Edexcel C4 (Core Mathematics 4) 2017 June

1. The curve \(C\) has parametric equations

$$x = 3 t - 4 , y = 5 - \frac { 6 } { t } , \quad t > 0$$

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\) The point \(P\) lies on \(C\) where \(t = \frac { 1 } { 2 }\)
2. Find the equation of the tangent to \(C\) at the point \(P\). Give your answer in the form \(y = p x + q\), where \(p\) and \(q\) are integers to be determined.
3. Show that the cartesian equation for \(C\) can be written in the form $$y = \frac { a x + b } { x + 4 } , \quad x > - 4$$ where \(a\) and \(b\) are integers to be determined.

2. \(\quad \mathrm { f } ( x ) = ( 2 + k x ) ^ { - 3 } , \quad | k x | < 2\), where \(k\) is a positive constant The binomial expansion of \(\mathrm { f } ( x )\), in ascending powers of \(x\), up to and including the term in \(x ^ { 2 }\) is $$A + B x + \frac { 243 } { 16 } x ^ { 2 }$$ where \(A\) and \(B\) are constants.

1. Write down the value of \(A\).
2. Find the value of \(k\).
3. Find the value of \(B\).

3. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation \(y = \frac { 6 } { \left( \mathrm { e } ^ { x } + 2 \right) } , x \in \mathbb { R }\)
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 \(x = 1\) The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 6 } { \left( \mathrm { e } ^ { x } + 2 \right) }\)

1. Complete the table above by giving the missing value of \(y\) to 5 decimal places.
2. 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.
3. Use the substitution \(u = \mathrm { e } ^ { x }\) to show that the area of \(R\) can be given by $$\int _ { a } ^ { b } \frac { 6 } { u ( u + 2 ) } \mathrm { d } u$$ where \(a\) and \(b\) are constants to be determined.
4. Hence use calculus to find the exact area of \(R\). [Solutions based entirely on graphical or numerical methods are not acceptable.]

4. The curve \(C\) has equation $$4 x ^ { 2 } - y ^ { 3 } - 4 x y + 2 ^ { y } = 0$$ The point \(P\) with coordinates \(( - 2,4 )\) lies on \(C\).

1. Find the exact value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at the point \(P\). The normal to \(C\) at \(P\) meets the \(y\)-axis at the point \(A\).
2. Find the \(y\) coordinate of \(A\), giving your answer in the form \(p + q \ln 2\), where \(p\) and \(q\) are constants to be determined.
   (3)

5. \begin{figure}[h] \end{figure} Diagram not drawn to scale The finite region \(S\), shown shaded in Figure 2, is bounded by the \(y\)-axis, the \(x\)-axis, the line with equation \(x = \ln 4\) and the curve with equation $$y = \mathrm { e } ^ { x } + 2 \mathrm { e } ^ { - x } , \quad x \geqslant 0$$ The region \(S\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
Use integration to find the exact value of the volume of the solid generated. Give your answer in its simplest form.
[0pt] [Solutions based entirely on graphical or numerical methods are not acceptable.]

6. With respect to a fixed origin \(O\), the lines \(l _ { 1 }\) and \(l _ { 2 }\) are given by the equations $$l _ { 1 } : \mathbf { r } = \left( \begin{array} { r } 4
28
4 \end{array} \right) + \lambda \left( \begin{array} { r } - 1
- 5
1 \end{array} \right) , \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { l } 5
3
1 \end{array} \right) + \mu \left( \begin{array} { r } 3
0
- 4 \end{array} \right)$$ where \(\lambda\) and \(\mu\) are scalar parameters. The lines \(l _ { 1 }\) and \(l _ { 2 }\) intersect at the point \(X\).

1. Find the coordinates of the point \(X\).
2. Find the size of the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\), giving your answer in degrees to 2 decimal places. The point \(A\) lies on \(l _ { 1 }\) and has position vector \(\left( \begin{array} { r } 2
   18
   6 \end{array} \right)\)
3. Find the distance \(A X\), giving your answer as a surd in its simplest form. The point \(Y\) lies on \(l _ { 2 }\). Given that the vector \(\overrightarrow { Y A }\) is perpendicular to the line \(l _ { 1 }\)
4. find the distance \(Y A\), giving your answer to one decimal place. The point \(B\) lies on \(l _ { 1 }\) where \(| \overrightarrow { A X } | = 2 | \overrightarrow { A B } |\).
5. Find the two possible position vectors of \(B\).

7. \begin{figure}[h] \end{figure} Figure 3 shows a vertical cylindrical tank of height 200 cm containing water. Water is leaking from a hole \(P\) on the side of the tank. At time \(t\) minutes after the leaking starts, the height of water in the tank is \(h \mathrm {~cm}\). The height \(h \mathrm {~cm}\) of the water in the tank satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = k ( h - 9 ) ^ { \frac { 1 } { 2 } } , \quad 9 < h \leqslant 200$$ where \(k\) is a constant. Given that, when \(h = 130\), the height of the water is falling at a rate of 1.1 cm per minute,

1. find the value of \(k\). Given that the tank was full of water when the leaking started,
2. solve the differential equation with your value of \(k\), to find the value of \(t\) when \(h = 50\)

8. \begin{figure}[h] \end{figure} Diagram not drawn to scale Figure 4 shows a sketch of part of the curve \(C\) with parametric equations $$x = 3 \theta \sin \theta , \quad y = \sec ^ { 3 } \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$ The point \(P ( k , 8 )\) lies on \(C\), where \(k\) is a constant.

1. Find the exact value of \(k\). The finite region \(R\), shown shaded in Figure 4, is bounded by the curve \(C\), the \(y\)-axis, the \(x\)-axis and the line with equation \(x = k\).
2. Show that the area of \(R\) can be expressed in the form $$\lambda \int _ { \alpha } ^ { \beta } \left( \theta \sec ^ { 2 } \theta + \tan \theta \sec ^ { 2 } \theta \right) \mathrm { d } \theta$$ where \(\lambda , \alpha\) and \(\beta\) are constants to be determined.
3. Hence use integration to find the exact value of the area of \(R\).