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2015 Paper 1 Q7
Let \[ \f(x) = 3ax^2 - 6x^3\, \] and, for each real number \(a\), let \({\rm M}(a)\) be the greatest value of \(\f(x)\) in the interval \(-\frac13 \le x \le 1\). Determine \({\rm M} (a)\) for \(a\ge0\). [The formula for \({\rm M} (a)\) is different in different ranges of \(a\); you will need to identify three ranges.]
Solution: \(f'(x) = 6ax-18x^2\), therefore \(f\) has turning points at \(0\) and \(\frac{a}3\) (ie decreasing for \(x \leq 0\) and \(x \geq \frac{a}{3}\) and increasing otherwise). Therefore possible maxima are \(f(-\tfrac13), f(\frac{a}{3}), f(1)\) where we consider \(\frac{a}{3}\) if \(a \leq 3\) and \(1\) otherwise. \(f(-\frac13) = \frac{a}{3} + \frac{2}{9} = \frac{3a+2}{9}\) \(f(\frac{a}{3}) = \frac{a^3}{3} - \frac{2a^3}{9} = \frac{a^3}{9}\) \(f(1) = 3(a-2)\) Comparing \(\frac{a^3}{9}\) to \(\frac{3a+2}{9}\) we have a double root at \(a = -1\) and a single root at \(a = 2\), therefore \(\frac{a^3}9\) is larger if \(a \geq 2\) Comparing \(3(a-2)\) to \(\frac{3a+2}9\) we have a cross-over at \(a = \frac{7}3\). Therefore we have: \begin{align*} M(a) &= \begin{cases} \frac{3a+2}{9} & 0 \leq a \leq 2 \\ \frac{a^3}{9} & 2 \leq a \leq 3 \\ 3(a-2) & 3 \leq a \end{cases} \end{align*}
1997 Paper 1 Q7
Find constants \(a_{1}\), \(a_{2}\), \(u_{1}\) and \(u_{2}\) such that, whenever \({\mathrm P}\) is a cubic polynomial, \[\int_{-1}^{1}{\mathrm P}(t)\,{\mathrm d}t =a_{1}{\mathrm P}(u_{1})+a_{2}{\mathrm P}(u_{2}).\]
Solution: Since this is true for all cubic polynomials, it must be true in particular for \(1, x, x^2, x^3\), therefore: \begin{align*} \int_{-1}^{1} 1 {\mathrm d}t &=a_{1}+a_{2} &=2\\ \int_{-1}^{1} x {\mathrm d}t &=a_{1}u_1+a_{2}u_2 &= 0 \\ \int_{-1}^{1} x^2 {\mathrm d}t &=a_{1}u_1^2+a_{2}u_2^2 &= \frac23\\ \int_{-1}^{1} x^3 {\mathrm d}t &=a_{1}u_1^3+a_{2}u_2^3 &= 0\\ \end{align*} \begin{align*} && \begin{cases} a_{1}+a_{2} &=2 \\ a_{1}u_1+a_{2}u_2 &= 0 \\ a_{1}u_1^2+a_{2}u_2^2 &= \frac23\\ a_{1}u_1^3+a_{2}u_2^3 &= 0\\ \end{cases} \\ \Rightarrow && \begin{cases} a_{1}(u_1^2 - \frac13) + a_{2}(u_2^2 - \frac13) &= 0 \\ a_{1}u_1(u_1^2 - \frac13) + a_{2}u_2(u_2^2 - \frac13) &= 0 \end{cases} \\ \Rightarrow && \begin{cases} u_i = \pm \frac1{\sqrt{3}} \\ a_i = 1\end{cases} \end{align*} Therefore we have: \[\int_{-1}^{1}{\mathrm P}(t)\,{\mathrm d}t ={\mathrm P} \l \frac1{\sqrt{3}} \r+{\mathrm P}\l -\frac1{\sqrt{3}} \r \] [Note: this question is actually asking about Gauss-Legendre polynomials, and could be done directly by appealing to standard results]
1993 Paper 1 Q7
Sketch the curve \[ \mathrm{f}(x)=x^{3}+Ax^{2}+B \] first in the case \(A>0\) and \(B>0\), and then in the case \(A<0\) and \(B>0.\) Show that the equation \[ x^{3}+ax^{2}+b=0, \] where \(a\) and \(b\) are real, will have three distinct real roots if \[ 27b^{2}+3a^{3}b<0, \] but will have fewer than three if \[ 27b^{2}+4a^{3}b<0. \]
1990 Paper 2 Q3
Sketch the curves given by \[ y=x^{3}-2bx^{2}+c^{2}x, \] where \(b\) and \(c\) are non-negative, in the cases: \begin{questionparts} \item \(2b < c\sqrt{3}\) \item \(2b=c\sqrt{3}\neq0\) \item \(c\sqrt{3} < 2b < 2c\), \item \(b=c\neq0\) \item \(b > c > 0\), \item \(c=0,b\neq0\) \item \(c=b=0\). \end{questionpart} Sketch also the curves given by \(y^{2}=x^{3}-2bx^{2}+c^{2}x\) in the cases \((i), (v)\) and \((vii)\).
Solution:
