Part (a) of this question was reported to be wrongly worded such that the values of P, q and r could not be determined.  However, the marking scheme took care of it and awarded marks up to the points where candidates could derive the three simultaneous equations.  Thus candidates were not disadvantaged in any way.  The equation 2x2 – 7x + 3 could be factorize to have x = ½ or x = 3.  Substituting this into the polynomial 2x3 + (P + q) x + r gave the equations:    2p + 2q + 4r = -1 and 3p + 3q + r = -54.  Substituting x = -1 into f(x) gave
f(-1) = -36 = p + q – r = 34.

In part (b), factorizing x2 – 3x + 2 < 0 gave (x – 1)(x – 2) < 0.  For the equality to be satisfied, x < 1 and x > 2 or x > 1 and x < 2.  Thus the required inequality = 1< x <2.

Many candidates could not investigate correctly.