Standard +0.3 This is a straightforward rational inequality requiring standard algebraic manipulation: rearranging to a single fraction, finding critical points from numerator and denominator, and testing intervals. While it's FP2, this is a routine technique that's slightly above average difficulty only due to the sign analysis required, but involves no novel insight or complex multi-step reasoning.
\(3x - 5 - \frac{2}{x} = 0\), multiply through by \(x^2\)
\(\frac{3x^2-5x-2}{x} = 0\)
\(\frac{(3x+1)(x-2)}{x} = 0\) or \(x(3x+1)(x-2)=0\)
M1
Obtaining two non-zero CVs by any valid method (not calculator)
CVs \(x = -\frac{1}{3}, 2\)
A1
Non-zero CVs correct
\(x = 0\)
B1
\(x < -\frac{1}{3}\), \(0 < x < 2\)
M1A1
M1: deducing one appropriate range from CVs; A1: both ranges correct; use set notation with curved brackets for A1; if \(\leqslant\) used deduct final mark only
# Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3x - 5 - \frac{2}{x} = 0$, multiply through by $x^2$ | | |
| $\frac{3x^2-5x-2}{x} = 0$ | | |
| $\frac{(3x+1)(x-2)}{x} = 0$ or $x(3x+1)(x-2)=0$ | M1 | Obtaining two non-zero CVs by any valid method (not calculator) |
| CVs $x = -\frac{1}{3}, 2$ | A1 | Non-zero CVs correct |
| $x = 0$ | B1 | |
| $x < -\frac{1}{3}$, $0 < x < 2$ | M1A1 | M1: deducing one appropriate range from CVs; A1: both ranges correct; use set notation with curved brackets for A1; if $\leqslant$ used deduct final mark only |
---