Edexcel C34 (Core Mathematics 3 & 4) 2018 June

1. (i) Find

$$\int \frac { 2 x ^ { 2 } + 5 x + 1 } { x ^ { 2 } } \mathrm {~d} x , \quad x > 0$$ (ii) Find $$\int x \cos 2 x \mathrm {~d} x$$

2. A curve \(C\) has parametric equations $$x = \frac { 3 } { 2 } t - 5 , \quad y = 4 - \frac { 6 } { t } \quad t \neq 0$$

1. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at \(t = 3\), giving your answer as a fraction in its simplest form.
2. Show that a cartesian equation of \(C\) can be expressed in the form $$y = \frac { a x + b } { x + 5 } \quad x \neq k$$ where \(a , b\) and \(k\) are integers to be found.

3. $$f ( x ) = 2 ^ { x - 1 } - 4 + 1.5 x \quad x \in \mathbb { R }$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as $$x = \frac { 1 } { 3 } \left( 8 - 2 ^ { x } \right)$$ The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\), where \(\alpha = 1.6\) to one decimal place.
2. Starting with \(x _ { 0 } = 1.6\), use the iteration formula $$x _ { n + 1 } = \frac { 1 } { 3 } \left( 8 - 2 ^ { x _ { n } } \right)$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 3 decimal places.
3. By choosing a suitable interval, prove that \(\alpha = 1.633\) to 3 decimal places.

4. (a) Find the binomial expansion of $$( 1 + p x ) ^ { - 4 } , \quad | p x | < 1$$ in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\), giving each coefficient as simply as possible in terms of the constant \(p\). $$f ( x ) = \frac { 3 + 4 x } { ( 1 + p x ) ^ { 4 } } \quad | p x | < 1$$ where \(p\) is a positive constant. In the series expansion of \(\mathrm { f } ( x )\), the coefficient of \(x ^ { 2 }\) is twice the coefficient of \(x\).
(b) Find the value of \(p\).
(c) Hence find the coefficient of \(x ^ { 3 }\) in the series expansion of \(\mathrm { f } ( x )\), giving your answer as a simplified fraction.

1. 1. The functions \(f\) and \(g\) are defined by

$$\begin{array} { l l } \mathrm { f } : x \rightarrow \mathrm { e } ^ { 2 x } - 5 , & x \in \mathbb { R }
\mathrm {~g} : x \rightarrow \ln ( 3 x - 1 ) , & x \in \mathbb { R } , x > \frac { 1 } { 3 } \end{array}$$

1. Find \(\mathrm { f } ^ { - 1 }\) and state its domain.
2. Find \(\mathrm { fg } ( 3 )\), giving your answer in its simplest form.
   (ii) (a) Sketch the graph with equation $$y = | 4 x - a |$$ where \(a\) is a positive constant. State the coordinates of each point where the graph cuts or meets the coordinate axes. Given that $$| 4 x - a | = 9 a$$ where \(a\) is a positive constant,
3. find the possible values of $$| x - 6 a | + 3 | x |$$ giving your answers, in terms of \(a\), in their simplest form.

6. (a) Express \(\sqrt { 5 } \cos \theta - 2 \sin \theta\) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { \pi } { 2 }\) State the value of \(R\) and give the value of \(\alpha\) to 4 significant figures.
(b) Solve, for \(- \pi < \theta < \pi\), $$\sqrt { 5 } \cos \theta - 2 \sin \theta = 0.5$$ giving your answers to 3 significant figures. [Solutions based entirely on graphical or numerical methods are not acceptable.] $$\mathrm { f } ( x ) = A ( \sqrt { 5 } \cos \theta - 2 \sin \theta ) + B \quad \theta \in \mathbb { R }$$ where \(A\) and \(B\) are constants. Given that the range of f is $$- 15 \leqslant f ( x ) \leqslant 33$$ (c) find the value of \(B\) and the possible values of \(A\).

7. \begin{figure}[h] \end{figure} Figure 1 shows a hemispherical bowl.
Water is flowing into the bowl at a constant rate of \(180 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).
When the height of the water is \(h \mathrm {~cm}\), the volume of water \(V \mathrm {~cm} ^ { 3 }\) is given by $$V = \frac { 1 } { 3 } \pi h ^ { 2 } ( 90 - h ) , \quad 0 \leqslant h \leqslant 30$$ Find the rate of change of the height of the water, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), when \(h = 15\) Give your answer to 2 significant figures.

8. With respect to a fixed origin \(O\), the lines \(l _ { 1 }\) and \(l _ { 2 }\) are given by the equations $$l _ { 1 } : \mathbf { r } = \left( \begin{array} { r } 1
- 3
2 \end{array} \right) + \lambda \left( \begin{array} { l } 1
2
3 \end{array} \right) , \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { l } 6
4
1 \end{array} \right) + \mu \left( \begin{array} { r } 1
1
- 1 \end{array} \right)$$ where \(\lambda\) and \(\mu\) are scalar parameters.

1. 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 \(\mu = - 1\)
2. Find the acute angle between the line segment \(P Q\) and \(l _ { 1 }\), giving your answer in degrees to 2 decimal places.
3. Find the shortest distance from the point \(Q\) to the line \(l _ { 1 }\), giving your answer to 3 significant figures.

9. \begin{figure}[h] \end{figure} Diagram not drawn to scale

1. 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.
2. Find the exact value of the volume of the solid generated, giving your answer in its simplest form.
   \section*{Leave
   k}

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

11. \begin{figure}[h] \end{figure}

1. 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 \(x = 6\)
2. 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.]

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.]

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

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

1. 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 $$\mathrm { f } ( x ) = \frac { 24 \left( x ^ { 2 } - 4 \right) ^ { \frac { 1 } { 2 } } } { x ^ { 3 } } \quad x > 2$$
2. Use your answer to part (a) to find the range of f.
3. State a reason why f-1 does not exist.