Edexcel C34 (Core Mathematics 3 & 4) 2019 June

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

1. 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.
2. 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.
3. Using a suitable interval, show that 2.077 is a root of the equation \(\mathrm { f } ( x ) = 0\) correct to 3 decimal places.
4. Hence state a root, to 3 decimal places, of the equation $$2 ( x + 2 ) ^ { 3 } + x - 18 = 0$$

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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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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 $$y = \mathrm { f } ( x ) , \quad 0 \leqslant x \leqslant k , \quad \text { where } k \text { is a constant }$$

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

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

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). The point \(P\) on \(C\) has coordinates \(( 0,2 )\).
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)

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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 \(38 ^ { \circ } \mathrm { C }\),

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

6. \begin{figure}[h] \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. 1. State the coordinates of \(P\).
   2. State the coordinates of \(Q\).
2. 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.
3. 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}
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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.
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8. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = \frac { 6 x + 2 } { 3 x ^ { 2 } + 5 } , \quad x \in \mathbb { R }$$

1. 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.
2. Using part (a), find the coordinates of point \(A\) and the coordinates of point \(B\).
3. State the coordinates of the maximum turning point of the function with equation $$y = \mathrm { f } ( 2 x ) + 4 \quad x \in \mathbb { R }$$
4. Find the range of the function $$\operatorname { g } ( x ) = \frac { 6 x + 2 } { 3 x ^ { 2 } + 5 } , \quad x \leqslant 0$$

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.)

1. (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

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

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] \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.

1. 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 \(B\).

1. Find the vector \(\overrightarrow { A B }\).
2. Find a vector equation for the line \(l\).
3. Show that the size of the angle \(C A B\) is \(62.8 ^ { \circ }\), to one decimal place.
4. 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 \(C A B\),
5. find the two possible position vectors of point \(D\).
   

13. \begin{figure}[h] \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.

1. 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.
2. Use the substitution \(u = x ^ { 2 }\) to show that the area of \(R\) is given by $$\int _ { 1 } ^ { 4 } 6 u ^ { \frac { 1 } { 2 } } \ln ( 2 u ) \mathrm { d } u$$
3. 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.

14. \begin{figure}[h] \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,

1. find the exact distance \(P Q\).
   \(C _ { 1 }\) and \(C _ { 2 }\) intersect at the point \(R\).
2. Find the exact coordinates of \(R\).

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