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1999 Paper 1 Q8
The function \(\f\) satisfies \(0\leqslant\f(t)\leqslant K\) when \(0\leqslant t\leqslant x\). Explain by means of a sketch, or otherwise, why \[0\leqslant\int_{0}^{x} \f (t)\,{\mathrm d}t \leqslant Kx.\] By considering \(\displaystyle \int_{0}^{1}\frac{t}{n(n-t)}\,{\mathrm d}t\), or otherwise, show that, if \(n>1\), \[ 0\le \ln \left( \frac n{n-1}\right) -\frac 1n \le \frac 1 {n-1} - \frac 1n \] and deduce that \[ 0\le \ln N -\sum_{n=2}^N \frac1n \le 1. \] Deduce that as \(N\to \infty\) \[ \sum_{n=1}^N \frac1n \to\infty. \] Noting that \(2^{10}=1024\), show also that if \(N<10^{30}\) then \[ \sum_{n=1}^N \frac1n <101. \]
1999 Paper 2 Q6
Find \(\displaystyle \ \frac{\d y}{\d x} \ \) if $$ y = \frac{ax+b}{cx+d}. \tag{*} $$ By using changes of variable of the form \((*)\), or otherwise, show that \[ \int_0^1 \frac{1}{(x+3)^2} \; \ln \left(\frac{x+1}{x+3}\right)\d x = {\frac16} \ln3 - {\frac14}\ln 2 - \frac 1{12}, \] and evaluate the integrals \[ \int_0^1 \frac{1}{(x+3)^2} \; \ln \left(\frac{x^2+3x+2}{(x+3)^2}\right)\d x \mbox{ and } \int_0^1 \frac{1}{(x+3)^2} \; \ln\left(\frac{x+1}{x+2}\right)\d x . \] [Not on original paper:] By changing to the variable \(y\) defined by $$ y=\frac{2x-3}{x+1},$$ evaluate the integral $$ \int_2^4 \frac{2x-3}{(x+1)^3}\; \ln\!\left(\frac{2x-3}{x+1}\right)\d x.$$ Evaluate the integral $$ \int_9^{25} {\big({2z^{-3/2} -5z^{-2}}\big)}\ln{\big(2-5z^{-1/2}\big)}\; \d z.$$
Solution: \begin{align*} && y &= \frac{ax+b}{cx+d} \\ &&&= \frac{\frac{a}{c}(cx+d) - \frac{da}{c} + b}{cx+d} \\ \Rightarrow && y' &= \left (b - \frac{da}{c} \right)(-1)(cx+d)^{-2} \cdot c \\ &&&= (ad-bc)(cx+d)^{-2} \end{align*} \begin{align*} && y &= \frac{x+1}{x+3} \\ && \frac{\d y}{\d x} &= \frac{2}{(x+3)^2} \\ \Rightarrow && I &= \int_0^1 \frac{1}{(x+3)^2} \; \ln \left(\frac{x+1}{x+3}\right)\d x \\ &&&= \int_{y=1/3}^{y=1/2} \frac12 \ln y \, \d y \\ &&&= \frac12 \left [ y \ln y - y \right]_{1/3}^{1/2} \\ &&&= \frac12 \left ( \frac12\ln \frac12 - \frac12 - \frac13 \ln\frac13 + \frac13 \right) \\ &&&= \frac16 \ln 3 -\frac14 \ln 2 -\frac1{12} \end{align*} \begin{align*} && J &= \int_0^1 \frac1{(x+3)^2} \ln \left ( \frac{x^2+3x+2}{(x+3)^2} \right) \d x \\ &&&= \int_0^1 \frac1{(x+3)^2} \left ( \ln \frac{x+1}{x+3} + \ln \frac{x+2}{x+3} \right) \d x \\ &&&= I + \int_0^1 \frac1{(x+3)^2} \ln \left ( \frac{x+2}{x+3} \right) \d x \\ &&&= I + \int_{y=2/3}^{y=3/4} \ln y\, \d y \\ &&&= I + \left [ y \ln y- y\right]_{2/3}^{3/4} \\ &&&= I + \left ( \frac34 \ln \frac34 - \frac34 - \frac23 \ln \frac23 + \frac23 \right) \\ &&&= I + \left ( \frac34 \ln 3 - \frac32 \ln 2- \frac1{12} - \frac23 \ln 2 + \frac23 \ln 3\right) \\ &&&= I + \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\ &&&= \frac16 \ln 3 -\frac14 \ln 2 -\frac1{12} + \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\ &&&= \frac{19}{12} \ln 3 -\frac{29}{12}\ln 2 - \frac16 \end{align*} \begin{align*} && K &= \int_0^1 \frac{1}{(x+3)^2} \; \ln\left(\frac{x+1}{x+2}\right)\d x \\ &&&= \int_0^1 \frac{1}{(x+3)^2} \; \left ( \ln\left(\frac{x+1}{x+3}\right) - \ln \left ( \frac{x+3}{x+2} \right) \right)\d x \\ &&&= \frac16 \ln 3 -\frac14 \ln 2 -\frac1{12} - \left ( \frac{17}{12} \ln 3 - \frac{13}6 \ln 2- \frac1{12} \right) \\ &&&= -\frac54 \ln 3 +\frac{23}{12} \ln 2 \end{align*}
1999 Paper 3 Q3
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 $$ \lim_{n\rightarrow\infty}\left\{{n\over n^2+1} + {n\over n^2+4} + \cdots + {n\over n^2+n^2}\right\}\,. $$
Solution:
1999 Paper 3 Q6
A closed curve is given by the equation $$ x^{2/n} + y^{2/n} = a^{2/n} \eqno(*) $$ where \(n\) is an odd integer and \(a\) is a positive constant. Find a parametrization \(x=x(t)\), \(y=y(t)\) which describes the curve anticlockwise as \(t\) ranges from \(0\) to \(2\pi\). Sketch the curve in the case \(n=3\), justifying the main features of your sketch. The area \(A\) enclosed by such a curve is given by the formula $$ A= {1\over 2} \int_0^{2\pi} \left[ x(t) {\d y(t)\over \d t} - y(t) {\d x(t)\over \d t} \right] \,\d t \,. $$ Use this result to find the area enclosed by (\(*\)) for \(n=3\).
1999 Paper 3 Q8
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 \(n=0,\; 1,\;2,\; \ldots\;\).
- Find \(y(x)\) for \(0\le x \le 2\pi\).
- Show that \(y(x) = \frac12 \sin 2x \) for \(2\pi\le x\le 4\pi\), and find \(y(x)\) for all \(x\ge0\).
- Show that $$ \int_0^\infty y^2 \,\d x = \pi \sum_{n=1}^\infty {1\over n^2} \,. $$
1999 Paper 3 Q9
The gravitational force between two point particles of masses \(m\) and \(m'\) is mutually attractive and has magnitude $$ {G m m' \over r^2}\,, $$ where \(G\) is a constant and \(r\) is the distance between them. A particle of unit mass lies on the axis of a thin uniform circular ring of radius \(r\) and mass \(m\), at a distance \(x\) from its centre. Explain why the net force on the particle is directed towards the centre of the ring and show that its magnitude is $$ {G m x \over (x^2 + r^2)^{3/2}} \,. $$ The particle now lies inside a thin hollow spherical shell of uniform density, mass \(M\) and radius \(a\), at a distance \(b\) from its centre. Show that the particle experiences no gravitational force due to the shell. %Explain without calculation the effect on this result if %the shell has finite thickness \(x\).
1998 Paper 1 Q2
Show, by means of a suitable change of variable, or otherwise, that \[ \int_{0}^{\infty}\mathrm{f}((x^{2}+1)^{1/2}+x)\,{\mathrm d}x =\frac{1}{2} \int_{1}^{\infty}(1+t^{-2})\mathrm{f}(t)\,{\mathrm d}t. \] Hence, or otherwise, show that \[ \int_{0}^{\infty}((x^{2}+1)^{1/2}+x)^{-3}\,{\mathrm d}x =\frac{3}{8}. \]
Solution: \begin{align*} && t &= (x^2+1)^{1/2}+x \\ && 1&=t^2-2tx \\ && x &= \frac{t^2-1}{2t} = \frac12 \left (t - \frac1t\right) \\ && \frac{\d x}{\d t} &= \frac12 \left ( 1+ \frac{1}{t^2} \right) \\ \Rightarrow && \int_0^{\infty} f((x^2+1)^{1/2}+x) \d x &= \int_{t=1}^{t = \infty}f(t) \frac12(1 + t^{-2}) \d t\\ &&&= \frac12 \int_1^{\infty}(1+t^{-2})f(t) \d t \end{align*} \begin{align*} \int_{0}^{\infty}((x^{2}+1)^{1/2}+x)^{-3}\,{\mathrm d}x &= \frac12 \int_1^{\infty}(1+t^{-2})t^{-3} \d t \\ &= \frac12 \left [\frac{-1}{2}t^{-2}-\frac{1}{4}t^{-4} \right]_{1}^{\infty} \\ &= \frac12 \left ( \frac12 + \frac14\right) = \frac38 \end{align*}
1998 Paper 1 Q8
Fluid flows steadily under a constant pressure gradient along a straight tube of circular cross-section of radius \(a\). The velocity \(v\) of a particle of the fluid is parallel to the axis of the tube and depends only on the distance \(r\) from the axis. The equation satisfied by \(v\) is \[\frac{1}{r}\frac{{\mathrm d}\ }{{\mathrm d}r} \left(r\frac{{\mathrm d}v}{{\mathrm d}r}\right) =-k,\] where \(k\) is constant. Find the general solution for \(v\). Show that \(|v|\rightarrow\infty\) as \(r\rightarrow 0\) unless one of the constants in your solution is chosen to be~\(0\). Suppose that this constant is, in fact, \(0\) and that \(v=0\) when \(r=a\). Find \(v\) in terms of \(k\), \(a\) and \(r\). The volume \(F\) flowing through the tube per unit time is given by \[F=2\pi\int_{0}^{a}rv\,{\mathrm d}r. \] Find \(F\).
1998 Paper 2 Q4
The integral \(I_n\) is defined by $$I_n=\int_0^\pi(\pi/2-x)\sin(nx+x/2)\,{\rm cosec}\,(x/2)\,\d x,$$ where \(n\) is a positive integer. Evaluate \(I_n-I_{n-1}\), and hence evaluate \(I_n\) leaving your answer in the form of a sum.
Solution: \begin{align*} && I_n - I_{n-1} &= \int_0^\pi \left ( \frac{\pi}{2} - x \right) \left ( \sin\left(nx + \frac{x}{2}\right) - \sin \left ((n-1)x + \frac{x}{2} \right)\right) \cosec \frac{x}{2} \d x \\ &&&= \int_0^\pi \left ( \frac{\pi}{2} - x \right) \left (2 \sin \left ( \frac{nx + \frac{x}{2} - (n-1)x - \frac{x}{2} }{2}\right)\cos nx \right) \cosec \frac{x}{2} \d x \\ &&&= \int_0^\pi \left ( \frac{\pi}{2} - x \right) \left (2 \sin \left ( \frac{x}{2}\right)\cos nx \right) \cosec \frac{x}{2} \d x \\ &&&=2 \int_0^\pi \left ( \frac{\pi}{2} - x \right) \cos nx \d x \\ &&&=\pi \left [ \frac{\sin nx}{n}\right]_0^{\pi} - 2\int_0^\pi x \cos n x \d x \\ &&&= 0 - 2\left[ \frac{x \sin nx}{n} \right]_0^{\pi} + 2\int_0^\pi \frac{\sin nx}{n} \d x \\ &&&= 2\left[ -\frac{\cos nx}{n^2} \right]_0^{\pi} \\ &&&=2 \frac{1-(-1)^{n}}{n^2} \\ \\ && I_0 &= \int_0^\pi (\pi/2 - x) \d x =0 \\ \Rightarrow && I_{2k+2} = I_{2k+1} &= 4 \left (\frac{1}{1^2} + \frac{1}{3^2} + \cdots + \frac{1}{(2k+1)^2} \right) \end{align*}
1998 Paper 3 Q2
Let $$ {\rm I}(a,b) = \int_0^1 t^{a}(1-t)^{b} \, \d t \; \qquad (a\ge0,\ b\ge0) .$$
- Show that \({\rm I}(a,b)={\rm I}(b,a)\),
- Show that \({\rm I}(a,b)={\rm I}(a+1,b)+{\rm I}(a,b+1)\).
- Show that \((a+1){\rm I}(a,b)=b{\rm I}(a+1,b-1)\) when \(a\) and \(b\) are positive and hence calculate \({\rm I}(a,b)\) when \(a\) and \(b\) are positive integers.
Solution:
- Let \(u = 1-t, \d u = -\d t\), then: \begin{align*} \mathrm{I}(a,b) &= \int_0^1 t^a(1-t)^b \d t \\ &= \int_{u=1}^{u=0} -(1-u)^a u^b \d u \\ &= \int_0^1(1-u)^a u^b \d u \\ &= \mathrm{I}(b, a) \end{align*}
- \begin{align*} \mathrm{I}(a+1,b)+\mathrm{I}(a,b+1) &= \int_0^1 t^{a+1}(1-t)^b + t^a(1-t)^{b+1} \d t \\ &= \int_0^1 (t+(1-t))t^a(1-t)^b \d t \\ &= \int_0^1 t^a(1-t)^b \d t \\ &= \mathrm{I}(a,b) \end{align*}
- Integrating by parts with \(\frac{du}{dt} = t^a, v = (1-t)^{b}\)\begin{align*} \mathrm{I}(a,b) &= \int_0^1 t^a (1-t)^b \d t \\ &= \left [ \frac{t^{a+1}}{a+1} (1-t)^b \right ]_0^1 + \int_0^1 \frac{t^{a+1}}{a+1} b(1-t)^{b-1} \\ &= \frac{b}{a+1} \int_0^1 t^{a+1}(1-t)^{b-1} \d t \\ &= \frac{b}{a+1} \mathrm{I}(a+1,b-1) \end{align*} Claim: \(\mathrm{I}(a,b) = \frac{a!b!}{(a+b+1)!}\) Proof: Note that \(I(a,0) = \frac{1}{a+1}\) so the formula holds for this case. We will induct on \(b\). The base case is done. Suppose that for \(b = k\) our formula is true, ie: \(\mathrm{I}(a,k) = \frac{a!k!}{(a+k+1)!}\) for all \(a\) (and fixed \(k\)) \begin{align*} \mathrm{I}(a,k+1) &= \frac{k+1}{a+1} \mathrm{I}(a+1,k) \\ &= \frac{k+1}{a+1} \frac{(a+1)!k!}{(a+1+k+1)!} \\ &= \frac{a!(k+1)!}{(a+(k+1)+1)!} \end{align*} So the formula is true for \(b=k+1\). Therefore, since it is true if \(b=0\) and if \(b=k\) is true then \(b=k+1\) is true, it is true for all values of \(b\).
1998 Paper 3 Q4
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 \(0\le\theta\le 2\pi\).
1998 Paper 3 Q9
A uniform right circular cone of mass \(m\) has base of radius \(a\) and perpendicular height \(h\) from base to apex. Show that its moment of inertia about its axis is \({3\over 10} ma^2\), and calculate its moment of inertia about an axis through its apex parallel to its base. \newline[{\em Any theorems used should be stated clearly.}] The cone is now suspended from its apex and allowed to perform small oscillations. Show that their period is $$ 2\pi\sqrt{ 4h^2 + a^2\over 5gh} \,. $$ \newline[{\em You may assume that the centre of mass of the cone is a distance \({3\over 4}h\) from its apex.}]
1997 Paper 1 Q6
Find constants \(a_{0}\), \(a_{1}\), \(a_{2}\), \(a_{3}\), \(a_{4}\), \(a_{5}\), \(a_{6}\) and \(b\) such that \[x^{4}(1-x)^{4}=(a_{6}x^{6}+a_{5}x^{5}+a_{4}x^{4}+a_{3}x^{3}+ a_{2}x^{2}+a_{1}x+a_{0})(x^{2}+1)+b.\] Hence, or otherwise, prove that \[\int_{0}^{1}\frac{x^{4}(1-x)^{4}}{1+x^{2}}{\rm d}x =\frac{22}{7}-\pi.\] Evaluate \(\displaystyle{\int_{0}^{1}x^{4}(1-x)^{4}{\rm d}x}\) and deduce that \[\frac{22}{7}>\pi>\frac{22}{7}-\frac{1}{630}.\]
Solution: Plugging in \(x = i\) we obtain \((1-i)^4 = (-2i)^2 = -4 \Rightarrow b = -4\). \begin{align*} x^4(1-x)^4 &= x^4(1-4x+6x^2-4x^3+x^4) \\ &= x^8-4x^7+6x^6-4x^5+x^4 \\ &= x^6(x^2+1) - x^6 -4x^7+6x^6-4x^5+x^4 \\ &= x^6(x^2+1) -4x^5(x^2+1)+4x^5 +5x^6-4x^5+x^4 \\ &= (x^6-4x^5)(x^2+1) +5x^4(x^2+1)-5x^4+x^4 \\ &= (x^6-4x^5+5x^4)(x^2+1) -4x^2(x^2+1)+4x^2 \\ &= (x^6-4x^5+5x^4-4x^2)(x^2+1) +4(x^2+1)-4 \\ &= (x^6-4x^5+5x^4-4x^2+4)(x^2+1) -4 \\ \end{align*} So \begin{align*} \int_0^1 \frac{x^4(1-x)^4}{1+x^2} \d x &= \int_0^1 (x^6-4x^5+5x^4-4x^2+4) - \frac{4}{1+x^2} \d x \\ &= \frac17 - \frac46+1-\frac43+4 - \pi \\ &= \frac{22}7 - \pi \end{align*} \begin{align*} \int_0^1 x^4(1-x)^4 \d x &= B(5,5) \\ &= \frac{4!4!}{9!} \\ &= \frac1{630} \end{align*} Therefore since \(0 < \frac{x^4(1-x)^4}{1+x^2} < x^4(1-x)^4\) we must have that \begin{align*} && 0 &< \frac{22}7 - \pi \\ \Rightarrow && \pi & < \frac{22}{7} \\ && \frac{22}{7} - \pi &< \frac1{630} \\ \Rightarrow && \frac{22}{7} - \frac1{630} &< \pi \end{align*} which is what we wanted.
1997 Paper 1 Q7
Find constants \(a_{1}\), \(a_{2}\), \(u_{1}\) and \(u_{2}\) such that, whenever \({\mathrm P}\) is a cubic polynomial, \[\int_{-1}^{1}{\mathrm P}(t)\,{\mathrm d}t =a_{1}{\mathrm P}(u_{1})+a_{2}{\mathrm P}(u_{2}).\]
Solution: Since this is true for all cubic polynomials, it must be true in particular for \(1, x, x^2, x^3\), therefore: \begin{align*} \int_{-1}^{1} 1 {\mathrm d}t &=a_{1}+a_{2} &=2\\ \int_{-1}^{1} x {\mathrm d}t &=a_{1}u_1+a_{2}u_2 &= 0 \\ \int_{-1}^{1} x^2 {\mathrm d}t &=a_{1}u_1^2+a_{2}u_2^2 &= \frac23\\ \int_{-1}^{1} x^3 {\mathrm d}t &=a_{1}u_1^3+a_{2}u_2^3 &= 0\\ \end{align*} \begin{align*} && \begin{cases} a_{1}+a_{2} &=2 \\ a_{1}u_1+a_{2}u_2 &= 0 \\ a_{1}u_1^2+a_{2}u_2^2 &= \frac23\\ a_{1}u_1^3+a_{2}u_2^3 &= 0\\ \end{cases} \\ \Rightarrow && \begin{cases} a_{1}(u_1^2 - \frac13) + a_{2}(u_2^2 - \frac13) &= 0 \\ a_{1}u_1(u_1^2 - \frac13) + a_{2}u_2(u_2^2 - \frac13) &= 0 \end{cases} \\ \Rightarrow && \begin{cases} u_i = \pm \frac1{\sqrt{3}} \\ a_i = 1\end{cases} \end{align*} Therefore we have: \[\int_{-1}^{1}{\mathrm P}(t)\,{\mathrm d}t ={\mathrm P} \l \frac1{\sqrt{3}} \r+{\mathrm P}\l -\frac1{\sqrt{3}} \r \] [Note: this question is actually asking about Gauss-Legendre polynomials, and could be done directly by appealing to standard results]
1997 Paper 2 Q3
Find constants \(a,\,b,\,c\) and \(d\) such that $$\frac{ax+b}{ x^2+2x+2}+\frac{cx+d}{ x^2-2x+2}= \frac{1}{ x^4+4}.$$ Show that $$\int_0^1\frac {\d x}{ x^4+4}\;= \frac{1}{16} \ln 5 +\frac{1}{8} \tan^{-1}2 .$$
Solution: First notice that \((x^2+2+2x)(x^2+2-2x) = (x^2+2)^2-4x^2 = x^2+4\) and so \begin{align*} && \frac{1}{x^4+4} &= \frac{ax+b}{ x^2+2x+2}+\frac{cx+d}{ x^2-2x+2} \\ \Rightarrow && 1 &= (ax+b)(x^2-2x+2) + (cx+d)(x^2+2x+2) \\ &&&= (a+c)x^3+(b-2a+d+2c)x^2+(2a-2b+2c+2d)x+2b+2d \\ \Rightarrow && 0 &= a+c \\ && 2a &= b+d+2c \\ && b &= a+c+d \\ && \frac12 &= b+d \\ \Rightarrow && c &= -a \\ && 2a &= \frac12+2(-a) \\ \Rightarrow && a &= \frac18, c = -\frac18 \\ && b &= d \\ \Rightarrow && b &= \frac14, d = \frac14 \end{align*} Therefore \begin{align*} \int_0^1\frac {\d x}{ x^4+4} &= \int_0^1 \frac18\left ( \frac{x+2}{x^2+2x+2} -\frac{x-2}{x^2-2x+2}\right) \d x \\ &= \frac1{16}\int_0^1 \left ( \frac{2x+2+2}{x^2+2x+2} -\frac{2x-2-2}{x^2-2x+2}\right) \d x \\ &= \frac1{16} \left [ \ln(x^2+2x+2) -\ln(x^2-2x+2)\right]_0^1 + \frac1{16} \int_0^1 \left ( \frac{2}{(x+1)^2+1} + \frac{2}{(x-1)^2+1} \right) \d x \\ &= \frac1{16} \left (\ln 5 - \ln 1 -(\ln 2-\ln 2) \right) + \frac18 \left [ \tan^{-1} (x+1)+\tan^{-1}(x-1) \right]_0^1 \\ &= \frac1{16} \ln 5 + \frac18 \left (\tan^{-1} 2+\tan^{-1} 0 - \tan^{-1}1-\tan^{-1} (-1) \right) \\ &= \frac1{16} \ln 5+ \frac18 \tan^{-1} 2 \end{align*}