MARKING SCHEME

Paper 2 Higher Tier Autumn 2008 Mark Comments6. One correct (see note on the right for def n .)evaluation of x 3 + 3x − 8 for an x satisfying:1·5 ≤ x ≤ 1·6Watch for pupils who are trying to make x 3 +3x equalto 8 rather than x 3 + 3x − 8 equal to 0.B1Calculations should be accurate to 1 figure (including 0) roundedor truncated. Values like −0·0154 can be represented as −0 in thisquestion. By convention, 0 is taken as +0, that is, a small +venumber.If no calculations are given, accept use of “too low” or “too high”Two correct (see note on the right for def n .) evaluationsof x 3 + 3x − 8 for an x satisfying:1·505 ≤ x < 1·525which give opposite signs for f(x).Two correct (OR F.T.) evaluations (1 sig. fig.)1·505 ≤ x ≤ 1·515which give opposite signs for f(x).Thus solution is 1·51 correct to 2 decimal place.Candidates must give a method that proves that thesolution is 1·51 correct to 2 decimal places.1·505 1·525B1M1A14OR >0 and