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2024 Paper 2 Q5
- The functions \(\mathrm{f}_1\) and \(\mathrm{F}_1\), each with domain \(\mathbb{Z}\), are defined by \[ \mathrm{f}_1(n) = n^2 + 6n + 11, \] \[ \mathrm{F}_1(n) = n^2 + 2. \] Show that \(\mathrm{F}_1\) has the same range as \(\mathrm{f}_1\).
- The function \(\mathrm{g}_1\), with domain \(\mathbb{Z}\), is defined by \[ \mathrm{g}_1(n) = n^2 - 2n + 5. \] Show that the ranges of \(\mathrm{f}_1\) and \(\mathrm{g}_1\) have empty intersection.
- The functions \(\mathrm{f}_2\) and \(\mathrm{g}_2\), each with domain \(\mathbb{Z}\), are defined by \[ \mathrm{f}_2(n) = n^2 - 2n - 6, \] \[ \mathrm{g}_2(n) = n^2 - 4n + 2. \] Find any integers that lie in the intersection of the ranges of the two functions.
- Show that \(p^2 + pq + q^2 \geqslant 0\) for all real \(p\) and \(q\). The functions \(\mathrm{f}_3\) and \(\mathrm{g}_3\), each with domain \(\mathbb{Z}\), are defined by \[ \mathrm{f}_3(n) = n^3 - 3n^2 + 7n, \] \[ \mathrm{g}_3(n) = n^3 + 4n - 6. \] Find any integers that lie in the intersection of the ranges of the two functions.
Solution:
- \(\,\) \begin{align*} && f_1(n) &= n^2 + 6n + 11 \\ &&&= (n+3)^2 + 2 \\ &&&=F_1(n+3) \end{align*} Since \(n \mapsto n+3\) is a bijection on \(\mathbb{Z}\) both functions must have exactly the same range.
- \(g_1(n) = n^2-2n+5 = (n-1)^2 + 4\). Since squares are always \(0, 1 \pmod{4}\) it's impossible for \(f_1\) and \(g_1\) to take the same value therefore the ranges have empty intersection.
- \(\,\) \begin{align*} && f_2(n) &= n^2-2n - 6 \\ &&&= (n-1)^2-7 \\ && g_2(n) &= n^2-4n+2 \\ &&&= (n-2)^2 - 2 \end{align*} so suppose \(x^2 - 7 = y^2 - 2\) then \begin{align*} && x^2 - 7 &= y^2 -2 \\ \Rightarrow && 5 &= y^2 - x^2 \\ &&&= (y-x)(y+x) \end{align*} So we have cases: \(y-x = -5, y + x = -1 \Rightarrow y = -3\) and the output is \(7\) \(y-x=-1, y+x = -5 \Rightarrow y = -3\) same output \(y-x=1, y+x = 5 \Rightarrow y = 3\) same output \(y-x=5, y-x = 1 \Rightarrow y = 3\) same ouput.
- \begin{align*} && 0 &\leq \frac12(p^2+q^2)+\frac12(p+q)^2 \\ &&&= p^2 + q^2 + pq \end{align*} Looking at \(f_3\) we see \begin{align*} && f_3(n) &= n^3 - 3n^2 + 7n \\ &&&= (n-1)^3 -3n + 7n +1 \\ &&&= (n-1)^3 +4(n-1) -3 \\ &&&= g_3(n-1) + 3 \end{align*} So suppose we have two values which are equal, ie \begin{align*} && x^3 + 4x -3 &= y^3 +4y -6 \\ \Rightarrow && 3 &= y^3-x^3+4y-4x \\ &&&= (y-x)(y^2+xy+x^2+4) \end{align*} Since \(x^2+xy+y^2 \geq 0\) then the right hand factor is always a positive integer bigger than \(3\) and in particular there will be no solutions and hence no integers in the intersection of the ranges.
1989 Paper 2 Q4
The function \(\mathrm{f}\) is defined by \[ \mathrm{f}(x)=\frac{\left(x-a\right)\left(x-b\right)}{\left(x-c\right)\left(x-d\right)}\qquad\left(x\neq c,\ x\neq d\right), \] where \(a,b,c\) and \(d\) are real and distinct, and \(a+b\neq c+d\). Show that \[ \frac{x\mathrm{f}'(x)}{\mathrm{f}(x)}=\left(1-\frac{a}{x}\right)^{-1}+\left(1-\frac{b}{x}\right)^{-1}-\left(1-\frac{c}{x}\right)^{-1}-\left(1-\frac{d}{x}\right)^{-1}, \] \((x\neq0,x\neq a,x\neq b)\) and deduce that when \(\left|x\right|\) is much larger than each of \(\left|a\right|,\left|b\right|,\left|c\right|\) and \(\left|d\right|,\) the gradient of \(\mathrm{f}(x)\) has the same sign as \((a+b-c-d).\) It is given that there is a real value of real value of \(x\) for which \(\mathrm{f}(x)\) takes the real value \(z\) if and only if \[ [\left(c-d\right)^{2}z+\left(a-c\right)\left(b-d\right)+\left(a-d\right)\left(b-c\right)]^{2}\geqslant4\left(a-c\right)\left(b-d\right)\left(a-d\right)\left(b-c\right). \] Describe briefly a method by which this result could be proved, but do not attempt to prove it. Given that \(a < b\) and \(a < c < d\), make sketches of the graph of \(\mathrm{f}\) in the four distinct cases which arise, indicating the cases for which the range of \(\mathrm{f}\) is not the whole of \(\mathbb{R}.\)
Solution: Notice that \(\ln f(x) = \ln (x - a) + \ln (x-b) - \ln (x-c) - \ln (x-d)\) therefore: \begin{align*} \frac{\d}{\d x}: && \frac{f'(x)}{f(x)} &= (x-a)^{-1}+(x-b)^{-1}-(x-c)^{-1} - (x-d)^{-1} \\ &&&= \frac{1}{x} \left ( (1-\frac{a}{x})^{-1}+(1-\frac{b}{x})^{-1}-(1-\frac{c}{x})^{-1} - (1-\frac{d}{x})^{-1}\right) \end{align*} Multiplying by \(x\) gives the desired result. When \(|x|\) is very large then: \begin{align*} \frac{x f'(x)}{f(x)} &\approx 1 + \frac{a}{x} + o(\frac{1}{x^2})+ 1 + \frac{b}{x} + o(\frac{1}{x^2})-(1 + \frac{c}{x} + o(\frac{1}{x^2}))-(1 + \frac{d}{x} + o(\frac{1}{x^2})) \\ &= \frac{a+b-c-d}{x} + o(x^{-2}) \end{align*} Dividing by \(x\) we obtain \(\frac{f'(x)}{f(x)} \approx \frac{a+b-c-d}{x^2} + o(x^{-3})\) if \(|x|\) is sufficiently large this will be dominated by the \(\frac{a+b-c-d}{x^2}\) term which will have the same sign as \((a+b-c-d)\). When \(|x|\) is very large all of the brackets will have the same sign, and therefore \(f(x)\) will be positive, and so \(f'(x)\) must have the same sign as \(a+b-c-d\). To prove this result, we could set \(f(x) = k\) and rearrange to form a quadratic in \(x\). We could then check the discriminant is non-zero. Case 1: \(a < c < d < b\) and \(a+b > c+d \Rightarrow\) not all values reached and approx asymtope from below on the right and above on the left.
