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1993 Paper 3 Q3
The matrices \(\mathbf{A},\mathbf{B}\) and \(\mathbf{M}\) are given by \[ \mathbf{A}=\begin{pmatrix}a & 0 & 0\\ b & c & 0\\ d & e & f \end{pmatrix},\quad\mathbf{B}=\begin{pmatrix}1 & p & q\\ 0 & 1 & r\\ 0 & 0 & 1 \end{pmatrix},\quad\mathbf{M}=\begin{pmatrix}1 & 3 & 2\\ 4 & 13 & 5\\ 3 & 8 & 7 \end{pmatrix}, \] where \(a,b,\ldots,r\) are real numbers. Given that \(\mathbf{M=AB},\) show that \(a=1,b=4,c=1,d=3,e=1,f=-2,p=3,q=2\) and \(r=-3\) gives the unique solution for \(\mathbf{A}\) and \(\mathbf{B}.\) Evaluate \(\mathbf{A}^{-1}\) and \(\mathbf{B}^{-1},\) Hence, or otherwise, solve the simultaneous equations \begin{alignat*}{1} x+3y+2z & =7\\ 4x+13y+5z & =18\\ 3x+8y+7z & =25. \end{alignat*}
Solution: \begin{align*} && \begin{pmatrix}1 & 3 & 2\\ 4 & 13 & 5\\ 3 & 8 & 7 \end{pmatrix} &= \begin{pmatrix}a & 0 & 0\\ b & c & 0\\ d & e & f \end{pmatrix}\begin{pmatrix}1 & p & q\\ 0 & 1 & r\\ 0 & 0 & 1 \end{pmatrix} \\ &&&= \begin{pmatrix} a & ap & aq \\ b & pb + c & qb + cr\\ d & pd + e & qd + er +f \end{pmatrix} \\ \Rightarrow && a,b,d,p,q&=1,4,3,3,2\\ &&&= \begin{pmatrix} 1 & 3 & 2 \\ 4 & 12 + c & 8+ cr\\ 3 & 9 + e & 6 + er +f \end{pmatrix} \\ \Rightarrow && c, e&=1,-1\\ &&&= \begin{pmatrix} 1 & 3 & 2 \\ 4 & 13 & 8+ r\\ 3 & 8 & 6 -r +f \end{pmatrix} \\ \Rightarrow && r, f &= -3, -2 \end{align*} \begin{align*} \mathbf{A}^{-1} &= \begin{pmatrix} 1 & 0 & 0 \\ 4 & 1 & 0\\ 3 & -1 & -2 \end{pmatrix}^{-1} \\ &=\frac{1}{-2} \begin{pmatrix} -2 & 0 & 0 \\ 8 & -2 & 0\\ -7 & 1 & 1 \end{pmatrix} \\ \\ \mathbf{B}^{-1} &= \begin{pmatrix} 1 & 3 & 2 \\ 0 & 1 & -3\\ 0 & 0 & 1 \end{pmatrix}^{-1} \\ &= \begin{pmatrix} 1 & -3 & -11 \\ 0 & 1 & 3\\ 0 & 0 & 1 \end{pmatrix} \\ \end{align*} We want to solve \(\mathbf{M}\mathbf{v} = \begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix}\), ie \begin{align*} \mathbf{v} &= \mathbf{M}^{-1}\begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\ &= \mathbf{B}^{-1} \mathbf{A}^{-1}\begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\ &= \frac{1}{-2}\mathbf{B}^{-1} \begin{pmatrix} -2 & 0 & 0 \\ 8 & -2 & 0\\ -7 & 1 & 1 \end{pmatrix} \begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\ &= \frac{1}{-2}\mathbf{B}^{-1} \begin{pmatrix} -14 \\ 20 \\ -6 \end{pmatrix} \\ &= \begin{pmatrix} 1 & -3 & -11 \\ 0 & 1 & 3\\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 7 \\ -10 \\ 3 \end{pmatrix} \\ &= \begin{pmatrix} 4\\ -1 \\ 3 \end{pmatrix} \end{align*} This algorithm is called the "LU-decomposition"
1993 Paper 3 Q7
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.
Solution: Let \(t = \sinh^2 y\), then \begin{align*} && \sinh(2x) &= \cosh y \tag{1}\\ && \sinh(2y) &= 2 \cosh x \tag{2} \\ \\ && \cosh(2x) &= 2 \cosh^2 x -1 \\ (2): &&&= \frac12 \sinh^2(2y) -1 \\ && 1 &= \left (\frac12 \sinh^2(2y) -1 \right)^2 - \cosh^2 y \\ &&&= \frac14 \sinh^4(2y)-\sinh^2(2y)+1-\cosh^2 y \\ \Rightarrow && 0 &= \frac14 (\cosh^2 (2y)-1)^2- (\cosh^2 (2y)-1) - \cosh^2 y \\ &&&= \frac14 \left ( \left (1+2\sinh^2 y \right)^2-1 \right)^2 -\left ( \left (1+2\sinh^2 y \right)^2 -1\right) - (1 + \sinh^2 y ) \\ &&&= \frac14 \left ( 1 + 4t+4t^2 -1\right)^2 - \left ( 1+4t+4t^2-1\right) - (1 + t) \\ &&&= \frac14 (4t + 4t^2)^2 - (4t+4t^2)-1-t \\ &&&= 4(t+t^2)^2 - 4t^2-5t-1 \\ &&&= 4t^4+8t^3+4t^2-4t^2-5t-1 \\ &&& = 4t^4+8t^3-5t-1 \\ &&&= (t+1)(4t^3+4t^2-4t-1) \end{align*} Since \(\sinh^2 y\) is positive, we must be a root of the second cubic. Let \(f(t) = 4t^3+4t^2-4t-1\), then \(f(0) = -1\) and \(f'(t) = 12t^2+8t-4 = 4(t+1)(3t-1)\), so we have turning points at \(-1\) and \(\frac13\). Since \(f(-1) = 3 > 0\) and \(f(0) < 0\) we must have exactly one root larger than zero. Therefore there is a unique root. \(f(0.7) = -0.468 < 0\) \(f(0.8) = 0.408 > 0\) since \(f\) is continuous and changes sign, the root must fall in the interval \((0.7, 0.8)\). Let \(t_{n+1} = t_n - \frac{f(t_n)}{f'(t_n)}\), and \(t_0 = 0.75\), then \begin{align*} t_0 &= 0.75 \\ t_1 &= 0.7571428571 \\ t_2 &= 0.7570684728 \\ t_3 &= 0.7570684647 \end{align*} So \(t \approx 0.757068\), \(\sinh y \approx 0.870097\), \(y \approx 0.786474\), \(x \approx 0.546965\)
1992 Paper 1 Q11
Three light elastic strings \(AB,BC\) and \(CD\), each of natural length \(a\) and modulus of elasticity \(\lambda,\) are joined together as shown in the diagram. \noindent
1991 Paper 1 Q5
A set of \(n\) distinct vectors \(\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{n},\) where \(n\geqslant2\), is called regular if it satisfies the following two conditions:
- there are constants \(\alpha\) and \(\beta\), with \(\alpha>0\), such that for any \(i\) and \(j\), \[ \mathbf{a}_{i}\cdot\mathbf{a}_{j}=\begin{cases} \alpha^{2} & \mbox{ when }i=j\\ \beta & \mbox{ when }i\neq j, \end{cases} \]
- the centroid of \(\mathbf{a}_{1},\mathbf{a}_{2},\ldots,\mathbf{a}_{n}\) is the origin \(\mathbf{0}.\) {[}The centroid of vectors \(\mathbf{b}_{1},\mathbf{b}_{2},\ldots,\mathbf{b}_{m}\) is the vector \(\frac{1}{m}(\mathbf{b}_{1}+\mathbf{b}_{2}+\cdots+\mathbf{b}_{m}).\){]}
Solution: \begin{align*} && \mathbf{0} &= \sum_i \mathbf{a}_i \tag{ii} \\ \Rightarrow && 0 &= \mathbf{a}_i \cdot \mathbf{0} \\ &&&= \sum_j \mathbf{a}_i \cdot \mathbf{a}_j \\ &&&= (n-1)\beta + \alpha^2 \tag{i} \\ \Rightarrow && (n-1)\beta &= -\alpha^2 \end{align*} Suppose we have \(\mathbf{a}_j = \binom{x}{y}\), \(j \neq 1\) then \(x = \beta\). We also must have \(\beta^2 + y^2 = 1\), so there are at most two values for \(y\), ie two extra vectors. ie \(n = 2, 3\). If \(n = 2 \Rightarrow \mathbf{a}_2 = - \mathbf{a}_1\). If \(n = 3\) \begin{align*} && \mathbf{0} &= \binom{1}{0} + \binom{\beta}{y} + \binom{\beta}{-y} \\ \Rightarrow && \beta = -1/2 \\ \Rightarrow && y &= \pm \frac{\sqrt{3}}{2} \end{align*} Suppose $\mathbf{a}_{1}=\begin{pmatrix}1\\ 0\\ 0 \end{pmatrix}\(, \)\mathbf{a}_{2}=\begin{pmatrix}\cos \theta \\ \sin \theta \\ 0 \end{pmatrix}$ (since we need \(\mathbf{a}_2 \cdot \mathbf{a}_2 = 1\)). \(\beta = \cos \theta\)). We can have \(\cos \theta = - 1\). Suppose we have \(\mathbf{a}_j =\begin{pmatrix}x\\ y \\ z \end{pmatrix}\), so \(x = \cos \theta\), and \(y^2 + z^2 = \sin^2 \theta\), so we can write it as: \(\mathbf{a}_j =\begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi \\ \sin \theta \sin \phi \end{pmatrix}\). We must also have \(\beta = \begin{pmatrix} \cos \theta \\ \sin \theta \cos \phi \\ \sin \theta \sin \phi \end{pmatrix} \cdot \begin{pmatrix} \cos \theta \\ \sin \theta\\ 0 \end{pmatrix} = \cos^2 \theta + \sin^2 \theta \cos \phi = \cos \theta\), so \(\cos \phi = \frac{\cos \theta - \cos^2\theta}{1-\cos^2 \theta} = \frac{\cos \theta}{1+\cos \theta}\). Therefore there is one value for \(\cos \phi\), so at most two values for \(\sin \phi\), Therefore we can have either \(2, 3,4\) or \(5\) different values in the set. \(n = 2\), we've already handled. If \(n = 3\), then \(\beta = -\frac12\), \(\cos \phi = -1\), so we can only have two different values for \(\sin \theta\), ie: \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -1/2\\ \frac{\sqrt{3}}{2} \\ 0 \end{pmatrix}, \begin{pmatrix} -1/2\\ -\frac{\sqrt{3}}{2} \\ 0 \end{pmatrix} \right \}\) Finally, if \(n = 4\), we have \(\beta = -\frac13\), \(\cos \phi = \frac{-1/3}{2/3} = -\frac12\). \(\sin \theta = \pm \frac{\sqrt{3}}{2}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ \frac{2\sqrt{2}}{3} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ -\frac{\sqrt{2}}{3} \\ \frac{\sqrt{6}}{3} \end{pmatrix},\begin{pmatrix} -\frac13\\ -\frac{\sqrt{2}}{3} \\ -\frac{\sqrt{6}}{3} \end{pmatrix} \right \}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ -\frac{2\sqrt{2}}{3} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac13\\ \frac{\sqrt{2}}{3} \\ \frac{\sqrt{6}}{3} \end{pmatrix},\begin{pmatrix} -\frac13\\ \frac{\sqrt{2}}{3} \\ -\frac{\sqrt{6}}{3} \end{pmatrix} \right \}\) If \(n = 5\), then \(\beta = -\frac14\), \(\cos \phi = \frac{-1/4}{3/4} = -\frac13\). \(\sin \theta = \frac{\sqrt{15}}{4}\), \(\sin \phi = \frac{2\sqrt{2}}{3}\) \(\displaystyle \left \{\begin{pmatrix} 1\\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ \frac{\sqrt{15}}{4} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ -\frac{\sqrt{15}}{4} \\ 0 \end{pmatrix}, \begin{pmatrix} -\frac14\\ \frac{\sqrt{15}}{12} \\ \frac{\sqrt{30}}{6} \end{pmatrix}, \begin{pmatrix} -\frac14\\ -\frac{\sqrt{15}}{12} \\ -\frac{\sqrt{30}}{6} \end{pmatrix}, \right \}\)
1991 Paper 2 Q1
Let \(\mathrm{h}(x)=ax^{2}+bx+c,\) where \(a,b\) and \(c\) are constants, and \(a\neq0\). Give a condition which \(a,b\) and \(c\) must satisfy in order that \(\mathrm{h}(x)\) can be written in the form \[ a(x+k)^{2},\tag{*} \] where \(k\) is a constant. If \(\mathrm{f}(x)=3x^{2}+4x\) and \(\mathrm{g}(x)=x^{2}-2\), find the two constant values of \(\lambda\) such that \(\mathrm{f}(x)+\lambda\mathrm{g}(x)\) can be written in the form \((*)\). Hence, or otherwise, find constants \(A,B,C,D,m\) and \(n\) such that \begin{alignat*}{1} \mathrm{f}(x) & =A(x+m)^{2}+B(x+n)^{2}\\ \mathrm{g}(x) & =C(x+m)^{2}+D(x+n)^{2}. \end{alignat*} If \(\mathrm{f}(x)=3x^{2}+4x\) and \(\mathrm{g}(x)=x^{2}+\alpha\) and it is given by that there is only one value of \(\lambda\) for which \(\mathrm{f}(x)+\lambda\mathrm{g}(x)\) can be written in the form \((*)\), find \(\alpha\).
Solution: For \(h(x)\) to be written in this form \(b^2=4ac\). Suppose \(f(x) = 3x^2+4x\), \(g(x) = x^2-2\). then, \begin{align*} && f(x) + \lambda g(x) &= (3+\lambda)x^2+4x - 2 \lambda \\ \Rightarrow && 0 &= 16 + 8(3+\lambda) \lambda \\ \Rightarrow && 0 &= 2+ 3 \lambda + \lambda^2 \\ &&&= (\lambda +1)(\lambda + 2) \\ \Rightarrow && \lambda &= -1 , -2 \\ \end{align*} \begin{align*} && f(x) - g(x) &= 2(x+1)^2 \\ && f(x) -2g(x) &= (x+2)^2 \\ \Rightarrow && g(x) &= 2(x+1)^2 - (x+2)^2 \\ && f(x) &= 4(x+1)^2 - (x+2)^2 \end{align*} Suppose \(f(x) = 3x^2+4x, g(x) = x^2 + \alpha\), then \begin{align*} && f(x) + \lambda g(x) &= (3+\lambda)x^2+4x+\lambda \alpha \\ \Rightarrow && 0 &= 16 -2\lambda \alpha(\lambda + 3) \\ && 0 &= \alpha \lambda^2 +3\lambda-8 \\ \Rightarrow && 0 &= 9 +32 \alpha \\ \Rightarrow && \alpha &= -\frac{9}{32} \end{align*}
1991 Paper 2 Q3
It is given that \(x,y\) and \(z\) are distinct and non-zero, and that they satisfy \[ x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}. \] Show that \(x^{2}y^{2}z^{2}=1\) and that the value of \(x+\dfrac{1}{y}\) is either \(+1\) or \(-1\).
Solution: \begin{align*} && x-y &= \frac1z - \frac1y \\ && x-z &= \frac1x - \frac1y \\ && y-z &= \frac1x - \frac1z \\ \Rightarrow && (x-y)(x-z)(y-z) &= \frac{(y-z)(y-x)(z-x)}{x^2y^2z^2} \\ \Rightarrow && x^2y^2 z^2 &= 1 \\ \end{align*} Suppose \(x + \frac1{y} =k \Rightarrow xy + 1 = ky\) Therefore \(y + \frac{1}{z} = y \pm xy = k\) Therefore \(1 \mp y = k(y \mp 1) \Rightarrow k = \pm 1\)
1991 Paper 3 Q4
The point \(P\) moves on a straight line in three-dimensional space. The position of \(P\) is observed from the points \(O_{1}(0,0,0)\) and \(O_{2}(8a,0,0).\) At times \(t=t_{1}\) and \(t=t_{1}'\), the lines of sight from \(O_{1}\) are along the lines \[ \frac{x}{2}=\frac{z}{3},y=0\quad\mbox{ and }\quad x=0,\frac{y}{3}=\frac{z}{4} \] respectively. At times \(t=t_{2}\) and \(t=t_{2}'\), the lines of sight from \(O_{2}\) are \[ \frac{x-8a}{-3}=\frac{y}{1}=\frac{z}{3}\quad\mbox{ and }\quad\frac{x-8a}{-4}=\frac{y}{2}=\frac{z}{5} \] respectively. Find an equation or equations for the path of \(P\).
1989 Paper 1 Q9
Sketch the graph of \(8y=x^{3}-12x\) for \(-4\leqslant x\leqslant4\), marking the coordinates of the turning points. Similarly marking the turning points, sketch the corresponding graphs in the \((X,Y)\)-plane, if \begin{alignat*}{3} \rm{(a)} & \quad & & X=\tfrac{1}{2}x, & \qquad & Y=y,\\ \rm{(b)} & & & X=x, & & Y=\tfrac{1}{2}y,\\ \rm{(c)} & & & X=\tfrac{1}{2}x+1, & & Y=y,\\ \rm{(d)} & & & X=x, & & Y=\tfrac{1}{2}y+1. \end{alignat*} Find values for \(a,b,c,d\) such that, if \(X=ax+b,\) \(Y=cy+d\), then the graph in the \((X,Y)\)-plane corresponding to \(8y=x^{3}-12x\) has turning points at \((X,Y)=(0,0)\) and \((X,Y)=(1,1)\).
Solution: \(8\frac{\d y}{\d x} = 3(x^2-4)\) so the turning points are at \((\pm 2, \mp 2)\)
1989 Paper 2 Q10
State carefully the conditions which the fixed vectors \(\mathbf{a,b,u}\) and \(\mathbf{v}\) must satisfy in order to ensure that the line \(\mathbf{r=a+}\lambda\mathbf{u}\) intersects the line \(\mathbf{r=b+\mu}\mathbf{v}\) in exactly one point. Find the two values of the fixed scalar \(b\) for which the planes with equations \[ \left.\begin{array}{c} x+y+bz=b+2\\ bx+by+z=2b+1 \end{array}\right\} \tag{*} \] do not intersect in a line. For other values of \(b\), express the line of intersection of the two planes in the form \(\mathbf{r=a}+\lambda\mathbf{u},\) where \(\mathbf{a\cdot u}=0\). Find the conditions which \(b\) and the fixed scalars \(c\) and \(d\) must satisfy to ensure that there is exactly one point on the line \[ \mathbf{r=}\left(\begin{array}{c} 0\\ 0\\ c \end{array}\right)+\mu\left(\begin{array}{c} 1\\ d\\ 0 \end{array}\right) \] whose coordinates satisfy both equations \((*)\).
Solution: There are two requirements (assuming they are lines not fixed points): 1. They cannot be parallel, ie \(\mathbf{u} \neq \lambda \mathbf{v}\) for any \(\lambda\) 2. They must lie in the same plane, ie \((\mathbf{b}-\mathbf{a})\cdot (\mathbf{u} \times \mathbf{v}) = 0\) The planes will not intersect in a line if they are either parallel and separate or parallel and the same. If \(b = 1\) or \(b=-1\) the planes are parallel. \begin{align*} && (x+y) + b z &= b+ 2\\ &&b(x+y) + z &= 2b + 1 \\ \Rightarrow && (1-b^2)z &= 2b+1 - b^2 -2b \\ &&&= 1-b^2 \\ \Rightarrow && z &= 1 \\ && x+ y &= 2 \\ \end{align*} Therefore our line is \(\mathbf{r} = \begin{pmatrix} 1+t \\ 1-t \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \) We must have: \(d \neq -1\) to ensure that the lines aren't parallel. We must also have: \begin{align*} 0 &= \left ( \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} -\begin{pmatrix} 0 \\ 0 \\ c \end{pmatrix}\right) \cdot \left ( \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ d \\ 0 \end{pmatrix} \right) \\ &= \begin{pmatrix} 1 \\ 1 \\ 1-c \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ d+1 \end{pmatrix} \\ &= (1-c)(d+1) \end{align*} So \(c =1\)
1988 Paper 2 Q2
The numbers \(x,y\) and \(z\) are non-zero, and satisfy \[ 2a-3y=\frac{\left(z-x\right)^{2}}{y}\quad\mbox{ and }\quad2a-3z=\frac{\left(x-y\right)^{2}}{z}, \] for some number \(a\). If \(y\neq z\), prove that \[ x+y+z=a, \] and that \[ 2a-3x=\frac{\left(y-z\right)^{2}}{x}. \] Determine whether this last equation holds only if \(y\neq z\).
Solution: \begin{align*} && \begin{cases} 2a-3y=\frac{\left(z-x\right)^{2}}{y} \\ 2a-3z=\frac{\left(x-y\right)^{2}}{z} \end{cases} \\ \Rightarrow && \begin{cases} 2ay-3y^2=\left(z-x\right)^{2} \\ 2az-3z^2=\left(x-y\right)^{2} \end{cases} \\ \Rightarrow && 2a(y-z)-3(y+z)(y-z) &= (z-x+x-y)(z-x-x+y) \\ \Rightarrow && (y-z)(2a-3y-3z) &= (z-y)(z-2x+y) \\ \Rightarrow && 2a-3y-3z &= 2x-y-z \tag{\(y \neq z\)} \\ \Rightarrow && a &= x+y+z \\ \end{align*} This is is our first result. \begin{align*} && 2a-3y-3z &= 2x-y-z \\ \Rightarrow && 2a-3y-3x &= 3z-y-x \\ \Rightarrow && (y-x)2a-3(y-x)(y+x) &= (y-x)(2z-x-y) \\ \Rightarrow && 2a(y-x)-3(y^2-x^2) &= (z-y)^2-(x-z)^2 \\ \Rightarrow && 2ax - 3x^2 &= (y-z)^2 \\ \Rightarrow && 2a - 3x &= \frac{(y-z)^2}{x} \end{align*} Suppose \(x = \frac23 a, y = z = \frac16 a\) then all equations are satisfied, but \(y = z\).