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2005 Paper 3 Q3
D: 1700.0 B: 1500.0
quadratics composition of functions quartic necessary and sufficient condition completing the square polynomial roots algebraic manipulation

Let \(\f(x)=x^2+px+q\) and \(\g(x)=x^2+rx+s\,\). Find an expression for \(\f ( \g (x))\) and hence find a necessary and sufficient condition on \(a\), \(b\) and \(c\) for it to be possible to write the quartic expression \(x^4+ax^3+bx^2+cx+d\) in the form \(\f ( \g (x))\), for some choice of values of \(p\), \(q\), \(r\) and \(s\). Show further that this condition holds if and only if it is possible to write the quartic expression \(x^4+ax^3+bx^2+cx+d\) in the form \((x^2+vx+w)^2-k\), for some choice of values of \(v\), \(w\) and \(k\). Find the roots of the quartic equation \(x^4-4x^3+10x^2-12x+4=0\,\).

Solution: \begin{align*} && f(g(x)) &= (g(x))^2 + p(g(x)) + q \\ &&&= (x^2+rx+s)^2 + p(x^2+rx+s) + q \\ &&&= x^4 + 2rx^3 + (2s+r^2+p)x^2 +(2rs+pr)x + (s^2+ps+q) \end{align*} So we need \(2r=a ,2s+r^2+p = b, r(2s+p) = c\). (We have full control over \(d\) since we can always chance \(q\) only affecting \(d\). \begin{align*} && r &= \frac{a}{2} \\ && b-r^2 & =rc \\ && b - \frac{a^2}{4} & =\frac{ac}{2} \\ \Rightarrow && 4b-a^2&= 2ac \end{align*} Clearly this condition is necessary. It is sufficient since if it is true the equations are solveable. \((x^2+vx+w)^2 = x^4 + 2vx^3 + (2vw+v^2)x^2+2vw x + w^2\). We don't care about the constant term since we can control this with \(k\), so we just need to check \(4(2vw+v^2) - (2v)^2 = 8wv\) so this does satisfy the condition. The reverse is also clear. \begin{align*} && 0 &= x^4-4x^3+10x^2-12x+4 \\ &&&= (x^2-2x+3)^2-5 \\ \Rightarrow && 0 &= x^2 - 2x+3 \pm \sqrt{5} \\ && x &= \frac{2 \pm \sqrt{4 - 4(3 \pm \sqrt{5})}}{2} \\ &&&= 1 \pm \sqrt{\mp \sqrt{5} -2} \\ &&& = 1 \pm \sqrt{\sqrt{5}-2}, 1 \pm i\sqrt{\sqrt{5}+2} \end{align*}

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2000 Paper 3 Q5
D: 1700.0 B: 1500.0
vectors matrix linear transformation 2x2 determinant cross product 2D necessary and sufficient condition geometric proof ratio

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}\( define \)\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 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 \[ \Delta\!\! \left( \bf a, \bf b \right) : \Delta\!\! \left( \bf b, \bf c \right) : \Delta\!\! \left( \bf c, \bf a \right) = \Delta\!\! \left( \bf p, \bf q \right) : \Delta\!\! \left( \bf q, \bf r \right) : \Delta\!\! \left( \bf r, \bf p \right). \]

Solution:

  1. First notice that there is a matrix taking \((1,0)\) and \((0,1)\) to \(P\) and \(Q\). Notice there is also a matrix taking \((1,0)\) and \((0,1)\) to \(A\) and \(B\). Since \(A\) and \(B\) are not parallel, this map is invertible. Then we must be able to compose this inverse with the second map to obtain a matrix \(\mathbf{M}\) satisfying our conditions.
  2. \(\,\) \begin{align*} && LHS &= \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 \\ &&&= (a_1b_2-a_2b_1) \binom{c_1}{c_2} + (c_1a_2-c_2a_1)\binom{b_1}{b_2} + (b_1c_2-b_2c_1)\binom{a_1}{a_2} \\ &&&= \binom{a_1b_2c_1-a_2b_1c_1+c_1a_2b_1-c_2a_1b_1+b_1c_2a_1-b_2c_1a_1}{a_1b_2c_2-a_2b_1c_2+c_1a_2b_2-c_2a_1b_2+b_1c_1a_2-b_2c_1a_2} \\ &&&= \binom{0}{0} \\ &&&= \mathbf{0} \end{align*} First note that the matrix taking \(P\), \(Q\) to \(A\), \(B\) is unique. (\(\Rightarrow\)) Suppose \(\mathbf{Ma} = \mathbf{p}\) and \(\mathbf{Mb} = \mathbf{q}\) and \(\mathbf{Mc} = \mathbf{r}\). Then notice that \begin{align*} && \mathbf{0} &= \mathbf{M0} \\ &&&= \mathbf{M}\left ( \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\right) \\ &&&= \Delta\!\! \left( \bf a, \bf b \right)\mathbf{M} \bf c + \Delta\!\! \left( \bf c, \bf a \right) \mathbf{M}\bf b + \Delta\!\! \left( \bf b, \bf c \right) \mathbf{M}\bf a\\ &&&= \Delta\!\! \left( \bf a, \bf b \right)\bf r + \Delta\!\! \left( \bf c, \bf a \right)\bf q + \Delta\!\! \left( \bf b, \bf c \right) \bf p\\ \end{align*} However, since \(\mathbf{p}, \mathbf{q}, \mathbf{r}\) are not parallel, then these coefficients must be a scalar multiples of \(\Delta(\mathbf{p}, \mathbf{q}), \cdots\) as required. \((\Leftarrow)\) Suppose we have this relationship, and \(\mathbf{Ma} = \mathbf{p}\) and \(\mathbf{Mb} = \mathbf{q}\), then \begin{align*} && \mathbf{0} &= \mathbf{M0} \\ &&&= \mathbf{M}\left ( \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\right) \\ &&&= \Delta\!\! \left( \bf a, \bf b \right)\mathbf{M} \bf c + \Delta\!\! \left( \bf c, \bf a \right) \mathbf{M}\bf b + \Delta\!\! \left( \bf b, \bf c \right) \mathbf{M}\bf a\\ &&&= \Delta\!\! \left( \bf a, \bf b \right)\mathbf{Mc} + \Delta\!\! \left( \bf c, \bf a \right)\bf q + \Delta\!\! \left( \bf b, \bf c \right) \bf p\\ \end{align*} Since these are scalar multiples of \(\Delta(\mathbf{p}, \mathbf{q}), \cdots\) and we write this as \begin{align*} && \mathbf{0} &= \Delta(\mathbf{p}, \mathbf{q})\mathbf{Mc} + \Delta(\mathbf{r}, \mathbf{p})\mathbf{q} + \Delta (\mathbf{q}, \mathbf{r})\mathbf{p} \end{align*} But since \(\mathbf{p}, \mathbf{q}, \mathbf{r}\) are not parallel, this means that \(\mathbf{Mc}\) is uniquely defined to be \(\mathbf{r}\) as required.

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1999 Paper 3 Q1
D: 1700.0 B: 1500.0
roots of polynomials symmetric functions cubic equation geometric progression arithmetic progression Vieta's formulas necessary and sufficient condition

Consider the cubic equation \[ x^3-px^2+qx-r=0\;, \] where \(p\ne0\) and \(r\ne 0\).

  1. If the three roots can be written in the form \(ak^{-1}\), \(a\) and \(ak\) for some constants \(a\) and \(k\), show that one root is \(q/p\) and that \(q^3 -rp^3=0\;.\)
  2. If \(r=q^3/p^3\;\), show that \(q/p\) is a root and that the product of the other two roots is \((q/p)^2\). Deduce that the roots are in geometric progression.
  3. Find a necessary and sufficient condition involving \(p\), \(q\) and \(r\) for the roots to be in arithmetic progression.

Solution:

  1. If the roots are \(ak^{-1}, a, ak\) then we must have that \(p = a(k^{-1}+1+k)\), \(q = a^2(k^{-1}+k+1)\) and \(r = a^3\), therefore \(a = \frac{q}{p}\) (ie one of the roots is \(\frac q p\) and \(r = \left ( \frac{q}{p} \right)^3 \Rightarrow q^3 =rp^3 \Rightarrow q^3-rp^3 = 0\)
  2. Suppose \(r = q^3/p^3\) then \(\left (\frac{q}{p} \right)^3 - p\left (\frac{q}{p} \right)^2+q\left (\frac{q}{p} \right) - r = \frac{--pq^2+pq^2}{p^2} =0 \), therefore \(q/p\) is a root by the factor theorem. We must also have the product of the three roots is \(q^3/p^3\) but one of the roots is \(q/p\) therefore the product of the other two roots is \(q^2/p^2\), but the condition \(ac = b^2\) is precisely the condition that \(a,b,c\) is a geometric progression.
  3. If the three roots are \(a-d, a, a+d\) then \(p = 3a\), \(q = a^2-da+a^2+da+a^2-d^2 = 3a^2-d^2\), \(r = a(a^2-q^2)\), therefore \(\frac{p}{3}\left (q-\frac{2p^2}9 \right) = r\) Similarly, suppose \(\frac{p}{3}\) is a root, then the other two roots must sum to twice this and therefore they are in arithmetic progression. The condition \(\frac{p}{3}\) is a root is equivalent to: \(\frac{p^3}{27} - \frac{p^3}{9} + \frac{qp}{3} - r = 0\), ie exactly \(\frac{p}{3}\left (q-\frac{2p^2}9 \right) = r\), therefore this condition is both necessary and sufficient.

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1994 Paper 3 Q6
D: 1700.0 B: 1501.5
complex numbers rotation Argand diagram geometric transformation composition of transformations unit circle necessary and sufficient condition

The four points \(A,B,C,D\) in the Argand diagram (complex plane) correspond to the complex numbers \(a,b,c,d\) respectively. The point \(P_{1}\) is mapped to \(P_{2}\) by rotating about \(A\) through \(\pi/2\) radians. Then \(P_{2}\) is mapped to \(P_{3}\) by rotating about \(B\) through \(\pi/2\) radians, \(P_{3}\) is mapped to \(P_{4}\) by rotating about \(C\) through \(\pi/2\) radians and \(P_{4}\) is mapped to \(P_{5}\) by rotating about \(D\) through \(\pi/2\) radians, each rotation being in the positive sense. If \(z_{i}\) is the complex number corresponding to \(P_{i},\) find \(z_{5}\) in terms of \(a,b,c,d\) and \(z_{1}.\) Show that \(P_{5}\) will coincide with \(P_{1},\) irrespective of the choice of the latter if, and only if \[a-c=\mathrm{i}(b-d)\] and interpret this condition geometrically. The points \(A,B\) and \(C\) are now chosen to be distinct points on the unit circle and the angle of rotation is changed to \(\theta,\) where \(\theta\neq0,\) on each occasion. Find the necessary and sufficient condition on \(\theta\) and the points \(A,B\) and \(C\) for \(P_{4}\) always to coincide with \(P_{1}.\)

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