Questions — Edexcel (9670 questions)

7. The function \(f\) is defined by $$f : x \wp \rightarrow | 2 x - a | , x \in \mathbb { R }$$ where \(a\) is a positive constant.

1. Sketch the graph of \(y = \mathrm { f } ( x )\), showing the coordinates of the points where the graph cuts the axes.
2. On a separate diagram, sketch the graph of \(y = \mathrm { f } ( 2 x )\), showing the coordinates of the points where the graph cuts the axes.
3. Given that a solution of the equation \(\mathrm { f } ( x ) = \frac { 1 } { 2 } x\) is \(x = 4\), find the two possible values of \(a\).

8. (a) Prove that $$\frac { 1 - \cos 2 \theta } { \sin 2 \theta } \equiv \tan \theta , \theta \neq \frac { n \pi } { 2 } , \quad n \in \mathbb { Z }$$ (b) Solve, giving exact answers in terms of \(\pi\), $$2 ( 1 - \cos 2 \theta ) = \tan \theta , \quad 0 < \theta < \pi$$ [P2 January 2002 Question 6]

9. \begin{figure}[h] \end{figure} Figure 2 shows part of the curve \(C\) with equation \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = 0.5 \mathrm { e } ^ { x } - x ^ { 2 } .$$ The curve \(C\) cuts the \(y\)-axis at \(A\) and there is a minimum at the point \(B\).

1. Find an equation of the tangent to \(C\) at \(A\). The \(x\)-coordinate of \(B\) is approximately 2.15 . A more exact estimate is to be made of this coordinate using iterations \(x _ { n + 1 } = \ln \mathrm { g } \left( x _ { n } \right)\).
2. Show that a possible form for \(\mathrm { g } ( x )\) is \(\mathrm { g } ( x ) = 4 x\).
3. Using \(x _ { n + 1 } = \ln 4 x _ { n }\), with \(x _ { 0 } = 2.15\), calculate \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\). Give the value of \(x _ { 3 }\) to 4 decimal places.

10. $$\mathrm { f } ( x ) = \frac { 2 } { x - 1 } - \frac { 6 } { ( x - 1 ) ( 2 x + 1 ) } , x > 1$$

1. Prove that \(\mathrm { f } ( x ) = \frac { 4 } { 2 x + 1 }\).
2. Find the range of f.
3. Find \(\mathrm { f } ^ { - 1 } ( x )\).
4. Find the range of \(\mathrm { f } ^ { - 1 } ( x )\).

11. Use the derivatives of \(\sin x\) and \(\cos x\) to prove that the derivative of \(\tan x\) is \(\sec ^ { 2 } x\).

12. Express \(\frac { 3 } { x ^ { 2 } + 2 x } + \frac { x - 4 } { x ^ { 2 } - 4 }\) as a single fraction in its simplest form.
[0pt] [P2 June 2002 Question 2]

13. Figure 1
\includegraphics[max width=\textwidth, alt={}, center]{d0c23635-3b9b-4666-9cb4-21b931fb3719-06_626_759_313_537} Figure 1 shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = 10 + \ln ( 3 x ) - \frac { 1 } { 2 } \mathrm { e } ^ { x } , 0.1 \leq x \leq 3.3$$ Given that \(\mathrm { f } ( k ) = 0\),

1. show, by calculation, that \(3.1 < k < 3.2\).
2. Find \(\mathrm { f } ^ { \prime } ( x )\). The tangent to the graph at \(x = 1\) intersects the \(y\)-axis at the point \(P\).
3. 1. Find an equation of this tangent.
   2. Find the exact \(y\)-coordinate of \(P\), giving your answer in the form \(a + \ln b\).

14. $$\mathrm { f } ( x ) = x ^ { 2 } - 2 x - 3 , x \in \mathbb { R } , x \geq 1 .$$

1. Find the range of f .
2. Write down the domain and range of \(\mathrm { f } ^ { - 1 }\).
3. Sketch the graph of \(\mathrm { f } ^ { - 1 }\), indicating clearly the coordinates of any point at which the graph intersects the coordinate axes. Given that \(\mathrm { g } ( x ) = | x - 4 | , x \in \mathbb { R }\),
4. find an expression for \(\operatorname { gf } ( x )\).
5. Solve \(\operatorname { gf } ( x ) = 8\).

15. Express \(\frac { y + 3 } { ( y + 1 ) ( y + 2 ) } - \frac { y + 1 } { ( y + 2 ) ( y + 3 ) }\) as a single fraction in its simplest form.

16. (a) Express \(1.5 \sin 2 x + 2 \cos 2 x\) in the form \(R \sin ( 2 x + \alpha )\), where \(R > 0\) and \(0 < \alpha < \frac { 1 } { 2 } \pi\), giving your values of \(R\) and \(\alpha\) to 3 decimal places where appropriate.
(b) Express \(3 \sin x \cos x + 4 \cos ^ { 2 } x\) in the form \(a \cos 2 x + b \sin 2 x + c\), where \(a , b\) and \(c\) are constants to be found.
(c) Hence, using your answer to part (a), deduce the maximum value of \(3 \sin x \cos x + 4 \cos ^ { 2 } x\).

17. The curve \(C\) with equation \(y = p + q \mathrm { e } ^ { x }\), where \(p\) and \(q\) are constants, passes through the point \(( 0,2 )\). At the point \(P ( \ln 2 , p + 2 q )\) on \(C\), the gradient is 5 .

1. Find the value of \(p\) and the value of \(q\). The normal to \(C\) at \(P\) crosses the \(x\)-axis at \(L\) and the \(y\)-axis at \(M\).
2. Show that the area of \(\triangle O L M\), where \(O\) is the origin, is approximately 53.8. \section*{18.} \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{d0c23635-3b9b-4666-9cb4-21b931fb3719-08_487_695_259_683}
   \end{figure} Figure 1 shows a sketch of the curve with equation \(y = \mathrm { e } ^ { - x } - 1\).
3. Copy Fig. 1 and on the same axes sketch the graph of \(y = \frac { 1 } { 2 } | x - 1 |\). Show the coordinates of the points where the graph meets the axes. The \(x\)-coordinate of the point of intersection of the graph is \(\alpha\).
4. Show that \(x = \alpha\) is a root of the equation \(x + 2 \mathrm { e } ^ { - x } - 3 = 0\).
5. Show that \(- 1 < \alpha < 0\). The iterative formula \(x _ { \mathrm { n } + 1 } = - \ln \left[ \frac { 1 } { 2 } \left( 3 - x _ { n } \right) \right]\) is used to solve the equation \(x + 2 \mathrm { e } ^ { - x } - 3 = 0\).
6. Starting with \(x _ { 0 } = - 1\), find the values of \(x _ { 1 }\) and \(x _ { 2 }\).
7. Show that, to 2 decimal places, \(\alpha = - 0.58\).

19. The function f is defined by \(\mathrm { f } : x \mapsto \frac { 3 x - 1 } { x - 3 } , x \in \mathbb { R } , x \neq 3\).

1. Prove that \(\mathrm { f } ^ { - 1 } ( x ) = \mathrm { f } ( x )\) for all \(x \in \mathbb { R } , x \neq 3\).
2. Hence find, in terms of \(k , \operatorname { ff } ( k )\), where \(x \neq 3\). \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 3} \includegraphics[alt={},max width=\textwidth]{d0c23635-3b9b-4666-9cb4-21b931fb3719-09_817_1139_623_406}
   \end{figure} Figure 3 shows a sketch of the one-one function g , defined over the domain \(- 2 \leq x \leq 2\).
3. Find the value of \(\mathrm { fg } ( - 2 )\).
4. Sketch the graph of the inverse function \(\mathrm { g } ^ { - 1 }\) and state its domain. The function h is defined by \(\mathrm { h } : x \mapsto 2 \mathrm {~g} ( x - 1 )\).
5. Sketch the graph of the function h and state its range.

20. Express \(\frac { x } { ( x + 1 ) ( x + 3 ) } + \frac { x + 12 } { x ^ { 2 } - 9 }\) as a single fraction in its simplest form.
21. (a) Sketch the graph of \(y = | 2 x + a | , a > 0\), showing the coordinates of the points where the graph meets the coordinate axes.
(b) On the same axes, sketch the graph of \(y = \frac { 1 } { x }\).
(c) Explain how your graphs show that there is only one solution of the equation $$x | 2 x + a | - 1 = 0$$ (d) Find, using algebra, the value of \(x\) for which \(x | 2 x + 1 | - 1 = 0\).
22. The curve with equation \(y = \ln 3 x\) crosses the \(x\)-axis at the point \(P ( p , 0 )\).
(a) Sketch the graph of \(y = \ln 3 x\), showing the exact value of \(p\). The normal to the curve at the point \(Q\), with \(x\)-coordinate \(q\), passes through the origin.
(b) Show that \(x = q\) is a solution of the equation \(x ^ { 2 } + \ln 3 x = 0\).
(c) Show that the equation in part (b) can be rearranged in the form \(x = \frac { 1 } { 3 } \mathrm { e } ^ { - x ^ { 2 } }\).
(d) Use the iteration formula \(x _ { n + 1 } = \frac { 1 } { 3 } \mathrm { e } ^ { - x _ { n } ^ { 2 } }\), with \(x _ { 0 } = \frac { 1 } { 3 }\), to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\). Hence write down, to 3 decimal places, an approximation for \(q\).
23. (a) Express \(\sin x + \sqrt { 3 } \cos x\) in the form \(R \sin ( x + \alpha )\), where \(R > 0\) and \(0 < \alpha < 90 ^ { \circ }\).
(b) Show that the equation \(\sec x + \sqrt { 3 } \operatorname { cosec } x = 4\) can be written in the form $$\sin x + \sqrt { 3 } \cos x = 2 \sin 2 x$$ (c) Deduce from parts (a) and (b) that sec \(x + \sqrt { 3 } \operatorname { cosec } x = 4\) can be written in the form $$\sin 2 x - \sin \left( x + 60 ^ { \circ } \right) = 0$$ 24. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of the curve with equation \(y = \mathrm { f } ( x ) , x \geq 0\). The curve meets the coordinate axes at the points \(( 0 , c )\) and \(( d , 0 )\). In separate diagrams sketch the curve with equation
(a) \(y = \mathrm { f } ^ { - 1 } ( x )\),
(b) \(y = 3 \mathrm { f } ( 2 x )\). Indicate clearly on each sketch the coordinates, in terms of \(c\) or \(d\), of any point where the curve meets the coordinate axes. Given that f is defined by $$\mathrm { f } : x \mapsto 3 \left( 2 ^ { - x } \right) - 1 , x \in \mathbb { R } , x \geq 0 ,$$ (c) state
(i) the value of \(c\),
(ii) the range of \(f\).
(d) Find the value of \(d\), giving your answer to 3 decimal places. The function g is defined by $$\mathrm { g } : x \rightarrow \log _ { 2 } x , x \in \mathbb { R } , x \geq 1 .$$ (e) Find \(\mathrm { fg } ( x )\), giving your answer in its simplest form.
25. (a) Simplify \(\frac { x ^ { 2 } + 4 x + 3 } { x ^ { 2 } + x }\).
(b) Find the value of \(x\) for which \(\log _ { 2 } \left( x ^ { 2 } + 4 x + 3 \right) - \log _ { 2 } \left( x ^ { 2 } + x \right) = 4\).
26. The functions \(f\) and \(g\) are defined by $$\begin{aligned} & \mathrm { f } : x \mapsto x ^ { 2 } - 2 x + 3 , x \in \mathbb { R } , 0 \leq x \leq 4
& \mathrm {~g} : x \mapsto \lambda x ^ { 2 } + 1 , \text { where } \lambda \text { is a constant, } x \in \mathbb { R } . \end{aligned}$$ (a) Find the range of f .
(b) Given that \(\operatorname { gf } ( 2 ) = 16\), find the value of \(\lambda\).
27. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve with equation \(y = \mathrm { f } ( x ) , - 1 \leq x \leq 3\). The curve touches the \(x\)-axis at the origin \(O\), crosses the \(x\)-axis at the point \(A ( 2,0 )\) and has a maximum at the point \(B \left( \frac { 4 } { 3 } , 1 \right)\). In separate diagrams, show a sketch of the curve with equation
(a) \(y = \mathrm { f } ( x + 1 )\),
(b) \(y = | \mathrm { f } ( x ) |\),
(c) \(y = \mathrm { f } ( | x | )\),
marking on each sketch the coordinates of points at which the curve
(i) has a turning point,
(ii) meets the \(x\)-axis.
28. (a) Sketch, on the same set of axes, the graphs of $$y = 2 - \mathrm { e } ^ { - x } \text { and } y = \sqrt { } x$$ [It is not necessary to find the coordinates of any points of intersection with the axes.]
Given that \(\mathrm { f } ( x ) = \mathrm { e } ^ { - x } + \sqrt { } x - 2 , x \geq 0\),
(b) explain how your graphs show that the equation \(\mathrm { f } ( x ) = 0\) has only one solution,
(c) show that the solution of \(\mathrm { f } ( x ) = 0\) lies between \(x = 3\) and \(x = 4\). The iterative formula \(x _ { n + 1 } = \left( 2 - \mathrm { e } ^ { - x _ { n } } \right) ^ { 2 }\) is used to solve the equation \(\mathrm { f } ( x ) = 0\).
(d) Taking \(x _ { 0 } = 4\), write down the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), and hence find an approximation to the solution of \(\mathrm { f } ( x ) = 0\), giving your answer to 3 decimal places.
[0pt] [P2 June 2003 Question 5] 28a. (i) Given that \(\cos ( x + 30 ) ^ { \circ } = 3 \cos ( x - 30 ) ^ { \circ }\), prove that \(\tan x ^ { \circ } = - \frac { \sqrt { 3 } } { 2 }\).
(ii) (a) Prove that \(\frac { 1 - \cos 2 \theta } { \sin 2 \theta } \equiv \tan \theta\).
(b) Verify that \(\theta = 180 ^ { \circ }\) is a solution of the equation \(\sin 2 \theta = 2 - 2 \cos 2 \theta\).
(c) Using the result in part (a), or otherwise, find the other two solutions, \(0 < \theta < 360 ^ { \circ }\), of the equation using \(\sin 2 \theta = 2 - 2 \cos 2 \theta\).
[0pt] [P2 June 2003 Question 8]
29. (a) Express as a fraction in its simplest form $$\frac { 2 } { x - 3 } + \frac { 13 } { x ^ { 2 } + 4 x - 21 }$$ (b) Hence solve $$\frac { 2 } { x - 3 } + \frac { 13 } { x ^ { 2 } + 4 x - 21 } = 1$$ [P2 November 2003 Question 1]
30. Prove that $$\frac { 1 - \tan ^ { 2 } \theta } { 1 + \tan ^ { 2 } \theta } \equiv \cos 2 \theta$$

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

$$\begin{aligned} & \mathrm { f } : x \mapsto | x - a | + a , x \in \mathbb { R } ,
& \mathrm {~g} : x \mapsto 4 x + a , \quad x \in \mathbb { R } . \end{aligned}$$ where \(a\) is a positive constant.
(a) On the same diagram, sketch the graphs of f and g , showing clearly the coordinates of any points at which your graphs meet the axes.
(b) Use algebra to find, in terms of \(a\), the coordinates of the point at which the graphs of f and g intersect.
(c) Find an expression for \(\mathrm { fg } ( x )\).
(d) Solve, for \(x\) in terms of \(a\), the equation $$\mathrm { fg } ( x ) = 3 a$$

1. The curve \(C\) has equation \(y = \mathrm { f } ( x )\), where

$$\mathrm { f } ( x ) = 3 \ln x + \frac { 1 } { x } , \quad x > 0$$ The point \(P\) is a stationary point on \(C\).
(a) Calculate the \(x\)-coordinate of \(P\).
(b) Show that the \(y\)-coordinate of \(P\) may be expressed in the form \(k - k \ln k\), where \(k\) is a constant to be found. The point \(Q\) on \(C\) has \(x\)-coordinate 1 .
(c) Find an equation for the normal to \(C\) at \(Q\). The normal to \(C\) at \(Q\) meets \(C\) again at the point \(R\).
(d) Show that the \(x\)-coordinate of \(R\)
(i) satisfies the equation \(6 \ln x + x + \frac { 2 } { x } - 3 = 0\),
(ii) lies between 0.13 and 0.14 .
33. The function f is given by \(\mathrm { f } : x \mapsto 2 + \frac { 3 } { x + 2 } , x \in \mathbb { R } , x \neq - 2\).
(a) Express \(2 + \frac { 3 } { x + 2 }\) as a single fraction.
(b) Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
(c) Write down the domain of \(\mathrm { f } ^ { - 1 }\).
34. The function f is even and has domain \(\mathbb { R }\). For \(x \geq 0 , \mathrm { f } ( x ) = x ^ { 2 } - 4 a x\), where \(a\) is a positive constant.
(a) In the space below, sketch the curve with equation \(y = \mathrm { f } ( x )\), showing the coordinates of all the points at which the curve meets the axes.
(b) Find, in terms of \(a\), the value of \(\mathrm { f } ( 2 a )\) and the value of \(\mathrm { f } ( - 2 a )\). Given that \(a = 3\),
(c) use algebra to find the values of \(x\) for which \(\mathrm { f } ( x ) = 45\).
35. Given that \(y = \log _ { a } x , x > 0\), where \(a\) is a positive constant,
(a) (i) express \(x\) in terms of \(a\) and \(y\),
(ii) deduce that \(\ln x = y \ln a\).
(b) Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { x \ln a }\). The curve \(C\) has equation \(y = \log _ { 10 } x , x > 0\). The point \(A\) on \(C\) has \(x\)-coordinate 10 . Using the result in part (b),
(c) find an equation for the tangent to \(C\) at \(A\). The tangent to \(C\) at \(A\) crosses the \(x\)-axis at the point \(B\).
(d) Find the exact \(x\)-coordinate of \(B\).
36. (i) (a) Express ( \(12 \cos \theta - 5 \sin \theta\) ) in the form \(R \cos ( \theta + \alpha )\), where \(R > 0\) and \(0 < \alpha < 90 ^ { \circ }\).
(b) Hence solve the equation $$12 \cos \theta - 5 \sin \theta = 4$$ for \(0 < \theta < 90 ^ { \circ }\), giving your answer to 1 decimal place.
(ii) Solve $$8 \cot \theta - 3 \tan \theta = 2 ,$$ for \(0 < \theta < 90 ^ { \circ }\), giving your answer to 1 decimal place.
37. Express as a single fraction in its simplest form $$\frac { x ^ { 2 } - 8 x + 15 } { x ^ { 2 } - 9 } \times \frac { 2 x ^ { 2 } + 6 x } { ( x - 5 ) ^ { 2 } }$$ [P2 June 2004 Question 1]
38. (i) Given that \(\sin x = \frac { 3 } { 5 }\), use an appropriate double angle formula to find the exact value of \(\sec 2 x\).
(ii) Prove that $$\cot 2 x + \operatorname { cosec } 2 x \equiv \cot x , \quad \left( x \neq \frac { n \pi } { 2 } , n \in \mathrm { Z } \right)$$ [P2 June 2004 Question 2]
39. $$\mathrm { f } ( x ) = x ^ { 3 } + x ^ { 2 } - 4 x - 1$$ The equation \(\mathrm { f } ( x ) = 0\) has only one positive root, \(\alpha\).
(a) Show that \(\mathrm { f } ( x ) = 0\) can be rearranged as $$x = \sqrt { \left( \frac { 4 x + 1 } { x + 1 } \right) } , x \neq - 1$$ The iterative formula \(x _ { n + 1 } = \sqrt { \left( \frac { 4 x _ { n } + 1 } { x _ { n } + 1 } \right) }\) is used to find an approximation to \(\alpha\).
(b) Taking \(x _ { 1 } = 1\), find, to 2 decimal places, the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\).
(c) By choosing values of \(x\) in a suitable interval, prove that \(\alpha = 1.70\), correct to 2 decimal places.
(d) Write down a value of \(x _ { 1 }\) for which the iteration formula \(x _ { n + 1 } = \sqrt { \left( \frac { 4 x _ { n } + 1 } { x _ { n } + 1 } \right) }\) does not produce a valid value for \(x _ { 2 }\). Justify your answer.
40. $$\mathrm { f } ( x ) = x + \frac { \mathrm { e } ^ { x } } { 5 } , \quad x \in \mathbb { R }$$ (a) Find \(\mathrm { f } ^ { \prime } ( x )\). The curve \(C\), with equation \(y = \mathrm { f } ( x )\), crosses the \(y\)-axis at the point \(A\).
(b) Find an equation for the tangent to \(C\) at \(A\).
(c) Complete the table, giving the values of \(\sqrt { \left( x + \frac { \mathrm { e } ^ { x } } { 5 } \right) }\) to 2 decimal places.

1. The function f is given by

$$f : x \mapsto \ln ( 3 x - 6 ) , \quad x \in \mathbb { R } , \quad x > 2 .$$ (a) Find \(\mathrm { f } ^ { - 1 } ( x )\).
(b) Write down the domain of \(\mathrm { f } ^ { - 1 }\) and the range of \(\mathrm { f } ^ { - 1 }\).
(c) Find, to 3 significant figures, the value of \(x\) for which \(\mathrm { f } ( x ) = 3\). The function g is given by $$\mathrm { g } : x \mapsto \ln | 3 x - 6 | , \quad x \in \mathbb { R } , \quad x \neq 2 .$$ (d) Sketch the graph of \(y = \mathrm { g } ( x )\).
(e) Find the exact coordinates of all the points at which the graph of \(y = \mathrm { g } ( x )\) meets the coordinate axes.

1. Use the binomial theorem to expand

$$\sqrt { } ( 4 - 9 x ) , \quad | x | < \frac { 4 } { 9 } ,$$ in ascending powers of \(x\), up to and including the term in \(x ^ { 3 }\), simplifying each term.

3. (a) Express \(\frac { 5 x + 3 } { ( 2 x - 3 ) ( x + 2 ) }\) in partial fractions.
(b) Hence find the exact value of \(\int _ { 2 } ^ { 6 } \frac { 5 x + 3 } { ( 2 x - 3 ) ( x + 2 ) } \mathrm { d } x\), giving your answer as a single logarithm.

5. \begin{figure}[h] \end{figure} Figure 1 shows the graph of the curve with equation $$y = x \mathrm { e } ^ { 2 x } , \quad x \geq 0 .$$ The finite region \(R\) bounded by the lines \(x = 1\), the \(x\)-axis and the curve is shown shaded in Figure 1.

1. Use integration to find the exact value of the area for \(R\).
2. Complete the table with the values of \(y\) corresponding to \(x = 0.4\) and 0.8 .

   \(x\)	0	0.2	0.4	0.6	0.8	1
   \(y = x \mathrm { e } ^ { 2 x }\)	0	0.29836		1.99207		7.38906

3. Use the trapezium rule with all the values in the table to find an approximate value for this area, giving your answer to 4 significant figures.

6. A curve has parametric equations $$x = 2 \cot t , \quad y = 2 \sin ^ { 2 } t , \quad 0 < t \leq \frac { \pi } { 2 } .$$

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of the parameter \(t\).
2. Find an equation of the tangent to the curve at the point where \(t = \frac { \pi } { 4 }\).
3. Find a cartesian equation of the curve in the form \(y = \mathrm { f } ( x )\). State the domain on which the curve is defined.

7. The line \(l _ { 1 }\) has vector equation $$\mathbf { r } = \left( \begin{array} { l } 3
1
2 \end{array} \right) + \lambda \left( \begin{array} { r } 1
- 1
4 \end{array} \right)$$ and the line \(l _ { 2 }\) has vector equation $$\mathbf { r } = \left( \begin{array} { r } 0
4
- 2 \end{array} \right) + \mu \left( \begin{array} { r } 1
- 1
0 \end{array} \right) ,$$ where \(\lambda\) and \(\mu\) are parameters.
The lines \(l _ { 1 }\) and \(l _ { 2 }\) intersect at the point \(B\) and the acute angle between \(l _ { 1 }\) and \(l _ { 2 }\) is \(\theta\).

1. Find the coordinates of \(B\).
2. Find the value of \(\cos \theta\), giving your answer as a simplified fraction. The point \(A\), which lies on \(l _ { 1 }\), has position vector \(\mathbf { a } = 3 \mathbf { i } + \mathbf { j } + 2 \mathbf { k }\).
   The point \(C\), which lies on \(l _ { 2 }\), has position vector \(\mathbf { c } = 5 \mathbf { i } - \mathbf { j } - 2 \mathbf { k }\).
   The point \(D\) is such that \(A B C D\) is a parallelogram.
3. Show that \(| \overrightarrow { A B } | = | \overrightarrow { B C } |\).
4. Find the position vector of the point \(D\).

8. Liquid is pouring into a container at a constant rate of \(20 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) and is leaking out at a rate proportional to the volume of the liquid already in the container.

1. Explain why, at time \(t\) seconds, the volume, \(V \mathrm {~cm} ^ { 3 }\), of liquid in the container satisfies the differential equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = 20 - k V$$ where \(k\) is a positive constant. The container is initially empty.
2. By solving the differential equation, show that $$V = A + B \mathrm { e } ^ { - k t }$$ giving the values of \(A\) and \(B\) in terms of \(k\). Given also that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = 10\) when \(t = 5\),
3. find the volume of liquid in the container at 10 s after the start. Materials required for examination
   Mathematical Formulae (Green) Items included with question papers
   Nil Paper Reference(s)
   6666 \section*{Advanced Level} \section*{Monday 23 January 2006 - Afternoon} Candidates may use any calculator EXCEPT those with the facility for symbolic algebra, differentiation and/or integration. Thus candidates may NOT use calculators such as the Texas Instruments TI 89, TI 92, Casio CFX 9970G, Hewlett Packard HP 48G.

1. A measure of the effective voltage, \(M\) volts, in an electrical circuit is given by

$$M ^ { 2 } = \int _ { 0 } ^ { 1 } V ^ { 2 } \mathrm {~d} t$$ where \(V\) volts is the voltage at time \(t\) seconds. Pairs of values of \(V\) and \(t\) are given in the following table. Use the trapezium rule with five values of \(V ^ { 2 }\) to estimate the value of \(M\). \begin{figure}[h] \end{figure} Figure 1 shows part of a curve \(C\) with equation \(y = x ^ { 2 } + 3\). The shaded region is bounded by \(C\), the \(x\)-axis and the lines \(x = 1\) and \(x = 3\). The shaded region is rotated \(360 ^ { \circ }\) about the \(x\)-axis. Using calculus, calculate the volume of the solid generated. Give your answer as an exact multiple of \(\pi\).

3. (a) Given that \(a ^ { x } = \mathrm { e } ^ { k x }\), where \(a\) and \(k\) are constants, \(a > 0\) and \(x \in \mathbb { R }\), prove that \(k = \ln a\).
(b) Hence, using the derivative of \(\mathrm { e } ^ { k x }\), prove that when \(y = 2 ^ { x }\), $$\frac { \mathrm { d } y } { \mathrm { dx } } = 2 ^ { x } \ln 2 .$$ (c) Hence deduce that the gradient of the curve with equation \(y = 2 ^ { x }\) at the point \(( 2,4 )\) is \(\ln 16\).

4. $$\mathrm { f } ( x ) = ( 1 + 3 x ) ^ { - 1 } , | x | < \frac { 1 } { 3 }$$

1. Expand \(\mathrm { f } ( x )\) in ascending powers of \(x\) up to and including the term in \(x ^ { 3 }\).
2. Hence show that, for small \(x\), $$\frac { 1 + x } { 1 + 3 x } \approx 1 - 2 x + 6 x ^ { 2 } - 18 x ^ { 3 }$$
3. Taking a suitable value for \(x\), which should be stated, use the series expansion in part (b) to find an approximate value for \(\frac { 101 } { 103 }\), giving your answer to 5 decimal places.

5. (a) Use integration by parts to show that $$\int _ { 0 } ^ { \frac { \pi } { 4 } } x \sec ^ { 2 } x \mathrm {~d} x = \frac { 1 } { 4 } \pi - \frac { 1 } { 2 } \ln 2$$ \begin{figure}[h] \end{figure} The finite region \(R\), bounded by the equation \(y = x ^ { \frac { 1 } { 2 } } \sec x\), the line \(x = \frac { \pi } { 4 }\) and the \(x\)-axis is shown in Fig. 1. The region \(R\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
(b) Find the volume of the solid of revolution generated.
(c) Find the gradient of the curve with equation \(y = x ^ { \frac { 1 } { 2 } } \sec x\) at the point where \(x = \frac { \pi } { 4 }\).

6. Two submarines are travelling in straight lines through the ocean. Relative to a fixed origin, the vector equations of the two lines, \(l _ { 1 }\) and \(l _ { 2 }\), along which they travel are $$\begin{aligned} \mathbf { r } & = 3 \mathbf { i } + 4 \mathbf { j } - 5 \mathbf { k } + \lambda ( \mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k } )
\text { and } \mathbf { r } & = 9 \mathbf { i } + \mathbf { j } - 2 \mathbf { k } + \mu ( 4 \mathbf { i } + \mathbf { j } - \mathbf { k } ) , \end{aligned}$$ where \(\lambda\) and \(\mu\) are scalars.

1. Show that the submarines are moving in perpendicular directions.
2. Given that \(l _ { 1 }\) and \(l _ { 2 }\) intersect at the point \(A\), find the position vector of \(A\). The point \(b\) has position vector \(10 \mathbf { j } - 11 \mathbf { k }\).
3. Show that only one of the submarines passes through the point \(B\).
4. Given that 1 unit on each coordinate axis represents 100 m , find, in km , the distance \(A B\).

7. In a chemical reaction two substances combine to form a third substance. At time \(t , t \geq 0\), the concentration of this third substance is \(x\) and the reaction is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = k ( 1 - 2 x ) ( 1 - 4 x ) , \text { where } k \text { is a positive constant. }$$

1. Solve this differential equation and hence show that $$\ln \left| \frac { 1 - 2 x } { 1 - 4 x } \right| = 2 k t + c , \text { where } c \text { is an arbitrary constant. }$$
2. Given that \(x = 0\) when \(t = 0\), find an expression for \(x\) in terms of \(k\) and \(t\).
3. Find the limiting value of the concentration \(x\) as \(t\) becomes very large. \section*{8.} \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 2} \includegraphics[alt={},max width=\textwidth]{615ec68b-3a32-4309-bb54-acf39ed09f96-05_716_1026_326_468}
   \end{figure} Part of the design of a stained glass window is shown in Fig. 2. The two loops enclose an area of blue glass. The remaining area within the rectangle \(A B C D\) is red glass. The loops are described by the curve with parametric equations $$x = 3 \cos t , \quad y = 9 \sin 2 t , \quad 0 \leq t < 2 \pi .$$
4. Find the cartesian equation of the curve in the form \(y ^ { 2 } = \mathrm { f } ( x )\).
5. Show that the shaded area in Fig. 2, enclosed by the curve and the \(x\)-axis, is given by $$\int _ { 0 } ^ { \frac { \pi } { 2 } } A \sin 2 t \sin t \mathrm {~d} t , \text { stating the value of the constant } A \text {. }$$
6. Find the value of this integral. The sides of the rectangle \(A B C D\), in Fig. 2, are the tangents to the curve that are parallel to the coordinate axes. Given that 1 unit on each axis represents 1 cm ,
7. find the total area of the red glass.