6 } { t } \quad t > 3$$ The finite region $R$, shown shaded in Figure 3, is bounded by the curve $C$, the $y$-axis, the $x$-axis and the line with equation $x = 2 \ln 9$ The region $R$ is rotated $360 ^ { \circ }$ about the $x$-axis to form a solid of revolution.
(c) Show that the exact volume of the solid generated is $$k \times I$$ where $k$ is a constant to be found.

Edexcel C34 2019 June Q12

Relative to a fixed origin $O$,
the point $A$ has position vector $( 2 \mathbf { i } - 3 \mathbf { j } - 2 \mathbf { k } )$
the point $B$ has position vector $( 3 \mathbf { i } + 2 \mathbf { j } + 5 \mathbf { k } )$
the point $C$ has position vector ( $2 \mathbf { i } + 4 \mathbf { j } - 3 \mathbf { k }$ )

The line $l$ passes through the points $A$ and $B$.

Find the vector $\overrightarrow { A B }$.
Find a vector equation for the line $l$.
Show that the size of the angle $C A B$ is $62.8 ^ { \circ }$, to one decimal place.
Hence find the area of triangle $C A B$, giving your answer to 3 significant figures. The point $D$ lies on the line $l$. Given that the area of triangle $C A D$ is twice the area of triangle $C A B$,
find the two possible position vectors of point $D$.

Edexcel C34 2019 June Q13

13. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{a9870c94-0910-46ec-a54a-44a431cb324e-42_649_709_242_614} \captionsetup{labelformat=empty} \caption{Figure 4}

\end{figure} Figure 4 shows a sketch of the curve with equation $y = 12 x ^ { 2 } \ln \left( 2 x ^ { 2 } \right) , x > 0$ The finite region $R$, shown shaded in Figure 4, is bounded by the curve, the line with equation $x = 1$, the $x$-axis and the line with equation $x = 2$ The table below shows corresponding values of $x$ and $y$ for $y = 12 x ^ { 2 } \ln \left( 2 x ^ { 2 } \right)$, with the values of $y$ given to 3 significant figures.

$x$	1	1.25	1.5	1.75	2
$y$	8.32	21.4	40.6	66.6	99.8

Use the trapezium rule, with all the values of $y$, to obtain an estimate for the area of $R$, giving your answer to 2 significant figures.
Use the substitution $u = x ^ { 2 }$ to show that the area of $R$ is given by $$\int _ { 1 } ^ { 4 } 6 u ^ { \frac { 1 } { 2 } } \ln ( 2 u ) \mathrm { d } u$$
Hence, using calculus, find the exact area of $R$, writing your answer in the form $a + b \ln 2$, where $a$ and $b$ are constants to be found.

Edexcel C34 2019 June Q14

14. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{a9870c94-0910-46ec-a54a-44a431cb324e-46_524_855_255_539} \captionsetup{labelformat=empty} \caption{Figure 5}

\end{figure} Figure 5 shows a sketch of the curves $C _ { 1 }$ and $C _ { 2 }$ $$\begin{aligned} & C _ { 1 } \text { has equation } y = 3 + \mathrm { e } ^ { x + 1 } \quad x \in \mathbb { R }
& C _ { 2 } \text { has equation } y = 10 - \mathrm { e } ^ { x } \quad x \in \mathbb { R } \end{aligned}$$ Given that $C _ { 1 }$ and $C _ { 2 }$ cut the $y$-axis at the points $P$ and $Q$ respectively,

find the exact distance $P Q$.
$C _ { 1 }$ and $C _ { 2 }$ intersect at the point $R$.
Find the exact coordinates of $R$.
VIIIV SIHI NI IAIUM ION OC VIIV SIHI NI JIIIM ION OC VIIV SIHI NI JIIYM ION OC

Edexcel C34 2017 October Q1

1. $$f ( x ) = x ^ { 5 } + x ^ { 3 } - 12 x ^ { 2 } - 8 , \quad x \in \mathbb { R }$$

Show that the equation $\mathrm { f } ( x ) = 0$ can be written as $$x = \sqrt [ 3 ] { \frac { 4 \left( 3 x ^ { 2 } + 2 \right) } { x ^ { 2 } + 1 } }$$
Use the iterative formula $$x _ { n + 1 } = \sqrt [ 3 ] { \frac { 4 \left( 3 x _ { n } ^ { 2 } + 2 \right) } { x _ { n } ^ { 2 } + 1 } }$$ with $x _ { 0 } = 2$, to find $x _ { 1 } , x _ { 2 }$ and $x _ { 3 }$ giving your answers to 3 decimal places. The equation $\mathrm { f } ( x ) = 0$ has a single root, $\alpha$.
By choosing a suitable interval, prove that $\alpha = 2.247$ to 3 decimal places.

Edexcel C34 2017 October Q2

2. The curve $C$ has equation $$y ^ { 3 } + x ^ { 2 } y - 6 x = 0$$

Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
Hence find the exact coordinates of the points on $C$ for which $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$

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