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2022 Paper 3 Q3
D: 1500.0 B: 1500.0
implicit differentiation stationary points discriminant curve sketching simultaneous equations polynomial equations

  1. The curve \(C_1\) has equation \[ ax^2 + bxy + cy^2 = 1 \] where \(abc \neq 0\) and \(a > 0\). Show that, if the curve has two stationary points, then \(b^2 < 4ac\).
  2. The curve \(C_2\) has equation \[ ay^3 + bx^2y + cx = 1 \] where \(abc \neq 0\) and \(b > 0\). Show that the \(x\)-coordinates of stationary points on this curve satisfy \[ 4cb^3 x^4 - 8b^3 x^3 - ac^3 = 0\,. \] Show that, if the curve has two stationary points, then \(4ac^6 + 27b^3 > 0\).
  3. Consider the simultaneous equations \begin{align*} ay^3 + bx^2 y + cx &= 1 \\ 2bxy + c &= 0 \\ 3ay^2 + bx^2 &= 0 \end{align*} where \(abc \neq 0\) and \(b > 0\). Show that, if these simultaneous equations have a solution, then \(4ac^6 + 27b^3 = 0\).

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2014 Paper 1 Q3
D: 1500.0 B: 1484.0
integration definite integrals polynomial equations curve sketching inequality substitution optimisation algebraic manipulation

The numbers \(a\) and \(b\), where \(b > a\ge0\), are such that \[ \int_a^b x^2 \d x = \left ( \int_a^b x \d x\right)^{\!\!2}\,. \]

  1. In the case \(a=0\) and \(b>0\), find the value of \(b\).
  2. In the case \(a=1\), show that \(b\) satisfies \[ 3b^3 -b^2-7b -7 =0\,. \] Show further, with the help of a sketch, that there is only one (real) value of \(b\) that satisfies this equation and that it lies between \(2\) and \(3\).
  3. Show that \(3p^2 + q^2 = 3p^2q\), where \(p=b+a\) and \(q=b-a\), and express \(p^2\) in terms of \(q\). Deduce that \(1< b-a\le\frac43\).

Solution:

  1. \(\,\) \begin{align*} && \int_0^b x^2 \d x &= \left ( \int_0^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} &= \left ( \frac{b^2}{2} \right)^2 \\ \Rightarrow && b &= \frac{4}{3} \end{align*}
  2. \(\,\) \begin{align*} && \int_1^b x^2 \d x &= \left ( \int_1^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} - \frac{1}{3} &= \left ( \frac{b^2}{2} - \frac{1}{2} \right)^2 \\ \Rightarrow && 4(b^3 - 1) &= 3(b^2-1)^2 \\ \Rightarrow && 4(b^3-1) &= 3(b^4-2b^2+1) \\ \Rightarrow && 0 &= 3b^4-4b^3-6b^2+7 \\ &&&= (b-1)(3b^3-b^2-7b-7) \\ \Rightarrow && 0 &= 3b^3-b^2-7b-7 \end{align*}
    TikZ diagram
    Let \(f(x) = 3x^3-x^2-7x-7\) then \(f(2) = 3 \cdot 8 - 4 - 14 - 7 = -1 < 0\), \(f(3) = 3 \cdot 27 - 9 - 21 - 7 = 44 > 0\) therefore the root must lie between \(2\) and \(3\).
  3. \(,\) \begin{align*} && \int_a^b x^2 \d x &= \left ( \int_a^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} - \frac{a^3}{3} &= \left ( \frac{b^2}{2} - \frac{a^2}{2} \right)^2 \\ \Rightarrow && 4(b^3 - a^3) &= 3(b^2-a^2)^2 \\ \Rightarrow && 4(b^2+ab+a^2) &= 3(b-a)(b+a)^2 \\ \Rightarrow && 4 \left ( \left ( \frac{p+q}{2}\right)^2+\left ( \frac{p+q}{2}\right)\left ( \frac{p-q}{2}\right)+\left ( \frac{p-q}{2}\right)^2\right) &= 3qp^2 \\ \Rightarrow && 3p^2 + q^2 &= 3qp^2 \\ \Rightarrow && 3p^2(q-1) &= q^2 \\ \Rightarrow && p^2 &= \frac{q^2}{3(q-1)} \\ \Rightarrow && 1 &\leq \frac{1}{3(q-1)} \\ \Rightarrow && 3(q-1) &\leq 1 \\ \Rightarrow && q & \leq \frac{4}{3} \\ \end{align*}

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