CAIE Further Paper 2 (Further Paper 2) 2024 November

2 It is given that $$x = 1 + \frac { 1 } { t } \quad \text { and } \quad y = \cos ^ { - 1 } t \quad \text { for } 0 < t < 1$$

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { t ^ { 2 } } { \sqrt { 1 - t ^ { 2 } } }\).
   \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-05_2723_33_99_22}
2. Show that \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = - t ^ { a } \left( 1 - t ^ { 2 } \right) ^ { b } \left( 2 - t ^ { 2 } \right)\), where \(a\) and \(b\) are constants to be determined.

3 A curve has equation \(y = \mathrm { e } ^ { x }\) for \(\ln \frac { 4 } { 3 } \leqslant x \leqslant \ln \frac { 12 } { 5 }\). The area of the surface generated when the curve is rotated through \(2 \pi\) radians about the \(x\)-axis is denoted by \(A\).

1. Use the substitution \(u = \mathrm { e } ^ { x }\) to show that $$A = 2 \pi \int _ { \frac { 4 } { 3 } } ^ { \frac { 12 } { 5 } } \sqrt { 1 + u ^ { 2 } } \mathrm {~d} u$$
2. Use the substitution \(u = \sinh v\) to show that $$A = \pi \left( \frac { 904 } { 225 } + \ln \frac { 5 } { 3 } \right) .$$ \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-06_2716_38_109_2012}
   \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-07_2726_35_97_20}

4 The matrix \(\mathbf { A }\) is given by $$\mathbf { A } = \left( \begin{array} { r r r } - 11 & 1 & 8
0 & - 2 & 0
- 16 & 1 & 13 \end{array} \right)$$

1. Show that \(\left( \begin{array} { l } 1
   1
   1 \end{array} \right)\) is an eigenvector of \(\mathbf { A }\) and state the corresponding eigenvalue.
2. Show that the characteristic equation of \(\mathbf { A }\) is \(\lambda ^ { 3 } - 19 \lambda - 30 = 0\) and hence find the other eigenvalues of \(\mathbf { A }\).
   \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-08_2715_44_110_2006}
   \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-09_2726_33_97_22}
3. Use the characteristic equation of \(\mathbf { A }\) to find \(\mathbf { A } ^ { - 1 }\).

5 Find the particular solution of the differential equation $$6 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } - 5 \frac { \mathrm {~d} x } { \mathrm {~d} t } + x = t ^ { 2 } + t + 1$$ given that, when \(t = 0 , x = 12\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - 6\).
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\includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-10_2715_40_110_2007}
\includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-11_2726_35_97_20}

6
\includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-12_533_1532_278_264} The diagram shows the curve with equation \(y = \left( \frac { 1 } { 2 } \right) ^ { x }\) for \(0 \leqslant x \leqslant 1\), together with a set of \(N\) rectangles each of width \(\frac { 1 } { N }\).

1. By considering the sum of the areas of these rectangles, show that \(\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x > L _ { N }\), where $$L _ { N } = \frac { 1 } { 2 N \left( 2 ^ { \frac { 1 } { N } } - 1 \right) }$$ \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-12_2717_38_109_2009}
2. Use a similar method to find, in terms of \(N\), an upper bound \(U _ { N }\) for \(\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x\).
3. Find the least value of \(N\) such that \(U _ { N } - L _ { N } \leqslant 10 ^ { - 3 }\).
4. Given that \(\int _ { 0 } ^ { 1 } \left( \frac { 1 } { 2 } \right) ^ { x } \mathrm {~d} x = \frac { 1 } { 2 \ln 2 }\) ,use the value of \(N\) found in part(c)to find upper and lower bounds for \(\ln 2\) .

7

1. Show that an appropriate integrating factor for $$\sqrt { x ^ { 2 } + 16 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = x \sqrt { x ^ { 2 } + 16 }$$ is \(\frac { 1 } { 4 } x + \frac { 1 } { 4 } \sqrt { x ^ { 2 } + 16 }\) .
   \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-15_2723_33_99_22}
2. Hence find the solution of the differential equation $$\sqrt { x ^ { 2 } + 16 } \frac { \mathrm {~d} y } { \mathrm {~d} x } + y = x \sqrt { x ^ { 2 } + 16 }$$ for which \(y = 6\) when \(x = 3\).

8

1. By considering the binomial expansion of \(\left( z + \frac { 1 } { z } \right) ^ { 7 }\), where \(z = \cos \theta + \mathrm { i } \sin \theta\), use de Moivre's theorem to show that $$\cos ^ { 7 } \theta = a \cos 7 \theta + b \cos 5 \theta + c \cos 3 \theta + d \cos \theta$$ where \(a , b , c\) and \(d\) are constants to be determined.
   Let \(I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \cos ^ { n } \theta \mathrm {~d} \theta\).
2. Show that $$n I _ { n } = 2 ^ { - \frac { 1 } { 2 } n } + ( n - 1 ) I _ { n - 2 }$$ \includegraphics[max width=\textwidth, alt={}, center]{4af32247-c1f9-4c1f-bdf8-bafe17aca1dc-18_2716_40_109_2009}
3. Using the results given in parts (a) and (b), find the exact value of \(I _ { 9 }\).
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