Standard +0.3 This is a standard modulus inequality requiring consideration of critical points (x = 1.5 and x = -1) and testing regions, but the algebraic manipulation is straightforward once cases are identified. Slightly above average difficulty as it requires systematic case analysis rather than just routine manipulation, but this is a well-practiced technique in C3.
implied by correct answer or plausible incorrect answer
Obtain \(\frac{2}{3} < x < 4\)
A1 5
or correctly expressed equiv; allow \(\leq\) signs
Or: Attempt solution of two linear equations or inequalities
M1
one eqn with signs of \(2x\) and \(x\) the same, second eqn with signs different
Obtain value \(\frac{2}{3}\)
A1
Obtain value \(4\)
B1
Attempt valid method for solving inequality
M1
implied by correct answer or plausible incorrect answer
Obtain \(\frac{2}{3} < x < 4\)
A1 (5)
or correctly expressed equiv; allow \(\leq\) signs
**Either:** Attempt to square both sides | M1 | producing 3 terms on each side
Obtain $3x^2 - 14x + 8 = 0$ | A1 | or inequality involving $<$ or $>$
Obtain correct values $\frac{2}{3}$ and $4$ | A1 |
Attempt valid method for solving inequality | M1 | implied by correct answer or plausible incorrect answer
Obtain $\frac{2}{3} < x < 4$ | A1 5 | or correctly expressed equiv; allow $\leq$ signs
**Or:** Attempt solution of two linear equations or inequalities | M1 | one eqn with signs of $2x$ and $x$ the same, second eqn with signs different
Obtain value $\frac{2}{3}$ | A1 |
Obtain value $4$ | B1 |
Attempt valid method for solving inequality | M1 | implied by correct answer or plausible incorrect answer
Obtain $\frac{2}{3} < x < 4$ | A1 (5) | or correctly expressed equiv; allow $\leq$ signs