Problems

Filters
Clear Filters

1531 problems found

2020 Paper 3 Q4
D: 1500.0 B: 1500.0

The plane \(\Pi\) has equation \(\mathbf{r} \cdot \mathbf{n} = 0\) where \(\mathbf{n}\) is a unit vector. Let \(P\) be a point with position vector \(\mathbf{x}\) which does not lie on the plane \(\Pi\). Show that the point \(Q\) with position vector \(\mathbf{x} - (\mathbf{x} \cdot \mathbf{n})\mathbf{n}\) lies on \(\Pi\) and that \(PQ\) is perpendicular to \(\Pi\).

  1. Let transformation \(T\) be a reflection in the plane \(ax+by+cz=0\), where \(a^2+b^2+c^2=1\). Show that the image of \(\mathbf{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\) under \(T\) is \(\begin{pmatrix} b^2+c^2-a^2 \\ -2ab \\ -2ac \end{pmatrix}\), and find the images of \(\mathbf{j}\) and \(\mathbf{k}\) under \(T\). Write down the matrix \(\mathbf{M}\) which represents transformation \(T\).
  2. The matrix \[ \begin{pmatrix} 0.64 & 0.48 & 0.6 \\ 0.48 & 0.36 & -0.8 \\ 0.6 & -0.8 & 0 \end{pmatrix} \] represents a reflection in a plane. Find the cartesian equation of the plane.
  3. The matrix \(\mathbf{N}\) represents a rotation through angle \(\pi\) about the line through the origin parallel to \(\begin{pmatrix} a \\ b \\ c \end{pmatrix}\), where \(a^2+b^2+c^2=1\). Find the matrix \(\mathbf{N}\).
  4. Identify the single transformation which is represented by the matrix \(\mathbf{NM}\).

2020 Paper 3 Q5
D: 1500.0 B: 1500.0

Show that for positive integer \(n\), \(x^n - y^n = (x-y)\displaystyle\sum_{r=1}^{n} x^{n-r} y^{r-1}\).

  1. Let \(\mathrm{F}\) be defined by \[ \mathrm{F}(x) = \frac{1}{x^n(x-k)} \quad \text{for } x \neq 0,\, k \] where \(n\) is a positive integer and \(k \neq 0\).
    1. Given that \[ \mathrm{F}(x) = \frac{A}{x-k} + \frac{\mathrm{f}(x)}{x^n}, \] where \(A\) is a constant and \(\mathrm{f}(x)\) is a polynomial, show that \[ \mathrm{f}(x) = \frac{1}{x-k}\left(1 - \left(\frac{x}{k}\right)^n\right). \] Deduce that \[ \mathrm{F}(x) = \frac{1}{k^n(x-k)} - \frac{1}{k}\sum_{r=1}^{n} \frac{1}{k^{n-r}x^r}. \]
    2. By differentiating \(x^n \mathrm{F}(x)\), prove that \[ \frac{1}{x^n(x-k)^2} = \frac{1}{k^n(x-k)^2} - \frac{n}{xk^n(x-k)} + \sum_{r=1}^{n} \frac{n-r}{k^{n+1-r}x^{r+1}}. \]
  2. Hence evaluate the limit of \[ \int_2^N \frac{1}{x^3(x-1)^2} \; \mathrm{d}x \] as \(N \to \infty\), justifying your answer.

2020 Paper 3 Q6
D: 1500.0 B: 1500.0

  1. Sketch the curve \(y = \cos x + \sqrt{\cos 2x}\) for \(-\frac{1}{4}\pi \leqslant x \leqslant \frac{1}{4}\pi\).
  2. The equation of curve \(C_1\) in polar co-ordinates is \[ r = \cos\theta + \sqrt{\cos 2\theta} \qquad -\tfrac{1}{4}\pi \leqslant \theta \leqslant \tfrac{1}{4}\pi. \] Sketch the curve \(C_1\).
  3. The equation of curve \(C_2\) in polar co-ordinates is \[ r^2 - 2r\cos\theta + \sin^2\theta = 0 \qquad -\tfrac{1}{4}\pi \leqslant \theta \leqslant \tfrac{1}{4}\pi. \] Find the value of \(r\) when \(\theta = \pm\frac{1}{4}\pi\). Show that, when \(r\) is small, \(r \approx \frac{1}{2}\theta^2\). Sketch the curve \(C_2\), indicating clearly the behaviour of the curve near \(r=0\) and near \(\theta = \pm\frac{1}{4}\pi\). Show that the area enclosed by curve \(C_2\) and above the line \(\theta = 0\) is \(\dfrac{\pi}{2\sqrt{2}}\).

2020 Paper 3 Q7
D: 1500.0 B: 1500.0

  1. Given that the variables \(x\), \(y\) and \(u\) are connected by the differential equations \[ \frac{\mathrm{d}u}{\mathrm{d}x} + \mathrm{f}(x)u = \mathrm{h}(x) \quad \text{and} \quad \frac{\mathrm{d}y}{\mathrm{d}x} + \mathrm{g}(x)y = u, \] show that \[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (\mathrm{g}(x) + \mathrm{f}(x))\frac{\mathrm{d}y}{\mathrm{d}x} + (\mathrm{g}'(x) + \mathrm{f}(x)\mathrm{g}(x))y = \mathrm{h}(x). \tag{1} \]
  2. Given that the differential equation \[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(1 + \frac{4}{x}\right)\frac{\mathrm{d}y}{\mathrm{d}x} + \left(\frac{2}{x} + \frac{2}{x^2}\right)y = 4x + 12 \tag{2} \] can be written in the same form as (1), find a first order differential equation which is satisfied by \(\mathrm{g}(x)\). If \(\mathrm{g}(x) = kx^n\), find a possible value of \(n\) and the corresponding value of \(k\). Hence find a solution of (2) with \(y = 5\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -3\) at \(x = 1\).

2020 Paper 3 Q8
D: 1500.0 B: 1500.0

A sequence \(u_k\), for integer \(k \geqslant 1\), is defined as follows. \[ u_1 = 1 \] \[ u_{2k} = u_k \text{ for } k \geqslant 1 \] \[ u_{2k+1} = u_k + u_{k+1} \text{ for } k \geqslant 1 \]

  1. Show that, for every pair of consecutive terms of this sequence, except the first pair, the term with odd subscript is larger than the term with even subscript.
  2. Suppose that two consecutive terms in this sequence have a common factor greater than one. Show that there are then two consecutive terms earlier in the sequence which have the same common factor. Deduce that any two consecutive terms in this sequence are co-prime (do not have a common factor greater than one).
  3. Prove that it is not possible for two positive integers to appear consecutively in the same order in two different places in the sequence.
  4. Suppose that \(a\) and \(b\) are two co-prime positive integers which do not occur consecutively in the sequence with \(b\) following \(a\). If \(a > b\), show that \(a-b\) and \(b\) are two co-prime positive integers which do not occur consecutively in the sequence with \(b\) following \(a-b\), and whose sum is smaller than \(a+b\). Find a similar result for \(a < b\).
  5. For each integer \(n \geqslant 1\), define the function \(\mathrm{f}\) from the positive integers to the positive rational numbers by \(\mathrm{f}(n) = \dfrac{u_n}{u_{n+1}}\). Show that the range of \(\mathrm{f}\) is all the positive rational numbers, and that \(\mathrm{f}\) has an inverse.

2020 Paper 3 Q9
D: 1500.0 B: 1500.0

Two inclined planes \(\Pi_1\) and \(\Pi_2\) meet in a horizontal line at the lowest points of both planes and lie on either side of this line. \(\Pi_1\) and \(\Pi_2\) make angles of \(\alpha\) and \(\beta\), respectively, to the horizontal, where \(0 < \beta < \alpha < \frac{1}{2}\pi\). A uniform rigid rod \(PQ\) of mass \(m\) rests with \(P\) lying on \(\Pi_1\) and \(Q\) lying on \(\Pi_2\) so that the rod lies in a vertical plane perpendicular to \(\Pi_1\) and \(\Pi_2\) with \(P\) higher than \(Q\).

  1. It is given that both planes are smooth and that the rod makes an angle \(\theta\) with the horizontal. Show that \(2\tan\theta = \cot\beta - \cot\alpha\).
  2. It is given instead that \(\Pi_1\) is smooth, that \(\Pi_2\) is rough with coefficient of friction \(\mu\) and that the rod makes an angle \(\phi\) with the horizontal. Given that the rod is in limiting equilibrium, with \(P\) about to slip down the plane \(\Pi_1\), show that \[ \tan\theta - \tan\phi = \frac{\mu}{(\mu + \tan\beta)\sin 2\beta} \] where \(\theta\) is the angle satisfying \(2\tan\theta = \cot\beta - \cot\alpha\).

2020 Paper 3 Q10
D: 1500.0 B: 1500.0

A light elastic spring \(AB\), of natural length \(a\) and modulus of elasticity \(kmg\), hangs vertically with one end \(A\) attached to a fixed point. A particle of mass \(m\) is attached to the other end \(B\). The particle is held at rest so that \(AB > a\) and is released. Find the equation of motion of the particle and deduce that the particle oscillates vertically. If the period of oscillation is \(\dfrac{2\pi}{\Omega}\), show that \(kg = a\Omega^2\). Suppose instead that the particle, still attached to \(B\), lies on a horizontal platform which performs simple harmonic motion vertically with amplitude \(b\) and period \(\dfrac{2\pi}{\omega}\). At the lowest point of its oscillation, the platform is a distance \(h\) below \(A\). Let \(x\) be the distance of the particle above the lowest point of the oscillation of the platform. When the particle is in contact with the platform, show that the upward force on the particle from the platform is \[ mg + m\Omega^2(a + x - h) + m\omega^2(b - x). \] Given that \(\omega < \Omega\), show that, if the particle remains in contact with the platform throughout its motion, \[ h \leqslant a\left(1 + \frac{1}{k}\right) + \frac{\omega^2 b}{\Omega^2}. \] Find the corresponding inequality if \(\omega > \Omega\). Hence show that, if the particle remains in contact with the platform throughout its motion, it is necessary that \[ h \leqslant a\left(1 + \frac{1}{k}\right) + b, \] whatever the value of \(\omega\).

2020 Paper 3 Q11
D: 1500.0 B: 1500.0

The continuous random variable \(X\) is uniformly distributed on \([a,b]\) where \(0 < a < b\).

  1. Let \(\mathrm{f}\) be a function defined for all \(x \in [a,b]\)
    • with \(\mathrm{f}(a) = b\) and \(\mathrm{f}(b) = a\),
    • which is strictly decreasing on \([a,b]\),
    • for which \(\mathrm{f}(x) = \mathrm{f}^{-1}(x)\) for all \(x \in [a,b]\).
    The random variable \(Y\) is defined by \(Y = \mathrm{f}(X)\). Show that \[ \mathrm{P}(Y \leqslant y) = \frac{b - \mathrm{f}(y)}{b - a} \quad \text{for } y \in [a,b]. \] Find the probability density function for \(Y\) and hence show that \[ \mathrm{E}(Y^2) = -ab + \int_a^b \frac{2x\,\mathrm{f}(x)}{b-a} \; \mathrm{d}x. \]
  2. The random variable \(Z\) is defined by \(\dfrac{1}{Z} + \dfrac{1}{X} = \dfrac{1}{c}\) where \(\dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{b}\). By finding the variance of \(Z\), show that \[ \ln\left(\frac{b-c}{a-c}\right) < \frac{b-a}{c}. \]

2020 Paper 3 Q12
D: 1500.0 B: 1500.0

\(A\) and \(B\) both toss the same biased coin. The probability that the coin shows heads is \(p\), where \(0 < p < 1\), and the probability that it shows tails is \(q = 1 - p\). Let \(X\) be the number of times \(A\) tosses the coin until it shows heads. Let \(Y\) be the number of times \(B\) tosses the coin until it shows heads.

  1. The random variable \(S\) is defined by \(S = X + Y\) and the random variable \(T\) is the maximum of \(X\) and \(Y\). Find an expression for \(\mathrm{P}(S = s)\) and show that \[ \mathrm{P}(T = t) = pq^{t-1}(2 - q^{t-1} - q^t). \]
  2. The random variable \(U\) is defined by \(U = |X - Y|\), and the random variable \(W\) is the minimum of \(X\) and \(Y\). Find expressions for \(\mathrm{P}(U = u)\) and \(\mathrm{P}(W = w)\).
  3. Show that \(\mathrm{P}(S = 2 \text{ and } T = 3) \neq \mathrm{P}(S = 2) \times \mathrm{P}(T = 3)\).
  4. Show that \(U\) and \(W\) are independent, and show that no other pair of the four variables \(S\), \(T\), \(U\) and \(W\) are independent.

2019 Paper 1 Q1
D: 1500.0 B: 1500.0

A straight line passes through the fixed point \((1 , k)\) and has gradient \(- \tan \theta\), where \(k > 0\) and \(0 < \theta < \frac{1}{2}\pi\). Find, in terms of \(\theta\) and \(k\), the coordinates of the points \(X\) and \(Y\) where the line meets the \(x\)-axis and the \(y\)-axis respectively.

  1. Find an expression for the area \(A\) of triangle \(OXY\) in terms of \(k\) and \(\theta\). (The point \(O\) is the origin.) You are given that, as \(\theta\) varies, \(A\) has a minimum value. Find an expression in terms of \(k\) for this minimum value.
  2. Show that the length \(L\) of the perimeter of triangle \(OXY\) is given by $$L = 1 + \tan \theta + \sec \theta + k(1 + \cot \theta + \cosec \theta).$$ You are given that, as \(\theta\) varies, \(L\) has a minimum value. Show that this minimum value occurs when \(\theta = \alpha\) where $$\frac{1 - \cos \alpha}{1 - \sin \alpha} = k.$$ Find and simplify an expression for the minimum value of \(L\) in terms of \(\alpha\).


Solution: \(y = (-\tan \theta)(x-1)+k\) so when \(x = 0\), \(y = k + \tan \theta\), so \(Y = (0, k+\tan \theta)\). When \(y = 0\), \(x = 1 + \frac{k}{\tan \theta}\)

  1. \(A = \frac12 (k+\tan \theta)\left ( 1 + \frac{k}{\tan \theta} \right) = k + \frac12 \left (\tan \theta + \frac{k^2}{\tan \theta} \right)\) Notice that \(x + \frac{k^2}{x} \geq 2 k\) by AM-GM, so the minimum is \(k + \frac12 \cdot 2k = 2k\)
  2. \(\,\) \begin{align*} L &= k + \tan \theta + 1 + k \cot \theta + \sqrt{(k + \tan \theta)^2 + \left (1 + \frac{k}{\tan \theta} \right)^2} \\ &= k + \tan \theta + 1 + k \cot \theta + \sqrt{k^2 + 2 k \tan \theta +\tan^2 \theta + 1 + 2k \cot \theta + k^2\cot^2 \theta} \\ &= k + \tan \theta + 1 + k \cot \theta + \sqrt{\sec^2 \theta+ 2k \sec\theta\cosec \theta + k^2\cosec^2 \theta} \\ &= k + \tan \theta + 1 + k \cot \theta +\sec \theta + k\cosec \theta\\ &= 1 + \tan \theta + \sec \theta + k (1 + \cot \theta + \cosec \theta) \end{align*} \begin{align*} && \frac{\d L}{\d \theta} &= \sec^2 \theta + \tan \theta \sec \theta + k(-\cosec^2 \theta - \cot \theta \cosec \theta ) \\ \Rightarrow && 0 &=\sec^2 \alpha+ \tan \theta \sec \alpha+ k(-\cosec^2 \alpha- \cot \alpha\cosec \alpha) \\ \Rightarrow && k &= \frac{\sec^2 \alpha+ \tan \alpha\sec \alpha}{\cosec^2 \alpha+ \cot \alpha\cosec \alpha} \\ &&&= \frac{\sin^2 \alpha(1 + \sin \alpha)}{\cos^2 \alpha (1+ \cos \alpha)} \\ &&&= \frac{(1-\cos^2 \alpha)(1 + \sin \alpha)}{(1-\sin^2 \alpha )(1+ \cos \alpha)} \\ &&&= \frac{1-\cos \alpha}{1-\sin \alpha} \\ \Rightarrow && L &= 1 + \tan \alpha + \sec \alpha + \frac{1-\cos \alpha}{1-\sin \alpha} \left (1 + \cot \alpha + \cosec \alpha \right) \\ &&&= \frac{1+\tan \alpha + \sec \alpha -\sin \alpha-\sin \alpha \tan \alpha-\tan \alpha}{1-\sin \alpha} + \\ &&&\quad \quad \frac{1+\cot \alpha + \cosec \alpha-\cos \alpha-\cos \alpha \cot \alpha -\cot \alpha}{1-\sin \alpha} \\ &&&= \frac{2+\sec \alpha(1-\sin^2 \alpha)-\sin \alpha + \cosec \alpha(1-\cos^2 \alpha)-\cos \alpha}{1-\sin \alpha} \\ &&&= \frac{2+\cos\alpha-\sin \alpha + \sin\alpha-\cos \alpha}{1-\sin \alpha} \\ &&&= \frac{2}{1-\sin \alpha} \end{align*}

2019 Paper 1 Q2
D: 1500.0 B: 1500.0

The curve \(C\) is given parametrically by the equations \(x = 3t^2\), \(y = 2t^3\). Show that the equation of the tangent to \(C\) at the point \((3p^2 , 2p^3)\) is \(y = px - p^3\). Find the point of intersection of the tangents to \(C\) at the distinct points \((3p^2 , 2p^3)\) and \((3q^2 , 2q^3)\). Hence show that, if these two tangents are perpendicular, their point of intersection is \((u^2 + 1 , -u)\), where \(u = p + q\). The curve \(\tilde{C}\) is given parametrically by the equations \(x = u^2 + 1\), \(y = -u\). Find the coordinates of the points that lie on both \(C\) and \(\tilde{C}\). Sketch \(C\) and \(\tilde{C}\) on the same axes.


Solution: \begin{align*} && \frac{\d y}{\d x} &= \frac{\frac{\d y}{\d t}}{\frac{\d x}{\d t}} \\ &&&= \frac{6t^2}{6t} = t \\ \Rightarrow && \frac{y-2p^3}{x - 3p^2} &= p \\ \Rightarrow && y &= px-3p^3+2p^3 \\ && y &= px - p^3 \end{align*} The two lines will be \begin{align*} && y &= px - p^3 \\ && y &= qx - q^3 \\ \Rightarrow && p^3-q^3 &= (p-q)x \\ \Rightarrow && x &= p^2+pq+q^2 \\ && y &= p(p^2+pq+q^2)-p^3 \\ &&&= pq(p+q) \\ && (x,y) &= (p^2+pq+q^2,pq(p+q)) \\ \end{align*} If the tangents are \(\perp\) then \(pq=-1\), so we have \begin{align*} && (x,y) &= (p^2+2pq+q^2-pq, pq(p+q)) \\ &&&= ((p+q)^2-1, -(p+q)) \\ &&&= (u^2-1, -u) \end{align*} We have \(x = y^2+1\) and \(\left ( \frac{x}{3} \right)^3 = \left ( \frac{y}{2}\right)^2 \Rightarrow y^2 = \frac{4}{27}x^3\) so \begin{align*} && 0 &= \frac{4}{27}x^3-x+1 \\ &&0&=4x^3-27x+27 \\ &&&= (x+3)(2x-3)^2 \end{align*} So we have the points \((x,y) = \left (\frac32, \pm\frac{1}{\sqrt{2}}\right)\)

TikZ diagram

2019 Paper 1 Q3
D: 1500.0 B: 1500.0

By first multiplying the numerator and the denominator of the integrand by \((1 - \sin x)\), evaluate $$\int_0^{\frac{1}{4}\pi} \frac{1}{1 + \sin x} dx.$$ Evaluate also: $$\int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \frac{1}{1 + \sec x} dx \quad \text{and} \quad \int_0^{\frac{1}{3}\pi} \frac{1}{(1 + \sin x)^2} dx.$$


Solution: \begin{align*} \int_0^{\frac{1}{4}\pi} \frac{1}{1 + \sin x} dx &= \int_0^{\frac{1}{4}\pi} \frac{1-\sin x}{1 - \sin^2 x} dx \\ &= \int_0^{\frac{1}{4}\pi} \frac{1-\sin x}{\cos^2 x} dx \\ &= \int_0^{\frac{1}{4}\pi} \sec^2 x - \sec x \tan x dx \\ &= \left [\tan x-\sec x \right]_0^{\frac{1}{4}\pi} \\ &= 2 - \frac{1}{\sqrt{2}} \end{align*} \begin{align*} \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \frac{1}{1 + \sec x} \d x &= \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \frac{1-\sec x}{1 - \sec^2 x} \d x \\ &= \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \frac{\sec x-1}{\tan^2 x} \d x \\ &= \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \cot x \cosec x-\cot^2 x\d x \\ &= \left [ -\cosec x +x+\cot x\right]_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \\ &= \l -\frac{2}{\sqrt3}+\frac{\pi}{3}+\frac{1}{\sqrt{3}}\r - \l-\sqrt{2}+\frac{\pi}{4}+1 \r \\ &= \frac{\pi}{12}-\frac{1}{\sqrt{3}}+\sqrt{2}-1 \end{align*} \begin{align*} \int_0^{\frac{1}{3}\pi} \frac{1}{(1 + \sin x)^2} \d x &= \int_0^{\frac{1}{3}\pi} \frac{1-2\sin x+\sin^2x}{(1-\sin^2 x)^2} \d x \\ &= \int_0^{\frac{1}{3}\pi} \frac{1-2\sin x+\sin^2x}{\cos^4 x} \d x \\ \end{align*} Splitting this up into: \begin{align*} \int_0^{\frac{1}{3}\pi} \frac{-2\sin x}{\cos^4 x} \d x &= -\frac23 \left [ \frac{1}{\cos^3 x}\right]_0^{\frac{1}{3}\pi} \\ &= -\frac{16}3+\frac23 \\ &= -\frac{14}3 \end{align*} and \begin{align*} && \int_0^{\frac{1}{3}\pi} \frac{1+\sin^2x}{\cos^4 x} \d x &= \int_0^{\frac{1}{3}\pi} (\sec^2 x + \tan^2 x) \sec^2 x \d x \\ &&&= \int_0^{\frac{1}{3}\pi} (1+ 2\tan^2 x) \sec^2 x \d x \\ u = \tan x, \d u = \sec^2 x \d x&&&= \int_0^{\sqrt{3}}(1+2u^2) \d u \\ &&&= \left [u + \frac23 u^3 \right]_0^{\sqrt{3}} \\ &&&= \sqrt{3} + 2\sqrt{3} \\ &&&= 3\sqrt{3} \end{align*} And so our complete integral is: \[ \int_0^{\frac{1}{3}\pi} \frac{1}{(1 + \sin x)^2} \d x = 3\sqrt{3} - \frac{14}3\]

2019 Paper 1 Q4
D: 1500.0 B: 1500.0

  1. Find integers \(m\) and \(n\) such that $$\sqrt{3+2\sqrt{2}} = m + n\sqrt{2}.$$
  2. Let \(f(x) = x^4 - 10x^2 + 12x - 2\). Given that the equation \(f(x) = 0\) has four real roots, explain why \(f(x)\) can be written in the form $$f(x)=(x^2 + sx + p)(x^2 - sx + q)$$ for some real constants \(s\), \(p\) and \(q\), and find three equations for \(s\), \(p\) and \(q\). Show that $$s^2(s^2 - 10)^2 + 8s^2 - 144 = 0$$ and find the three possible values of \(s^2\). Use the smallest of these values of \(s^2\) to solve completely the equation \(f(x) = 0\), simplifying your answers as far as you can.


Solution:

  1. \((1+\sqrt{2})^2 = 3 + 2\sqrt{2}\) so \(\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}\)
  2. We can always factorise any quartic in the form \((x^2+ax+b)(x^2+cx+d)\), since \(x^3\) has a coefficient of \(a+b\) we must have \(a = -b\), ie the form in the question. \begin{align*} && 0 &= (x^2+sx+p)(x^2-sx+q) \\ &&&= x^4+(p+q-s^2)x^2+s(q-p)x+pq \\ \Rightarrow && pq &= -2 \\ && s(q-p) &= 12 \\ && p+q-s^2 &= -10 \\ \\ && p+q &= s^2-10 \\ && (p+q)^2 &= (s^2-10)^2 \\ && (q-p)^2 &= \frac{12}{s^2} \\ \Rightarrow && (s^2-10)^2 &= \frac{12}{s^2} + 4pq \\ && (s^2-10)^2 &= \frac{144}{s^2} -8 \\ && 0 &= s^2(s^2-10)^2+8s^2-144 \\ &&&= s^6-20s^4+108s^2-144 \\ &&&= (s^2-2)(s^2-6)(s^2-12) \end{align*} Suppose \(s = \sqrt{2}\), and we have \begin{align*} && q-p &= 6\sqrt{2} \\ && p+q &= -8 \\ \Rightarrow && q &= 3\sqrt{2}-4 \\ && p &= -4-3\sqrt{2} \end{align*} Solving our quadratic equations, we have \begin{align*} && 0 &= x^2-\sqrt{2}x-4+3\sqrt{2} \\ \Rightarrow && x &= \frac{\sqrt{2}\pm \sqrt{2-4\cdot(-4+3\sqrt{2})}}{2} \\ &&&= \frac{\sqrt{2}\pm \sqrt{18-12\sqrt{2}}}{2} \\ &&&= \frac{\sqrt{2}\pm (2\sqrt{3}-\sqrt{6})}{2} \\ \\ && 0 &= x^2+\sqrt{2}x-3\sqrt{2}-4 \\ && x &= \frac{-\sqrt{2} \pm \sqrt{2-4\cdot(3\sqrt{3}-4)}}{2}\\ && &= \frac{-\sqrt{2} \pm \sqrt{18+12\sqrt{2}}}{2}\\ && &= \frac{-\sqrt{2} \pm (\sqrt{6}+2\sqrt{3})}{2}\\ \end{align*}

2019 Paper 1 Q5
D: 1500.0 B: 1500.0

  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 \(\frac{21}{20}\).


Solution:

  1. If \(\vec{PQ} = \vec{SR}\) we have a parallelogram. \(\vec{PQ} = \vec{SR}\) and \(|\vec{PQ}| = |\vec{PS}|\) then we have a rhombus.
  2. If the four points lie in a plane then \((\vec{RS} \times \vec{RP}) \cdot \vec{RQ} =0\), so \begin{align*} && 0 &=\left ( \begin{pmatrix}-r\\ s-1 \\ 0 \end{pmatrix} \times \begin{pmatrix}p-r\\ -1 \\ -1 \end{pmatrix}\right) \cdot \begin{pmatrix}1-r\\ q-1 \\ -1 \end{pmatrix} \\ && &= \begin{pmatrix}1-s \\ -r \\r+(p-r)(1-s) \end{pmatrix} \cdot \begin{pmatrix}1-r\\ q-1 \\ -1 \end{pmatrix} \\ &&&= (1-s)(1-r)-r(q-1)-r-(p-r)(1-s) \\ &&&=(1-s)(1-r-p+r)-rq \\ \Rightarrow && rq &= (1-s)(1-p) \end{align*}
    1. \(\,\) \begin{align*} && \vec{PQ} &= \vec{SR} \\ \Leftrightarrow && \begin{pmatrix}1-p\\q \\ 0 \end{pmatrix} &= \begin{pmatrix}r\\1-s \\ 0 \end{pmatrix} \\ \Leftrightarrow && 1-p = r & \quad ; \quad q = 1-s\\ \Leftrightarrow && 1= r+p & \quad ; \quad 1 = q+s\\ \end{align*} The centroid is \(\frac14 (p+1+r, q+s+1, 2)\) which is clearly \(\frac12(1,1,1)\) iff those equations are true.
    2. \(\,\) \begin{align*} && |\vec{PQ}| &= |\vec{PS}| \\ \Leftrightarrow && (1-p)^2+q^2+ 0^2 &= p^2+s^2+1)\\ \Leftrightarrow && 1-2p+p^2+q^2 &= p^2 + s^2 + 1 \\ \Leftrightarrow && -2p+q^2 &= s^2 \end{align*} From the previous equations we have \(r = 1-p\), and \(-2p+(1-s)^2 = s^2 \Rightarrow -2p + 1 -2s = 0 \Rightarrow s = \frac12 - p\) and \(q = \frac12 + p\) \begin{align*} && \cos PQR &= \frac{\vec{QP}\cdot \vec{QR}}{|\vec{QP}||\vec{QR}|} \\ &&&= \frac{ \begin{pmatrix}p-1\\ -q \\ 0 \end{pmatrix} \cdot \begin{pmatrix}r-1\\ 1-q \\ 1 \end{pmatrix}}{\sqrt{(p-1)^2+q^2}\sqrt{(r-1)^2+(1-q)^2+1^2}} \\ &&&= \frac{ \begin{pmatrix}p-1\\ -\frac12-p \\ 0 \end{pmatrix} \cdot \begin{pmatrix}-p\\ \frac12-p \\ 1 \end{pmatrix}}{\sqrt{(p-1)^2+(-\frac12-p)^2}\sqrt{p^2+(\frac12-p)^2+1^2}} \\ &&&= \frac{ p-p^2-\frac14+p^2}{\sqrt{p^2-2p+1+\frac14+p+p^2}\sqrt{p^2+\frac14-p+p^2+1}} \\ &&&= \frac{4p-1}{\sqrt{8p^2-4p+5}\sqrt{8p^2-4p+5}}\\ &&&= \frac{4p-1}{8p^2-4p+5}\\ \end{align*} For \(PQRS\) to be a square \(\cos PQR = 0\), ie \(p = \frac14\) and so \((p,q,r,s) = (\frac14, \frac34, \frac34, \frac14)\) and \(|PQ| = \sqrt{(1-p)^2+q^2} = \sqrt{\left ( \frac34 \right)^2 + \left ( \frac34 \right)^2 } = \frac{3\sqrt{2}}4\), notice that \(\left ( \frac{21}{20} \right)^2 = \frac{441}{400} < \frac{9}{8}\) (\(441 < 450\)) therefore the sides are at least as long as \(\frac{21}{20}\)

2019 Paper 1 Q6
D: 1500.0 B: 1518.2

In both parts of this question, \(x\) is real and \(0 < \theta < \pi\).

  1. By completing the square, find in terms of \(\theta\) the minimum value as \(x\) varies of $$9x^2 - 12x \cos \theta + 4.$$ Find also the maximum value as \(x\) varies of \(12x^2 \sin \theta - 9x^4\). Hence determine the values of \(x\) and \(\theta\) that satisfy the equation $$9x^4 + (9 - 12 \sin \theta)x^2 - 12x \cos \theta + 4 = 0.$$
  2. Sketch the curve $$y = \frac{x^2}{x - \theta},$$ where \(\theta\) is a constant. Deduce that either \(\frac{x^2}{x - \theta} \leq 0\) or \(\frac{x^2}{x - \theta} \geq 4\theta\). By considering the numerator and denominator separately, or otherwise, show that $$\frac{\sin^2 \theta \cos^2 x}{1 + \cos^2 \theta \sin^2 x} \leq 1.$$ Hence determine the values of \(x\) and \(\theta\) that satisfy the equation $$\frac{x^2}{4\theta(x - \theta)} = \frac{\sin^2 \theta \cos^2 x}{1 + \cos^2 \theta \sin^2 x}.$$


Solution:

  1. \(\,\) \begin{align*} && y &= 9x^2 - 12x \cos \theta + 4 \\ &&&= (3x-2\cos \theta)^2+4-4\cos^2 \theta \\ &&&= (3x-2\cos \theta)^2 + 4 \sin^2 \theta \end{align*} Therefore the minimum is \(4\sin^2 \theta\) when \(x = \frac23 \cos \theta\). \begin{align*} && y &= 12x^2 \sin \theta - 9x^4 \\ &&&=4\sin^2 \theta -(3x^2-2\sin\theta)^2 \end{align*} Therefore the maximum is \(4\sin^2 \theta\) when \(x^2 = \frac23\sin \theta\) Therefore \begin{align*} && 0 &= 9x^4 + (9 - 12 \sin \theta)x^2 - 12x \cos \theta + 4 \\ && \underbrace{-9x^4+12x^2\sin \theta}_{\leq 4\sin^2 \theta } &= \underbrace{9x^2 - 12x \cos \theta + 4 }_{\geq 4 \sin^2 \theta} \end{align*} Therefore the equality cases must be achieved in both cases, ie \(x = \frac23 \cos \theta\) and \(x^2 = \frac23 \sin \theta\) \begin{align*} && x^2 &= \frac49\cos^2 \theta \\ &&&= \frac49(1-\sin^2 \theta) \\ &&&= \frac49(1-\frac94 x^2) \\ \Rightarrow && 2x^2 &= \frac49 \\ \Rightarrow && x &= \pm \frac{\sqrt{2}}3\\ \Rightarrow && \cos \theta &=\pm \frac32 \frac{\sqrt{2}}3 \\ &&&= \pm \frac{1}{\sqrt{2}} \\ \Rightarrow && \theta &= \frac{\pi}{4}, \frac{3\pi}{4} \\ \Rightarrow && (x, \theta) &= \left (\frac{\sqrt{2}}{3}, \frac{\pi}{4} \right), \left (-\frac{\sqrt{2}}{3}, \frac{3\pi}{4} \right) \end{align*}
  2. Sketching we obtain, noticing we can find the turning point by: \begin{align*} && \frac{x^2}{x-\theta} &= \lambda \\ \Leftrightarrow && x^2 - \lambda x +\theta \lambda &= 0 \\ \Leftrightarrow && 0 &\leq \Delta = \lambda^2 -4\lambda \theta \\ \Leftrightarrow && \lambda &\geq 4 \theta, \lambda \leq 0 \end{align*}
    TikZ diagram
    Notice that \(\sin^2 \theta \cos^2 x \leq 1\) and \(1 + cos^2 \theta \sin^2 x \geq 1\) and therefore we must have the inequality desired. \begin{align*} && \underbrace{\frac{x^2}{4\theta(x - \theta)}}_{\geq 1 \text{ or } \leq 0} &= \underbrace{\frac{\sin^2 \theta \cos^2 x}{1 + \cos^2 \theta \sin^2 x}}_{\in [0,1]} \\ \text{both}=0: && x = 0 &, \sin \theta = 0 \\ \text{both}=1: && x = 2\theta &, \sin^2 \theta = 1,\cos^2 x = 1 \\ && 1 &= \cos^2 2 \theta \\ &&&= (1-2 \sin^2 \theta)^2 \\ &&&= 1 \\ \Rightarrow && (x, \theta) &= \left(\frac{\pi}{2}, \pi\right) \end{align*}