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2024 Paper 3 Q3
D: 1500.0 B: 1500.0
differentiation curve sketching logarithmic functions inequalities monotonic functions turning points gradient analysis

Throughout this question, consider only \(x > 0\).

  1. Let \[\mathrm{g}(x) = \ln\left(1 + \frac{1}{x}\right) - \frac{x+c}{x(x+1)}\] where \(c \geqslant 0\).
    1. Show that \(y = \mathrm{g}(x)\) has positive gradient for all \(x > 0\) when \(c \geqslant \frac{1}{2}\).
    2. Find the values of \(x\) for which \(y = \mathrm{g}(x)\) has negative gradient when \(0 \leqslant c < \frac{1}{2}\).
  2. It is given that, for all \(c > 0\), \(\mathrm{g}(x) \to -\infty\) as \(x \to 0\). Sketch, for \(x > 0\), the graphs of \[y = \mathrm{g}(x)\] in the cases
    1. \(c = \frac{3}{4}\),
    2. \(c = \frac{1}{4}\).
  3. The function \(\mathrm{f}\) is defined as \[\mathrm{f}(x) = \left(1 + \frac{1}{x}\right)^{x+c}.\] Show that, for \(x > 0\),
    1. \(\mathrm{f}\) is a decreasing function when \(c \geqslant \frac{1}{2}\);
    2. \(\mathrm{f}\) has a turning point when \(0 < c < \frac{1}{2}\);
    3. \(\mathrm{f}\) is an increasing function when \(c = 0\).

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2009 Paper 1 Q3
D: 1516.0 B: 1468.7
quadratics inequalities number of solutions square root equation cube root equation discriminant analysis case analysis

  1. By considering the equation \(x^2+x-a=0\,\), show that the equation \(x={(a-x)\vphantom M}^{\frac12}\) has one real solution when \(a\ge0\) and no real solutions when \(a<0\,\). Find the number of distinct real solutions of the equation \[ x={\big((1+a)x-a\big)}^{\!\frac13} \] in the cases that arise according to the value of \(a\).
  2. Find the number of distinct real solutions of the equation \[ x={(b+x)\vphantom M}^{\frac12} \] in the cases that arise according to the value of \(b\,\).

Solution:

  1. \(\,\) \begin{align*} && x &= (a-x)^{\frac12} \\ \Rightarrow && x^2 &= a - x \\ \Rightarrow && 0 &= x^2 + x - a \end{align*} This has a roots if \(\Delta = 1 + 4a \geq 0 \Rightarrow a \geq -\frac14\). These roots also need to be positive (since \(x \geq 0\)). Since \(f(0) = -a\) we have one positive root if \(a \geq 0\). If \(a \leq 0\) then since the roots are symmetric about \(x = -\frac12\), both roots are negative and there are no positive roots. Therefore we have on real solution if \(a \geq 0\) and non otherwise. \begin{align*} && x & = \left ( (1+a)x - a \right)^{\frac13} \\ \Leftrightarrow && x^ 3 &= (1+a)x - a \\ \Leftrightarrow && 0 &= x^3- (1+a)x + a \\ \Leftrightarrow && 0 &= (x-1)(x^2+x-a) \\ \end{align*} Since every solution to the first equation is a solution to the second, we have \(x = 1\) always works, and there is an additional two solutions if \(a > -\frac14\) and a single extra solution if \(a = -\frac14\). We can also repeat solutions if \(1\) is a root of \(x^2+x -a\), ie when \(a = 2\) Therefore: One solution if \(a < -\frac14\) Two solutions if \(a = -\frac14, 2\) Three solutions if \(a > -\frac14, a \neq 2\)
  2. \(\,\) \begin{align*} && x &= (b+x)^{\frac12} \\ \Rightarrow && x^2 &= b + x \\ \Rightarrow && 0 &= x^2 - x - b \end{align*} This has a positive root if \(\frac14 - \frac12 - b \leq 0 \rightarrow b \geq \frac14\). It has two positive roots if \(b \geq 0\). Therefore two solutions if \(b > \frac14\) and one solution if \(b = \frac14\)

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2002 Paper 3 Q3
D: 1700.0 B: 1516.0
curve sketching quadratics inequalities surds square root equations optimisation discriminant number of solutions

Let \[\f(x) = a \sqrt{x} - \sqrt{x - b}\;, \] where \(x\ge b >0\) and \(a>1\,\). Sketch the graph of \(\f(x)\,\). Hence show that the equation \(\f(x) = c\), where \(c>0\), has no solution when \(c^2 < b \l a^2 - 1 \r\,\). Find conditions on \(c^2\) in terms of \(a\) and \(b\) for the equation to have exactly one or exactly two solutions. Solve the equations

  1. \(3 \sqrt{x} - \sqrt{x - 2} = 4\, ,\)
  2. \(3 \sqrt{x} - \sqrt{x - 3} = 5\;\).

Solution: \begin{align*} && f'(x) &= \frac12 ax^{-1/2}-\frac12(x-b)^{-1/2} \\ \Rightarrow f'(x) = 0: && 0 &= \frac{a\sqrt{x-b}-\sqrt{x}}{\sqrt{x(x-b)}} \\ \Rightarrow && x &= a^2(x-b)\\ \Rightarrow && x &= \frac{a^2b}{a^2-1} \\ && f(x) &= a^2 \sqrt{\frac{b}{a^2-1}} - \sqrt{\frac{a^2b}{a^2-1}-b} \\ &&&= a^2 \sqrt{\frac{b}{a^2-1}} - \sqrt{\frac{b}{a^2-1}} \\ &&&= \sqrt{b(a^2-1)} \end{align*}

TikZ diagram
If \(c\) is below the turning point, ie \(c^2 < b(a^2-1)\) there is no solution. If \(c^2 = b(a^2-1)\) there is exactly one solution. If \(b(a^2-1) < c^2 < (f(b))^2 = a^2b\) then there are two solutions, otherwise there is exactly one solution.
  1. \(c^2 = 16\), \(2 \cdot (3^2-1) = 16\), so we should have exactly one solution at \(x = \frac{3^2 \cdot 2}{3^2 -1 } = \frac{9}{4}\)
  2. \(c^2 = 25\) and \(3 \cdot (3^2 - 1) = 24, 3 \cdot (3^2) = 27\), so we look for two solutions. \begin{align*} && 5 & = 3 \sqrt{x} - \sqrt{x-3} \\ \Rightarrow && 25 &= 9x+x-3-6\sqrt{x(x-3)} \\ \Rightarrow && 3\sqrt{x(x-3)} &= 5x-14 \\ \Rightarrow && 9x(x-3) &= 25x^2-140x+196 \\ \Rightarrow && 0 &= 16x^2-113x+196 \\ &&&= (x-4)(16x-49) \\ \Rightarrow && x &= 4, \frac{49}{16} \end{align*}

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2001 Paper 1 Q2
D: 1500.0 B: 1484.0
inequalities quadratics rational inequality surds squaring method sign analysis domain restrictions

Solve the inequalities

  1. \(1+2x-x^2 >2/x \quad (x\ne0)\) ,
  2. \(\sqrt{3x+10} > 2+\sqrt{x+4} \quad (x\ge -10/3)\).

Solution:

  1. \(\,\)
    TikZ diagram
    \begin{align*} && 1+2x-x^2 = 2/x \\ \Rightarrow && 0 &= x^3-2x^2-x+2 \\ &&&= (x+1)(x^2-3x+2) \\ &&&= (x+1)(x-1)(x-2) \end{align*} Therefore the inequality is satisfied on \((1,2)\) and \((-1,0)\)
  2. \(\,\)
    TikZ diagram
    \begin{align*} && \sqrt{3x+10} &= 2+\sqrt{x+4} \\ && 3x+10 &= x+8 + 4\sqrt{x+4} \\ && 16(x+4) &= 4(x+1)^2 \\ && 4x+16 &= x^2+2x+1 \\ \Rightarrow && 0 &= x^2-2x-15 \\ &&&= (x-5)(x+3) \end{align*} Therefore \(x > 5\)

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2000 Paper 1 Q6
D: 1516.0 B: 1500.0
quadratics inequalities factorisation curve sketching linear factors region sketching simultaneous inequalities

Show that \[ x^2-y^2 +x+3y-2 = (x-y+2)(x+y-1) \] and hence, or otherwise, indicate by means of a sketch the region of the \(x\)-\(y\) plane for which $$ x^2-y^2 +x+3y>2. $$ Sketch also the region of the \(x\)-\(y\) plane for which $$ x^2-4y^2 +3x-2y<-2. $$ Give the coordinates of a point for which both inequalities are satisfied or explain why no such point exists.

Solution: \begin{align*} && (x-y+2)(x+y-1) &= (x-y)(x+y)-(x-y)+2(x+y)-2 \\ &&&= x^2-y^2+x+3y-2 \end{align*}

TikZ diagram
\begin{align*} x^2-4y^2 +3x-2y+2 &= (x - 2 y + 1) (x + 2 y + 2) \end{align*}
TikZ diagram
TikZ diagram
Consider the point \(x = 0, y = \frac32\), then \(\frac92 - \frac94 = \frac94 > 2\) and \(-4\cdot\frac94-2\cdot \frac32 = -12 < -2\) so this is an example of a point in both regions

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1999 Paper 1 Q3
D: 1500.0 B: 1500.0
inequalities sequences recurrence relation proof by contradiction quadratics cyclic system golden ratio

The \(n\) positive numbers \(x_{1},x_{2},\dots,x_{n}\), where \(n\ge3\), satisfy $$ x_{1}=1+\frac{1}{x_{2}}\, ,\ \ \ x_{2}=1+\frac{1}{x_{3}}\, , \ \ \ \dots\; , \ \ \ x_{n-1}=1+\frac{1}{x_{n}}\, , $$ and also $$ \ x_{n}=1+\frac{1}{x_{1}}\, . $$ Show that

  1. \(x_{1},x_{2},\dots,x_{n}>1\),
  2. \({\displaystyle x_{1}-x_{2}=-\frac{x_{2}-x_{3}}{x_{2}x_{3}}}\),
  3. \(x_{1}=x_{2}=\cdots=x_{n}\).
Hence find the value of \(x_1\).

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1999 Paper 1 Q4
D: 1500.0 B: 1484.0
modulus function curve sketching inequalities regions in plane absolute value geometric transformation translation

Sketch the following subsets of the \(x\)-\(y\) plane:

  1. \(|x|+|y|\le 1\) ;
  2. \(|x-1|+|y-1|\le 1 \) ;
  3. \(|x-1|-|y+1|\le 1 \) ;
  4. \(|x|\, |y-2|\le 1\) .

Solution:

  1. TikZ diagram
  2. TikZ diagram
  3. TikZ diagram
  4. TikZ diagram

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1995 Paper 1 Q1
D: 1484.0 B: 1484.0
inequalities polynomial factorisation cubic homogeneous equation curve sketching regions in plane factoring by grouping

  1. Find the real values of \(x\) for which \[ x^{3}-4x^{2}-x+4\geqslant0. \]
  2. Find the three lines in the \((x,y)\) plane on which \[ x^{3}-4x^{2}y-xy^{2}+4y^{3}=0. \]
  3. On a sketch shade the regions of the \((x,y)\) plane for which \[ x^{3}-4x^{2}y-xy^{2}+4y^{3}\geqslant0. \]

Solution:

  1. \(\,\) \begin{align*} && 0 & \leq x^3 - 4x^2 - x + 4 \\ &&&= (x-1)(x^2-3x-4) \\ &&&= (x-1)(x-4)(x+1) \\ \Leftrightarrow && x &\in [-1, 1] \cup [4, \infty) \end{align*}
  2. \(\,\) \begin{align*} && 0 &= x^{3}-4x^{2}y-xy^{2}+4y^{3} \\ && 0 &= (x-y)(x-4y)(x+y) \end{align*} Therefore the lines are \(y = x, 4y = x, y=-x\).
  3. TikZ diagram
    (quickest way to see this is to check the \(x\) or \(y\)-axis)

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