Problems

Filters
Clear Filters

1 problem found

2001 Paper 1 Q2
D: 1500.0 B: 1484.0

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 \(x > 5\)