UFM Pure

Prove that \[ \tan^{-1}t=t-\frac{t^{3}}{3}+\frac{t^{5}}{5}-\cdots+\frac{(-1)^{n}t^{2n+1}}{2n+1}+(-1)^{n+1}\int_{0}^{t}\frac{x^{2n+2}}{1+x^{2}}\,\mathrm{d}x. \] Hence show that, if \(0\leqslant t\leqslant1,\) then \[ \frac{t^{2n+3}}{2(2n+3)}\leqslant\left|\tan^{-1}t-\sum_{r=0}^{n}\frac{(-1)^{r}t^{2r+1}}{2r+1}\right|\leqslant\frac{t^{2n+3}}{2n+3}. \] Show that, as \(n\rightarrow\infty,\) \[ 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)}\rightarrow\pi, \] but that the error in approximating \(\pi\) by \({\displaystyle 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)}}\) is at least \(10^{-2}\) if \(n\) is less than or equal to \(98\).

If \(y=\mathrm{f}(x)\), then the inverse of \(\mathrm{f}\) (when it exists) can be obtained from Lagrange's identity. This identity, which you may use without proof, is \[ \mathrm{f}^{-1}(y)=y+\sum_{n=1}^{\infty}\frac{1}{n!}\frac{\mathrm{d}^{n-1}}{\mathrm{d}y^{n-1}}\left[y-\mathrm{f}\left(y\right)\right]^{n}, \] provided the series converges.

1. Verify Lagrange's identity when \(\mathrm{f}(x)=\alpha x\), \((0<\alpha<2)\).
2. Show that one root of the equation \[ \tfrac{1}{2}=x-\tfrac{1}{4}x^{3} \] is \[ x=\sum_{n=0}^{\infty}\frac{\left(3n\right)!}{n!\left(2n+1\right)!2^{4n+1}} \]
3. Find a solution for \(x\), as a series in \(\lambda,\) of the equation \[ x=\mathrm{e}^{\lambda x}. \]

Show that the sum of the infinite series \[ \log_{2}\mathrm{e}-\log_{4}\mathrm{e}+\log_{16}\mathrm{e}-\ldots+(-1)^{n}\log_{2^{2^{n}}}\mathrm{e}+\ldots \] is \[ \frac{1}{\ln(2\sqrt{2})}. \] {[}\(\log_{a}b=c\) is equivalent to \(a^{c}=b\).{]}

Prove that, for any numbers \(a_1\), \(a_2\), \(\ldots\)\,, and \(b_1\), \(b_2\), \(\ldots\)\,, and for \(n\ge1\), \[ \sum_{m=1}^n a_m(b_{m+1} -b_m) = a_{n+1}b_{n+1} -a_1b_1 -\sum_{m=1}^n b_{m+1}(a_{m+1} -a_m) \,. \]

1. By setting \(b_m = \sin mx\), show that \[ \sum_{m=1}^n \cos (m+\tfrac12)x = \tfrac12 \big(\sin (n+1)x - \sin x \big) \cosec \tfrac12 x \,. \] {\bf Note:} $\sin A - \sin B = \displaystyle 2 \cos \big( \tfrac{{\displaystyle A+B\vphantom{_1}}} {\displaystyle 2\vphantom{^1}} \big)\, \sin\big( \tfrac{{\displaystyle A-B\vphantom{_1}}}{\displaystyle 2\vphantom{^1}} \big)\, $.
2. Show that \[ \sum_{m=1}^n m\sin mx = \big (p \sin(n+1)x +q \sin nx\big) \cosec^2 \tfrac12 x \,, \] where \(p\) and \(q\) are to be determined in terms of \(n\). \vspace{3mm} {\bf Note:} \(2\sin A \sin B = \cos (A-B) - \cos (A+B)\,\); \\[2mm] \phantom {\bf Note:} \(2\cos A \sin B = \sin (A+B) - \sin (A-B)\,\).

1. Prove that, for any positive integers \(n\) and \(r\), \[ \frac{1}{^{n+r}\C_{r+1}} =\frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right). \] Hence determine \[ \sum_{n=1}^{\infty}{\frac{1}{^{n+r}\C_{r+1}}} \,, \] and deduce that \ \(\displaystyle \sum_{n=2}^\infty \frac 1 {^{n+2}\C_3} = \frac12\,\).
2. Show that, for \(n \ge 3\,\), \[ \frac{3!}{n^3} < \frac{1}{^{n+1}\C_{3}} \ \ \ \ \ \text{and} \ \ \ \ \ \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} < \frac{5!}{n^3} \,. \] By summing these inequalities for \(n \ge 3\,\), show that \[ \frac{115}{96} < \sum_{n=1}^{\infty}{\frac{1}{n^3}} < \frac{116}{96} \, . \]

1. By considering \ $\displaystyle \frac1 {1+ x^r} - \frac1 {1+ x^{r +1}} $ \ for \(\vert x \vert \ne 1\), simplify \[ \sum_{r=1}^N \frac{x^r}{(1+x^r)(1+x^{r+1})} \,. \] Show that, for \(\vert x \vert <1\), \[ \sum_{r=1}^\infty \frac{x^r}{(1+x^r)(1+x^{r+1})} = \frac x {1-x^2} \,. \]
2. Deduce that \[ \sum_{r=1}^\infty \sech(ry)\sech((r + 1)y) = 2\e^{-y} \cosech (2 y) \] for \(y > 0\). Hence simplify \[ \sum_{r=-\infty}^\infty \sech(ry) \sech((r + 1)y) \,,\] for \(y>0\).

Evaluate the integral \[ \hphantom{ \ \ \ \ \ \ \ \ \ (m> \tfrac12)\,.} \int_{m-\frac12} ^\infty \frac 1{x^2}\, \d x { \ \ \ \ \ \ \ \ \ (m> \tfrac12)\,.} \] Show by means of a sketch that \[ \sum_{r=m}^n \frac 1 {r^2} \approx \int_{m-\frac12}^{n+\frac12} \frac1 {x^2} \, \d x \,, \tag{\(*\)} \] where \(m\) and \(n\) are positive integers with \(m

The numbers \(\.f(r)\) satisfy \(\.f(r)>\.f(r+1)\) for \(r=1\), \(2\), \dots. Show that, for any non-negative integer \(n\), \[ k^n(k-1) \, \.f(k^{n+1}) \le \sum_{r=k^n}^{k^{n+1}-1}\.f(r) \le k^n(k-1)\, \.f(k^n)\, \] where \(k\) is an integer greater than 1.

1. By taking \(\.f(r) = 1/r\), show that \[ \frac{N+1}2 \le \sum_{r=1}^{2^{N+1}-1} \frac1r \le N+1 \,. \] Deduce that the sum \(\sum\limits_{r=1}^\infty \frac1r\) does not converge.
2. By taking \(\.f(r)= 1/r^3\), show that \[ \sum_{r=1}^\infty \frac1 {r^3} \le 1 \tfrac 13 \,. \]
3. Let \(S(n)\) be the set of positive integers less than \(n\) which do not have a \(2\) in their decimal representation and let \(\sigma(n)\) be the sum of the reciprocals of the numbers in \(S(n)\), so for example \(\sigma(5) = 1+\frac13+\frac14\). Show that \(S(1000)\) contains \(9^3-1\) distinct numbers. Show that \(\sigma (n) < 80\) for all \(n\).

In this question, the following theorem may be used.\newline {\sl Let \(u_1\), \(u_2\), \(\ldots\) be a sequence of (real) numbers. If the sequence is bounded above (that is, \(u_n\le b\) for all \(n\), where \(b\) is some fixed number) and increasing (that is, \(u_n\ge u_{n-1}\) for all \(n\)), then the sequence tends to a limit (that is, converges).} The sequence \(u_1\), \(u_2\), \(\ldots\) is defined by \(u_1=1\) and \[ u_{n+1} = 1+\frac 1{u_n} \ \ \ \ \ \ \ \ \ \ (n\ge1)\,. \tag{\(*\)} \]

1. Show that, for \(n\ge3\), \[ u_{n+2}-u_n = \frac{u_{n} - u_{n-2}}{(1+u_n)(1+u_{n-2})} . \]
2. Prove, by induction or otherwise, that \(1\le u_n \le 2\) for all \(n\).
3. Show that the sequence \(u_1\), \(u_3\), \(u_5\), \(\ldots\) tends to a limit, and that the sequence \(u_2\), \(u_4\), \(u_6\), \(\ldots\) tends to a limit. Find these limits and deduce that the sequence \(u_1\), \(u_2\), \(u_3\), \(\ldots\,\) tends to a limit. Would this conclusion change if the sequence were defined by \((*)\) and \(u_1=3\)?

The sequence \(F_0\), \(F_1\), \(F_2\), \(\ldots\,\) is defined by \(F_0=0\), \(F_1=1\) and, for \(n\ge0\), \[ F_{n+2} = F_{n+1} + F_n \,. \]

1. Show that \(F_0F_3-F_1F_2 = F_2F_5- F_3F_4\,\).
2. Find the values of \(F_nF_{n+3} - F_{n+1}F_{n+2}\) in the two cases that arise.
3. Prove that, for \(r=1\), \(2\), \(3\), \(\ldots\,\), \[ \arctan \left( \frac 1{F_{2r}}\right) =\arctan \left( \frac 1{F_{2r+1}}\right)+ \arctan \left( \frac 1{F_{2r+2}}\right) \] and hence evaluate the following sum (which you may assume converges): \[ \sum_{r=1}^\infty \arctan \left( \frac 1{F_{2r+1}}\right) \,. \]

In this question, \(\vert x \vert <1\) and you may ignore issues of convergence.

1. Simplify \[ (1-x)(1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^n})\,, \] where \(n\) is a positive integer, and deduce that \[ \frac1{1-x} = (1+x)(1+x^2)(1+x^4) \cdots (1+x^{2^n}) + \frac {x^{2^{n+1}}}{1-x}\,. \] Deduce further that \[ \ln(1-x) = - \sum_{r=0}^\infty \ln \left (1+ x ^{2^r}\right) \,, \] and hence that \[ \frac1 {1-x} = \frac 1 {1+x} + \frac {2x}{1+x^2} + \frac {4x^3}{1+x^4} +\cdots\,. \]
2. Show that \[ \frac{1+2x}{1+x+x^2} = \frac{1-2x}{1-x+x^2} + \frac{2x-4x^3}{1-x^2+x^4} + \frac {4x^3-8x^7}{1-x^4+x^8} + \cdots\,. \]

The positive numbers \(\alpha\), \(\beta\) and \(q\) satisfy \(\beta-\alpha >q\). Show that \[ \frac{\alpha^2+\beta^2 -q^2}{\alpha\beta}-2> 0\,. \] The sequence \(u_0\), \(u_1\), \(\ldots\) is defined by \(u_0=\alpha\), \(u_1=\beta\) and \[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u_{n+1} = \frac {u_{n}^2 -q^2}{u_{n-1}} \ \ \ \ \ \ \ \ \ \ \ (n\ge1), \] where \(\alpha\), \(\beta\) and \(q\) are given positive numbers (and \(\alpha\) and \(\beta\) are such that no term in the sequence is zero). Prove that \(u_n(u_n+u_{n+2}) = u_{n+1}(u_{n-1}+u_{n+1})\,\). Prove also that \[ u_{n+1} -pu_n + u_{n-1}=0 \] for some number \(p\) which you should express in terms of \(\alpha\), \(\beta\) and \(q\). Hence, or otherwise, show that if \(\beta> \alpha+q\), the sequence is strictly increasing (that is, \(u_{n+1}-u_n > 0\) for all \(n\)). Comment on the case \(\beta =\alpha +q\).

A sequence of numbers \(t_0\), \(t_1\), \(t_2\), \(\ldots\,\) satisfies \[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t_{n+2} = p t_{n+1}+qt_{n} \ \ \ \ \ \ \ \ \ \ (n\ge0), \] where \(p\) and \(q\) are real. Throughout this question, \(x\), \(y\) and \(z\) are non-zero real numbers.

1. Show that, if \(t_n=x\) for all values of \(n\), then \(p+q=1\) and \(x\) can be any (non-zero) real number.
2. Show that, if \(t_{2n} = x\) and \(t_{2n+1}=y\) for all values of \(n\), then \(q\pm p=1\). Deduce that either \(x=y\) or \(x=-y\), unless \(p\) and \(q\) take certain values that you should identify.
3. Show that, if \(t_{3n} = x\), \(t_{3n+1}=y\) and \(t_{3n+2}=z\) for all values of \(n\), then \[ p^3+q^3 +3pq-1=0\,. \] Deduce that either \(p+q=1\) or \((p-q)^2 +(p+1)^2+(q+1)^2=0\). Hence show that either \(x=y=z\) or \(x+y+z=0\).

Given that \(y = \cos(m \arcsin x)\), for \(\vert x \vert <1\), prove that \[ (1-x^2) \frac {\d^2 y}{\d x^2} -x \frac {\d y}{\d x} +m^2y=0\,. \] Obtain a similar equation relating \(\dfrac{\d^3y}{\d x^3}\,\),\; \(\dfrac{\d^2y}{\d x^2}\, \) and \(\, \dfrac{\d y}{\d x}\,\), and a similar equation relating \(\dfrac{\d^4y}{\d x^4}\,\),~~\(\dfrac{\d^3y}{\d x^3}\,\) and \(\,\dfrac{\d^2 y}{\d x^2}\,\). Conjecture and prove a relation between \(\dfrac{\d^{n+2}y}{\d x^{n+2}}\,\), \ \(\dfrac{\d^{n+1}y}{\d x^{n+1}}\;\) and \(\;\dfrac{\d^n y}{\d x^n}\,\). Obtain the first three non-zero terms of the Maclaurin series for \(y\). Show that, if \(m\) is an even integer, \(\cos m\theta\) may be written as a polynomial in \(\sin\theta\) beginning \[ 1 - \frac{m^2\sin^2\theta}{2!}+ \frac{m^2(m^2-2^2)\sin^4\theta}{4!} -\cdots \,. \, \tag{\(\vert\theta\vert < \tfrac12 \pi\)} \] State the degree of the polynomial.

Let \(S_k(n) \equiv \sum\limits_{r=0}^n r^k\,\), where \(k\) is a positive integer, so that \[ S_1(n) \equiv \tfrac12 n(n+1) \text{ and } S_2(n) \equiv \tfrac16 n(n+1)(2n+1)\,. \]

1. By considering \(\sum\limits_{r=0}^n \left[ (r+1)^k-r^k\right]\, \), show that \[ kS_{k-1}(n)=(n+1)^k -(n+1) - \binom{k}{2} S_{k-2}(n) - \binom {k}{3} S_{k-3}(n) - \cdots - \binom{k}{k-1} S_{1}(n) \;. \tag{\(*\)} \] Obtain simplified expressions for \(S_3(n)\) and \(S_4(n)\).
2. Explain, using \((*)\), why \(S_k(n)\) is a polynomial of degree \(k+1\) in \(n\). Show that in this polynomial the constant term is zero and the sum of the coefficients is 1.

The sequence of real numbers \(u_1\), \(u_2\), \(u_3\), \(\ldots\) is defined by \begin{equation*} u_1=2 \,, \qquad\text{and} \qquad u_{n+1} = k - \frac{36}{u_n} \quad \text{for } n\ge1, \tag{\(*\)} \end{equation*} where \(k\) is a constant.

1. Determine the values of \(k\) for which the sequence \((*)\) is: (a) constant; (b) periodic with period 2; (c) periodic with period 4.
2. In the case \(k=37\), show that \(u_n\ge 2\) for all \(n\). Given that in this case the sequence \((*)\) converges to a limit \(\ell\), find the value of \(\ell\).

The sequence \(u_n\) (\(n= 1, 2, \ldots\)) satisfies the recurrence relation \[ u_{n+2}= \frac{u_{n+1}}{u_n}(ku_n-u_{n+1}) \] where \(k\) is a constant. If \(u_1=a\) and \(u_2=b\,\), where \(a\) and \(b\) are non-zero and \(b \ne ka\,\), prove by induction that \[ u_{2n}=\Big(\frac b a \Big) u_{2n-1} \] \[ u_{2n+1}= c u_{2n} \] for \(n \ge 1\), where \(c\) is a constant to be found in terms of \(k\), \(a\) and \(b\). Hence express \(u_{2n}\) and \(u_{2n-1}\) in terms of \(a\), \(b\), \(c\) and \(n\). Find conditions on \(a\), \(b\) and \(k\) in the three cases:

1. the sequence \(u_n\) is geometric;
2. \(u_n\) has period 2;
3. the sequence \(u_n\) has period 4.

Given a sequence \(w_0\), \(w_1\), \(w_2\), \(\ldots\,\), the sequence \(F_1\), \(F_2\), \(\ldots\) is defined by $$F_n = w_n^2 + w_{n-1}^2 - 4w_nw_{n-1} \,.$$ Show that $\; F_{n}-F_{n-1} = \l w_n-w_{n-2} \r \l w_n+w_{n-2}-4w_{n-1} \r \; \( for \)n \ge 2\,$.

1. The sequence \(u_0\), \(u_1\), \(u_2\), \(\ldots\) has \(u_0 = 1\), and \(u_1 = 2\) and satisfies \[ u_n = 4u_{n-1} -u_{n-2} \quad (n \ge 2)\;. \] Prove that \ $ u_n^2 + u_{n-1}^2 = 4u_nu_{n-1}-3 \; $ for \(n \ge 1\,\).
2. A sequence \(v_0\), \(v_1\), \(v_2\), \(\ldots\,\) has \(v_0=1\) and satisfies \begin{equation*} v_n^2 + v_{n-1}^2 = 4v_nv_{n-1}-3 \quad (n \ge 1). \tag{\(\ast\)} \end{equation*} \makebox[7mm]{(a) \hfill}Find \(v_1\) and prove that, for each \(n\ge2\,\), either \(v_n= 4v_{n-1} -v_{n-2}\) or \(v_n=v_{n-2}\,\). \makebox[7mm]{(b) \hfill}Show that the sequence, with period 2, defined by \begin{equation*} v_n = \begin{cases} 1 & \mbox{for \(n\) even} \\ 2 & \mbox{for \(n\) odd} \end{cases} \end{equation*} \makebox[7mm]{\hfill}satisfies \((\ast)\). \makebox[7mm]{(c) \hfill}Find a sequence \(v_n\) with period 4 which has \(v_0=1\,\), and satisfies~\((\ast)\).

Given that \(\f''(x) > 0\) when \(a \le x \le b\,\), explain with the aid of a sketch why \[ (b-a) \, \f \Big( {a+b \over 2} \Big) < \int^b_a \f(x) \, \mathrm{d}x < (b-a) \, \displaystyle \frac{\f(a) + \f(b)}{2} \;. \] By choosing suitable \(a\), \(b\) and \(\f(x)\,\), show that \[ {4 \over (2n-1)^2} < {1 \over n-1} - {1 \over n} < {1 \over 2} \l {1 \over n^2} + {1 \over (n-1)^2}\r \,, \] where \(n\) is an integer greater than 1. Deduce that \[ 4 \l {1 \over 3^2} +{1 \over 5^2} + {1 \over 7^2} + \cdots \r < 1 < {1 \over 2} + \left( {1 \over 2^2} +{1 \over 3^2} + {1 \over 4^2} + \cdots \right)\,. \] Show that \[ {1 \over 2} \l {1 \over 3^2} + {1 \over 4^2} + {1 \over 5^2} + \frac 1 {6^2} + \cdots \r < {1 \over 3^2} +{1 \over 5^2} + {1 \over 7^2} + \cdots \] and hence show that \[ {3 \over 2} \displaystyle < \sum_{n=1}^\infty {1 \over n^2} <{7 \over 4}\;. \]

A sequence \(t_0\), \(t_1\), \(t_2\), \(...\) is said to be {\sl strictly increasing} if \(t_{n+1} > t_n\) for all \(n\ge{0}\,\).

1. The terms of the sequence \(x_0\,\), \(x_1\,\), \(x_2\,\), \(\ldots\) satisfy $$ \ds x_{n+1}=\frac{x_n^2 +6}{5} $$ for \(n\ge{0}\,\). Prove that if \(x_0 > 3\) then the sequence is strictly increasing.
2. The terms of the sequence \(y_0\,\), \(y_1\,\), \(y_2\,\), \(\ldots\) satisfy $$ \ds y_{n+1}= 5-\frac 6 {y_n} $$ for \(n\ge{0}\,\). Prove that if \(2 < y_0 < 3\) then the sequence is strictly increasing but that \(y_n<3\) for all \(n\,\).

Show that \[ 2\sin \frac12 \theta \, \cos r\theta = \sin\big(r+\frac12\big)\theta - \sin\big(r-\frac12\big)\theta \;. \] Hence, or otherwise, find all solutions of the equation \[ \cos a\theta + \cos (a + 1) \theta + \dots + \cos(b-2)\theta+\cos (b - 1 ) \theta = 0 \;, \] where \(a\) and \(b\) are positive integers with \(a < b-1\,\).

Show that, if \(n>0\,\), then $$ \int_{e^{1/n}}^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x = {2 \over {n^2\e}}\;. $$ You may assume that \(\ds \frac{\ln x} x \to 0\;\) as \(x\to\infty\,\). Explain why, if \(1 < a < b\,\), then $$ \int_b^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x < \int_a^\infty\,{{\ln x} \over {x^{n+1}}}\,\d x\;. $$ Deduce that $$ \sum_{n=1}^{N}{1 \over n^2} < {\e \over 2}\int_{\e^{1/N}}^{\infty} \left({1-x^{-N}} \over {x^2-x}\right) \ln x\,\d x\;, $$ where \(N\,\) is any integer greater than \(1\).

Given that $$\e = 1 + {1 \over 1 !} + {1 \over 2 !} + {1 \over 3 !} + \cdots + {1 \over r !} + \cdots \; ,$$ use the binomial theorem to show that $$ {\left( 1 + {1 \over n} \right)}^{\!n} < \e $$ for any positive integer \(n\). The product \({\rm P }( n )\) is defined, for any positive integer \(n\), by $$ {\rm P} ( n ) = {3 \over 2} \cdot {5 \over 4} \cdot {9 \over 8} \cdot \ldots \cdot {2^n + 1 \over 2^n} . $$ Use the arithmetic-geometric mean inequality, $$ {a_1 + a_2 + \cdots + a_n \over n} \ge \ {\left( a_1 \cdot a_2 \cdot \ldots \cdot a_n \right)}^{1 \over n}\,, $$ to show that \({\rm P }( n ) < \e\) for all \(n\) . Explain briefly why \({\rm P} ( n )\) tends to a limit as \(n\to\infty\). Show that this limit, \(L\), satisfies \(2 < L\le\e\).

The sequence \(u_0\), \(u_1\), \(u_2\), ... is defined by $$ u_0=1,\hspace{0.2in} u_1=1, \quad u_{n+1}=u_n+u_{n-1} \hspace{0.2in}{\rm for}\hspace{0.1in}n \ge 1\,. $$ Prove that $$ u^2_{n+2} + u^2_{n-1} = 2( u^2_{n+1} + u^2_n ) \,. $$ Using induction, or otherwise, prove the following result: \[ u_{2n} = u^2_n + u^2_{n-1} \quad \mbox{ and }\quad u_{2n+1} = u^2_{n+1} - u^2_{n-1} \] for any positive integer \(n\).

Justify, by means of a sketch, the formula $$ \lim_{n\rightarrow\infty}\left\{{1\over n}\sum_{m=1}^n \f(1+m/n)\right\} = \int_1^2 \f(x)\,\d x \,. $$ Show that $$ \lim_{n\rightarrow\infty}\left\{{1\over n+1} + {1\over n+2} + \cdots + {1\over n+n}\right\} = \ln 2 \,. $$ Evaluate $$ \lim_{n\rightarrow\infty}\left\{{n\over n^2+1} + {n\over n^2+4} + \cdots + {n\over n^2+n^2}\right\}\,. $$

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\begin{align*} && V &= \lim_{n\rightarrow\infty}\left\{{1\over n+1} + {1\over n+2} + \cdots + {1\over n+n}\right\} \\ && &= \lim_{n\rightarrow\infty}\left\{\sum_{m=1}^n \frac{1}{n+m}\right\} \\ && &= \lim_{n\rightarrow\infty}\left\{\frac1n\sum_{m=1}^n \frac{1}{1+\frac{m}{n}}\right\} \\ &&&=\int_1^2 \frac{1}{x} \d x \\ &&&= \left [\ln x \right]_1^2 = \ln 2 \end{align*} \begin{align*} V &= \lim_{n\rightarrow\infty}\left\{{n\over n^2+1} + {n\over n^2+4} + \cdots + {n\over n^2+n^2}\right\} \\ &= \lim_{n\rightarrow\infty}\left\{\sum_{m=1}^n \frac{n}{n^2+m^2}\right\} \\ &= \lim_{n\rightarrow\infty}\left\{\frac{1}{n}\sum_{m=1}^n \frac{1}{1+\left (\frac{m}{n} \right)^2}\right\} \\ &= \int_0^1 \frac{1}{1+x^2} \d x \\ &= \left [\tan^{-1} x \right]_0^1 \\ &= \frac{\pi}4 \end{align*}

1. Let \(a\), \(b\) and \(c\) be three non-zero complex numbers with the properties \(a + b + c = 0\) and \(a^2 + b^2 + c^2 = 0\). Show that \(a\), \(b\) and \(c\) cannot all be real. Show further that \(a\), \(b\) and \(c\) all have the same modulus.
2. Show that it is not possible to find three non-zero complex numbers \(a\), \(b\) and \(c\) with the properties \(a + b + c = 0\) and \(a^3 + b^3 + c^3 = 0\).
3. Show that if any four non-zero complex numbers \(a\), \(b\), \(c\) and \(d\) have the properties \(a + b + c + d = 0\) and \(a^3 + b^3 + c^3 + d^3 = 0\), then at least two of them must have the same modulus.
4. Show, by taking \(c = 1\), \(d = -2\) and \(e = 3\) that it is possible to find five real numbers \(a\), \(b\), \(c\), \(d\) and \(e\) with distinct magnitudes and with the properties \(a + b + c + d + e = 0\) and \(a^3 + b^3 + c^3 + d^3 + e^3 = 0\).

The \(n\)th degree polynomial P\((x)\) is said to be reflexive if:

1. [(a)] P\((x)\) is of the form \(x^n - a_1x^{n-1} + a_2x^{n-2} - \cdots + (-1)^na_n\) where \(n \geq 1\);
2. [(b)] \(a_1, a_2, \ldots, a_n\) are real;
3. [(c)] the \(n\) (not necessarily distinct) roots of the equation P\((x) = 0\) are \(a_1, a_2, \ldots, a_n\).

1. Find all reflexive polynomials of degree less than or equal to 3.
2. For a reflexive polynomial with \(n > 3\), show that $$2a_2 = -a_2^2 - a_3^2 - \cdots - a_n^2.$$ Deduce that, if all the coefficients of a reflexive polynomial of degree \(n\) are integers and \(a_n \neq 0\), then \(n \leq 3\).
3. Determine all reflexive polynomials with integer coefficients.

1. The function \(\f\) is given by \[ \f(\beta)=\beta - \frac 1 \beta - \frac 1 {\beta^2} \ \ \ \ \ \ \ \ (\beta\ne0) \,. \] Find the stationary point of the curve \(y=\f(\beta)\,\) and sketch the curve. Sketch also the curve \(y=\g(\beta)\,\), where \[ \g(\beta) = \beta + \frac 3 \beta - \frac 1 {\beta^2} \ \ \ \ \ \ \ \ (\beta\ne0)\,. \]
2. Let \(u\) and \(v\) be the roots of the equation \[ x^2 +\alpha x +\beta = 0 \,, \] where \(\beta\ne0\,\). Obtain expressions in terms of \(\alpha\) and \(\beta\) for \(\displaystyle u+v + \frac 1 {uv}\) and \( \displaystyle \frac 1 u + \frac 1 v + uv\,\).
3. Given that \(\displaystyle u+v + \frac 1 {uv} = -1\,\), and that \(u\) and \(v\) are real, show that \(\displaystyle \frac 1 u+ \frac 1 v + {uv} \le -1\;\).
4. Given instead that \(\displaystyle u+v + \frac 1 {uv} = 3 \;\), and that \(u\) and \(v\) are real, find the greatest value of \(\displaystyle \frac 1 u+ \frac 1v + {uv}\,\).

Let \(\alpha\), \(\beta\), \(\gamma\) and \(\delta\) be the roots of the quartic equation \[ x^4 +px^3 +qx^2 +r x +s =0 \,. \] You are given that, for any such equation, \(\,\alpha \beta + \gamma\delta\,\), \(\alpha\gamma+\beta\delta\,\) and \(\,\alpha \delta + \beta\gamma\,\) satisfy a cubic equation of the form \[ y^3+Ay^2+ (pr-4s)y+ (4qs-p^2s -r^2) =0 \,. \] Determine \(A\). Now consider the quartic equation given by \(p=0\,\), \(q= 3\,\), \(r=-6\,\) and \(s=10\,\).

1. Find the value of \(\alpha\beta + \gamma \delta\), given that it is the largest root of the corresponding cubic equation.
2. Hence, using the values of \(q\) and \(s\), find the value of \((\alpha +\beta)(\gamma+\delta)\,\) and the value of \(\alpha\beta\) given that \(\alpha\beta >\gamma\delta\,\).
3. Using these results, and the values of \(p\) and \(r\), solve the quartic equation.

1. Let \(w\) and \(z\) be complex numbers, and let \(u= w+z\) and \(v=w^2+z^2\). Prove that \(w\) and \(z\) are real if and only if \(u\) and \(v\) are real and \(u^2\le2v\).
2. The complex numbers \(u\), \(w\) and \(z\) satisfy the equations \begin{align*} w+z-u&=0 \\ w^2+z^2 -u^2 &= - \tfrac 23 \\ w^3+z^3 -\lambda u &= -\lambda\, \end{align*} where \(\lambda \) is a positive real number. Show that for all values of \(\lambda\) except one (which you should find) there are three possible values of \(u\), all real. Are \(w\) and \(z\) necessarily real? Give a proof or counterexample.

Let \(a\), \(b\) and \(c\) be real numbers such that \(a+b+c=0\) and let \[(1+ax)(1+bx)(1+cx) = 1+qx^2 +rx^3\,\] for all real \(x\). Show that \(q = bc+ca+ab\) and \(r= abc\).

1. Show that the coefficient of \(x^n\) in the series expansion (in ascending powers of \(x\)) of \(\ln (1+qx^2+rx^3)\) is \((-1)^{n+1} S_n\) where \[S_n = \frac{a^n+b^n+c^n}{n} \,, \ \ \ \ \ \ \ \ (n\ge1).\]
2. Find, in terms of \(q\) and \(r\), the coefficients of \(x^2\), \(x^3\) and \(x^5\) in the series expansion (in ascending powers of \(x\)) of \(\ln (1+qx^2+rx^3)\) and hence show that \(S_2S_3 =S_5\).
3. Show that \(S_2S_5 =S_7\).
4. Give a proof of, or find a counterexample to, the claim that \(S_2S_7=S_9\).

The numbers \(x\), \(y\) and \(z\) satisfy \begin{align*} x+y+z&= 1\\ x^2+y^2+z^2&=2\\ x^3+y^3+z^3&=3\,. \end{align*} Show that \[ yz+zx+xy=-\frac12 \,.\] Show also that \(x^2y+x^2z+y^2z+y^2x+z^2x+z^2y=-1\,\), and hence that \[ xyz=\frac16 \,.\] Let \(S_n=x^n+y^n+z^n\,\). Use the above results to find numbers \(a\), \(b\) and \(c\) such that the relation \[ S_{n+1}=aS_{n}+bS_{n-1}+cS_{n-2}\,, \] holds for all \(n\).

Find all values of \(a\), \(b\), \(x\) and \(y\) that satisfy the simultaneous equations \begin{alignat*}{3} a&+b & &=1 &\\ ax&+by & &= \tfrac13& \\ ax^2&+by^2& &=\tfrac15& \\ ax^3 &+by^3& &=\tfrac17\,.& \end{alignat*} \noindent{\bf [} {\bf Hint}: you may wish to start by multiplying the second equation by \(x+y\). {\bf ]}

In this question, do not consider the special cases in which the denominators of any of your expressions are zero. Express \(\tan(\theta_1+\theta_2+\theta_3+\theta_4)\) in terms of \(t_i\), where \(t_1=\tan\theta_1\,\), etc. Given that \(\tan\theta_1\), \(\tan\theta_2\), \(\tan\theta_3\) and \(\tan\theta_4\) are the four roots of the equation \[at^4+bt^3+ct^2+dt+e=0 \] (where \(a\ne0\)), find an expression in terms of \(a\), \(b\), \(c\), \(d\) and \(e\) for \(\tan(\theta_1+\theta_2+\theta_3+\theta_4)\). The four real numbers \(\theta_1\), \(\theta_2\), \(\theta_3\) and \(\theta_4\) lie in the range \(0\le \theta_i<2\pi\) and satisfy the equation \[ p\cos2\theta+\cos(\theta-\alpha)+p=0\,,\] where \(p\) and \(\alpha\) are independent of \(\theta\). Show that \(\theta_1+\theta_2+\theta_3+\theta_4=n\pi\) for some integer \(n\).

In this question, you may assume that if \(k_1,\dots,k_n\) are distinct positive real numbers, then \[\frac1n\sum_{r=1}^nk_r>\left({\prod\limits_{r=1}^n} k_r\right )^{\!\! \frac1n},\] i.e. their arithmetic mean is greater than their geometric mean. Suppose that \(a\), \(b\), \(c\) and \(d\) are positive real numbers such that the polynomial \[{\rm f}(x)=x^4-4ax^3+6b^2x^2-4c^3x+d^4\] has four distinct positive roots.

1. Show that \(pqr,qrs,rsp\) and \(spq\) are distinct, where \(p,q,r\) and \(s\) are the roots of the polynomial \(\mathrm{f}\).
2. By considering the relationship between the coefficients of \(\mathrm{f}\) and its roots, show that \(c > d\).
3. Explain why the polynomial \(\mathrm{f}'(x)\) must have three distinct roots.
4. By differentiating \(\mathrm{f}\), show that \(b > c\).
5. Show that \(a > b\).

1. If \(x+y+z=\alpha,\) \(xy+yz+zx=\beta\) and \(xyz=\gamma,\) find numbers \(A,B\) and \(C\) such that \[ x^{3}+y^{3}+z^{3}=A\alpha^{3}+B\alpha\beta+C\gamma. \] Solve the equations \begin{alignat*}{1} x+y+z & =1\\ x^{2}+y^{2}+z^{2} & =3\\ x^{3}+y^{3}+z^{3} & =4. \end{alignat*}
2. The area of a triangle whose sides are \(a,b\) and \(c\) is given by the formula \[ \mathrm{area}=\sqrt{s(s-a)(s-b)(s-c)} \] where \(s\) is the semi-perimeter \(\frac{1}{2}(a+b+c).\) If \(a,b\) and \(c\) are the roots of the equation \[ x^{3}-16x^{2}+81x-128=0, \] find the area of the triangle.

The cubic equation \[ x^{3}-px^{2}+qx-r=0 \] has roots \(a,b\) and \(c\). Express \(p,q\) and \(r\) in terms of \(a,b\) and \(c\).

1. If \(p=0\) and two of the roots are equal to each other, show that \[ 4q^{3}+27r^{2}=0. \]
2. Show that, if two of the roots of the original equation are equal to each other, then \[ 4\left(q-\frac{p^{2}}{3}\right)^{3}+27\left(\frac{2p^{3}}{27}-\frac{pq}{3}+r\right)^{2}=0. \]

The equation \[ x^{n}-qx^{n-1}+r=0, \] where \(n\geqslant5\) and \(q\) and \(r\) are real constants, has roots \(\alpha_{1},\alpha_{2},\ldots,\alpha_{n}.\) The sum of the products of \(m\) distinct roots is denoted by \(\Sigma_{m}\) (so that, for example, \(\Sigma_{3}=\sum\alpha_{i}\alpha_{j}\alpha_{k}\) where the sum runs over the values of \(i,j\) and \(k\) with \(n\geqslant i>j>k\geqslant1\)). The sum of \(m\)th powers of the roots is denoted by \(S_{m}\) (so that, for example, \(S_{3}=\sum\limits_{i=1}^{n}\alpha_{i}^{3}\)). Prove that \(S_{p}=q^{p}\) for \(1\leqslant p\leqslant n-1.\) You may assume that for any \(n\)th degree equation and \(1\leqslant p\leqslant n\) \[ S_{p}-S_{p-1}\Sigma_{1}+S_{p-2}\Sigma_{2}-\cdots+(-1)^{p-1}S_{1}\Sigma_{p-1}+(-1)^{p}p\Sigma_{p}=0.] \] Find expressions for \(S_{n},\) \(S_{n+1}\) and \(S_{n+2}\) in terms of \(q,r\) and \(n\). Suggest an expression for \(S_{n+m},\) where \(m < n\), and prove its validity by induction.

The point \(P(a\sec \theta, b\tan \theta )\) lies on the hyperbola \[ \dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1\,, \] where \(a>b>0\,\). Show that the equation of the tangent to the hyperbola at \(P\) can be written as \[ bx- ay \sin\theta = ab \cos\theta \,. \]

1. This tangent meets the lines \(\dfrac x a = \dfrac yb\) and \(\dfrac x a =- \dfrac y b\) at \(S\) and \(T\), respectively. How is the mid-point of \(ST\) related to \(P\)?
2. The point \(Q(a\sec \phi, b\tan \phi)\) also lies on the hyperbola and the tangents to the hyperbola at \(P\) and \(Q\) are perpendicular. These two tangents intersect at \((x,y)\). Obtain expressions for \(x^2\) and \(y^2\) in terms of \(a\), \(\theta\) and \(\phi\). Hence, or otherwise, show that \(x^2+y^2 = a^2 -b^2\).

The point with cartesian coordinates \((x,y)\) lies on a curve with polar equation \(r=\f(\theta)\,\). Find an expression for \(\dfrac{\d y}{\d x}\) in terms of \(\f(\theta)\), \(\f'(\theta)\) and \(\tan\theta\,\). Two curves, with polar equations \(r=\f(\theta)\) and \(r=\g(\theta)\), meet at right angles. Show that where they meet \[ \f'(\theta) \g'(\theta) +\f(\theta)\g(\theta) = 0 \,. \] The curve \(C\) has polar equation \(r=\f(\theta)\) and passes through the point given by \(r=4\), \(\theta = - \frac12\pi\). For each positive value of \(a\), the curve with polar equation \(r= a(1+\sin\theta)\) meets~\(C\) at right angles. Find \(\f(\theta)\,\). Sketch on a single diagram the three curves with polar equations \(r= 1+\sin\theta\,\), \ \(r= 4(1+\sin\theta)\) and \(r=\f(\theta)\,\).

Show Solution

\((x, y) = (f(\theta)\cos(\theta), f(\theta)\sin(\theta))\) so \begin{align*} \frac{dy}{d\theta} &= -f(\theta)\sin(\theta) + f'(\theta)\cos(\theta) \\ \frac{dx}{d\theta} &= f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) \\ \frac{dy}{dx} &= \frac{-f(\theta)\sin(\theta) + f'(\theta)\cos(\theta)}{f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) } \\ &= \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \end{align*} If the curves meet at right angles then the product of their gradients is \(-1\), ie \begin{align*} \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \cdot \frac{-g(\theta)\tan(\theta) + g'(\theta)}{g(\theta) + g'(\theta)\tan(\theta) } &= -1 \\ f(\theta)g(\theta)\tan^2 \theta - f(\theta)g'(\theta)\tan \theta - f'(\theta)g(\theta)\tan \theta + f'(\theta)g'(\theta) &= \\ \quad - \l f(\theta)g(\theta) + f(\theta)g'(\theta)\tan(\theta) + f'(\theta)g(\theta)\tan(\theta) + f'(\theta)g'(\theta)\tan^2 \theta \r \\ \tan^2\theta \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r + f'(\theta)g'(\theta) + f(\theta)g(\theta) &= 0 \\ (\tan^2\theta + 1) \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r &= 0 \\ f(\theta)g(\theta) + f'(\theta)g'(\theta) &= 0 \end{align*} \(g(\theta) = a(1+\sin\theta), g'(\theta) = a\cos\theta\) Therefore \(f'(\theta)a\cos \theta+f(\theta)a(1+\sin(\theta)) = 0\) \begin{align*} && \frac{f'(\theta)}{f(\theta)} &= -\sec(\theta) - \tan(\theta) \\ \Rightarrow && \ln(f(\theta)) &= -\ln |\tan(\theta) + \sec(\theta)| + \ln |\cos(\theta)| + C \\ \Rightarrow && f(\theta) &= A \frac{\cos \theta}{\tan \theta + \sec \theta} \\ &&&= A \frac{\cos^2 \theta}{\sin \theta + 1} \\ &&&= A \frac{1-\sin^2 \theta}{\sin \theta + 1} \\ &&&= A (1-\sin \theta) \end{align*} When \(\theta = -\frac12 \pi, r = 4\), so \(A = 2\).

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1. Show that under the changes of variable \(x= r\cos\theta\) and \(y = r\sin\theta\), where \(r\) is a function of \(\theta\) with \(r>0\), the differential equation \[ (y+x)\frac{\d y}{\d x} = y-x \] becomes \[ \frac{\d r}{\d\theta} + r=0 \,. \] Sketch a solution in the \(x\)-\(y\) plane.
2. Show that the solutions of \[ \left( y+x -x(x^2+y^2) \right) \, \frac{\d y }{\d x} = y-x - y(x^2+y^2) \] can be written in the form \\ \[ r^2 = \dfrac 1 {1+A\e^{2\theta}}\, \]\\ and sketch the different forms of solution that arise according to the value of \(A\).

In this question, \(r\) and \(\theta\) are polar coordinates with \(r \ge0\) and \(- \pi < \theta\le \pi\), and \(a\) and \(b\) are positive constants. Let \(L\) be a fixed line and let \(A\) be a fixed point not lying on \(L\). Then the locus of points that are a fixed distance (call it \(d\)) from \(L\) measured along lines through \(A\) is called a conchoid of Nicomedes.

1. Show that if \[ \vert r- a \sec\theta \vert = b\,, \tag{\(*\)} \] where \(a>b\), then \(\sec\theta >0\). Show that all points with coordinates satisfying (\(*\)) lie on a certain conchoid of Nicomedes (you should identify \(L\), \(d\) and \(A\)). Sketch the locus of these points.
2. In the case \(a < b\), sketch the curve (including the loop for which \(\sec\theta<0\)) given by \[ \vert r- a \sec\theta \vert = b\, . \] Find the area of the loop in the case \(a=1\) and \(b=2\). [Note: $ %\displaystyle \int \! \sec\theta \,\d \theta = \ln \vert \sec\theta + \tan\theta \vert + C \,. $]

A movable point \(P\) has cartesian coordinates \((x,y)\), where \(x\) and \(y\) are functions of \(t\). The polar coordinates of \(P\) with respect to the origin \(O\) are \(r\) and \(\theta\). Starting with the expression \[ \tfrac12 \int r^2 \, \d \theta \] for the area swept out by \(OP\), obtain the equivalent expression \[ \tfrac12 \int \left( x\frac{\d y}{\d t} - y \frac{\d x}{\d t}\right)\d t \,. \tag{\(*\)} \] The ends of a thin straight rod \(AB\) lie on a closed convex curve \(\cal C\). The point \(P\) on the rod is a fixed distance \(a\) from \(A\) and a fixed distance \(b\) from \(B\). The angle between \(AB\) and the positive \(x\) direction is \(t\). As \(A\) and \(B\) move anticlockwise round \(\cal C\), the angle \(t\) increases from \(0\) to \(2\pi\) and \(P\) traces a closed convex curve \(\cal D\) inside \(\cal C\), with the origin \(O\) lying inside \(\cal D\), as shown in the diagram.

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Show that in polar coordinates the gradient of any curve at the point \((r,\theta)\) is \[ \frac{ \ \ \dfrac{\d r }{\d\theta} \tan\theta + r \ \ } { \dfrac{\d r }{\d\theta} -r\tan\theta}\,. \] \noindent

Show that the equation (in plane polar coordinates) \(r=\cos\theta\), for $-\frac{1}{2}\pi \le \theta \le \frac{1}{2}\pi$, represents a circle. Sketch the curve \(r=\cos2\theta\) for \(0\le\theta\le 2\pi\), and describe the curves \(r=\cos2n\theta\), where \(n\) is an integer. Show that the area enclosed by such a curve is independent of \(n\). Sketch also the curve \(r=\cos3\theta\) for \(0\le\theta\le 2\pi\).

The curve \(C\) has the equation \(x^3+y^3 = 3xy\).

1. Show that there is no point of inflection on \(C\). You may assume that the origin is not a point of inflection.
2. The part of \(C\) which lies in the first quadrant is a closed loop touching the axes at the origin. By converting to polar coordinates, or otherwise, evaluate the area of this loop.

\noindent

Sketch the curve \(C\) whose polar equation is \[ r=4a\cos2\theta\qquad\mbox{ for }-\tfrac{1}{4}\pi<\theta<\tfrac{1}{4}\pi. \] The ellipse \(E\) has parametric equations \[ x=2a\cos\phi,\qquad y=a\sin\phi. \] Show, without evaluating the integrals, that the perimeters of \(C\) and \(E\) are equal. Show also that the areas of the regions enclosed by \(C\) and \(E\) are equal.

Show Solution

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\begin{align*} && \text{Perimeter}(C) &= \int_{-\pi/4}^{\pi/4} \sqrt{r^2 + \left ( \frac{\d r}{\d \theta} \right)^2} \d \theta \\ &&&= \int_{-\pi/4}^{\pi/4} \sqrt{16a^2 \cos^2 2 \theta + 64a^2 \sin^2 2 \theta } \d \theta \\ &&&= \int_{-\pi/4}^{\pi/4} 4a\sqrt{1 + 3 \sin^2 2 \theta } \d \theta \\ \\ \\ && \text{Perimeter}(D) &= \int_0^{2 \pi} \sqrt{\left ( \frac{\d x}{\d \phi}\right)^2+\left ( \frac{\d y}{\d \phi}\right)^2} \d \phi \\ &&&= \int_0^{2 \pi} \sqrt{ 4a^2 \sin^2 \phi+a^2 \cos^2 \phi} \d \phi \\ &&&= a^2\int_0^{2 \pi} \sqrt{ 1+3 \sin^2 \phi} \d \phi \\ \end{align*} But clearly these two integrals are equal. \begin{align*} && \text{A}(C) &= \frac12 \int_{-\pi/4}^{\pi/4} r^2 \d \theta \\ &&&= \frac12 \int_{-\pi/4}^{\pi/4} 16a^2 \cos^2 2 \theta \d \theta \\ &&&= 8a^2\int_{-\pi/4}^{\pi/4} \cos^2 2 \theta \d \theta \\ &&&= 8a^2 \frac{\pi}{4} = 2\pi a^2 \\ && \text{A}(D) &= 2\pi a^2 \end{align*}

The parametric equations \(E_{1}\) and \(E_{2}\) define the same ellipse, in terms of the parameters \(\theta_{1}\) and \(\theta_{2}\), (though not referred to the same coordinate axes). \begin{alignat*}{2} E_{1}:\qquad & x=a\cos\theta_{1}, & \quad & y=b\sin\theta_{1},\\ E_{2}:\qquad & x=\dfrac{k\cos\theta_{2}}{1+e\cos\theta_{2}}, & \quad & y=\dfrac{k\sin\theta_{2}}{1+e\cos\theta_{2}}, \end{alignat*} where \(0< b< a,\) \(0< e< 1\) and \(0< k\). Find the position of the axes for \(E_{2}\) relative to the axes for \(E_{1}\) and show that \(k=a(1-e^{2})\) and \(b^{2}=a^{2}(1-e^{2}).\) {[}The standard polar equation of an ellipse is \(r=\dfrac{\ell}{1+e\cos\theta}.]\) By considering expressions for the length of the perimeter of the ellipse, or otherwise, prove that \[ \int_{0}^{\pi}\sqrt{1-e^{2}\cos^{2}\theta}\,\mathrm{d}\theta=\int_{0}^{\pi}\frac{1-e^{2}}{(1+e\cos\theta)^{2}}\sqrt{1+e^{2}+2e\cos\theta}\,\mathrm{d}\theta. \] Given that \(e\) is so small that \(e^{6}\) may be neglected, show that the value of either integral is \[ \tfrac{1}{64}\pi(64-16e^{2}-3e^{4}). \]

The curve \(C\) has the differential equation in polar coordinates \[ \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r=5\sin3\theta,\qquad\text{for }\quad\frac{\pi}{5}\leqslant\theta\leqslant\frac{3\pi}{5}, \] and, when \(\theta=\dfrac{\pi}{2},\) \(r=1\) and \(\dfrac{\mathrm{d}r}{\mathrm{d}\theta}=-2.\) Show that \(C\) forms a closed loop and that the area of the region enclosed by \(C\) is \[ \frac{\pi}{5}+\frac{25}{48}\left[\sin\left(\frac{\pi}{5}\right)-\sin\left(\frac{2\pi}{5}\right)\right]. \]

Show by means of a sketch that the parabola \(r(1+\cos\theta)=1\) cuts the interior of the cardioid \(r=4(1+\cos\theta)\) into two parts. Show that the total length of the boundary of the part that includes the point \(r=1,\theta=0\) is \(18\sqrt{3}+\ln(2+\sqrt{3}).\)

Show Solution

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The curves will intersect when: \begin{align*} && \frac{1}{1+\cos \theta} &= 4 (1 + \cos \theta) \\ \Rightarrow && 1 + \cos \theta &= \pm \frac{1}{2} \\ \Rightarrow && \cos \theta &= -\frac12 \\ \Rightarrow && \theta &= \pm \frac{2\pi}{3}, \end{align*} Therefore we can measure the two sides of the boundaries. For the cardioid it will be: \begin{align*} s &= \int_{-2\pi/3}^{2 \pi /3} \sqrt{r^2 + \left ( \frac{\d r}{\d \theta} \right)^2} \d \theta \\ &= \int_{-2\pi/3}^{2 \pi /3}\sqrt{r^2 + \left ( \frac{\d r}{\d \theta} \right)^2} \d \theta \\ &= \int_{-2\pi/3}^{2 \pi /3}\sqrt{16(1 + \cos \theta)^2 + 16 \sin^2 \theta} \d \theta \\ &= 4\int_{-2\pi/3}^{2 \pi /3}\sqrt{2 + 2 \cos \theta} \d \theta \\ &= 8\int_{-2\pi/3}^{2 \pi /3}\sqrt{\cos^2 \frac{\theta}{2}} \d \theta \\ &= 8\int_{-2\pi/3}^{2 \pi /3}|\cos \frac{\theta}{2}| \d \theta \\ &= 16\int_{\pi}^{2\pi/3}(-\cos \frac{\theta}{2}) \d \theta + 8\int_{-\pi}^{\pi}\cos \frac{\theta}{2} \d \theta \\ &= 16 \cdot \left [ 2\sin \frac{\theta}{2}\right]_{\pi}^{2\pi/3}+ 8 \cdot 4 \\ &= 16 \cdot (\sqrt{3}-2) + 8 \cdot 4 \\ &= 16\sqrt{3} \end{align*} For the parabola we have that \(\sqrt{x^2+y^2} + x = 1 \Rightarrow x^2 + y^2 = 1 - 2x + x^2 \Rightarrow y^2 = 1-2x\). So we can parameterise our parabola as \(y = t, x = \frac{1-t^2}2\). And we are interested in the points \(t = -\sqrt{3}\) and \(t =\sqrt3\) \begin{align*} &&s &= \int_{-\sqrt3}^\sqrt3 \sqrt{\left ( \frac{\d x}{\d t} \right)^2 + \left ( \frac{\d y}{\d t} \right)^2 } \d t \\ &&&= \int_{-\sqrt3}^\sqrt3 \sqrt{t^2+1^2} \d t \\ \sinh u = t, \frac{\d t}{\d u} = \cosh u&&&= \int_{-\sinh^{-1} \sqrt3}^{\sinh^{-1}\sqrt3} \cosh^2 u \d u \\ &&&= \left [\frac12 u + \frac14 \sinh(2u) \right ]_{-\sinh^{-1} \sqrt3}^{\sinh^{-1}\sqrt3} \\ &&&= \sinh^{-1} \sqrt{3} + 2\sqrt{3} \\ &&&= \ln(2 + \sqrt{3}) + 2\sqrt{3} \end{align*} Therefore the total distance is as required.

Show that, for a given constant \(\gamma\) \((\sin\gamma\neq0)\) and with suitable choice of the constants \(A\) and \(B\), the line with cartesian equation \(lx+my=1\) has polar equations \[ \frac{1}{r}=A\cos\theta+B\cos(\theta-\gamma). \] The distinct points \(P\) and \(Q\) on the conic with polar equations \[ \frac{a}{r}=1+e\cos\theta \] correspond to \(\theta=\gamma-\delta\) and \(\theta=\gamma+\delta\) respectively, and \(\cos\delta\neq0.\) Obtain the polar equation of the chord \(PQ.\) Hence, or otherwise, obtain the equation of the tangent at the point where \(\theta=\gamma.\) The tangents at \(L\) and \(M\) to a conic with focus \(S\) meet at \(T.\) Show that \(ST\) bisects the angle \(LSM\) and find the position of the intersection of \(ST\) and \(LM\) in terms of your chosen parameters for \(L\) and \(M.\)

1. Show that in polar coordinates, the gradient of any curve at the point \((r,\theta)\) is \[ \left.\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\tan\theta+r\right)\right/\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}-r\tan\theta\right). \]
   

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2. A mirror is designed so that any ray of light which hits one side of the mirror and which is parallel to a certain fixed line \(L\) is reflected through a fixed point \(O\) on \(L\). For any ray hitting the mirror, the normal to the mirror at the point of reflection bisects the angle between the incident ray and the reflected ray, as shown in the figure. Prove that the mirror intersects any plane containing \(L\) in a parabola.

The distinct points \(P(ap^2 , 2ap)\), \(Q(aq^2 , 2aq)\) and \(R(ar^2,2ar)\) lie on the parabola \(y^2 = 4ax\), where \(a>0\). The points are such that the normal to the parabola at \(Q\) and the normal to the parabola at \(R\) both pass through \(P\).

1. Show that \(q^2 +qp + 2 = 0\).
2. Show that \(QR\) passes through a certain point that is independent of the choice of \(P\).
3. Let \(T\) be the point of intersection of \(OP\) and \(QR\), where \(O\) is the coordinate origin. Show that \(T\) lies on a line that is independent of the choice of \(P\). Show further that the distance from the \(x\)-axis to \(T\) is less than \(\dfrac {\;a}{\sqrt2}\,\).

1. The line \(L\) has equation \(y=mx+c\), where \(m>0\) and \(c>0\). Show that, in the case \(mc>a>0\), the shortest distance between \(L\) and the parabola \(y^2=4ax\) is \[ \frac{mc-a}{m\sqrt{m^2+1}}\,.\] What is the shortest distance in the case that \(mc\le a\)?
2. Find the shortest distance between the point \((p,0)\), where \(p>0\), and the parabola \(y^2=4ax\), where \(a>0\), in the different cases that arise according to the value of \(p/a\). [\textit{You may wish to use the parametric coordinates \((at^2, 2at)\) of points on the parabola.}] Hence find the shortest distance between the circle $(x-p)^2 + y^2 =b^2\(, where \)p>0\( and \)b>0\(, and the parabola \)y^2=4ax\(, where \)a>0$, in the different cases that arise according to the values of \(p\), \(a\) and~\(b\).

The point \(P(a\cos\theta\,,\, b\sin\theta)\), where \(a>b>0\), lies on the ellipse \[\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2}=1\,.\] The point \(S(-ea\,,\,0)\), where \(b^2=a^2(1-e^2)\,\), is a focus of the ellipse. The point \(N\) is the foot of the perpendicular from the origin, \(O\), to the tangent to the ellipse at \(P\). The lines \(SP\) and \(ON\) intersect at \(T\). Show that the \(y\)-coordinate of \(T\) is \[\dfrac{b\sin\theta}{1+e\cos\theta}\,.\] Show that \(T\) lies on the circle with centre \(S\) and radius \(a\).

Show Solution

Find the gradient of the tangent of the ellipse at \(P\): \begin{align*} && \frac{2x}{a^2} + \frac{2y}{b^2} \frac{\d y}{\d x} &= 0 \\ \Rightarrow && \frac{\d y}{\d x} &= - \frac{2xb^2}{2ya^2} \\ &&&=- \frac{a \cos \theta b^2}{b \sin \theta a^2} \\ &&&=-\frac{b}{a} \cot \theta \end{align*} Therefore the gradient of \(ON\) is \(\frac{a}{b} \tan \theta\). \begin{align*} && y &= \frac{a}{b} \tan \theta x \\ && \frac{y-0}{x-(-ea)} &= \frac{b\sin \theta-0}{a\cos \theta -(-ea)} \\ && y &= \frac{b \sin \theta}{a(e+\cos \theta)}(x+ea) \\ \Rightarrow && y &= \frac{b \sin \theta}{a(\cos \theta+e)}\frac{b}{a} \cot \theta y+ \frac{eb \sin \theta}{\cos \theta + e} \\ &&&= \frac{b^2 \cos \theta}{a^2(\cos \theta +e)}y + \frac{eb \sin \theta}{\cos \theta + e} \\ \Rightarrow && (\cos \theta+e)y &= (1-e^2)\cos \theta y +eb \sin \theta\\ && e(1+e\cos \theta)y &= eb \sin \theta \\ \Rightarrow && y &= \frac{b \sin \theta}{1+e\cos \theta} \\ && x &= \frac{b \sin \theta}{1+e\cos \theta} \frac{b}{a} \cot \theta \\ &&&= \frac{b^2 \cos \theta}{a(1+e\cos \theta)} \end{align*} Therefore \(\displaystyle T\left (\frac{b^2 \cos \theta}{a(1+e\cos \theta)}, \frac{b \sin \theta}{1+e\cos \theta} \right)\). Finally, we can look at the distance of \(T\) from \(S\) \begin{align*} && d^2 &= \left (\frac{b^2 \cos \theta}{a(1+e\cos \theta)}-(-ea) \right)^2 + \left (\frac{b \sin \theta}{1+e\cos \theta} -0\right)^2 \\ &&&= \frac{\left (b^2 \cos \theta+ea^2(1+e\cos\theta)\right)^2 + \left ( ab \sin \theta\right)^2}{a^2(1+e\cos \theta)^2} \\ &&&= \frac{b^4\cos^2\theta+e^2a^4(1+e\cos\theta)^2+2ea^2b^2(1+e\cos\theta)+a^2b^2\sin^2\theta}{a^2(1+e\cos\theta)^2} \\ &&&= \frac{a^4(1-e^2)^2\cos^2\theta+e^2a^4(1+e\cos\theta)^2+2ea^2a^2(1-e^2)(1+e\cos\theta)+a^4(1-e^2)\sin^2\theta}{a^2(1+e\cos\theta)^2} \\ &&&= a^2 \left ( \frac{(1-e^2)^2\cos^2\theta+e^2(1+e\cos\theta)^2+2e(1-e^2)(1+e\cos\theta)+(1-e^2)(1-\cos^2\theta)}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \left ( \frac{e^2(1+e\cos\theta)^2+(1-e^2)((1-e^2)\cos^2\theta+2e(1+e\cos\theta)+(1-\cos^2\theta))}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \left ( \frac{e^2(1+e\cos\theta)^2+(1-e^2)(1+e\cos\theta)^2}{(1+e\cos\theta)^2} \right) \\ &&&= a^2 \end{align*} Therefore a circle radius \(a\) centre \(S\).

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Let \(P\) be the point on the curve \(y=ax^2+bx+c\) (where \(a\) is non-zero) at which the gradient is \(m\). Show that the equation of the tangent at \(P\) is \[ y-mx=c-\frac{(m-b)^2}{4a}\;. \] Show that the curves \(y=a_1 x^2+b_1 x+c_1\) and \(y=a_2 x^2+b_2 x+c_2\) (where \(a_1\) and \(a_2\) are non-zero) have a common tangent with gradient \(m\) if and only if \[ (a_2 -a_1 )m^2 + 2(a_1 b_2-a_2 b_1)m + 4a_1 a_2(c_2-c_1)+ a_2 b_1^2-a_1 b_2 ^2=0\;. \] Show that, in the case \(a_1 \ne a_2 \,\), the two curves have exactly one common tangent if and only if they touch each other. In the case \(a_1 =a_2\,\), find a necessary and sufficient condition for the two curves to have exactly one common tangent.

In the \(x\)--\(y\) plane, the point \(A\) has coordinates \((a\,,0)\) and the point \(B\) has coordinates \((0\,,b)\,\), where \(a\) and \(b\) are positive. The point \(P\,\), which is distinct from \(A\) and \(B\), has coordinates~\((s,t)\,\). \(X\) and \(Y\) are the feet of the perpendiculars from \(P\) to the \(x\)--axis and \(y\)--axis respectively, and \(N\) is the foot of the perpendicular from \(P\) to the line \(AB\,\). Show that the coordinates \((x\,,y)\) of \(N\) are given by \[ x= \frac {ab^2 -a(bt-as)}{a^2+b^2} \;, \ \ \ y = \frac{a^2b +b(bt-as)}{a^2+b^2} \;. \] Show that, if $\ds \ \left( \frac{t-b} s\right)\left( \frac t {s-a}\right) = -1\;\(, then \)N$ lies on the line \(XY\,\). Give a geometrical interpretation of this result.

A parabola has the equation \(y=x^{2}.\) The points \(P\) and \(Q\) with coordinates \((p,p^{2})\) and \((q,q^{2})\) respectively move on the parabola in such a way that \(\angle POQ\) is always a right angle.

1. Find and sketch the locus of the midpoint \(R\) of the chord \(PQ.\)
2. Find and sketch the locus of the point \(T\) where the tangents to the parabola at \(P\) and \(Q\) intersect.

The straight line \(OSA,\) where \(O\) is the origin, bisects the angle between the positive \(x\) and \(y\) axes. The ellipse \(E\) has \(S\) as focus. In polar coordinates with \(S\) as pole and \(SA\) as the initial line, \(E\) has equation \(\ell=r(1+e\cos\theta).\) Show that, at the point on \(E\) given by \(\theta=\alpha,\) the gradient of the tangent to the ellipse is given by \[ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\sin\alpha-\cos\alpha-e}{\sin\alpha+\cos\alpha+e}. \] The points on \(E\) given by \(\theta=\alpha\) and \(\theta=\beta\) are the ends of a diameter of \(E\). Show that \[ \tan(\alpha/2)\tan(\beta/2)=-\frac{1+e}{1-e}. \] [Hint. A diameter of an ellipse is a chord through its centre.]

Show Solution

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\begin{align*} && \ell &= r(1 + e \cos \theta) \\ \Rightarrow && 0 &= \frac{\d r}{\d \theta}(1 + e \cos \theta) - re \sin \theta \\ \Rightarrow && \frac{\d r}{\d \theta} &= \frac{re \sin \theta}{1+e \cos \theta} \end{align*} Suppose we consider the \((x',y')\) plane, which is essentially the \(x-y\) plan rotated by \(45^\circ\), then we would have \begin{align*} && \frac{\d y'}{ \d x'} &= \frac{\frac{\d y'}{\d \theta}}{\frac{\d x'}{\d \theta}} \\ &&&= \frac{\frac{\d r}{\d \theta} \sin \theta + r\cos \theta}{\frac{\d r}{\d \theta} \cos \theta - r\sin \theta} \\ &&&= \frac{\frac{re \sin \theta}{1+e \cos \theta} \sin \theta + r\cos \theta}{\frac{re \sin \theta}{1+e \cos \theta} \cos\theta -r\sin \theta} \\ &&&= \frac{re\sin^2 \theta+r \cos \theta(1+e \cos \theta)}{re\sin \theta \cos \theta -r \sin \theta (1+e \cos \theta)} \\ &&&= \frac{\cos \theta + e \cos^2 \theta+e \sin^2 \theta}{-\sin \theta} \\ &&&= \frac{\cos \theta + e}{-\sin \theta} \end{align*} Since our frame is rotated by \(45^\circ\) we need to consider the appropriate gradient for this. We know that \(m = \tan \theta\) so \(m' = \tan (\theta+45^{\circ}) = \frac{1+m}{1-m}\) therefore we should have \begin{align*} && \frac{\d y}{ \d x} &= \frac{1+\frac{\cos \theta + e}{-\sin \theta}}{1-\frac{\cos \theta + e}{-\sin \theta}} \\ &&&= \frac{\cos \theta - \sin \theta + e}{-\sin \theta - \cos \theta-e} \\ &&&= \frac{\sin \theta - \cos \theta -e}{\sin \theta + \cos \theta +e} \end{align*} As required. The tangents at those points are parallel, therefore \begin{align*} && \frac{\cos \alpha+e}{\sin \alpha} &= \frac{\cos \beta+e}{\sin \beta} \\ \Rightarrow && \frac{\frac{1-\tan^2 \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}}+e}{\frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}}} &= \frac{\frac{1-\tan^2 \frac{\beta}{2}}{1+\tan^2 \frac{\beta}{2}}+e}{\frac{2\tan \frac{\beta}{2}}{1+\tan^2 \frac{\beta}{2}}} \\ && \frac{1-\tan^2 \frac{\alpha}{2}+e(1+\tan^2\frac{\alpha}{2})}{2\tan\frac{\alpha}{2}} &= \frac{1-\tan^2 \frac{\beta}{2}+e(1+\tan^2\frac{\beta}{2})}{2\tan\frac{\beta}{2}} \\ && \frac{(1+e)+(e-1)\tan^2 \frac{\alpha}{2}}{2\tan \frac{\alpha}{2}} &= \frac{(1+e)+(e-1)\tan^2 \frac{\beta}{2}}{2\tan \frac{\beta}{2}} \\ && \frac{(1+e)}{\tan\frac{\alpha}2} - (1-e)\tan\frac{\alpha}2 &= \frac{(1+e)}{\tan\frac{\beta}2} - (1-e)\tan\frac{\beta}2 \end{align*} ie both \(\tan \frac{\alpha}{2}\) and \(\tan \frac{\beta}{2}\) are roots of a quadratic of the form \((1-e)x^2-cx-(1+e)\) but this means the product of the roots is \(-\frac{1+e}{1-e}\)

Find the equations of the tangent and normal to the parabola \(y^{2}=4ax\) at the point \((at^{2},2at).\) For \(i=1,2,\) and 3, let \(P_{i}\) be the point \((at_{i}^{2},2at_{i}),\) where \(t_{1},t_{2}\) and \(t_{3}\) are all distinct. Let \(A_{1}\) be the area of the triangle formed by the tangents at \(P_{1},P_{2}\) and \(P_{3},\) and let \(A_{2}\) be the area of the triangle formed by the normals at \(P_{1},P_{2}\) and \(P_{3}.\) Using the fact that the area of the triangle with vertices at \((x_{1},y_{1}),(x_{2},y_{2})\) and \((x_{3},y_{3})\) is the absolute value of \[ \tfrac{1}{2}\det\begin{pmatrix}x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1\\ x_{3} & y_{3} & 1 \end{pmatrix}, \] show that \(A_{3}=(t_{1}+t_{2}+t_{3})^{2}A_{1}.\) Deduce a necessary and sufficient condition in terms of \(t_{1},t_{2}\) and \(t_{3}\) for the normals at \(P_{1},P_{2}\) and \(P_{3}\) to be concurrent.

In this question, you should ignore issues of convergence.

1. Let \[ I = \int_0^1 \frac{\f(x^{-1}) } {1+x} \, \d x \,, \] where \(\f(x)\) is a function for which the integral exists. Show that \[ I = \sum_{n=1}^\infty \int_n^{n+1} \frac{\f(y) } {y(1+y)}\, \d y \] and deduce that, if \(\f(x) = \f(x+1)\) for all \(x\), then \[ I= \int_0^1 \frac{\f(x)} {1+x} \, \d x \,. \]
2. The {\em fractional part}, \(\{x\}\), of a real number \(x\) is defined to be \(x-\lfloor x\rfloor\) where \(\lfloor x \rfloor\) is the largest integer less than or equal to \(x\). For example \(\{3.2\} = 0.2\) and \(\{3\}=0\,\). Use the result of part (i) to evaluate \[ \displaystyle \int _0^1 \frac { \{x^{-1}\}}{1+x}\, \d x \text{ and } \displaystyle \int _0^1 \frac { \{2x^{-1}\}}{1+x}\, \d x \,. \]
3. (Bonus) Use the same method to evaluate \[ \int_0^1 \frac {x\{x^{-1}\}}{1-x^2} \, \d x \,. \]
4. (Bonus - harder) Use the same method to evaluate \[ \int_0^1 \frac {x^2\{x^{-1}\}}{1-x^2} \, \d x \,. \]

In this question, you should ignore issues of convergence.

1. Write down the binomial expansion, for \(\vert x \vert<1\,\), of \(\;\dfrac{1}{1+x}\,\) and deduce that %. By considering %$ %\displaystyle \int \frac 1 {1+x} \, \d x %\,, %$ %show that \[ \displaystyle \ln (1+x) = -\sum_{n=1}^\infty \frac {(-x)^n}n \, \] for \(\vert x \vert <1 \,\).
2. Write down the series expansion in powers of \(x\) of \(\displaystyle \e^{-ax}\,\). Use this expansion to show that \[ \int_0^\infty \frac {\left(1- \e^{-ax}\right)\e^{-x}}x \,\d x = \ln(1+a) \ \ \ \ \ \ \ (\vert a \vert <1)\,. \]
3. Deduce the value of \[ \int_0^1 \frac{x^p - x^q}{\ln x} \, \d x \ \ \ \ \ \ (\vert p\vert <1, \ \vert q\vert <1) \,. \]

1. By use of calculus, show that \(x- \ln(1+x)\) is positive for all positive \(x\). Use this result to show that \[ \sum_{k=1}^n \frac 1 k > \ln (n+1) \,. \]
2. By considering \( x+\ln (1-x)\), show that \[ \sum_{k=1}^\infty \frac 1 {k^2} <1+ \ln 2 \,. \]

In this question, you may ignore questions of convergence. Let \(y= \dfrac {\arcsin x}{\sqrt{1-x^2}}\,\). Show that \[ (1-x^2)\frac {\d y}{\d x} -xy -1 =0 \] and prove that, for any positive integer \(n\), \[ (1-x^2) \frac{\d^{n+2}y}{\d x^{n+2}} - (2n+3)x \frac{\d ^{n+1}y}{\d x ^{n+1}} -(n+1)^2 \frac{\d^ny}{\d x^n}=0\, . \] Hence obtain the Maclaurin series for \( \dfrac {\arcsin x}{\sqrt{1-x^2}}\,\), giving the general term for odd and for even powers of \(x\). Evaluate the infinite sum \[ 1 + \frac 1 {3!} + \frac{2^2}{5!} + \frac {2^2\times 3^2}{7!}+\cdots + \frac {2^2\times 3^2\times \cdots \times n^2}{(2n+1)!} + \cdots\,. \]

1. Show that \[ \sum_{n=1} ^\infty \frac{n+1}{n!} = 2\e - 1 \] and \[ \sum _{n=1}^\infty \frac {(n+1)^2}{n!} = 5\e-1\,. \] Sum the series $\displaystyle \sum _{n=1}^\infty \frac {(2n-1)^3}{n!} \,.$
2. Sum the series $\displaystyle \sum_{n=0}^\infty \frac{(n^2+1)2^{-n}}{(n+1)(n+2)}\,$, giving your answer in terms of natural logarithms.

In this question, you may assume that the infinite series \[ \ln(1+x) = x-\frac{x^2}2 + \frac{x^3}{3} -\frac {x^4}4 +\cdots + (-1)^{n+1} \frac {x^n}{n} + \cdots \] is valid for \(\vert x \vert <1\).

1. Let \(n\) be an integer greater than 1. Show that, for any positive integer \(k\), \[ \frac1{(k+1)n^{k+1}} < \frac1{kn^{k}}\,. \] Hence show that \(\displaystyle \ln\! \left(1+\frac1n\right) <\frac1n\,\). Deduce that \[ \left(1+\frac1n\right)^{\!n}<\e\,. \]
2. Show, using an expansion in powers of \(\dfrac1y\,\), that $ \displaystyle \ln \! \left(\frac{2y+1}{2y-1}\right) > \frac 1y %= \sum _{r=0}^\infty \frac 1{(2r+1)(2y)^{2r}}\,. \( for \)y>\frac12$. Deduce that, for any positive integer \(n\), \[ \e < \left(1+\frac1n\right)^{\! n+\frac12}\,. \]
3. Use parts (i) and (ii) to show that as \(n\to\infty\) \[ \left(1+\frac1n\right)^{\!n} \to \e\,. \]

The function \(\f(t)\) is defined, for \(t\ne0\), by \[ \f(t) = \frac t {\e^t-1}\,. \] \begin{questionparts} \item By expanding \(\e^t\), show that \(\displaystyle \lim _{t\to0} \f(t) = 1\,\). Find \(\f'(t)\) and evaluate \(\displaystyle \lim _{t\to0} \f'(t)\,\). \item Show that \(\f(t) +\frac12 t\) is an even function. [{\bf Note:} A function \(\g(t)\) is said to be {\em even} if \(\g(t) \equiv \g(-t)\).] \item Show with the aid of a sketch that \( \e^t( 1-t)\le 1\,\) and deduce that \(\f'(t)\ne 0\) for \(t\ne0\). \end{questionpart} Sketch the graph of \(\f(t)\).

Show Solution

1. Claim \(f(t) + \frac12 t\) is an even function. Proof: Consider \(f(-t) - \frac12t\), then \begin{align*} f(-t) - \frac12t &= \frac{-t}{e^{-t}-1} - \frac12t \\ &= \frac{-te^t}{1-e^t} - \frac12 t \\ &= \frac{t(1-e^t) -t}{1-e^t} - \frac12 t \\ &= t - \frac{t}{1-e^t} - \frac12 t \\ &= \frac{t}{e^t-1} + \frac12 t \end{align*} So it is even.
2. 

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   Drawing the tangent to \(y = e^{-x}\) at \((0,1)\) we find that \(e^{-t} \geq (1-t)\) for all \(t\), in particular, \(e^t(1-t) \leq 1\) \(f'(t) = \frac{(e^t(1-t) -1}{(e^t-1)^2} \leq 0\) and \(f'(t) = -\frac12\) when \(t = 0\)

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[Note: This is the exponential generating function for the Bernoulli numbers]

The function \(f\) satisfies the identity \begin{equation} f(x) +f(y) \equiv f(x+y) \tag{\(*\)} \end{equation} for all \(x\) and \(y\). Show that \(2\f(x)\equiv \f(2x)\) and deduce that \(f''(0)=0\). By considering the Maclaurin series for \(\f(x)\), find the most general function that satisfies \((*)\). [{\it Do not consider issues of existence or convergence of Maclaurin series in this question.}]

1. By considering the function \(\G\), defined by \(\ln\big(\g(x)\big) =\G(x)\), find the most general function that, for all \(x\) and \(y\), satisfies the identity \[ \g(x) \g(y) \equiv \g(x+y)\,. \]
2. By considering the function \(H\), defined by \(\h(\e^u) =H(u)\), find the most general function that satisfies, for all positive \(x\) and \(y\), the identity \[ \h(x) +\h(y) \equiv \h(xy) \,. \]
3. Find the most general function \(t\) that, for all \(x\) and \(y\), satisfies the identity \begin{equation*} t(x) + t(y) \equiv t(z)\,, \end{equation*} where \(z= \dfrac{x+y}{1-xy}\,\).

1. Let \[ \tan x = \sum\limits_{n=0}^\infty a_n x^n \text{ and } \cot x = \dfrac 1 x +\sum\limits_{n=0}^\infty b_nx^n \] for \(0< x < \frac12\pi\,\). Explain why \(a_n=0\) for even \(n\). Prove the identity \[ \cot x - \tan x \equiv 2 \cot 2x\, \] and show that \[a_{n} = (1-2^{n+1})b_n\,.\]
2. Let $ \displaystyle {\rm cosec}\, x = \frac1x +\sum\limits _{n=0}^\infty c_n x^n\,$ for \(0< x < \frac12\pi\,\). By considering \(\cot x + \tan x\), or otherwise, show that \[ c_n = (2^{-n} -1)b_n \,. \]
3. Show that \[ \left(1+x{ \sum\limits_{n=0}^\infty} b_n x^n \right)^2 +x^2 = \left(1+x{ \sum\limits_{n=0} ^\infty} c_n x^n \right)^2\,. \] Deduce from this and the previous results that \(a_1=1\), and find \(a_3\).

Given that \(y = \ln ( x + \sqrt{x^2 + 1})\), show that \( \displaystyle \frac{\d y}{\d x} = \frac1 {\sqrt{x^2 + 1} }\;\). Prove by induction that, for \(n \ge 0\,\), \[ \l x^2 + 1 \r y^{\l n + 2 \r} + \l 2n + 1 \r x y^{\l n + 1 \r} + n^2 y^{\l n \r} = 0\;, \] where \(\displaystyle y^{(n)} = \frac{\d^n y}{\d x^n}\) and \(y^{(0)} =y\,\). Using this result in the case \(x = 0\,\), or otherwise, show that the Maclaurin series for \(y\) begins \[ x - {x^3 \over 6} +{3 x^5 \over 40} \] and find the next non-zero term.

Sketch the graph of \({\rm f}(s)={ \e}^s(s-3)+3\) for \(0\le s < \infty\). Taking \({\e\approx 2.7}\), find the smallest positive integer, \(m\), such that \({\rm f}(m) > 0\). Now let $$ {\rm b}(x) = {x^3 \over \e^{x/T} -1} \, $$ where \(T\) is a positive constant. Show that \({\rm b}(x)\) has a single turning point in \(0 < x < \infty\). By considering the behaviour for small \(x\) and for large \(x\), sketch \({\rm b}(x)\) for \(0\le x < \infty\). Let $$ \int_0^\infty {\rm b}(x)\,\d x =B, $$ which may be assumed to be finite. Show that \(B = K T^n\) where \(K\) is a constant, and \(n\) is an integer which you should determine. Given that \(\displaystyle{B \approx 2 \int_0^{Tm} {\rm b}(x) {\,\rm d }x}\), use your graph of \({\rm b}(x)\) to find a rough estimate for \(K\).

The exponential of a square matrix \({\bf A}\) is defined to be $$ \exp ({\bf A}) = \sum_{r=0}^\infty {1\over r!} {\bf A}^r \,, $$ where \({\bf A}^0={\bf I}\) and \(\bf I\) is the identity matrix. Let $$ {\bf M}=\left(\begin{array}{cc} 0 & -1 \\ 1 & \phantom{-} 0 \end{array} \right) \,. $$ Show that \({\bf M}^2=-{\bf I}\) and hence express \(\exp({\theta {\bf M}})\) as a single \(2\times 2\) matrix, where \(\theta\) is a real number. Explain the geometrical significance of \(\exp({\theta {\bf M}})\). Let $$ {\bf N}=\left(\begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array}\right) \,. $$ Express similarly \(\exp({s{\bf N}})\), where \(s\) is a real number, and explain the geometrical significance of \(\exp({s{\bf N}})\). For which values of \(\theta\) does $$ \exp({s{\bf N}})\; \exp({\theta {\bf M}})\, = \, \exp({\theta {\bf M}})\;\exp({s{\bf N}}) $$ for all \(s\)? Interpret this fact geometrically.

1. By considering the series expansion of \((x^2+5x+4){\rm \; e}^x\) show that \[10{\rm\, e}=4+\frac{3^2}{1!}+\frac{4^2}{2!}+\frac{5^2}{3!}+\cdots\;.\]
2. Show that \[5{\rm\, e}=1+\frac{2^2}{1!}+\frac{3^2}{2!}+\frac{4^2}{3!}+\cdots\;.\]
3. Evaluate \[1+\frac{2^3}{1!}+\frac{3^3}{2!}+\frac{4^3}{3!}+\cdots\;.\]

The function \(\mathrm{f}\) is given by \(\mathrm{f}(x)=\sin^{-1}x\) for \(-1 < x < 1.\) Prove that \[ (1-x^{2})\mathrm{f}''(x)-x\mathrm{f}'(x)=0. \] Prove also that \[ (1-x^{2})\mathrm{f}^{(n+2)}(x)-(2n+1)x\mathrm{f}^{(n+1)}(x)-n^{2}\mathrm{f}^{(n)}(x)=0, \] for all \(n>0\), where \(\mathrm{f}^{(n)}\) denotes the \(n\)th derivative of \(\mathrm{f}\). Hence express \(\mathrm{f}(x)\) as a Maclaurin series. The function \(\mathrm{g}\) is given by \[ \mathrm{g}(x)=\ln\sqrt{\frac{1+x}{1-x}}, \] for \(-1 < x < 1.\) Write down a power series expression for \(\mathrm{g}(x),\) and show that the coefficient of \(x^{2n+1}\) is greater than that in the expansion of \(\mathrm{f},\) for each \(n > 0\).

1. Evaluate \[ \sum_{r=1}^{n}\frac{6}{r(r+1)(r+3)}. \]
2. Expand \(\ln(1+x+x^{2}+x^{3})\) as a series in powers of \(x\), where \(\left|x\right|<1\), giving the first five non-zero terms and the general term.
3. Expand \(\mathrm{e}^{x\ln(1+x)}\) as a series in powers of \(x\), where \(-1 < x\leqslant1\), as far as the term in \(x^{4}\).

The points \(P\,(0,a),\) \(Q\,(a,0)\) and \(R\,(a,-a)\) lie on the curve \(C\) with cartesian equation \[ xy^{2}+x^{3}+a^{2}y-a^{3}=0,\qquad\mbox{ where }a>0. \] At each of \(P,Q\) and \(R\), express \(y\) as a Taylor series in \(h\), where \(h\) is a small increment in \(x\), as far as the term in \(h^{2}.\) Hence, or otherwise, sketch the shape of \(C\) near each of these points. Show that, if \((x,y)\) lies on \(C\), then \[ 4x^{4}-4a^{3}x-a^{4}\leqslant0. \] Sketch the graph of \(y=4x^{4}-4a^{3}x-a^{4}.\) Given that the \(y\)-axis is an asymptote to \(C\), sketch the curve \(C\).

Show Solution

\begin{align*} && 0 &= xy^{2}+x^{3}+a^{2}y-a^{3} \\ \frac{\d }{\d x} : && 0 &= y^2+2xyy' + 3x^2+a^2y' \\ \Rightarrow && y' &= -\frac{y^2+3x^2}{a^2+2xy} \\ \\ \frac{\d^2 }{\d x^2}: && 0 &= 2yy'+2yy'+2x(y')^2+2xyy''+6x+a^2y'' \\ \Rightarrow && y'' &= -\frac{4yy'+2x(y')^2+6x}{a^2+2xy} \\ \\ P: && y &= a \\ && y' &= -\frac{a^2}{a^2} = -1 \\ && y'' &= -\frac{-4a}{a^2} = \frac{4}{a} \\ \Rightarrow && y &\approx a - h+\frac{2}{a}h^2 \\ \\ Q: && y &= 0 \\ && y' &= -\frac{3a^2}{a^2} = -3 \\ && y'' &= -\frac{18a+6a}{a^2} = -\frac{24}{a} \\ \Rightarrow && y &\approx 0-3h-\frac{12}{a}h \\ \\ R: && y &= -a \\ && y' &= -\frac{a^2+3a^2}{a^2-2a^2} = 4 \\ && y'' &= -\frac{-16a+32a+6a}{a^2-2a^2} = \frac{22}{a} \\ \Rightarrow && y &\approx -a+4h + \frac{11}{a}h^2 \end{align*} Alternatively: \begin{align*} && 0 &= xy^{2}+x^{3}+a^{2}y-a^{3} \\ P(0,a): && y &\approx a + c_1h + c_2h^2 \\ && 0 &= h(a+c_1h)^2 + a^2(a + c_1h + c_2h^2)-a^3 \\ &&&= a^3-a^3 + (a^2+a^2c_1)h+(2ac_1+a^2c_2)h^2 \\ \Rightarrow && c_1 &=-1, c_2 =\frac{2}{a} \\ \Rightarrow && y &\approx a - h + \frac{2}{a}h^2 \\ \\ Q(a,0): && y &\approx c_1h + c_2h^2 \\ && 0 &= (a+h)(c_1h)^2+(a+h)^3+a^2(c_1h + c_2h^2 )-a^3 \\ &&&= a^3-a^3+(3a^2+a^2c_1)h + (ac_1^2+3a+a^2c_2)h^2 + \cdots \\ \Rightarrow && c_1 &=-3, c_2 = -\frac{12}{a} \\ \Rightarrow && y &\approx -3h -\frac{12}{a}h \\ \\ R(a,-a): && y &\approx -a + c_1h + c_2h^2 \\ && 0 &= (a+h)(-a + c_1h+c_2h^2)^2+(a+h)^3+a^2(-a + c_1h + c_2h^2)-a^3 \\ &&&= (a^2-2a^2c_1+3a^2+a^2c_1)h+(-2ac_1+c_1^2+\cdots)h^2 \\ \Rightarrow && c_1 &=4, c_2 = \frac{11}{a} \\ \Rightarrow && y &\approx -a + 4h + \frac{11}{a} \end{align*}

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If \((x,y)\) lies on the curve, then viewing it as a quadratic in \(y\) we must have \(\Delta = (a^2)^2-4\cdot x \cdot (x^3-a^3) \geq 0 \Rightarrow a^4-4x^4+4xa^3 \geq 0 \Rightarrow 4x^4-4a^3x-a^4 \leq 0\)

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Let \begin{alignat*}{2} \tan x & =\ \ \, \quad{\displaystyle \sum_{n=0}^{\infty}a_{n}x^{n}} & & \text{ for small }x,\\ x\cot x & =1+\sum_{n=1}^{\infty}b_{n}x^{n}\quad & & \text{ for small }x\text{ and not zero}. \end{alignat*} Using the relation \[ \cot x-\tan x=2\cot2x,\tag{*} \] or otherwise, prove that \(a_{n-1}=(1-2^{n})b_{n}\), for \(n\geqslant1\). Let \[ x\mathrm{cosec}x=1+{\displaystyle \sum_{n=1}^{\infty}c_{n}x^{n}\quad\text{ for small }x\neq0. \qquad \qquad \, } \] Using a relation similar to \((*)\) involving \(2\mathrm{cosec}2x\), or otherwise, prove that \[ c_{n}=\frac{2^{n-1}-1}{2^{n}-1}\frac{1}{2^{n-1}}a_{n-1}\qquad(n\geqslant1). \]

Show, by finding \(R\) and \(\gamma\), that \(A \sinh x + B\cosh x \) can be written in the form \(R\cosh (x+\gamma)\) if \(B>A>0\). Determine the corresponding forms in the other cases that arise, for \(A>0\), according to the value of \(B\). Two curves have equations \(y = \sech x\) and \(y = a\tanh x + b\,\), where \(a>0\).

1. In the case \(b>a\), show that if the curves intersect then the \(x\)-coordinates of the points of intersection can be written in the form \[ \pm\arcosh \left( \frac 1 {\sqrt{b^2-a^2}}\right) - {\rm artanh \,} \frac a b .\]
2. Find the corresponding result in the case \(a>b>0\,\).
3. Find necessary and sufficient conditions on \(a\) and \(b\) for the curves to intersect at two distinct points.
4. Find necessary and sufficient conditions on \(a\) and \(b\) for the curves to touch and, given that they touch, express the \(y\)-coordinate of the point of contact in terms of \(a\).

Starting from the result that \[ \.h(t) >0\ \mathrm{for}\ 0< t < x \Longrightarrow \int_0^x \.h(t)\ud t > 0 \,, \] show that, if \(\.f''(t) > 0\) for \(0 < t < x_0\) and \(\.f(0)=\.f'(0) =0\), then \(\.f(t)>0\) for \(0 < t < x_0\).

1. Show that, for \(0 < x < \frac12\pi\), \[ \cos x \cosh x <1 \,. \]
2. Show that, for \(0 < x < \frac12\pi\), \[ \frac 1 {\cosh x} < \frac {\sin x} x < \frac x {\sinh x} \,. \] %
3. Show that, for \(0 < x < \frac12\pi\), \(\tanh x < \tan x\).

Let \(y = \ln (x^2-1)\,\), where \(x >1\), and let \(r\) and \(\theta\) be functions of \(x\) determined by \(r= \sqrt{x^2-1}\) and \(\coth\theta= x\). Show that \[ \frac {\d y}{\d x} = \frac {2\cosh \theta}{r} \text{ and } \frac {\d^2 y}{\d x^2} = -\frac {2 \cosh 2\theta}{r^2}\,, \] and find an expression in terms of \(r\) and \(\theta\) for \(\dfrac {\d^3 y}{\d x^3}\,\). Find, with proof, a similar formula for \(\dfrac{\d^n y}{\d x^n}\) in terms of \(r\) and \(\theta\).

1. Solve the equation \(u^2+2u\sinh x -1=0\) giving \(u\) in terms of \(x\). Find the solution of the differential equation \[ \left( \frac{\d y}{\d x}\right)^{\!2} +2 \frac{\d y}{\d x} \sinh x -1 = 0 \] that satisfies \(y=0\) and \(\dfrac {\d y}{\d x} >0\) at \(x=0\).
2. Find the solution, not identically zero, of the differential equation \[ \sinh y \left( \frac{\d y}{\d x}\right)^{\!2} +2 \frac{\d y}{\d x} -\sinh y = 0 \] that satisfies \(y=0\) at \(x=0\), expressing your solution in the form \(\cosh y=\f(x)\). Show that the asymptotes to the solution curve are \(y=\pm(-x+\ln 4)\).

Define \(\cosh x\) and \(\sinh x\) in terms of exponentials and prove, from your definitions, that \[ \cosh^{4}x-\sinh^{4}x=\cosh2x \] and \[ \cosh^{4}x+\sinh^{4}x=\tfrac{1}{4}\cosh4x+\tfrac{3}{4}. \] Find \(a_{0},a_{1},\ldots,a_{n}\) in terms of \(n\) such that \[ \cosh^{n}x=a_{0}+a_{1}\cosh x+a_{2}\cosh2x+\cdots+a_{n}\cosh nx. \] Hence, or otherwise, find expressions for \(\cosh^{2m}x-\sinh^{2m}x\) and \(\cosh^{2m}x+\sinh^{2m}x,\) in terms of \(\cosh kx,\) where \(k=0,\ldots,2m.\)

The real numbers \(x\) and \(y\) satisfy the simultaneous equations $$ \sinh (2x) = \cosh y \qquad\hbox{and}\qquad \sinh(2y) = 2 \cosh x. $$ Show that \(\sinh^2 y\) is a root of the equation $$ 4t^3 + 4t^2 -4t -1=0 $$ and demonstrate that this gives at most one valid solution for \(y\). Show that the relevant value of \(t\) lies between \(0.7\) and \(0.8\), and use an iterative process to find \(t\) to 6 decimal places. Find \(y\) and hence find \(x\), checking your answers and stating the final answers to four decimal places.

1. Given that \[ \mathrm{f}(x)=\ln(1+\mathrm{e}^{x}), \] prove that \(\ln[\mathrm{f}'(x)]=x-\mathrm{f}(x)\) and that \(\mathrm{f}''(x)=\mathrm{f}'(x)-[\mathrm{f}'(x)]^{2}.\) Hence, or otherwise, expand \(\mathrm{f}(x)\) as a series in powers of \(x\) up to the term in \(x^{4}.\)
2. Given that \[ \mathrm{g}(x)=\frac{1}{\sinh x\cosh2x}, \] explain why \(\mathrm{g}(x)\) can not be expanded as a series of non-negative powers of \(x\) but that \(x\mathrm{g}(x)\) can be so expanded. Explain also why this latter expansion will consist of even powers of \(x\) only. Expand \(x\mathrm{g}(x)\) as a series as far as the term in \(x^{4}.\)

The transformation \(T\) from \(\binom{x}{y}\) to \(\binom{x'}{y'}\) in two-dimensional space is given by \[ \begin{pmatrix}x'\\ y' \end{pmatrix}=\begin{pmatrix}\cosh u & \sinh u\\ \sinh u & \cosh u \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix}, \] where \(u\) is a positive real constant. Show that the curve with equation \(x^{2}-y^{2}=1\) is transformed into itself. Find the equations of two straight lines through the origin which transform into themselves. A line, not necessary through the origin, which has gradient \(\tanh v\) transforms under \(T\) into a line with gradient \(\tanh v'\). Show that \(v'=v+u\). The lines \(\ell_{1}\) and \(\ell_{2}\) with gradients \(\tanh v_{1}\) and \(\tanh v_{2}\) transform under \(T\) into lines with gradients \(\tanh v_{1}'\) and \(\tanh v_{2}'\) respectively. Find the relation satisfied by \(v_{1}\) and \(v_{2}\) that is the necessary and sufficient for \(\ell_{1}\) and \(\ell_{2}\) to intersect at the same angle as their transforms. In the case when \(\ell_{1}\) and \(\ell_{2}\) meet at the origin, illustrate in a diagram the relation between \(\ell_{1}\), \(\ell_{2}\) and their transforms.

Solve the quadratic equation \(u^{2}+2u\sinh x-1=0\), giving \(u\) in terms of \(x\). Find the solution of the differential equation \[ \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}\sinh x-1=0 \] which satisfies \(y=0\) and \(y'>0\) at \(x=0\). Find the solution of the differential equation \[ \sinh x\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^{2}+2\frac{\mathrm{d}y}{\mathrm{d}x}-\sinh x=0 \] which satisfies \(y=0\) at \(x=0\).

The real variables \(\theta\) and \(u\) are related by the equation \(\tan\theta=\sinh u\) and \(0\leqslant\theta<\frac{1}{2}\pi.\) Let \(v=\mathrm{sech}u.\) Prove that

1. \(v=\cos\theta;\)
2. \(\dfrac{\mathrm{d}\theta}{\mathrm{d}u}=v;\)
3. \(\sin2\theta=-2\dfrac{\mathrm{d}v}{\mathrm{d}u}\quad\) and \(\quad\cos2\theta=-\cosh u\dfrac{\mathrm{d}^{2}v}{\mathrm{d}u^{2}};\)
4. \({\displaystyle \frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}v}{\mathrm{d}\theta^{2}}+\frac{\mathrm{d}v}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta^{2}}+\left(\frac{\mathrm{d}u}{\mathrm{d}\theta}\right)^{2}=0.}\)

Given that \(y=\cosh(n\cosh^{-1}x),\) for \(x\geqslant1,\) prove that \[ y=\frac{(x+\sqrt{x^{2}-1})^{n}+(x-\sqrt{x^{2}-1})^{n}}{2}. \] Explain why, when \(n=2k+1\) and \(k\in\mathbb{Z}^{+},\) \(y\) can also be expressed as the polynomial \[ a_{0}x+a_{1}x^{3}+a_{2}x^{5}+\cdots+a_{k}x^{2k+1}. \] Find \(a_{0},\) and show that

1. \(a_{1}=(-1)^{k-1}2k(k+1)(2k+1)/3\);
2. \(a_{2}=(-1)^{k}2(k-1)k(k+2)(2k+1)/15.\)

The real numbers \(x\) and \(y\) are related to the real numbers \(u\) and \(v\) by \[ 2(u+\mathrm{i}v)=\mathrm{e}^{x+\mathrm{i}y}-\mathrm{e}^{-x-\mathrm{i}y}. \] Show that the line in the \(x\)-\(y\) plane given by \(x=a\), where \(a\) is a positive constant, corresponds to the ellipse \[ \left(\frac{u}{\sinh a}\right)^{2}+\left(\frac{v}{\cosh a}\right)^{2}=1 \] in the \(u\)-\(v\) plane. Show also that the line given by \(y=b\), where \(b\) is a constant and \(0<\sin b<1,\) corresponds to one branch of a hyperbola in the \(u\)-\(v\) plane. Write down the \(u\) and \(v\) coordinates of one point of intersection of the ellipse and hyperbola branch, and show that the curves intersect at right-angles at this point. Make a sketch of the \(u\)-\(v\) plane showing the ellipse, the hyperbola branch and the line segments corresponding to:

1. \(x=0\);
2. \(y=\frac{1}{2}\pi,\) \(\quad 0\leqslant x\leqslant a.\)

Show Solution

\begin{align*} && 2(u+iv) &= e^{a+iy} - e^{-a-iy} \\ && &=(e^a \cos y - e^{-a} \cos y) + (e^a \sin y + e^{-a} \sin y)i \\ &&&= 2 \sinh a \cos y + 2\cosh a \sin y i\\ \Rightarrow && \frac{u}{\sinh a} &= \cos y \\ && \frac{v}{\cosh a} &= \sin y \\ \Rightarrow && 1 &= \left(\frac{u}{\sinh a}\right)^{2}+\left(\frac{v}{\cosh a}\right)^{2} \end{align*} \begin{align*} && 2(u+iv) &= e^{x+ib} - e^{-x-ib} \\ &&&= 2\sinh x \cos b + 2\cosh x \sin b i \\ \Rightarrow && \frac{u}{\cos b} &= \sinh x \\ && \frac{v}{\sin b} &= \cosh x \\ \Rightarrow && 1 &= \left (\frac{v}{\sin b} \right)^2 - \left (\frac{u}{\cos b} \right)^2 \end{align*} Therefore all the points lie of a hyperbola, and since \(\frac{v}{\sin b} > 0 \Rightarrow v > 0\) it's one branch of the hyperbola. (And all points on it are reachable as \(x\) varies from \(-\infty < x < \infty\). \begin{align*} 2(u+iv) &= e^{a+ib} - e^{-a-ib} \\ &= 2 \sinh a \cos b + 2 \cosh a \sin b i \end{align*} so we can take \(u = \sinh a \cos b, v = \cosh a \sin b\). \begin{align*} \frac{\d }{\d u} && 0 &= \frac{2 u}{\sinh^2 a} + \frac{2v}{\cosh^2 a} \frac{\d v}{\d u} \\ \Rightarrow && \frac{\d v}{\d u} &= -\frac{u}{v} \coth^2 a \\ \\ && \frac{\d v}{\d u} \rvert_{(u,v)} &= -\frac{\sinh a \cos b}{\cosh a \sin b} \coth^2 a \\ &&&= -\cot b \coth a \\ \frac{\d }{\d u} && 0 &= \frac{2 v}{\sin^2 b} \frac{\d v}{\d u} - \frac{2u}{\cos^2 b} \\ \Rightarrow && \frac{\d v}{\d u} &= \frac{u}{v} \tan^2 b \\ && \frac{\d v}{\d u} \rvert_{(u,v)} &= \frac{\sinh a \cos b}{\cosh a \sin b} \tan^2 b \\ &&&= \tanh a \tan b \end{align*} Therefore they are negative reciprocals and hence perpendicular.

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Show that the following functions are positive when \(x\) is positive:

1. [ \(x-\tanh x\)
2. \(x\sinh x-2\cosh x+2\)
3. \(2x\cosh2x-3\sinh2x+4x\).

Show Solution

1. Notice that \(f(x) = x - \tanh x\) has \(f'(x) = 1-\textrm{sech}^2 x = \tanh^2 x > 0\) so \(f(x)\) is strictly increasing on \((0, \infty)\) and \(f(0) = 0\) therefore \(f(x)\) is positive for all \(x\) positive
2. Let \(f(x) = x\sinh x-2\cosh x+2\) then \(f'(x) = \sinh x +x \cosh x - 2 \sinh x = x \cosh x -\sinh x = \cosh x ( x - \tanh x) > 0\) by the first part. \(f(0) = 0\) so \(f(x)\) is positive for all \(x\) positive.
3. Let \(f(x) = 2x\cosh2x-3\sinh2x+4x\) then \begin{align*} f'(x) &= 2\cosh 2x +4x\sinh 2x - 6 \cosh 2x + 4 \\ &= 4( x\sinh 2x-\cosh 2x +1) \\ &= 4(x2\cosh x \sinh x -2\cosh^2x ) \\ &= 8 \cosh^2 x (x - \tanh x) \end{align*} Which is always positive when \(x\) > 0, \(f(0) = 0\) so \(f(x) > 0\) for all positive \(x\).

Let \(f(x) = \frac{x(\cosh x)^{\frac{1}{3}}}{\sinh x}\) then \begin{align*} f'(x) &= \frac{(\cosh x)^{\frac13}\sinh x+\frac13 x \cosh^{-\frac23} x \sinh^2 x - x(\cosh x)^{\frac13} \cosh x}{\sinh^2 x} \\ &= \frac{\cosh x \sinh x + \frac13 x \sinh^2 x - x \cosh^2 x}{\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{3\cosh x \sinh x + x( \sinh^2 x - 3 \cosh^2 x)}{3\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{\frac32 \sinh 2x + x( -2\cosh 2x - 2)}{3\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{3 \sinh 2x -4x\cosh 2x - 4x}{6\cosh x^{\frac23} x \sinh^2 x} \\ \end{align*} which from the earlier part is always negative.

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Let \(f(x) = \sqrt{x^2 + 1} - x\).

1. Using a binomial series, or otherwise, show that, for large \(|x|\), \(\sqrt{x^2 + 1} \approx |x| + \frac{1}{2|x|}\). Sketch the graph \(y = f(x)\).
2. Let \(g(x) = \tan^{-1} f(x)\) and, for \(x \neq 0\), let \(k(x) = \frac{1}{2}\tan^{-1}\frac{1}{x}\).
   1. Show that \(g(x) + g(-x) = \frac{1}{2}\pi\).
   2. Show that \(k(x) + k(-x) = 0\).
   3. Show that \(\tan k(x) = \tan g(x)\) for \(x > 0\).
   4. Sketch the graphs \(y = g(x)\) and \(y = k(x)\) on the same axes.
   5. Evaluate \(\int_0^1 k(x) \, dx\) and hence write down the value of \(\int_{-1}^0 g(x) \, dx\).

1. Show, by means of the substitution \(u=\cosh x\,\), that \[ \int \frac{\sinh x}{\cosh 2x} \d x = \frac 1{2\sqrt2} \ln \left\vert \frac{\sqrt2 \cosh x - 1}{\sqrt2 \cosh x + 1 } \right\vert + C \,.\]
2. Use a similar substitution to find an expression for \[ \int \frac{\cosh x}{\cosh 2x} \d x \,.\]
3. Using parts (i) and (ii) above, show that \[ \int_0^1 \frac 1{1+u^4} \d u = \frac{\pi + 2\ln(\sqrt2 +1)}{4\sqrt2}\,. \]

The definite integrals \(T\), \(U\), \(V\) and \(X\) are defined by \begin{align*} T&= \int_{\frac13}^{\frac12} \frac{{\rm artanh}\, t}t \,\d t\,, & U&= \int _{\ln 2 }^{\ln 3 } \frac{u}{2\sinh u}\, \d u \,, \\[3mm] V&= - \int_{\frac13}^{\frac12} \frac{\ln v}{1-v^2} \,\d v \,, & X&= \int _{\frac12\ln2}^{\frac12\ln3} \ln ({\coth x})\, \d x\,. \end{align*} Show, without evaluating any of them, that \(T\), \(U\), \(V\) and \(X\) are all equal.

The following result applies to any function \(\f\) which is continuous, has positive gradient and satisfies \(\f(0)=0\,\): \[ ab\le \int_0^a\f(x)\, \d x + \int_0^b \f^{-1}(y)\, \d y\,, \tag{\(*\)}\] where \(\f^{-1}\) denotes the inverse function of \(\f\), and \(a\ge 0\) and \(b\ge 0\).

1. By considering the graph of \(y=\f(x)\), explain briefly why the inequality \((*)\) holds. In the case \(a>0\) and \(b>0\), state a condition on \(a\) and \(b\) under which equality holds.
2. By taking \(\f(x) = x^{p-1}\) in \((*)\), where \(p>1\), show that if \(\displaystyle \frac 1p + \frac 1q =1\) then \[ ab \le \frac{a^p}p + \frac{b^q}q\,. \] Verify that equality holds under the condition you stated above.
3. Show that, for \(0\le a \le \frac12 \pi\) and \(0\le b \le 1\), \[ ab \le b\arcsin b + \sqrt{1-b^2} \, - \cos a\,. \] Deduce that, for \(t\ge1\), \[ \arcsin (t^{-1}) \ge t - \sqrt{t^2-1}\,. \]

In this question, \(a\) is a positive constant.

1. Express \(\cosh a\) in terms of exponentials. By using partial fractions, prove that \[ \int_0^1 \frac 1{ x^2 +2x\cosh a +1} \, \d x = \frac a {2\sinh a}\,. \]
2. Find, expressing your answers in terms of hyperbolic functions, \[ \int_1^\infty \frac 1 {x^2 +2x \sinh a -1} \,\d x \, \] and \[ \int_0^\infty \frac 1 {x^4 +2x^2\cosh a +1} \,\d x \,.\]

1. Show, with the aid of a sketch, that \(y> \tanh (y/2)\) for \(y>0\) and deduce that \begin{equation} \textrm{arcosh} x > \dfrac{x-1}{\sqrt{x^2-1}} \text{ for } x>1. \tag{\(*\)} \end{equation}
2. By integrating \((*)\), show that $\textrm{arcosh} x > 2 \dfrac{{x-1}}{\sqrt{x^2-1}} \( for \)x>1$.
3. Show that $\textrm{arcosh} x >3 \dfrac{\sqrt{x^2-1}}{{x+2}} \( for \)x>1$.

Show that if \(\displaystyle \int\frac1{u \, \f(u)}\; \d u = \F(u) + c\;\), then \(\displaystyle \int\frac{m}{x \, \f(x^m)} \;\d x = \F(x^m) + c\;\), where \(m\ne0\). Find:

1. \(\displaystyle\int\frac1{x^n-x} \, \d x\,\);
2. \(\displaystyle\int\frac1 {\sqrt{x^n+x^2}}\, \d x\,\).

Show that \[ \int_0^a \frac{\sinh x}{2\cosh^2 x -1} \, \mathrm{d} x = \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}\cosh a -1}{\sqrt{2}\cosh a +1}\r + \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}+1}{\sqrt{2}-1}\r \] and find \[ \int_0^a \frac{\cosh x}{1+2\sinh^2 x} \, \mathrm{d} x \, . \] Hence show that \[ \int_0^\infty \frac{\cosh x - \sinh x}{1+2\sinh^2 x} \, \mathrm{d} x = \frac{\pi}{2\sqrt{2}} - \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}+1}{\sqrt{2}-1}\r \, . \] By substituting \(u = \e^x\) in this result, or otherwise, find \[ \int_1^\infty \frac{1}{1+u^4} \, \mathrm{d} u \, . \]

Given that \(x+a>0\) and \(x+b>0\,\), and that \(b>a\,\), show that \[ \frac{\mathrm{d} \ }{\mathrm{d} x} \arcsin \left ( \frac{x + a }{ \ x + b} \right) = \frac{ \sqrt{\;b - a\;}} {( x + b ) \sqrt{ a + b + 2x} \ \ } \] and find $\displaystyle \frac{\mathrm{d} \ }{ \mathrm{d} x} \; \mathrm{arcosh} \left ( \frac{x + b }{ \ x + a} \right)$. Hence, or otherwise, integrate, for \(x > -1\,\),

1. \[ \int \frac{1}{ ( x + 1) \sqrt{x + 3} } \mathrm{d} x \]
2. \[ \int \frac{1} {( x + 3 ) \sqrt{x + 1} } \mathrm{d} x \] .

Show that \( \cosh^{-1} x = \ln ( x + \sqrt{x^2-1})\). Show that the area of the region defined by the inequalities \(\displaystyle y^2 \ge x^2-8\) and \(\displaystyle x^2\ge 25y^2 -16 \) is \((72/5) \ln 2\).

Show Solution

\begin{align*} && x &= \cosh y \\ \Rightarrow && x &= \tfrac12 (e^y + e^{-y} ) \\ \Rightarrow && 0 &= e^{2y} - 2xe^y + 1 \\ \Rightarrow && e^y &= \frac{2x \pm \sqrt{4x^2-4}}{2} \\ &&&= x \pm \sqrt{x^2-1} \\ \Rightarrow &&e^y &= x + \sqrt{x^2-1} \tag{by convention \(\cosh^{-1} > 0\)} \\ \Rightarrow && y &= \ln (x + \sqrt{x^2-1}) \end{align*}

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\begin{align*} && A &= 4 \left ( \int_0^3 \frac15\sqrt{16+x^2} \d x - \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x \right) \\ \\ x = 4\sinh u: && \int_0^3 \sqrt{4^2+x^2} \d x &= \int_{u=0}^{u=\sinh^{-1}(3/4)} \sqrt{4^2 (1+\sinh^2 u)} 4 \cosh u \d u \\ &&&= \int_0^{\sinh^{-1}(3/4)} 16 \cosh^{2}u \d u \\ &&&= 8\int_0^{\sinh^{-1}(3/4)} (1+\cosh 2u) \d u \\ &&&= 8 \left[u + \frac12 \sinh 2u\right]_0^{\sinh^{-1}(3/4)} \\ &&&= 8 \left (\sinh^{-1}(3/4) + \frac12 \sinh \left ( 2 \sinh^{-1}(3/4) \right) \right) \\ \\ && \sinh^{-1}(3/4) &= \ln\left ( \frac34 + \sqrt{\left ( \frac{3}{4} \right)^2 + 1} \right) \\ &&&= \ln \left ( \frac34 +\frac{5}{4} \right) \\ &&&= \ln 2 \\ \\ \Rightarrow && \int_0^3 \sqrt{4^2+x^2} \d x &= 8 \ln 2 + 4 \left ( \frac{e^{2 \ln 2} - e^{-2\ln2}}{2} \right) \\ &&&= 8 \ln 2 + 2 \cdot 4 - 2\cdot \frac{1}{4} \\ &&&= 8 \ln 2 + \frac{15}{2} \end{align*} \begin{align*} x = 2\sqrt{2} \cosh u: && \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x &= \int_{u=0}^{u = \cosh^{-1} \frac{3}{2\sqrt{2}}} \sqrt{8(\cosh^2 u - 1)} 2 \sqrt{2} \sinh u \d u \\ &&&= \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} 8\sinh^2 u \d u \\ &&&= 4 \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} 2\sinh^2 u \d u \\ &&&= 4 \int_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} \cosh 2 u -1 \d u \\ &&&= 4 \left [\frac12 \sinh 2u - u \right]_0^{ \cosh^{-1} \frac{3}{2\sqrt{2}}} \\ \\ && \cosh^{-1} \frac{3}{2\sqrt{2}} &= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\left ( \frac{3}{2\sqrt{2}} \right)^2-1} \right) \\ &&&= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\frac{9}{8} - 1} \right) \\ &&&= \ln \left ( \frac{3}{2\sqrt{2}} + \sqrt{\frac{1}{8} } \right) \\ &&&= \ln \frac{4}{2\sqrt{2}} \\ &&&= \frac12 \ln 2 \\ \\ && \int_{2\sqrt{2}}^3\sqrt{x^2-8} \d x &= 4 \left ( \frac12 \frac{e^{\ln2} - e^{-\ln2}}{2} - \frac12 \ln 2\right) \\ &&&= 2 - \frac12 -2 \ln 2 \\ &&&= \frac32 - 2 \ln 2 \end{align*} \begin{align*} A &= 4 \left (\frac15\left(8\ln 2 + \frac{15}2 \right)- \left ( \frac32 - 2 \ln 2\right)\right) \\ &=4\cdot \left( \frac{8}{5} + 2 \right) \ln 2 \\ &= \frac{72}{5} \ln 2 \end{align*}

Calculate the following integrals

1. \({\displaystyle \int\frac{x}{(x-1)(x^{2}-1)}\,\mathrm{d}x}\);
2. \({\displaystyle \int\frac{1}{3\cos x+4\sin x}\,\mathrm{d}x}\);
3. \({\displaystyle \int\frac{1}{\sinh x}\,\mathrm{d}x}.\)

1. The four points \(P\), \(Q\), \(R\) and \(S\) are the vertices of a plane quadrilateral. What is the geometrical shape of \(PQRS\) if \(\vec{PQ} = \vec{SR}\)? What is the geometrical shape of \(PQRS\) if \(\vec{PQ} = \vec{SR}\) and \(|\vec{PQ}| = |\vec{PS}|\)?
2. A cube with edges of unit length has opposite vertices at \((0,0,0)\) and \((1,1,1)\). The points $$P(p,0,0), \quad Q(1,q,0), \quad R(r,1,1) \quad \text{and} \quad S(0,s,1)$$ lie on edges of the cube. Given that the four points lie in the same plane, show that $$rq = (1-s)(1-p).$$
   1. Show that \(\vec{PQ} = \vec{SR}\) if and only if the centroid of the quadrilateral \(PQRS\) is at the centre of the cube. Note: the centroid of the quadrilateral \(PQRS\) is the point with position vector $$\frac{1}{4}(\vec{OP} + \vec{OQ} + \vec{OR} + \vec{OS}),$$ where \(O\) is the origin.
   2. Given that \(\vec{PQ} = \vec{SR}\) and \(|\vec{PQ}| = |\vec{PS}|\), express \(q\), \(r\) and \(s\) in terms of \(p\). Show that $$\cos PQR = \frac{4p-1}{5-4p+8p^2}.$$ Write down the values of \(p\), \(q\), \(r\) and \(s\) if \(PQRS\) is a square, and show that the length of each side of this square is greater than \(\frac{21}{20}\).

The points \(O\), \(A\) and \(B\) are the vertices of an acute-angled triangle. The points \(M\) and \(N\) lie on the sides \(OA\) and \(OB\) respectively, and the lines \(AN\) and \(BM\) intersect at \(Q\). The position vector of \(A\) with respect to \(O\) is \(\bf a\), and the position vectors of the other points are labelled similarly. Given that \(\vert MQ \vert = \mu \vert QB\vert \), and that \(\vert NQ \vert = \nu \vert QA\vert \), where \(\mu\) and \(\nu\) are positive and \(\mu \nu <1\), show that \[ {\bf m} = \frac {(1+\mu)\nu}{1+\nu} \, {\bf a} \,. \] The point \(L\) lies on the side \(OB\), and \(\vert OL \vert = \lambda \vert OB \vert \,\). Given that \(ML\) is parallel to \(AN\), express~\(\lambda\) in terms of \(\mu\) and \(\nu\). What is the geometrical significance of the condition \(\mu\nu<1\,\)?

All vectors in this question lie in the same plane. The vertices of the non-right-angled triangle \(ABC\) have position vectors \(\bf a\), \(\bf b\) and \(\bf c\), respectively. The non-zero vectors \(\bf u\) and \(\bf v\) are perpendicular to \(BC\) and \(CA\), respectively. Write down the vector equation of the line through \(A\) perpendicular to \(BC\), in terms of \(\bf u\),~\(\bf a\) and a parameter \(\lambda \). The line through \(A\) perpendicular to \(BC\) intersects the line through \(B\) perpendicular to \(CA\) at \(P\). Find the position vector of \(P\) in terms of \(\bf a\),~\(\bf b\), \(\bf c\) and \(\bf u\). Hence show that the line \(CP\) is perpendicular to the line \(AB\).

The sides \(OA\) and \(CB\) of the quadrilateral \(OABC\) are parallel. The point \(X\) lies on \(OA\), between \(O\) and \(A\). The position vectors of \(A\), \(B\), \(C\) and \(X\) relative to the origin \(O\) are \(\bf a\), \(\bf b\), \(\bf c\) and \(\bf x\), respectively. Explain why \(\bf c\) and \(\bf x\) can be written in the form \[ {\bf c} = k {\bf a} + {\bf b} \text{ \ \ \ \ and \ \ \ \ } {\bf x} = m {\bf a}\,, \] where \(k\) and \(m\) are scalars, and state the range of values that each of \(k\) and \(m\) can take. %

1. The lines \(OB\) and \(AC\) intersect at \(D\), the lines \(XD\) and \(BC\) intersect at \(Y\) and the lines \(OY\) and \(AB\) intersect at \(Z\). Show that the position vector of \(Z\) relative to \(O\) can be written as \[ \frac{ {\bf b} + mk {\bf a}}{mk+1}\,. \] %

\noindent

1. Let \({\bf x}_1\) and \({\bf x}_2\) be the position vectors of the centres of \(C_1\) and \(C_2\), respectively. Show that the position vector of \(P\) is \[ \frac{r_1 {\bf x}_2- r_2 {\bf x}_1}{r_1-r_2} \,, \] where \(r_1\) and \(r_2\) are the radii of \(C_1\) and \(C_2\), respectively.
2. The circle \(C_3\) does not overlap either \(C_1\) or \(C_2\) and its radius, \(r_3\), satisfies \(r_1 \ne r_3 \ne r_2\). The focus of \(C_1\) and \(C_3\) is \(Q\), and the focus of \(C_2\) and \(C_3\) is \(R\). Show that \(P\), \(Q\) and~\(R\) lie on the same straight line.
3. Find a condition on \(r_1\), \(r_2\) and \(r_3\) for \(Q\) to lie half-way between \(P\) and \(R\).

The four distinct points \(P_i\) (\(i=1\), \(2\), \(3\), \(4\)) are the vertices, labelled anticlockwise, of a cyclic quadrilateral. The lines \(P_1P_3\) and \(P_2P_4\) intersect at \(Q\).

1. By considering the triangles \(P_1QP_4\) and \(P_2QP_3\) show that \((P_1Q)( QP_3) = (P_2Q) (QP_4)\,\).
2. Let \(\+p_i\) be the position vector of the point \(P_i\) (\(i=1\), \(2\), \(3\), \(4\)). Show that there exist numbers \(a_i\), not all zero, such that \begin{equation} \sum\limits_{i=1}^4 a_i =0 \qquad\text{and}\qquad \sum\limits_{i=1}^4 a_i \+p_i ={\bf 0} \,. \tag{\(*\)} \end{equation}
3. Let \(a_i\) (\(i=1\),~\(2\), \(3\),~\(4\)) be any numbers, not all zero, that satisfy~\((*)\). Show that \(a_1+a_3\ne 0\) and that the lines \(P_1P_3\) and \(P_2P_4\) intersect at the point with position vector \[ \frac{a_1 \+p_1 + a_3 \+p_3}{a_1+a_3} \,. \] Deduce that \(a_1a_3 (P_1P_3)^2 = a_2a_4 (P_2P_4)^2\,\).

In the triangle \(OAB\), the point \(D\) divides the side \(BO\) in the ratio \(r:1\) (so that \(BD = rDO\)), and the point \(E\) divides the side \(OA\) in the ratio \(s:1\) (so that \(OE =s EA\)), where \(r\) and \(s\) are both positive.

1. The lines \(AD\) and \(BE\) intersect at \(G\). Show that \[ \+g= \frac{rs}{1+r+rs} \, \+a + \frac 1 {1+r+rs} \, \+b \,, \] where \+a, \+b and \+g are the position vectors with respect to \(O\) of \(A\), \(B\) and \(G\), respectively.
2. The line through \(G\) and~\(O\) meets \(AB\) at \(F\). Given that \(F\) divides \(AB\) in the ratio \(t:1\), find an expression for~\(t\) in terms of \(r\) and~\(s\).

The four vertices \(P_i\) (\(i= 1, 2, 3, 4\)) of a regular tetrahedron lie on the surface of a sphere with centre at \(O\) and of radius 1. The position vector of \(P_i\) with respect to \(O\) is \({\bf p}_i\) (\(i= 1, 2, 3, 4\)). Use the fact that \({\bf p}_1+ {\bf p}_2+{\bf p}_3+{\bf p}_4={\bf 0}\,\) to show that \({\bf p}_i \,.\, {\bf p}_j =-\frac13\,\) for \(i\ne j\). Let \(X\) be any point on the surface of the sphere, and let \(XP_i\) denote the length of the line joining \(X\) and \(P_i\) (\(i= 1, 2, 3, 4\)).

1. By writing \((XP_i) ^2\) as \(({\bf p}_i- {\bf x)}\,.\,({\bf p}_i- {\bf x})\), where \({\bf x}\) is the position vector of \(X\) with respect to \(O\), show that \[ \sum_{i=1}^4(XP_i) ^2 =8\,. \]
2. Given that \(P_1\) has coordinates \((0,0,1)\) and that the coordinates of \(P_2\) are of the form \((a,0,b)\), where \(a>0\), show that \(a=2\sqrt2/3\) and \(b=-1/3\), and find the coordinates of \(P_3\) and \(P_4\).
3. Show that \[ \sum_{i=1}^4 (XP_i)^4 = 4 \sum_{i=1}^4 (1- {\bf x}\,.\,{\bf p}_i)^2\,. \] By letting the coordinates of \(X\) be \( (x,y,z)\), show further that $ \sum\limits_{i=1}^4 (XP_i)^4\( is independent of the position of \)X$.

Three distinct points, \(X_1\), \(X_2\) and \(X_3\), with position vectors \({\bf x}_1\), \({\bf x}_2\) and \({\bf x}_3\) respectively, lie on a circle of radius 1 with its centre at the origin \(O\). The point \(G\) has position vector \(\frac13({\bf x}_1+{\bf x}_2+{\bf x}_3)\). The line through \(X_1\) and \(G\) meets the circle again at the point \(Y_1\) and the points \(Y_2\) and \(Y_3\) are defined correspondingly. Given that \(\overrightarrow{GY_1} =-\lambda_1\overrightarrow{GX_1}\), where \(\lambda_1\) is a positive scalar, show that \[ \overrightarrow{OY_1}= \tfrac13 \big( (1-2\lambda_1){\bf x}_1 +(1+\lambda_1)({\bf x}_2+{\bf x}_3)\big) \] and hence that \[ \lambda_1 = \frac {3-\alpha-\beta-\gamma} {3+\alpha -2\beta-2\gamma} \,,\] where \(\alpha = {\bf x}_2 \,.\, {\bf x}_3\), \(\beta = {\bf x}_3\,.\, {\bf x}_1\) and \(\gamma = {\bf x}_1\,.\, {\bf x}_2\). Deduce that $\dfrac {GX_1}{GY_1} + \dfrac {GX_2}{GY_2} + \dfrac {GX_3}{GY_3} =3 \,$.

Show Solution

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\begin{align*} && \mathbf{y}_1 &= \overrightarrow{OG}+\overrightarrow{GY_1} \\ &&&= \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3) -\lambda_1 \left (\mathbf{x}_1 - \frac13(\mathbf{x}_1+\mathbf{x}_2+\mathbf{x}_3)\right) \\ &&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\ && 1 &= \mathbf{y}_1 \cdot \mathbf{y}_1 \\ &&&= \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \cdot \frac13 \left ( (1-2\lambda_1)\mathbf{x}_1+(1+\lambda_1)(\mathbf{x}_2+\mathbf{x}_3)\right) \\ &&&= \frac19\left ( (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\mathbf{x}_1 \cdot \mathbf{x}_2+\mathbf{x}_1 \cdot \mathbf{x}_3) + 2(1+\lambda_1)^2 \mathbf{x}_2 \cdot \mathbf{x}_3 \right) \\ \Rightarrow && 9 &= (1-2\lambda_1)^2+2(1+\lambda_1)^2 + 2(1-2\lambda_1)(1+\lambda_1)(\beta+\gamma) + 2(1+\lambda_1)^2 \alpha \\ &&&= 3+6\lambda_1^2+2(\beta+\gamma)-2(\beta+\gamma)\lambda_1 - 4\lambda_1^2(\beta+\gamma) + 2\alpha+4\lambda_1\alpha + 2\lambda_1^2 \alpha \\ && 0 &= (-6+2(\alpha+\beta+\gamma))+2(2\alpha-(\beta+\gamma))\lambda_1 + (6+2(\alpha-2(\beta+\gamma)))\lambda_1^2 \\ \Rightarrow && 0 &= ((\alpha+\beta+\gamma)-3)+(2\alpha-(\beta+\gamma))\lambda_1 + (3+\alpha-2(\beta+\gamma))\lambda_1^2 \\ &&&= (\lambda_1+1)((3+\alpha-2(\beta+\gamma))\lambda_1+ ((\alpha+\beta+\gamma)-3)) \\ \Rightarrow && \lambda_1 &= \frac{3-(\alpha+\beta+\gamma)}{3+\alpha-2(\beta+\gamma)} \end{align*} as required. Since \(\dfrac {GX_1}{GY_1} = \frac1{\lambda_1}\) we must have, \begin{align*} && \frac {GX_1}{GY_1} + \frac {GX_2}{GY_2} + \frac {GX_3}{GY_3} &= \frac1{\lambda_1}+\frac1{\lambda_2}+\frac1{\lambda_3} \\ &&&= \frac{(3+\alpha-2\beta-2\gamma)+(3+\beta-2\gamma-2\alpha)+3+\gamma-2\alpha-2\beta)}{3-\alpha-\beta-\gamma} \\ &&&= \frac{9-3(\alpha+\beta+\gamma)}{3-(\alpha+\beta+\gamma)} \\ &&&= 3 \end{align*}

The points \(A\) and \(B\) have position vectors \(\bf a \) and \(\bf b\) with respect to an origin \(O\), and \(O\), \(A\)~and~\(B\) are non-collinear. The point \(C\), with position vector \(\bf c\), is the reflection of \(B\) in the line through \(O\) and \(A\). Show that \(\bf c\) can be written in the form \[ \bf c = \lambda \bf a -\bf b \] where \(\displaystyle \lambda = \frac{2\,{\bf a .b}}{{\bf a.a}}\). The point \(D\), with position vector \(\bf d\), is the reflection of \(C\) in the line through \(O\) and \(B\). Show that \(\bf d\) can be written in the form \[ \bf d = \mu\bf b - \lambda \bf a \] for some scalar \(\mu\) to be determined. Given that \(A\), \(B\) and \(D\) are collinear, find the relationship between \(\lambda\) and \(\mu\). In the case \(\lambda = -\frac12\), determine the cosine of \(\angle AOB\) and describe the relative positions of \(A\), \(B\) and \(D\).

The points \(A\) and \(B\) have position vectors \(\bf i +j+k\) and \(5{\bf i} - {\bf j} -{\bf k}\), respectively, relative to the origin \(O\). Find \(\cos2\alpha\), where \(2\alpha\) is the angle \(\angle AOB\).

1. The line \(L _1\) has equation \({\bf r} =\lambda(m{\bf i}+n {\bf j} + p{\bf k})\). Given that \(L _1\) is inclined equally to \(OA\) and to \(OB\), determine a relationship between \(m\), \(n\) and~\(p\). Find also values of \(m\), \(n\) and~\(p\) for which \(L _1\) is the angle bisector of \(\angle AOB\).
2. The line \(L _2\) has equation \({\bf r} =\mu(u{\bf i}+v {\bf j} + w{\bf k})\). Given that \( L _2\) is inclined at an angle \(\alpha\) to \(OA\), where \(2\alpha = \angle AOB\), determine a relationship between \(u\), \(v\) and \(w\). Hence describe the surface with Cartesian equation \(x^2+y^2+z^2 =2(yz+zx+xy)\).

Relative to a fixed origin \(O\), the points \(A\) and \(B\) have position vectors \(\bf{a}\) and \(\bf{b}\), respectively. (The points \(O\), \(A\) and \(B\) are not collinear.) The point \(C\) has position vector \(\bf c\) given by \[ {\bf c} =\alpha {\bf a}+ \beta {\bf b}\,, \] where \(\alpha\) and \(\beta\) are positive constants with \(\alpha+\beta<1\,\). The lines \(OA\) and \(BC\) meet at the point \(P\) with position vector \(\bf p\) and the lines \(OB\) and \(AC\) meet at the point \(Q\) with position vector \(\bf q\). Show that \[ {\bf p} =\frac{\alpha {\bf a} }{1-\beta}\,, \] and write down \(\bf q\) in terms of \(\alpha,\ \beta\) and \(\bf {b}\). Show further that the point \(R\) with position vector \(\bf r\) given by \[ {\bf r} =\frac{\alpha {\bf a} + \beta {\bf b}}{\alpha + \beta}\,, \] lies on the lines \(OC\) and \(AB\). The lines \(OB\) and \(PR\) intersect at the point \(S\). Prove that $ \dfrac{OQ}{BQ} = \dfrac{OS}{BS}\,$.

The non-collinear points \(A\), \(B\) and \(C\) have position vectors \(\bf a\), \(\bf b\) and \(\bf c\), respectively. The points \(P\) and \(Q\) have position vectors \(\bf p\) and \(\bf q\), respectively, given by \[ {\bf p}= \lambda {\bf a} +(1-\lambda){\bf b} \text{ \ \ \ and \ \ \ } {\bf q}= \mu {\bf a} +(1-\mu){\bf c} \] where \(0<\lambda<1\) and \(\mu>1\). Draw a diagram showing \(A\), \(B\), \(C\), \(P\) and \(Q\). Given that \(CQ\times BP = AB\times AC\), find \(\mu\) in terms of \(\lambda\), and show that, for all values of \(\lambda\), the the line \(PQ\) passes through the fixed point \(D\), with position vector \({\bf d}\) given by \({\bf d= -a +b +c}\,\). What can be said about the quadrilateral \(ABDC\)?

The points \(A\) and \(B\) have position vectors \(\bf a\) and \(\bf b\), respectively, relative to the origin \(O\). The points \(A\), \(B\) and \(O\) are not collinear. The point \(P\) lies on \(AB\) between \(A\) and \(B\) such that \[ AP : PB = (1-\lambda):\lambda\,. \] Write down the position vector of \(P\) in terms of \(\bf a\), \(\bf b\) and \(\lambda\). Given that \(OP\) bisects \(\angle AOB\), determine \(\lambda\) in terms of \(a\) and \(b\), where \(a=\vert \bf a\vert\) and $b=\vert \bb\vert$. The point \(Q\) also lies on \(AB\) between \(A\) and \(B\), and is such that \(AP=BQ\). Prove that $$OQ^2-OP^2=(b-a)^2\,.$$

The points \(B\) and \(C\) have position vectors \(\mathbf{b}\) and \(\mathbf{c}\), respectively, relative to the origin \(A\), and \(A\), \(B\) and \(C\) are not collinear.

1. The point \(X\) has position vector \(s \mathbf{b}+t\mathbf{c}\). Describe the locus of \(X\) when \(s+t=1\).
2. The point \(P\) has position vector \(\beta \mathbf{b}+\gamma\mathbf{c}\), where \(\beta\) and \(\gamma\) are non-zero, and \(\beta+\gamma\ne1\). The line \(AP\) cuts the line \(BC\) at \(D\). Show that \(BD:DC=\gamma:\beta\).
3. The line \(BP\) cuts the line \(CA\) at \(E\), and the line \(CP\) cuts the line \(AB\) at \(F\). Show that \[ \frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA}=1\,. \]

1. The line \(L_1\) has vector equation $\displaystyle {\bf r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} \hphantom{-} 2 \\ \hphantom{-} 2 \\ -3 \end{pmatrix} $. The line \(L_2\) has vector equation $\displaystyle {\bf r} = \begin{pmatrix} \hphantom{-} 4 \\ -2 \\ \hphantom{-} 9 \end{pmatrix} + \mu \begin{pmatrix} \hphantom{-} 1 \\ \hphantom{-} 2 \\ -2 \end{pmatrix} . $ Show that the distance \(D\) between a point on \(L_1\) and a point on \(L_2\) can be expressed in the form \[ D^2 = \left(3\mu -4 \lambda-5 \right)^2 + \left( \lambda -1 \right)^2 + 36\,. \] Hence determine the minimum distance between these two lines and find the coordinates of the points on the two lines that are the minimum distance apart.
2. The line \(L_3\) has vector equation ${\bf r} = \begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix} + \alpha \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} . $ The line \(L_4\) has vector equation $ {\bf r} = \begin{pmatrix} \hphantom{-} 3 \\ \hphantom{-} 3 \\ -2 \end{pmatrix} + \beta \begin{pmatrix} \, 4k\\ 1-k \\ \!\!\! -3k \end{pmatrix} . $ Determine the minimum distance between these two lines, explaining geometrically the two different cases that arise according to the value of \(k\).

Show that the line through the points with position vectors \(\bf x\) and \(\bf y\) has equation \[{\bf r} = (1-\alpha){\bf x} +\alpha {\bf y}\,, \] where \(\alpha\) is a scalar parameter. The sides \(OA\) and \(CB\) of a trapezium \(OABC\) are parallel, and \(OA>CB\). The point \(E\) on \(OA\) is such that \(OE : EA = 1:2\), and \(F\) is the midpoint of \(CB\). The point \(D\) is the intersection of \(OC\) produced and \(AB\) produced; the point \(G\) is the intersection of \(OB\) and \(EF\); and the point \(H\) is the intersection of \(DG\) produced and \(OA\). Let \(\bf a\) and \(\bf c\) be the position vectors of the points \(A\) and \(C\), respectively, with respect to the origin \(O\).

1. Show that \(B\) has position vector \(\lambda {\bf a} + {\bf c}\) for some scalar parameter \(\lambda\).
2. Find, in terms of \(\bf a\), \(\bf c\) and \(\lambda\) only, the position vectors of \(D\), \(E\), \(F\), \(G\) and \(H\). Determine the ratio \(OH:HA\).

The position vectors, relative to an origin \(O\), at time \(t\) of the particles \(P\) and \(Q\) are $$\cos t \; {\bf i} + \sin t\;{\bf j} + 0 \; {\bf k} \text{ and } \cos (t+\tfrac14{\pi})\, \big[{\tfrac32}{\bf i} + { \tfrac {3\sqrt{3}}2} {\bf k}\big] + 3\sin(t+\tfrac14{\pi}) \; {\bf j}\;,$$ respectively, where \(0\le t \le 2\pi\,\).

1. Give a geometrical description of the motion of \(P\) and \(Q\).
2. Let \(\theta\) be the angle \(POQ\) at time \(t\) that satisfies \(0\le\theta\le\pi\,\). Show that \[ \cos\theta = \tfrac{3\surd2}{8} -\tfrac14 \cos( 2t +\tfrac14 \pi)\;. \]
3. Show that the total time for which \(\theta \ge \frac14 \pi\) is \(\tfrac32 \pi\,\).

Given two non-zero vectors $\mathbf{a}=\begin{pmatrix}a_{1}\\ a_{2} \end{pmatrix}\( and \)\mathbf{b}=\begin{pmatrix}b_{1}\\ b_{2} \end{pmatrix}$ \mbox{define \(\Delta\!\! \left( \bf a, \bf b \right)\) by \(\Delta\!\! \left( \bf a, \bf b \right) = a_1 b_2 - a_2 b_1\).} Let \(A\), \(B\) and \(C\) be points with position vectors \(\bf a\), \(\bf b\) and \(\bf c\), respectively, no two of which are parallel. Let \(P\), \(Q\) and \(R\) be points with position vectors \(\bf p\), \(\bf q\) and \(\bf r\), respectively, none of which are parallel.

1. Show %, by considering first the case %\(\displaystyle \bf a = \pmatrix{\!a_1\!\cr 0 \cr}\), %or otherwise, that there exists a \(2 \times 2\) matrix \(\bf M\) such that \(P\) and \(Q\) are the images of \(A\) and \(B\) under the transformation represented by \(\bf M\).
2. Show that $ \Delta\!\! \left( \bf a, \bf b \right) \bf c + \Delta\!\! \left( \bf c, \bf a \right) \bf b + \Delta\!\! \left( \bf b, \bf c \right) \bf a = 0. $ Hence, or otherwise, prove that a necessary and sufficient condition for the points \(P\), \(Q\), and \(R\) to be the images of points \(A\), \(B\) and \(C\) under the transformation represented by some \(2 \times 2\) matrix \(\bf M\) is that \[ \Delta\!\! \left( \bf a, \bf b \right) : \Delta\!\! \left( \bf b, \bf c \right) : \Delta\!\! \left( \bf c, \bf a \right) = \Delta\!\! \left( \bf p, \bf q \right) : \Delta\!\! \left( \bf q, \bf r \right) : \Delta\!\! \left( \bf r, \bf p \right). \]

The line \(l\) has vector equation \({\bf r} = \lambda {\bf s}\), where \[ {\bf s} = (\cos\theta+\sqrt3\,) \; {\bf i} +(\surd2\;\sin\theta)\;{\bf j} +(\cos\theta-\sqrt3\,)\;{\bf k} \] and \(\lambda\) is a scalar parameter. Find an expression for the angle between \(l\) and the line \mbox{\({\bf r} = \mu(a\, {\bf i} + b\,{\bf j} +c\, {\bf k})\)}. Show that there is a line \(m\) through the origin such that, whatever the value of \(\theta\), the acute angle between \(l\) and \(m\) is \(\pi/6\). A plane has equation \(x-z=4\sqrt3\). The line \(l\) meets this plane at \(P\). Show that, as \(\theta\) varies, \(P\) describes a circle, with its centre on \(m\). Find the radius of this circle.

1. [(i)] Consider the sphere of radius \(a\) and centre the origin. %Show that the line through the point with position vector %\({\bf b}\) and parallel to a unit %vector \({\bf m}\) intersects the sphere at two points if %$$ %a^2 > {\bf b}.{\bf b} -({\bf b}.{\bf m})^2 \,. %$$ %What is the corresponding condition for there to be precisely one %point of intersection? %If this point has position vector \({\bf p}\), show that the line %is perpendicular to \({\bf p}\).
2. Show that the line \({\bf r} ={\bf b} + \lambda {\bf m}\), where \(\bf m\) is a unit vector, intersects the sphere \({\bf r}\cdot {\bf r} = a^2\) at two points if $$ a^2 > {\bf b}\cdot{\bf b} -({\bf b}\cdot{\bf m})^2 \,. $$ Write down the corresponding condition for there to be precisely one point of intersection. If this point has position vector \({\bf p}\), show that \({\bf m}\cdot{\bf p}=0\).
3. Now consider a second sphere of radius \(a\) and a plane perpendicular to a unit vector~\({\bf n}\). The centre of the sphere has position vector \({\bf d}\) and the minimum distance from the origin to the plane is \(l\). What is the condition for the plane to be tangential to this second sphere?
4. Show that the first and second spheres intersect at right angles ({\em i.e.\ }the two radii to each point of intersection are perpendicular) if $$ {\bf d}\cdot{\bf d} = 2 a^2 \,. $$

A plane \(\pi\) in 3-dimensional space is given by the vector equation \(\mathbf{r}\cdot\mathbf{n}=p,\) where \(\mathbf{n}\) is a unit vector and \(p\) is a non-negative real number. If \(\mathbf{x}\) is the position vector of a general point \(X\), find the equation of the normal to \(\pi\) through \(X\) and the perpendicular distance of \(X\) from \(\pi\). The unit circles \(C_{i},\) \(i=1,2,\) with centres \(\mathbf{r}_{i},\) lie in the planes \(\pi_{i}\) given by \(\mathbf{r}\cdot\mathbf{n}_{i}=p_{i},\) where the \(\mathbf{n}_{i}\) are unit vectors, and \(p_{i}\) are non-negative real numbers. Prove that there is a sphere whose surface contains both circles only if there is a real number \(\lambda\) such that \[ \mathbf{r}_{1}+\lambda\mathbf{n}_{1}=\mathbf{r}_{2}\pm\lambda\mathbf{n}_{2}. \] Hence, or otherwise, deduce the necessary conditions that \[ (\mathbf{r}_{1}-\mathbf{r}_{2})\cdot(\mathbf{n}_{1}\times\mathbf{n}_{2})=0 \] and that \[ (p_{1}-\mathbf{n}_{1}\cdot\mathbf{r}_{2})^{2}=(p_{2}-\mathbf{n}_{2}\cdot\mathbf{r}_{1})^{2}. \] Interpret each of these two conditions geometrically.

Show Solution

The equation of the normal to \(\pi\) through \(X\) is \(\mathbf{x} + \lambda \mathbf{n}\). The distance is \(|\mathbf{x}\cdot \mathbf{n}-p|\) We know that the centre of the sphere must lie above the centre of the circle following the normal, ie \(\mathbf{c} = \mathbf{r}_1+\lambda_1 \mathbf{n}_1 = \mathbf{r}_2+\lambda_2 \mathbf{n}_2\)

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We can also see that \(R^2 = 1 + \lambda_1^2 = 1 + \lambda_2^2 \Rightarrow \lambda_1 = \pm \lambda_2 \), from which we obtain the desired result. Therefore the condition is \begin{align*} && \mathbf{r}_1+\lambda \mathbf{n}_1 &= \mathbf{r}_2 \pm \lambda \mathbf{n}_2 \tag{1}\\ && \mathbf{r}_1 - \mathbf{r}_2 &= \lambda(\pm \mathbf{n}_1 - \mathbf{n}_2) \\ \Rightarrow && (\mathbf{r}_{1}-\mathbf{r}_{2})\cdot(\mathbf{n}_{1}\times\mathbf{n}_{2}) &= (\lambda(\pm \mathbf{n}_1 - \mathbf{n}_2))\cdot(\mathbf{n}_{1}\times\mathbf{n}_{2}) \\ &&&= \lambda \left (\pm \mathbf{n}_1 \cdot ( \mathbf{n}_{1}\times\mathbf{n}_{2}) - \mathbf{n}_2 \cdot (\mathbf{n}_{1}\times\mathbf{n}_{2})\right) \\ &&&= 0 \\ \\ \mathbf{n}_1 \cdot (1)&& \mathbf{r}_1 \cdot \mathbf{n}_1+\lambda \mathbf{n}_1 \cdot \mathbf{n}_1 &= \mathbf{r}_2 \cdot \mathbf{n}_1 \pm \lambda \mathbf{n}_2 \cdot \mathbf{n}_1 \\ && p_1 + \lambda &= \mathbf{r}_2 \cdot \mathbf{n}_1 \pm \lambda \mathbf{n}_2 \cdot \mathbf{n}_1 \\ \\ \mathbf{n}_2 \cdot (1)&& \mathbf{r}_1 \cdot \mathbf{n}_2+\lambda \mathbf{n}_1 \cdot \mathbf{n}_2 &= \mathbf{r}_2 \cdot \mathbf{n}_2 \pm \lambda \mathbf{n}_2 \cdot \mathbf{n}_2 \\ && \mathbf{r}_1 \cdot \mathbf{n}_2+\lambda \mathbf{n}_1 \cdot \mathbf{n}_2 &= p_2 \pm \lambda \\ && \pm \lambda -\lambda \mathbf{n}_1\cdot\mathbf{n}_2 &= \mathbf{r}_1 \cdot \mathbf{n}_2 - p_2 \\ &&&= \pm (\mathbf{r}_2\cdot \mathbf{n}_1 - p_1) \\ \Rightarrow && (p_{1}-\mathbf{n}_{1}\cdot\mathbf{r}_{2})^{2}&=(p_{2}-\mathbf{n}_{2}\cdot\mathbf{r}_{1})^{2} \end{align*} The first condition means the line between the centres lies in the plane spanned by the normal of the two planes \(\pi_1\) and \(\pi_2\). The second condition means that the distance of the center to the other plane is the same for both centres/planes.

Two non-parallel lines in 3-dimensional space are given by \(\mathbf{r}=\mathbf{p}_{1}+t_{1}\mathbf{m}_{1}\) and \(\mathbf{r}=\mathbf{p}_{2}+t_{2}\mathbf{m}_{2}\) respectively, where \(\mathbf{m}_{1}\) and \(\mathbf{m}_{2}\) are unit vectors. Explain by means of a sketch why the shortest distance between the two lines is \[ \frac{\left|(\mathbf{p}_{1}-\mathbf{p}_{2})\cdot(\mathbf{m}_{1}\times\mathbf{m}_{2})\right|}{\left|(\mathbf{m}_{1}\times\mathbf{m}_{2})\right|}. \]

1. Find the shortest distance between the lines in the case \[ \mathbf{p}_{1}=(2,1,-1)\qquad\mathbf{p}_{2}=(1,0,-2)\qquad\mathbf{m}_{1}=\tfrac{1}{5}(4,3,0)\qquad\mathbf{m}_{2}=\tfrac{1}{\sqrt{10}}(0,-3,1). \]
2. Two aircraft, \(A_{1}\) and \(A_{2},\) are flying in the directions given by the unit vectors \(\mathbf{m}_{1}\) and \(\mathbf{m}_{2}\) at constant speeds \(v_{1}\) and \(v_{2}.\) At time \(t=0\) they pass the points \(\mathbf{p}_{1}\) and \(\mathbf{p}_{2}\), respectively. If \(d\) is the shortest distance between the two aircraft during the flight, show that \[ d^{2}=\frac{\left|\mathbf{p}_{1}-\mathbf{p}_{2}\right|^{2}\left|v_{1}\mathbf{m}_{1}-v_{2}\mathbf{m}_{2}\right|^{2}-[(\mathbf{p}_{1}-\mathbf{p}_{2})\cdot(v_{1}\mathbf{m}_{1}-v_{2}\mathbf{m}_{2})]^{2}}{\left|v_{1}\mathbf{m}_{1}-v_{2}\mathbf{m}_{2}\right|^{2}}. \]
3. Suppose that \(v_{1}\) is fixed. The pilot of \(A_{2}\) has chosen \(v_{2}\) so that \(A_{2}\) comes as close as possible to \(A_{1}.\) How close is that, if \(\mathbf{p}_{1},\mathbf{p}_{2},\mathbf{m}_{1}\) and \(\mathbf{m}_{2}\) are as in (i)?

Let \(\mathbf{a},\mathbf{b}\) and \(\mathbf{c}\) be the position vectors of points \(A,B\) and \(C\) in three-dimensional space. Suppose that \(A,B,C\) and the origin \(O\) are not all in the same plane. Describe the locus of the point whose position vector \(\mathbf{r}\) is given by \[ \mathbf{r}=(1-\lambda-\mu)\mathbf{a}+\lambda\mathbf{b}+\mu\mathbf{c}, \] where \(\lambda\) and \(\mu\) are scalar parameters. By writing this equation in the form \(\mathbf{r}\cdot\mathbf{n}=p\) for a suitable vector \(\mathbf{n}\) and scalar \(p\), show that \[ -(\lambda+\mu)\mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})+\lambda\mathbf{b}\cdot(\mathbf{c}\times\mathbf{a})+\mu\mathbf{c}\cdot(\mathbf{a}\times\mathbf{b})=0 \] for all scalars \(\lambda,\mu.\) Deduce that \[ \mathbf{a}\cdot(\mathbf{b}\times\mathbf{c})=\mathbf{b}\cdot(\mathbf{c}\times\mathbf{a})=\mathbf{c}\cdot(\mathbf{a}\times\mathbf{b}). \] Say briefly what happens if \(A,B,C\) and \(O\) are all in the same plane.

The points \(A,B\) and \(C\) lie on the surface of the ground, which is an inclined plane. The point \(B\) is 100m due north of \(A,\) and \(C\) is 60m due east of \(B\). The vertical displacements from \(A\) to \(B,\) and from \(B\) to \(C\), are each 5m downwards. A plane coal seam lies below the surface and is to be located by making vertical bore-holes at \(A,B\) and \(C\). The bore-holes strike the coal seam at 95m, 45m and 76m below \(A,B\) and \(C\) respectively. Show that the coal seam is inclined at \(\cos^{-1}(\frac{4}{5})\) to the horizontal. The coal seam comes to the surface along a line. Find the bearing of this line.

By substituting \(y(x)=xv(x)\) in the differential equation \[ x^{3}\frac{\mathrm{d}v}{\mathrm{d}x}+x^{2}v=\frac{1+x^{2}v^{2}}{\left(1+x^{2}\right)v}, \] or otherwise, find the solution \(v(x)\) that satisfies \(v=1\) when \(x=1\). What value does this solution approach when \(x\) becomes large?

1. Use the substitution \(v= \sqrt y\) to solve the differential equation \[ \frac{\d y}{\d t} = \alpha y^{\frac12} - \beta y \ \ \ \ \ \ \ \ \ \ (y\ge0, \ \ t\ge0) \,, \] where \(\alpha\) and \(\beta\) are positive constants. Find the non-constant solution \(y_1(x)\) that satisfies \(y_1(0)=0\,\).
2. Solve the differential equation \[ \frac{\d y}{\d t} = \alpha y^{\frac23} - \beta y \ \ \ \ \ \ \ \ \ \ (y\ge0, \ \ t\ge0) \,, \] where \(\alpha\) and \(\beta\) are positive constants. Find the non-constant solution \(y_2(x)\) that satisfies \(y_2(0)=0\,\).
3. In the case \(\alpha=\beta\), sketch \(y_1(x)\) and \(y_2(x)\) on the same axes, indicating clearly which is \(y_1(x)\) and which is \(y_2(x)\). You should explain how you determined the positions of the curves relative to each other.

Given that \(y=xu\), where \(u\) is a function of \(x\), write down an expression for \(\dfrac {\d y}{\d x}\).

1. Use the substitution \(y=xu\) to solve \[ \frac {\d y}{\d x} = \frac {2y+x}{y-2x} \] given that the solution curve passes through the point \((1,1)\). Give your answer in the form of a quadratic in \(x\) and \(y\).
2. Using the substitutions \(x=X+a\) and \(y=Y+b\) for appropriate values of \(a\) and \(b\), or otherwise, solve \[ \frac {\d y}{\d x} = \frac {x-2y-4} {2x+y-3}\,, \] given that the solution curve passes through the point \((1,1)\).

1. Show that substituting \(y=xv\), where \(v\) is a function of \(x\), in the differential equation \[ xy \frac{\d y}{\d x} +y^2- 2x^2 =0 \quad (x\ne0) \] leads to the differential equation \[ xv\frac{\d v}{\d x} +2v^2 -2=0\,. \] Hence show that the general solution can be written in the form \[ x^2(y^2 -x^2) = C \,,\] where \(C\) is a constant.
2. Find the general solution of the differential equation \[ y \frac{\d y}{\d x} +6x +5y=0\, \quad (x\ne0). \]

Show that if \[ {\mathrm{d}y \over \mathrm{d} x}=\f(x)y + {\g(x) \over y} \] then the substitution \(u = y^2\) gives a linear differential equation for \(u(x)\,\). Hence or otherwise solve the differential equation \[ {\mathrm{d}y \over \mathrm{d} x}={y \over x} - {1 \over y}\;. \] Determine the solution curves of this equation which pass through \((1 \,, 1)\,\), \((2\, , 2)\) and \((4 \, , 4)\) and sketch graphs of all three curves on the same axes.

Show Solution

\begin{align*} && \frac{\d y}{\d x} &= f(x) y + \frac{g(x)}{y} \\ && y \frac{\d y}{\d x} &= f(x) y^2 + g(x) \\ u = y^2: && \frac12 \frac{\d u}{\d x} &= f(x) u + g(x) \end{align*} Which is a linear differential equation for \(u\). \begin{align*} && \frac12 u' &= \frac1x u -1 \\ \Rightarrow && u' - \frac2xu &= -1 \\ \Rightarrow && \frac{1}{x^2} u' - \frac{2}{x^3} u &= -\frac{1}{x^2} \\ \Rightarrow && (\frac{u}{x^2})' &= - \frac{1}{x^2} \\ \Rightarrow && \frac{u}{x^2} &= \frac1x + C \\ \Rightarrow && u &= Cx^2 + x \\ \Rightarrow && y^2 &= Cx^2 + x \end{align*} If \((1,1)\) is on the curve then \(1 = C + 1 \Rightarrow C = 0 \Rightarrow y^2 = x\). If \((2,2)\) is on the curve then \(4 = 4C + 2 \Rightarrow C = \frac12 \Rightarrow y^2 = x + \frac12 x^2\). If \((3,3)\) is on the curve then \(9 = 9C + 3 \Rightarrow C = \frac23 \Rightarrow y^2 = x + \frac23 x^2\)

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1. Show that the gradient at a point \(\l x\,, \, y \r\) on the curve \[ \l y + 2x \r^3 \l y - 4x \r = c\;, \] where \(c\) is a constant, is given by \[ \frac{\d y}{\d x} = \frac{16 x -y}{2y-5x} \;. \]
2. By considering the derivative with respect to \(x\) of \(\l y + ax \r^n \l y + bx \r\,\), or otherwise, find the general solution of the differential equation \[ \frac{\mathrm{d}y}{ \mathrm{d}x} = \frac{10x - 4y}{ 3x - y}\;. \]

1. Let \(y\) be the solution of the differential equation \[ \frac{\d y}{\d x} + 4x\e^{-x^2} {(y+3)}^{\frac12} = 0 \qquad (x \ge 0), \] that satisfies the condition \(y=6\) when \(x=0\). Find \(y\) in terms of \(x\) and show that \(y\to1\) as \(x \to \infty\).
2. Let \(y\) be any solution of the differential equation \[ \frac{\d y}{\d x} -x\e^{6 x^2} (y+3)^{1-k} = 0 \qquad (x \ge 0). \] %that satisfies the condition \(y=6\) %when \(x=0\). Find a value of \(k\) such that, as \(x \to \infty\), \(\e^{-3x^2}y\) tends to a finite non-zero limit, which you should determine.

If there are \(x\) micrograms of bacteria in a nutrient medium, the population of bacteria will grow at the rate \((2K-x)x\) micrograms per hour. Show that, if \(x=K\) when \(t=0\), the population at time \(t\) is given by \[ x(t)=K+K\frac{1-\mathrm{e}^{-2Kt}}{1+\mathrm{e}^{-2Kt}}. \] Sketch, for \(t\geqslant0\), the graph of \(x\) against \(t\). What happens to \(x(t)\) as \(t\rightarrow\infty\)? Now suppose that the situation is as described in the first paragraph, except that we remove the bacteria from the nutrient medium at a rate \(L\) micrograms per hour where \(K^{2}>L\). We set \(\alpha=\sqrt{K^{2}-L}.\) Write down the new differential equation for \(x\). By considering a new variable \(y=x-K+\alpha,\) or otherwise, show that, if \(x(0)=K\) then \(x(t)\rightarrow K+\alpha\) as \(t\rightarrow\infty\).

Let \(P,Q\) and \(R\) be functions of \(x\). Prove that, for any function \(y\) of \(x\), the function \[ Py''+Qy'+Ry \] can be written in the form \(\dfrac{\mathrm{d}}{\mathrm{d}x}(py'+qy),\) where \(p\) and \(q\) are functions of \(x\), if and only if \(P''-Q'+R=0.\) Solve the differential equation \[ (x-x^{4})y''+(1-7x^{3})y'-9x^{2}y=(x^{3}+3x^2)\mathrm{e}^{x}, \] given that when \(x=2,y=2\mathrm{e}^{2}\) and \(y'=0.\)

Let \(y,u,v,P\) and \(Q\) all be functions of \(x\). Show that the substitution \(y=uv\) in the differential equation \[ \frac{\mathrm{d}y}{\mathrm{d}x}+Py=Q \] leads to an equation for \(\dfrac{\mathrm{d}v}{\mathrm{d}x}\) in terms of \(x,Q\) and \(u\), provided that \(u\) satisfies a suitable first order differential equation. Hence or otherwise solve \[ \frac{\mathrm{d}y}{\mathrm{d}x}-\frac{2y}{x+1}=\left(x+1\right)^{\frac{5}{2}}, \] given that \(y(1)=0\). For what set of values of \(x\) is the solution valid?

The matrix \(\mathbf{F}\) is defined by \[ \mathbf{F}=\mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n}, \] where $\mathbf{A}=\begin{pmatrix}-3 & -1\\ 8 & 3 \end{pmatrix} \( and \) t \( is a variable scalar. Evaluate \)\mathbf{A}^{2}$, and show that \[ \mathbf{F}=\mathbf{I}\cosh t+\mathbf{A}\sinh t. \] Show also that \(\mathbf{F}^{-1}=\mathbf{I}\cosh t-\mathbf{A}\sinh t\), and that \(\dfrac{\mathrm{d}\mathbf{F}}{\mathrm{d}t}=\mathbf{FA}\). The vector $\mathbf{r}=\begin{pmatrix}x(t)\\ y(t) \end{pmatrix}$ satisfies the differential equation \[ \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}+\mathbf{A}\mathbf{r}=\mathbf{0}, \] with \(x=\alpha\) and \(y=\beta\) at \(t=0.\) Solve this equation by means of a suitable matrix integrating factor, and hence show that \begin{alignat*}{1} x(t) & =\alpha\cosh t+(3\alpha+\beta)\sinh t\\ y(t) & =\beta\cosh t-(8\alpha+3\beta)\sinh t. \end{alignat*}

1. Show that $$z^{m+1} - \frac{1}{z^{m+1}} = \left(z - \frac{1}{z}\right)\left(z^m + \frac{1}{z^m}\right) + \left(z^{m-1} - \frac{1}{z^{m-1}}\right)$$ Hence prove by induction that, for \(n \geq 1\), $$z^{2n} - \frac{1}{z^{2n}} = \left(z - \frac{1}{z}\right)\sum_{r=1}^n \left(z^{2r-1} + \frac{1}{z^{2r-1}}\right)$$ Find similarly \(z^{2n} - \frac{1}{z^{2n}}\) as a product of \((z + \frac{1}{z})\) and a sum.
2. 1. By choosing \(z = e^{i\theta}\), show that $$\sin 2n\theta = 2\sin\theta \sum_{r=1}^n \cos(2r-1)\theta$$
   2. Use this result, with \(n = 2\), to show that \(\cos\frac{2\pi}{5} = \cos\frac{\pi}{5} - \frac{1}{2}\).
   3. Use this result, with \(n = 7\), to show that \(\cos\frac{2\pi}{15} + \cos\frac{4\pi}{15} + \cos\frac{8\pi}{15} + \cos\frac{16\pi}{15} = \frac{1}{2}\).
3. Show that \(\sin\frac{\pi}{14} - \sin\frac{3\pi}{14} + \sin\frac{5\pi}{14} = \frac{1}{2}\).

1. If \(z=x+\mathrm{i}y,\) with \(x,y\) real, show that \[ \left|x\right|\cos\alpha+\left|y\right|\sin\alpha\leqslant\left|z\right|\leqslant\left|x\right|+\left|y\right| \] for all real \(\alpha.\)
2. By considering \((5-\mathrm{i})^{4}(1+\mathrm{i}),\) show that \[ \frac{\pi}{4}=4\tan^{-1}\left(\frac{1}{5}\right)-\tan^{-1}\left(\frac{1}{239}\right). \] Prove similarly that \[ \frac{\pi}{4}=3\tan^{-1}\left(\frac{1}{4}\right)+\tan^{-1}\left(\frac{1}{20}\right)+\tan^{-1}\left(\frac{1}{1985}\right). \]

Explain the geometrical relationship between the points in the Argand diagram represented by the complex numbers \(z\) and \(z\mathrm{e}^{\mathrm{i}\theta}.\) Write down necessary and sufficient conditions that the distinct complex numbers \(\alpha,\beta\) and \(\gamma\) represent the vertices of an equilateral triangle taken in anticlockwise order. Show that \(\alpha,\beta\) and \(\gamma\) represent the vertices of an equilateral triangle (taken in any order) if and only if \[ \alpha^{2}+\beta^{2}+\gamma^{2}-\beta\gamma-\gamma\alpha-\alpha\beta=0. \] Find necessary and sufficient conditions on the complex coefficients \(a,b\) and \(c\) for the roots of the equation \[ z^{3}+az^{2}+bz+c=0 \] to lie at the vertices of an equilateral triangle in the Argand digram.

Sum each of the series \[ \sin\left(\frac{2\pi}{23}\right)+\sin\left(\frac{6\pi}{23}\right)+\sin\left(\frac{10\pi}{23}\right)+\cdots+\sin\left(\frac{38\pi}{23}\right)+\sin\left(\frac{42\pi}{23}\right) \] and \[ \sin\left(\frac{2\pi}{23}\right)-\sin\left(\frac{6\pi}{23}\right)+\sin\left(\frac{10\pi}{23}\right)-\cdots-\sin\left(\frac{38\pi}{23}\right)+\sin\left(\frac{42\pi}{23}\right), \] giving each answer in terms of the tangent of a single angle. {[}No credit will be given for a numerical answer obtained purely by use of a calculator.{]}

1. Use De Moivre's theorem to show that, if \(\sin\theta\ne0\)\,, then \[ \frac{ \left(\cot \theta + \rm{i}\right)^{2n+1} -\left(\cot \theta - \rm{i}\right)^{2n+1}}{2\rm{i}} = \frac{\sin \left(2n+1\right)\theta} {\sin^{2n+1}\theta} \,, \] for any positive integer \(n\). Deduce that the solutions of the equation \[ \binom{2n+1}{1}x^{n}-\binom{2n+1}{3}x^{n-1} +\cdots + \left(-1\right)^{n}=0 \] are \[x=\cot^{2}\left(\frac{m\pi}{2n+1}\right) \] where \( m=1\), \(2\), \(\ldots\) , \(n\,\).
2. Hence show that \[ \sum_{m=1}^n \cot^{2}\left(\frac{m\pi}{2n+1}\right) =\frac{n\left(2n-1\right)}{3}. \]
3. Given that \(0<\sin \theta <\theta <\tan \theta\) for \(0 < \theta < \frac{1}{2}\pi\), show that \[ \cot^{2}\theta<\frac{1}{\theta^{2}}<1+\cot^{2}\theta. \] Hence show that \[ \sum^\infty_{m=1} \frac{1}{m^2}= \frac{\pi^2}{6}\,.\]

The transformation \(R\) in the complex plane is a rotation (anticlockwise) by an angle \(\theta\) about the point represented by the complex number \(a\). The transformation \(S\) in the complex plane is a rotation (anticlockwise) by an angle \(\phi\) about the point represented by the complex number \(b\).

1. The point \(P\) is represented by the complex number~\(z\). Show that the image of \(P\) under \(R\) is represented by \[ \e^{{\mathrm i} \theta}z + a(1-\e^{{\rm i} \theta})\,. \]
2. Show that the transformation \(SR\) (equivalent to \(R\) followed by \(S\)) is a rotation about the point represented by \(c\), where \[ %\textstyle c\,\sin \tfrac12 (\theta+\phi) = a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta + b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi \,, \] provided \(\theta+\phi \ne 2n\pi\) for any integer \(n\). What is the transformation \(SR\) if \(\theta +\phi = 2\pi\)?
3. Under what circumstances is \(RS =SR\)?

Let \(\omega = \e^{2\pi {\rm i}/n}\), where \(n\) is a positive integer. Show that, for any complex number \(z\), \[ (z-1)(z-\omega) \cdots (z - \omega^{n-1}) = z^n -1\,. \] The points \(X_0\), \(X_1\), \ldots\,, \(X_{n-1}\) lie on a circle with centre \(O\) and radius 1, and are the vertices of a regular polygon.

1. The point \(P\) is equidistant from \(X_0\) and \(X_1\). Show that, if \(n\) is even, \[ |PX_0| \times |PX_1 |\times \,\cdots\, \times |PX_{n-1}| = |OP|^n +1\, ,\] where \(|PX_ k|\) denotes the distance from \(P\) to \(X_k\). Give the corresponding result when \(n\) is odd. (There are two cases to consider.)
2. Show that \[ |X_0 X_1|\times |X_0 X_2|\times \,\cdots\, \times |X_0 X_{n-1}| =n\,. \]

1. If \(a\), \(b\) and \(c\) are all real, show that the equation \[ z^3+az^2+bz+c=0 \tag{\(*\)} \] has at least one real root.
2. Let \[ S_1= z_1+z_2+z_3, \ \ \ \ S_2= z_1^2 + z_2^2 + z_3^2, \ \ \ \ S_3= z_1^3 + z_2^3 + z_3^3\,, \] where \(z_1\), \(z_2\) and \(z_3\) are the roots of the equation \((*)\). Express \(a\) and \(b\) in terms of \(S_1\) and \(S_2\), and show that \[ 6c =- S_1^3 + 3 S_1S_2 - 2S_3\,. \]
3. The six real numbers \(r_k\) and \(\theta_k\) (\(k=1, \ 2, \ 3\)), where \(r_k>0\) and \(-\pi < \theta_k <\pi\), satisfy \[ \textstyle \sum\limits _{k=1}^3 r_k \sin (\theta_k) = 0\,, \ \ \ \ \textstyle \sum\limits _{k=1}^3 r_k^2 \sin (2\theta_k) = 0\,, \ \ \ \ \ \textstyle \sum\limits _{k=1}^3 r_k^3 \sin (3\theta_k) = 0\, . \] Show that \(\theta_k=0\) for at least one value of \(k\). Show further that if \(\theta_1=0\) then \(\theta_2 = - \theta_3\,\).

Evaluate \(\displaystyle \sum_{r=0}^{n-1} \e^{2i(\alpha + r\pi/n)}\) where \(\alpha\) is a fixed angle and \(n\ge2\). The fixed point \(O\) is a distance \(d\) from a fixed line \(D\). For any point \(P\), let \(s\) be the distance from \(P\) to \(D\) and let \(r\) be the distance from \(P\) to \(O\). Write down an expression for \(s\) in terms of \(d\), \(r\) and the angle \(\theta\), where \(\theta\) is as shown in the diagram below.

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Show that \((z-\e^{i\theta})(z-\e^{-i\theta})=z^2 -2z\cos\theta +1\,\). Write down the \((2n)\)th roots of \(-1\) in the form \(\e^{i\theta}\), where \(-\pi <\theta \le \pi\), and deduce that \[ z^{2n} +1 = \prod_{k=1}^n \left(z^2-2z \cos\left( \tfrac{(2k-1)\pi}{2n}\right) +1\right) \,. \] Here, \(n\) is a positive integer, and the \(\prod\) notation denotes the product.

1. By substituting \(z=i\) show that, when \(n\) is even, \[ \cos \left(\tfrac \pi {2n}\right) \cos \left(\tfrac {3\pi} {2n}\right) \cos \left(\tfrac {5\pi} {2n}\right) \cdots \cos \left(\tfrac{(2n-1) \pi} {2n}\right) = {(-1\vphantom{\dot A})}^{\frac12 n} 2^{1-n} \,. \]
2. Show that, when \(n\) is odd, \[ \cos^2 \left(\tfrac \pi {2n}\right) \cos ^2 \left(\tfrac {3\pi} {2n}\right) \cos ^2 \left(\tfrac {5\pi} {2n}\right) \cdots \cos ^2 \left(\tfrac{(n-2) \pi} {2n}\right) = n 2^{1-n} \,. \] You may use without proof the fact that \(1+z^{2n}= (1+z^2)(1-z^2+z^4 - \cdots + z^{2n-2})\,\) when \(n\) is odd.

Show that, provided \(q^2\ne 4p^3\), the polynomial \[ \hphantom{(p\ne0, \ q\ne0)\hspace{2cm}} x^3-3px +q \hspace {2cm} (p\ne0, \ q\ne0) \] can be written in the form \[ a(x-\alpha)^3 + b(x-\beta)^3\,, \] where \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(pt^2 -qt +p^2=0\), and \(a\) and \(b\) are constants which you should express in terms of \(\alpha\) and \(\beta\). Hence show that one solution of the equation \(x^3-24x+48=0\,\) is \[ x= \frac{2 (2-2^{\frac13})}{1-2^{\frac13}} \] and obtain similar expressions for the other two solutions in terms of \(\omega\), where \(\omega = \mathrm{e}^{2\pi\mathrm{i}/3}\,\). Find also the roots of \(x^3-3px +q=0\) when \(p=r^2\) and \(q= 2r^3\) for some non-zero constant \(r\).

For any given positive integer \(n\), a number \(a\) (which may be complex) is said to be a primitive \(n\)th root of unity if \(a^n=1\) and there is no integer \(m\) such that \(0 < m < n\) and \(a^m = 1\). Write down the two primitive 4th roots of unity. Let \({\rm C}_n(x)\) be the polynomial such that the roots of the equation \({\rm C}_n(x)=0\) are the primitive \(n\)th roots of unity, the coefficient of the highest power of \(x\) is one and the equation has no repeated roots. Show that \({\rm C}_4(x) = x^2+1\,\).

1. Find \({\rm C}_1(x)\), \({\rm C}_2(x)\), \({\rm C}_3(x)\), \({\rm C}_5(x)\) and \({\rm C}_6(x)\), giving your answers as unfactorised polynomials.
2. Find the value of \(n\) for which \({\rm C}_n(x) = x^4 + 1\).
3. Given that \(p\) is prime, find an expression for \({\rm C}_p(x)\), giving your answer as an unfactorised polynomial.
4. Prove that there are no positive integers \(q\), \(r\) and \(s\) such that \({\rm C}_q(x) \equiv {\rm C}_r(x) {\rm C}_s(x)\,\).

Show that $\big\vert \e^{\i\beta} -\e^{\i\alpha}\big\vert = 2\sin\frac12 (\beta-\alpha)\,\( for \)0<\alpha<\beta<2\pi\,$. Hence show that \[ \big\vert \e^{\i\alpha} -\e^{\i\beta}\big\vert \; \big\vert \e^{\i\gamma} -\e^{\i\delta}\big\vert + \big\vert \e^{\i\beta} -\e^{\i\gamma}\big\vert \; \big\vert \e^{\i\alpha} -\e^{\i\delta}\big\vert = \big\vert \e^{\i\alpha} -\e^{\i\gamma}\big\vert \; \big\vert \e^{\i\beta} -\e^{\i\delta}\big\vert \,, \] where \(0<\alpha<\beta<\gamma<\delta<2\pi\). Interpret this result as a theorem about cyclic quadrilaterals.

In this question, you may use without proof the results \[ 4 \cosh^3 y - 3 \cosh y = \cosh (3y) \ \ \ \ \text{and} \ \ \ \ \mathrm{arcosh} \, y = \ln ( y+\sqrt{y^2-1}). \] \noindent[ {\bf Note: } \(\mathrm{arcosh}y\) is another notation for \(\cosh^{-1}y\,\)] Show that the equation \(x^3 - 3a^2x = 2a^3 \cosh T\) is satisfied by \( 2a \cosh \l \frac13 T \r\) and hence that, if \(c^2\ge b^3>0\), one of the roots of the equation \(x^3-3bx=2c\) is \(\ds u+\frac{b}{u}\), where \(u = (c+\sqrt{c^2-b^3})^{\frac13}\;\). Show that the other two roots of the equation \(x^3-3bx=2c\) are the roots of the quadratic equation \[\ds x^2 + \Big( u+\frac{b}{u}\Big) x + u^2+\frac{b^2}{u^2}-b=0\, ,\] and find these roots in terms of \(u\), \(b\) and \(\omega\), where \(\omega = \frac{1}{2}(-1 + \mathrm{i}\sqrt{3})\). Solve completely the equation \(x^3-6x=6\,\).

Given that \(\alpha = \e^{\mathrm{i} \pi/3}\) , prove that \(1 + \alpha^2 = \alpha\). A triangle in the Argand plane has vertices \(A\), \(B\), and \(C\) represented by the complex numbers \(p\), \(q\alpha^2\) and \(- r\alpha\) respectively, where \(p\), \(q\) and \(r\) are positive real numbers. Sketch the triangle~\(ABC\). Three equilateral triangles \(ABL\), \(BCM\) and \(CAN\) (each lettered clockwise) are erected on sides \(AB\), \(BC\) and \(CA\) respectively. Show that the complex number representing \(N\) is \mbox{\(( 1 - \alpha) p- \alpha^2 r\)} and find similar expressions for the complex numbers representing \(L\) and \(M\). Show that lines \(LC\), \(MA\) and \(NB\) all meet at the origin, and that these three line segments have the common length \(p+q+r\).

Prove that \[ (\cos\theta +\mathrm{i}\sin\theta) (\cos\phi +\mathrm{i}\sin\phi) = \cos(\theta+\phi) +\mathrm{i}\sin(\theta+\phi) \] and that, for every positive integer \(n\), $$ {(\cos {\theta} + \mathrm{i}\sin {\theta})}^n = \cos{n{\theta}} + \mathrm{i}\sin{n{\theta}}. $$ By considering \((5-\mathrm{i})^2(1+\mathrm{i})\), or otherwise, prove that \[ \arctan\left(\frac{7}{17}\right)+2\arctan\left(\frac{1}{5}\right)=\frac{\pi}{4}\,. \] Prove also that \[ 3\arctan\left(\frac{1}{4}\right)+\arctan\left(\frac{1}{20}\right)+\arctan\left(\frac{1}{1985}\right)=\frac{\pi}{4}\,. \] [Note that \(\arctan\theta\) is another notation for \(\tan^{-1}\theta\).]

By considering the solutions of the equation \(z^n-1=0\), or otherwise, show that \[(z-\omega)(z-\omega^2)\dots(z-\omega^{n-1})=1+z+z^2+\dots+z^{n-1},\] where \(z\) is any complex number and \(\omega={\rm e}^{2\pi i/n}\). Let \(A_1,A_2,A_3,\dots,A_n\) be points equally spaced around a circle of radius \(r\) centred at \(O\) (so that they are the vertices of a regular \(n\)-sided polygon). Show that \[\overrightarrow{OA_1}+\overrightarrow{OA_2}+\overrightarrow{OA_3} +\dots+\overrightarrow{OA_n}=\mathbf0.\] Deduce, or prove otherwise, that \[\sum_{k=1}^n|A_1A_k|^2=2r^2n.\]

Show, using de Moivre's theorem, or otherwise, that \[ \tan7\theta=\frac{t(t^{6}-21t^{4}+35t^{2}-7)}{7t^{6}-35t^{4}+21t^{2}-1}\,, \] where \(t=\tan\theta.\)

1. By considering the equation \(\tan7\theta=0,\) or otherwise, obtain a cubic equation with integer coefficients whose roots are \[ \tan^{2}\left(\frac{\pi}{7}\right),\ \tan^{2}\left(\frac{2\pi}{7}\right)\ \mbox{ and }\tan^{2}\left(\frac{3\pi}{7}\right) \] and deduce the value of \[ \tan\left(\frac{\pi}{7}\right)\tan\left(\frac{2\pi}{7}\right)\tan\left(\frac{3\pi}{7}\right)\,. \]
2. Find, without using a calculator, the value of \[ \tan^{2}\left(\frac{\pi}{14}\right)+\tan^{2}\left(\frac{3\pi}{14}\right)+\tan^{2}\left(\frac{5\pi}{14}\right)\,. \]

If \(u\) and \(v\) are the two roots of \(z^{2}+az+b=0,\) show that \(a=-u-v\) and \(b=uv.\) Let \(\alpha=\cos(2\pi/7)+\mathrm{i}\sin(2\pi/7).\) Show that \(\alpha\) is a root of \(z^{6}-1=0\) and express the roots in terms of \(\alpha.\) The number \(\alpha+\alpha^{2}+\alpha^{4}\) is a root of a quadratic equation \[ z^{2}+Az+B=0 \] where \(A\) and \(B\) are real. By guessing the other root, or otherwise, find the numerical values of \(A\) and \(B\). Show that \[ \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}=-\frac{1}{2}, \] and evaluate \[ \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}, \] making it clear how you determine the sign of your answer.

By applying de Moivre's theorem to \(\cos5\theta+\mathrm{i}\sin5\theta,\) expanding the result using the binomial theorem, and then equating imaginary parts, show that \[ \sin5\theta=\sin\theta\left(16\cos^{4}\theta-12\cos^{2}\theta+1\right). \] Use this identity to evaluate \(\cos^{2}\frac{1}{5}\pi\), and deduce that \(\cos\frac{1}{5}\pi=\frac{1}{4}(1+\sqrt{5}).\)

Show that \[ \sin(2n+1)\theta=\sin^{2n+1}\theta\sum_{r=0}^{n}(-1)^{n-r}\binom{2n+1}{2r}\cot^{2r}\theta, \] where \(n\) is a positive integer. Deduce that the equation \[ \sum_{r=0}^{n}(-1)^{r}\binom{2n+1}{2r}x^{r}=0 \] has roots \(\cot^{2}(k\pi/(2n+1))\) for \(k=1,2,\ldots,n\). Show that

• sep}{3mm}
• [\bf (i)] \({\displaystyle \sum_{k=1}^{n}\cot^{2}\left(\frac{k\pi}{2n+1}\right)=\frac{n(2n-1)}{3}},\)
• [\bf (ii)] \({\displaystyle \sum_{k=1}^{n}\tan^{2}\left(\frac{k\pi}{2n+1}\right)=n(2n+1)},\)
• [\bf (iii)] \({\displaystyle \sum_{k=1}^{n}\mathrm{cosec}^{2}\left(\frac{k\pi}{2n+1}\right)=\frac{2n(n+1)}{3}}.\)

A path is made up in the Argand diagram of a series of straight line segments \(P_{1}P_{2},\) \(P_{2}P_{3},\) \(P_{3}P_{4},\ldots\) such that each segment is \(d\) times as long as the previous one, \((d\neq1)\), and the angle between one segment and the next is always \(\theta\) (where the segments are directed from \(P_{j}\) towards \(P_{j+1}\), and all angles are measured in the anticlockwise direction). If \(P_{j}\) represents the complex number \(z_{j},\) express \[ \frac{z_{n+1}-z_{n}}{z_{n}-z_{n-1}} \] as a complex number (for each \(n\geqslant2\)), briefly justifying your answer. If \(z_{1}=0\) and \(z_{2}=1\), obtain an expression for \(z_{n+1}\) when \(n\geqslant2\). By considering its imaginary part, or otherwise, show that if \(\theta=\frac{1}{3}\pi\) and \(d=2\), then the path crosses the real axis infinitely often.

Given that \(\sin\beta\neq0,\) sum the series \[ \cos\alpha+\cos(\alpha+2\beta)+\cdots+\cos(\alpha+2r\beta)+\cdots+\cos(\alpha+2n\beta) \] and \[ \cos\alpha+\binom{n}{1}\cos(\alpha+2\beta)+\cdots+\binom{n}{r}\cos(\alpha+2r\beta)+\cdots+\cos(\alpha+2n\beta). \] Given that \(\sin\theta\neq0,\) prove that \[ 1+\cos\theta\sec\theta+\cos2\theta\sec^{2}\theta+\cdots+\cos r\theta\sec^{r}\theta+\cdots+\cos n\theta\sec^{n}\theta=\frac{\sin(n+1)\theta\sec^{n}\theta}{\sin\theta}. \]

Show, using de Moivre's theorem, or otherwise, that \[ \tan9\theta=\frac{t(t^{2}-3)(t^{6}-33t^{4}+27t^{2}-3)}{(3t^{2}-1)(3t^{6}-27t^{4}+33t^{2}-1)},\qquad\mbox{ where }t=\tan\theta. \] By considering the equation \(\tan9\theta=0,\) or otherwise, obtain a cubic equation with integer coefficients whose roots are \[ \tan^{2}\left(\frac{\pi}{9}\right),\qquad\tan^{2}\left(\frac{2\pi}{9}\right)\qquad\mbox{ and }\qquad\tan^{2}\left(\frac{4\pi}{9}\right). \] Deduce the value of \[ \tan\left(\frac{\pi}{9}\right)\tan\left(\frac{2\pi}{9}\right)\tan\left(\frac{4\pi}{9}\right). \] Show that \[ \tan^{6}\left(\frac{\pi}{9}\right)+\tan^{6}\left(\frac{2\pi}{9}\right)+\tan^{6}\left(\frac{4\pi}{9}\right)=33273. \]

Let \(\omega=\mathrm{e}^{2\pi\mathrm{i}/3}.\) Show that \(1+\omega+\omega^{2}=0\) and calculate the modulus and argument of \(1+\omega^{2}.\) Let \(n\) be a positive integer. By evaluating \((1+\omega^{r})^{n}\) in two ways, taking \(r=1,2\) and \(3\), or otherwise, prove that \[ \binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\cdots+\binom{n}{k}=\frac{1}{3}\left(2^{n}+2\cos\left(\frac{n\pi}{3}\right)\right), \] where \(k\) is the largest multiple of \(3\) less than or equal to \(n\). Without using a calculator, evaluate \[ \binom{25}{0}+\binom{25}{3}+\cdots+\binom{25}{24} \] and \[ \binom{24}{2}+\binom{24}{5}+\cdots+\binom{24}{23}\,. \] {[}\(2^{25}=33554432.\){]}

The differential equation \[\frac{d^2x}{dt^2} = 2x\frac{dx}{dt}\] describes the motion of a particle with position \(x(t)\) at time \(t\). At \(t = 0\), \(x = a\), where \(a > 0\).

1. Solve the differential equation in the case where \(\frac{dx}{dt} = a^2\) when \(t = 0\). What happens to the particle as \(t\) increases from 0?
2. Solve the differential equation in the case where \(\frac{dx}{dt} = a^2 + p\) when \(t = 0\), where \(p > 0\). What happens to the particle as \(t\) increases from 0?
3. Solve the differential equation in the case where \(\frac{dx}{dt} = a^2 - q^2\) when \(t = 0\), where \(q > 0\). What happens to the particle as \(t\) increases from 0? Give conditions on \(a\) and \(q\) for the different cases which arise.

The coordinates of a particle at time \(t\) are \(x\) and \(y\). For \(t \geq 0\), they satisfy the pair of coupled differential equations \[ \begin{cases} \dot{x} &= -x -ky \\ \dot{y} &= x - y \end{cases}\] where \(k\) is a constant. When \(t = 0\), \(x = 1\) and \(y = 0\).

1. Let \(k = 1\). Find \(x\) and \(y\) in terms of \(t\) and sketch \(y\) as a function of \(t\). Sketch the path of the particle in the \(x\)-\(y\) plane, giving the coordinates of the point at which \(y\) is greatest and the coordinates of the point at which \(x\) is least.
2. Instead, let \(k = 0\). Find \(x\) and \(y\) in terms of \(t\) and sketch the path of the particle in the \(x\)-\(y\) plane.

The functions \(x(t)\) and \(y(t)\) satisfy the simultaneous differential equations \begin{alignat*}{1} \dfrac{\mathrm{d}x}{\mathrm{d}t}+2x-5y & =0\\ \frac{\mathrm{d}y}{\mathrm{d}t}+ax-2y & =2\cos t, \end{alignat*} subject to \(x=0,\) \(\dfrac{\mathrm{d}y}{\mathrm{d}t}=0\) at \(t=0.\) Solve these equations for \(x\) and \(y\) in the case when \(a=1\). Without solving the equations explicitly, state briefly how the form of the solutions for \(x\) and \(y\) if \(a>1\) would differ from the form when \(a=1.\)

Show that the second-order differential equation \[ x^2y''+(1-2p) x\, y' + (p^2-q^2) \, y= \f(x) \,, \] where \(p\) and \(q\) are constants, can be written in the form \[ x^a \big(x^b (x^cy)'\big)' = \f(x) \,, \tag{\(*\)} \] where \(a\), \(b\) and \(c\) are constants.

1. Use \((*)\) to derive the general solution of the equation \[ x^{2}y''+(1-2p)xy'+(p^2-q^{2})y=0 \] in the different cases that arise according to the values of \(p\) and \(q\).
2. Use \((*)\) to derive the general solution of the equation \[ x^{2}y''+(1-2p)xy'+p^2y=x^{n} \] in the different cases that arise according to the values of \(p\) and \(n\).

Two particles \(X\) and \(Y\), of equal mass \(m\), lie on a smooth horizontal table and are connected by a light elastic spring of natural length \(a\) and modulus of elasticity \(\lambda\). Two more springs, identical to the first, connect \(X\) to a point \(P\) on the table and \(Y\) to a point \(Q\) on the table. The distance between \(P\) and \(Q\) is \(3a\). Initially, the particles are held so that \(XP=a\), \(YQ= \frac12 a\,\), and \(PXYQ\) is a straight line. The particles are then released. At time \(t\), the particle \(X\) is a distance \(a+x\) from \(P\) and the particle \(Y\) is a distance \(a+y\) from \(Q\). Show that \[ m \frac{\.d ^2 x}{\.d t^2} = -\frac\lambda a (2x+y) \] and find a similar expression involving \(\dfrac{\.d^2 y}{\.d t^2}\). Deduce that \[ x-y = A\cos \omega t +B \sin\omega t \] where \(A\) and \(B\) are constants to be determined and \(ma\omega^2=\lambda\). Find a similar expression for \(x+y\). Show that \(Y\) will never return to its initial position.

A sphere of radius \(R\) and uniform density \(\rho_{\text{s}}\) is floating in a large tank of liquid of uniform density \(\rho\). Given that the centre of the sphere is a distance \(x\) above the level of the liquid, where \(x < R\), show that the volume of liquid displaced is \[ \frac \pi 3 (2R^3-3R^2x +x^3)\,. \] The sphere is acted upon by two forces only: its weight and an upward force equal in magnitude to the weight of the liquid it has displaced. Show that \[ 4 R^3\rho_{\text{s}} (g+\ddot x) = (2R^3 -3R^2x +x^3)\rho g\,. \] Given that the sphere is in equilibrium when \(x=\frac12 R\), find \(\rho_{\text{s}}\) in terms of \(\rho\). Find, in terms of \(R\) and \(g\), the period of small oscillations about this equilibrium position.

1. Let \(y(x)\) be a solution of the differential equation \( \dfrac {\d^2 y}{\d x^2}+y^3=0\) with \(y = 1\) and \(\dfrac{\d y}{\d x} =0\) at \(x=0\), and let \[ {\rm E} (x)= \left ( \frac {\d y}{\d x}\right)^{\!\!2} + \tfrac 12 y^4\,. \] Show by differentiation that \({\rm E}\) is constant and deduce that \( \vert y(x) \vert \le 1\) for all \(x\).
2. Let \(v(x)\) be a solution of the differential equation $ \dfrac{\d^2 v}{\d x^2} + x \dfrac {\d v}{\d x} +\sinh v =0 $ with \(v = \ln 3\) and \(\dfrac{\d v}{\d x} =0\) at \(x=0\), and let \[ {\rm E} (x)= \left ( \frac {\d v}{\d x}\right)^{\!\!2} + 2 \cosh v\,. \] Show that \(\dfrac{\d{\rm E}}{\d x}\le 0\) for \(x\ge0\) and deduce that \(\cosh v(x) \le \frac53\) for \(x\ge0\).
3. Let \(w(x)\) be a solution of the differential equation \[ \frac{\d^2 w}{\d x^2} + (5\cosh x - 4 \sinh x -3) \frac{\d w}{\d x} + (w\cosh w + 2 \sinh w) =0 \] with \(\dfrac{\d w }{\d x}=\dfrac 1 { \sqrt 2 }\) and \(w=0\) at \(x=0\). Show that \(\cosh w(x) \le \frac54\) for \(x\ge0\).

A pain-killing drug is injected into the bloodstream. It then diffuses into the brain, where it is absorbed. The quantities at time \(t\) of the drug in the blood and the brain respectively are \(y(t)\) and \(z(t)\). These satisfy \[ \dot y = - 2(y-z)\,, \ \ \ \ \ \ \ \dot z = - \dot y -3z\, , \] where the dot denotes differentiation with respect to \(t\). Obtain a second order differential equation for \(y\) and hence derive the solution \[ y= A\e^{-t} + B\e ^{-6t}\,, \ \ \ \ \ \ \ z= \tfrac12 A \e^{-t} - 2 B \e^{-6t}\,, \] where \(A\) and \(B\) are arbitrary constants. \begin{questionparts} \item Obtain the solution that satisfies \(z(0)=0\) and \(y(0)= 5\). The quantity of the drug in the brain for this solution is denoted by \(z_1(t)\). \item Obtain the solution that satisfies $ z(0)=z(1)= c$, where \(c\) is a given constant. %\[ %C=2(1-\e^{-1})^{-1} - 2(1-\e^{-6})^{-1}\,. %\] The quantity of the drug in the brain for this solution is denoted by \(z_2(t)\). \item Show that for \(0\le t \le 1\), \[ z_2(t) = \sum _{n=-\infty}^{0} z_1(t-n)\,, \] provided \(c\) takes a particular value that you should find. \end {questionparts}

Given that \(\displaystyle z = y^n \left( \frac{\d y}{\d x}\right)^{\!2}\), show that \[ \frac{\d z}{\d x} = y^{n-1} \frac{\d y}{\d x} \left( n \left(\frac{\d y}{\d x}\right)^{\!2} + 2y \frac{\d^2y}{\d x^2}\right) . \]

1. Use the above result to show that the solution to the equation \[ \left(\frac{\d y}{\d x}\right)^{\!2} + 2y \frac{\d^2y}{\d x^2} = \sqrt y \ \ \ \ \ \ \ \ \ \ (y>0) \] that satisfies \(y=1\) and \(\dfrac{\d y}{\d x}=0\) when \(x=0\) is \(y= \big ( \frac38 x^2+1\big)^{\frac23}\).
2. Find the solution to the equation \[ \left(\frac{\d y}{\d x}\right)^{\!2} -y \frac{\d^2y}{\d x^2} + y^2=0 \] that satisfies \(y=1\) and \(\dfrac{\d y}{\d x}=0\) when \(x=0\).

1. Find the general solution of the differential equation \[ \frac{\d u}{\d x} - \left(\frac { x +2}{x+1}\right)u =0\,. \]
2. Show that substituting\(y=z\e^{-x}\) (where \(z\) is a function of \(x\)) into the second order differential equation \[ (x+1) \frac{\d ^2 y}{\d x^2} + x \frac{\d y}{\d x} -y = 0 \tag{\(*\)} \] leads to a first order differential equation for \(\dfrac{\d z}{\d x}\,\). Find \(z\) and hence show that the general solution of \((*)\) is \[ y= Ax + B\e^{-x}\,, \] where \(A\) and \(B\) are arbitrary constants.
3. Find the general solution of the differential equation \[ (x+1) \frac{\d ^2 y}{\d x^2} + x \frac{\d y}{\d x} -y = (x+1)^2 . \]

A small bead \(B\), of mass \(m\), slides without friction on a fixed horizontal ring of radius \(a\). The centre of the ring is at \(O\). The bead is attached by a light elastic string to a fixed point \(P\) in the plane of the ring such that \(OP = b\), where \(b > a\). The natural length of the elastic string is \(c\), where \(c < b - a\), and its modulus of elasticity is \(\lambda\). Show that the equation of motion of the bead is \[ ma\ddot \phi = -\lambda\left( \frac{a\sin\phi}{c\sin\theta}-1\right)\sin(\theta+\phi) \,, \] where \(\theta=\angle BPO\) and \(\phi=\angle BOP\). Given that \(\theta\) and \(\phi\) are small, show that $a(\theta+\phi)\approx b\theta$. Hence find the period of small oscillations about the equilibrium position \(\theta=\phi =0\).

Show that, if \(y=\e^x\), then \[ (x-1) \frac{\d^2 y}{\d x^2} -x \frac{\d y}{\d x} +y=0\,. \tag{\(*\)} \] In order to find other solutions of this differential equation, now let \(y=u\e^x\), where \(u\) is a function of \(x\). By substituting this into \((*)\), show that \[ (x-1) \frac{\d^2 u}{\d x^2} + (x-2) \frac{\d u}{\d x} =0\,. \tag{\(**\)} \] By setting \( \dfrac {\d u}{\d x}= v\) in \((**)\) and solving the resulting first order differential equation for \(v\), find~\(u\) in terms of \(x\). Hence show that \(y=Ax + B\e^x\) satisfies \((*)\), where \(A\) and \(B\) are any constants.

A light spring is fixed at its lower end and its axis is vertical. When a certain particle \(P\) rests on the top of the spring, the compression is \(d\). When, instead, \(P\) is dropped onto the top of the spring from a height \(h\) above it, the compression at time \(t\) after \(P\) hits the top of the spring is \(x\). Obtain a second-order differential equation relating \(x\) and \(t\) for \(0\le t \le T\), where \(T\) is the time at which \(P\) first loses contact with the spring. Find the solution of this equation in the form \[ x= A + B\cos (\omega t) + C\sin(\omega t)\,, \] where the constants \(A\), \(B\), \(C\) and \(\omega\) are to be given in terms of \(d\), \(g\) and \(h\) as appropriate. Show that \[ T = \sqrt{d/g\;} \left (2 \pi - 2 \arctan \sqrt{2h/d\;}\;\right)\,. \]

1. The functions \(\f_n(x)\) are defined for \(n=0\), \(1\), \(2\), \(\ldots\)\, , by \[ \f_0(x) = \frac 1 {1+x^2}\, \qquad \text{and}\qquad \f_{n+1}(x) =\frac{\d \f_n(x)}{\d x}\,. \] Prove, for \(n\ge1\), that \[ (1+x^2)\f_{n+1}(x) + 2(n+1)x\f_n(x) + n(n+1)\f_{n-1}(x)=0\,. \]
2. The functions \(\P_n(x)\) are defined for \(n=0\), \(1\), \(2\), \(\ldots\)\, , by \[ \P_n(x) = (1+x^2)^{n+1}\f_n(x)\,. \] Find expressions for \(\P_0(x)\), \(\P_1(x)\) and \(\P_2(x)\). Prove, for \(n\ge0\), that \[ \P_{n+1}(x) -(1+x^2)\frac {\d \P_n(x)}{\d x}+ 2(n+1)x \P_n(x)=0\,, \] and that \(\P_n(x)\) is a polynomial of degree \(n\).

1. Let \(\displaystyle y= \sum_{n=0}^\infty a_n x^n\,\), where the coefficients \(a_n\) are independent of \(x\) and are such that this series and all others in this question converge. Show that \[ \displaystyle y'= \sum_{n=1}^\infty na_n x^{n-1}\,, \] and write down a similar expression for \(y''\). Write out explicitly each of the three series as far as the term containing \(a_3\).
2. It is given that \(y\) satisfies the differential equation \[ xy''-y'+4x^3y =0\,. \] By substituting the series of part (i) into the differential equation and comparing coefficients, show that \(a_1=0\). Show that, for \(n\ge4\), \[ a_n =- \frac{4}{n(n-2)}\, a_{n-4}\,, \] and that, if \(a_0=1\) and \(a_2=0\), then \( y=\cos (x^2)\,\). Find the corresponding result when \(a_0=0\) and \(a_2=1\).

In this question, \(p\) denotes \(\dfrac{\d y}{\d x}\,\).

1. Given that \[ y=p^2 +2 xp\,, \] show by differentiating with respect to \(x\) that \[ \frac{\d x}{\d p} = -2 - \frac {2x} p . \] Hence show that \(x = -\frac23p +Ap^{-2}\,,\) where \(A\) is an arbitrary constant. Find \(y\) in terms of \(x\) if \(p=-3\) when \(x=2\).
2. Given instead that \[ y=2xp +p \ln p\,,\] and that \(p=1\) when \(x=-\frac14\), show that \(x=-\frac12 \ln p - \frac14\,\) and find \(y\) in terms of \(x\).

1. Find functions \({\rm a}(x)\) and \({\rm b}(x)\) such that \(u=x\) and \(u=\e^{-x}\) both satisfy the equation $$\dfrac{\d^2u}{\d x^2} +{\rm a}(x) \dfrac{\d u}{\d x} + {\rm b} (x)u=0\,.$$ For these functions \({\rm a}(x)\) and \({\rm b}(x)\), write down the general solution of the equation. Show that the substitution \(y= \dfrac 1 {3u} \dfrac {\d u}{\d x}\) transforms the equation \[ \frac{\d y}{\d x} +3y^2 + \frac {x} {1+x} y = \frac 1 {3(1+x)} \tag{\(*\)} \] into \[ \frac{\d^2 u}{\d x^2} +\frac x{1+x} \frac{\d u}{\d x} - \frac 1 {1+x} u=0 \] and hence show that the solution of equation (\(*\)) that satisfies \(y=0\) at \(x=0\) is given by \(y = \dfrac{1-\e^{-x}}{3(x+\e^{-x})}\).
2. Find the solution of the equation $$ \frac{\d y}{\d x} +y^2 + \frac x {1-x} y = \frac 1 {1-x} $$ that satisfies \(y=2\) at \(x=0\).

Given that \(y=x\) and \(y=1-x^2\) satisfy the differential equation $$ \frac{\d^2 {y}}{\d x^2} + \p(x) \frac{\d {y}}{\d x} + \q(x) {y}=0\;, \tag{*} $$ show that \(\p(x)= -2x(1+x^2)^{-1}\) and \(\q(x) = 2(1+x^2)^{-1}\). Show also that \(ax+b(1-x^2)\) satisfies the differential equation for any constants \(a\) and \(b\). Given instead that \(y=\cos^2(\frac{1}{2}x^2)\) and \(y=\sin^2(\frac{1}{2}x^2)\) satisfy the equation \((*)\), find \(\p(x)\) and \(\q(x)\).

The function \(y(x)\) is defined for \(x\ge0\) and satisfies the conditions \[ y=0 \mbox{ \ \ and \ \ } \frac{\d y}{\d x}=1 \mbox{ \ \ at \(x=0\)}. \] When \(x\) is in the range \(2(n-1)\pi< x <2n\pi\), where \(n\) is a positive integer, \(y(t)\) satisfies the differential equation $$ {\d^2y \over \d x^2} + n^2 y=0. $$ Both \(y\) and \(\displaystyle \frac{\d y}{\d x} \) are continuous at \(x=2n\pi\) for \(n=0,\; 1,\;2,\; \ldots\;\).

1. Find \(y(x)\) for \(0\le x \le 2\pi\).
2. Show that \(y(x) = \frac12 \sin 2x \) for \(2\pi\le x\le 4\pi\), and find \(y(x)\) for all \(x\ge0\).
3. Show that $$ \int_0^\infty y^2 \,\d x = \pi \sum_{n=1}^\infty {1\over n^2} \,. $$

Show that \(\sin(k\sin^{-1} x)\), where \(k\) is a constant, satisfies the differential equation $$ (1-x^{2})\frac {\d^2 y}{\d x^2} -x\frac{\d y}{\d x} +k^{2}y=0. \eqno(*) $$ In the particular case when \(k=3\), find the solution of equation \((*)\) of the form \[ y=Ax^{3}+Bx^{2}+Cx+D, \] that satisfies \(y=0\) and \(\displaystyle \frac{\d y}{\d x}=3\) at \(x=0\). Use this result to express \(\sin 3\theta\) in terms of powers of \(\sin\theta\).

Suppose that \(y_n\) satisfies the equations \[(1-x^2)\frac{{\rm d}^2y_n}{{\rm d}x^2}-x\frac{{\rm d}y_n}{{\rm d}x}+n^2y_n=0,\] \[y_n(1)=1,\quad y_n(x)=(-1)^ny_n(-x).\] If \(x=\cos\theta\), show that \[\frac{{\rm d}^2y_n}{{\rm d}\theta^2}+n^2y_n=0,\] and hence obtain \(y_n\) as a function of \(\theta\). Deduce that for \(|x|\leqslant1\) \[y_0=1,\quad y_1=x,\] \[y_{n+1}-2xy_n+y_{n-1}=0.\]

Suppose that \[{\rm f}''(x)+{\rm f}(-x)=x+3\cos 2x\] and \({\rm f}(0)=1\), \({\rm f}'(0)=-1\). If \({\rm g}(x)={\rm f}(x)+{\rm f}(-x)\), find \({\rm g}(0)\) and show that \({\rm g}'(0)=0\). Show that \[{\rm g}''(x)+{\rm g}(x)=6\cos 2x,\] and hence find \({\rm g}(x)\). Similarly, if \({\rm h}(x)={\rm f}(x)-{\rm f}(-x)\), find \({\rm h}(x)\) and show that \[{\rm f}(x)=2\cos x -\cos2x-x.\]

Show that \(y=\sin^{2}(m\sin^{-1}x)\) satisfies the differential equation \[ (1-x^{2})y^{(2)}=xy^{(1)}+2m^{2}(1-2y), \] and deduce that, for all \(n\geqslant1,\) \[ (1-x^{2})y^{(n+2)}=(2n+1)xy^{(n+1)}+(n^{2}-4m^{2})y^{(n)}, \] where \(y^{(n)}\) denotes the \(n\)th derivative of \(y\). Derive the Maclaurin series for \(y\), making it clear what the general term is.

What is the general solution of the differential equation \[ \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+2k\frac{\mathrm{d}x}{\mathrm{d}t}+x=0 \] for each of the cases: (i) \(k>1;\) (ii) \(k=1\); (iii) \(0 < x < 1\)? In case (iii) the equation represents damped simple harmonic motion with damping factor \(k\). Let \(x(0)=0\) and let \(x_{1},x_{2},\ldots,x_{n},\ldots\) be the sequence of successive maxima and minima, so that if \(x_{n}\) is a maximum then \(x_{n+1}\) is the next minimum. Show that \(\left|x_{n+1}/x_{n}\right|\) takes a value \(\alpha\) which is independent of \(n\), and that \[ k^{2}=\frac{(\ln\alpha)^{2}}{\pi^{2}+(\ln\alpha)^{2}}. \]

Find functions \(\mathrm{f,g}\) and \(\mathrm{h}\) such that \[ \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}+\mathrm{f}(x)\frac{\mathrm{d}y}{\mathrm{d}x}+\mathrm{g}(x)y=\mathrm{h}(x)\tag{\ensuremath{*}} \] is satisfied by all three of the solutions \(y=x,y=1\) and \(y=x^{-1}\) for \(0 < x < 1.\) If \(\mathrm{f,g}\) and \(\mathrm{h}\) are the functions you have found in the first paragraph, what condition must the real numbers \(a,b\) and \(c\) satisfy in order that \[ y=ax+b+\frac{c}{x} \] should be a solution of \((*)\)?