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2011 Paper 2 Q1
D: 1600.0 B: 1500.0
curve sketching square root functions graphical solution simultaneous equations algebraic manipulation domain restrictions intersection of curves

  1. Sketch the curve \(y=\sqrt{1-x} + \sqrt{3+x}\;\). Use your sketch to show that only one real value of \(x\) satisfies \[ \sqrt{1-x} + \sqrt{3+x} = x+1\,, \] and give this value.
  2. Determine graphically the number of real values of \(x\) that satisfy \[ 2\sqrt{1-x} = \sqrt{3+x} + \sqrt{3-x}\;. \] Solve this equation.

Solution:

  1. TikZ diagram
    Clearly the only solution is \(x = 1\)
  2. TikZ diagram
    There is clearly only one solution, with \(x \approx -2\) \begin{align*} && 4(1-x) &= 6+2\sqrt{9-x^2} \\ && -2x-1 &=\sqrt{9-x^2} \\ \Rightarrow && 4x^2+4x+1 &= 9-x^2 \\ \Rightarrow && 0 &= 5x^2+4x-8 \\ &&x&= \frac{-2\pm 2\sqrt{11}}{5} \\ \Rightarrow && x &= -\left ( \frac{2+2\sqrt{11}}{5} \right) \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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