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2016 Paper 2 Q6
D: 1600.0 B: 1484.0
differential equation first order ODE substitution composition of functions chain rule Chebyshev polynomials verification functional equation

This question concerns solutions of the differential equation \[ (1-x^2) \left(\frac{\d y}{\d x}\right)^2 + k^2 y^2 = k^2\, \tag{\(*\)} \] where \(k\) is a positive integer. For each value of \(k\), let \(y_k(x)\) be the solution of \((*)\) that satisfies \(y_k(1)=1\); you may assume that there is only one such solution for each value of \(k\).

  1. Write down the differential equation satisfied by \(y_1(x)\) and verify that \(y_1(x) = x\,\).
  2. Write down the differential equation satisfied by \(y_2(x)\) and verify that \(y_2(x) = 2x^2-1\,\).
  3. Let \(z(x) = 2\big(y_n(x)\big)^2 -1\). Show that \[ (1-x^2) \left(\frac{\d z}{\d x}\right)^2 +4n^2 z^2 = 4n^2\, \] and hence obtain an expression for \(y_{2n}(x)\) in terms of \(y_n(x)\).
  4. Let \(v(x) = y_n\big(y_m(x)\big)\,\). Show that \(v(x) = y_{mn}(x)\,\).

Solution:

  1. When \(k =1\), we have \((1-x^2)(y')^2 + y^2 = 1\). Notice that if \(y_1 = x\) we have \(y_1' = 1\) and \((1-x^2) \cdot 1 + x^2 = 1\) so \(y_1\) is a solution, and we are allowed to assume this is the only solution. And notice that \(y_1(1) = 1\).
  2. When \(k = 2\) we have \((1-x^2)(y')^2 + 4y^2 = 4\). Trying \(y_2 = 2x^2-1\) we see that \(y_2' = 4x\) and \((1-x^2)(4x)^2 + 4(2x^2-1)^2 = 16x^2-16x^4+16x^4-16x^2+4 = 4\). We can also check that \(y(1) = 2 \cdot 1^2 - 1 = 1\)
  3. Let \(z(x) = 2(y_n(x))^2-1\), then \begin{align*} && \frac{\d z}{\d x} &= 4y'_n(x)y_n(x) \\ \Rightarrow && LHS &= (1-x^2)\left ( \frac{\d z}{\d x} \right)^2 + 4n^2 z^2 \\ &&&= (1-x^2)16(y'_n(x))^2(y_n(x))^2 + 4n^2(2(y_n(x))^2-1)^2 \\ &&&= 16y_n^2(1-x^2) \left [\frac{n^2-n^2y_n^2}{(1-x^2)} \right] + 16n^2y_n^4-16n^2y_n^2+4n^2 \\ &&&= 4n^2 = RHS \end{align*} Therefore \(y_{2n}(x) = 2(y_n(x))^2-1 = y_2(y_n(x))\) (notice also that \(z(1) = 2(y_n(1))^2-1 = 2-1 = 1\)).
  4. Let \(v(x) = y_n(y_m(x))\) so \begin{align*} && (y_m')^2 &= \frac{m^2(1-y_m^2)}{1-x^2} \\ && (y_n')^2 &= \frac{n^2(1-y_n^2)}{1-x^2} \\ \\ && \frac{\d v}{\d x} &= y_n'(y_m(x)) \cdot y_m'(x) \\ && (1-x^2)(v')^2 &= (1-x^2) \cdot (y_n'(y_m(x)))^2 (y_m')^2 \\ &&&= (1-x^2) \cdot (y_n'(y_m(x)))^2 \left ( \frac{m^2(1-y_m^2)}{1-x^2} \right) \\ &&&= (y_n'(y_m(x)))^2 m^2(1-y_m^2) \\ &&&= \frac{n^2(1-(y_n(y_m(x)))^2)}{1-y_m^2}m^2(1-y_m^2) \\ &&&= n^2m^2(1-v^2) \end{align*} Therefore \(v\) satisfies our differential equation and \(v(1) = y_n(y_m(1)) = y_n(1) = 1\) so it must be our desired solution.
[Note: this is another question about Chebyshev polynomials, and we have proven that we can compose them nicely. This might be more easily proven as \(T_n(x) = \cos(n \cos^{-1} x)\) and so \(T_n(T_m(x)) = \cos (n \cos^{-1}( \cos (m \cos^{-1} x))) = \cos(nm \cos^{-1}x) = T_{nm}(x)\)]

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2012 Paper 2 Q2
D: 1600.0 B: 1530.0
polynomials composition of functions degree of polynomial functional equation simultaneous equations proof by contradiction quadratic

If \(\p(x)\) and \(\q(x)\) are polynomials of degree \(m\) and \(n\), respectively, what is the degree of \(\p(\q(x))\)?

  1. The polynomial \(\p(x)\) satisfies \[ \p(\p(\p(x)))- 3 \p(x)= -2x\, \] for all \(x\). Explain carefully why \(\p(x)\) must be of degree 1, and find all polynomials that satisfy this equation.
  2. Find all polynomials that satisfy \[ 2\p(\p(x)) +3 [\p(x)]^2 -4\p(x) =x^4 \] for all \(x\).

Solution: If \(\p(x)\) and \(\q(x)\) are polynomials of degree \(m\) and \(n\), \(\p(\q(x))\) has degree \(mn\).

  1. Suppose \(\p(\p(\p(x)))- 3 \p(x)= -2x\), and suppose \(p(x)\) has degree \(n = \geq 2\), then \(\p(\p(\p(x)))\) has degree \(n^3\) and so the left hand side has degree higher than \(1\) and the right hand side is degree \(1\). Therefore \(\p(x)\) is degree \(1\) or \(0\). If \(p(x) = c\) then \(c^3-3c = -2x\) but the LHS doesn't depend on \(x\) which is also a contradiction. Therefore \(\p(x)\) is degree \(1\). Suppose \(\p(x) = ax+b\) then: \begin{align*} && -2x &= \p(\p(\p(x))) - 3\p(x) \\ &&&= \p(\p(ax+b)) - 3(ax+b) \\ &&&= \p(a(ax+b)+b) - 3ax -3b \\ &&&= a(a^2x+ab+b) + b - 3ax - 3b \\ &&&= (a^3-3a)x + b(a^2+a-2) \\ \Rightarrow &&& \begin{cases} a^3-3a&=-2 \\ b(a^2+a-2) &= 0\end{cases} \\ \Rightarrow &&& \begin{cases} a^3-3a+2 = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow &&& \begin{cases} (a-1)(a^2+a-2) = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow && (a,b) &= (1, b), (-2,b) \end{align*}
  2. Suppose \(2\p(\p(x)) +3 [\p(x)]^2 -4\p(x) =x^4\) and let \(\deg \p(x) = n\), then LHS has degree \(\max(n^2,2n,n)\) and the right hand side has degree \(4\). Therefore \(\p(x)\) must have degree \(2\). Let \(\p(x) = ax^2 + bx + c\), then, considering the coefficient of \(x^4\) in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) we will have \(2a^3+3a^2=1 \Rightarrow 2a^3+3a^2-1 = (a+1)^2(2a-1) \Rightarrow a = -1, a=\frac12\). Consider the coefficient of \(x^3\) in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) we have \(4a^2b+6ab = 0 \Rightarrow 2ab(2a+3) = 0\) Since \(a = -1, \frac12\) this means \(b = 0\). Consider the constant coefficient in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) (using \(b = 0\)). \(2ac^2+c+3c^2-4c = 0 \Rightarrow c(2ac+3c-3) = 0\). Therefore \(c = 0\) or \(a = -1, c = 3, a = \frac12, c = \frac34\), so our possible polynomials are: \(\p(x) = -x^2, \frac12x^2, -x^2+3, \frac12x^2+\frac34\)

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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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1996 Paper 3 Q8
D: 1700.0 B: 1516.0
simultaneous equations Möbius transformation inverse function composition of functions fixed points quadratic equation rational function domain and range

A transformation \(T\) of the real numbers is defined by \[ y=T(x)=\frac{ax-b}{cx-d}\,, \] where \(a,b,c\), \(d\) are real numbers such that \(ad\neq bc\). Find all numbers \(x\) such that \(T(x)=x.\) Show that the inverse operation, \(x=T^{-1}(y)\) expressing \(x\) in terms of \(y\) is of the same form as \(T\) and find corresponding numbers \(a',b',c'\),\(d'\). Let \(S_{r}\) denote the set of all real numbers excluding \(r\). Show that, if \(c\neq0,\) there is a value of \(r\) such that \(T\) is defined for all \(x\in S_{r}\) and find the image \(T(S_{r}).\) What is the corresponding result if \(c=0\)? If \(T_{1},\) given by numbers \(a_{1},b_{1},c_{1},d_{1},\) and \(T_{2},\) given by numbers \(a_{2},b_{2},c_{2},d_{2}\) are two such transformations, show that their composition \(T_{3},\) defined by \(T_{3}(x)=T_{2}(T_{1}(x)),\) is of the same form. Find necessary and sufficient conditions on the numbers \(a,b,c,d\) for \(T^{2}\), the composition of \(T\) with itself, to be the identity. Hence, or otherwise, find transformations \(T_{1},T_{2}\) and their composition \(T_{3}\) such that \(T_{1}^{2}\) and \(T_{2}^{2}\) are each the identity but \(T_{3}^{2}\) is not.

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