23 (a) shown M1 method to find at least one root in [0, 1], e.g. 2x3 + 4x (= 0) and f(0) (= f(1) (= 3) oe or f(0) = 0 and f(1) = 6 –3 –3), since there is a change in sign there must be at least one root in 0 < < 1 (as f is x continuous), or 0 and 6 are either side of 3 4x = 3 2x3 (b) – shown C1 for correct steps leading to rearranged equation = x = x (c) = 0.75 0.671677351 M1 for one correct iteration x1 = 0.5390625 x2 M1 for two further iterations = x3 A1 for 0.671(677351) 0.671677351
23 (a) shown M1 method to find at least one root in [0, 1], e.g. 2x3 + 4x –3 (= 0) and f(0) (= –3), f(1) (= 3) oe or f(0) = 0 and f(1) = 6 since there is a change in sign there must be at least one root in 0 < x < 1 (as f is continuous), or 0 and 6 are either side of 3 (b) 4x = 3 – 2x3 x = x = shown C1 for correct steps leading to rearranged equation (c) x1 = 0.75 x2 = 0.5390625 x3 = 0.671677351 0.671677351 M1 for one correct iteration M1 for two further iterations A1 for 0.671(677351)