2. (1) = 12 3b + + 2 = 0 (M1) f – c = 2x 3b, (x) f – = 6 3b = 0 (M1) (3) f – 3b = 6, = 2 (A1) b 1 3(2) + + 2 = 0, = 3 (A1) c c – Note: In the event of no working shown, award (C2) for 1 correct answer. [4] 3. (a) = ex/2 at = 0 = e0 = 1 1) (A1)(A1) 2 y x y P(0, (b) =  (A4) 4 V Notes: Award (A1) for  (A1) for each limit (A1) for (ex/2)2. (c) = (A1) V = (A1)  = e0] (A1) [eln2 – = 1] = (A1)(A1) [2  – = (AG) 5  [11] 4. (a) (seen anywhere) A1A1 attempt to substitute into the quotient rule (do not accept product rule) M1 e.g. correct manipulation that clearly leads to result A1 e.g. AG N0 4 (b) evidence of setting the derivative equal to zero (M1) = 0, 1– 2ln = 0 e.g. g′(x) x A1 A1 N2 3 [7] 5. (a) substituting into the second derivative M1 3 × e.g. = A1 f″ –5 since the second derivative is negative, B is a maximum R1 N0 (b) setting equal to zero (M1) f′(x) evidence of substituting = 2 (M1) x e.g. f′(2) correct substitution A1 e.g. correct simplification 6 2 + = 0, + = 0, 4 + = 0 A1 e.g. p p p –
2. f (1) = 12 – 3b + c + 2 = 0 (M1) f (x) = 2x – 3b, f (3) = 6 – 3b = 0 (M1) 3b = 6, b = 2 (A1) 1 – 3(2) + c + 2 = 0, c = 3 (A1) Note: In the event of no working shown, award (C2) for 1 correct answer. [4] 3. (a) y = ex/2 at x = 0 y = e0 = 1 P(0, 1) (A1)(A1) 2 (b) V =  (A4) 4 Notes: Award (A1) for  (A1) for each limit (A1) for (ex/2)2. (c) V = (A1) =  (A1) = [eln2 – e0] (A1) = [2 – 1] =  (A1)(A1) =  (AG) 5 [11] 4. (a) (seen anywhere) A1A1 attempt to substitute into the quotient rule (do not accept product rule) M1 e.g. correct manipulation that clearly leads to result A1 e.g. AG N0 4 (b) evidence of setting the derivative equal to zero (M1) e.g. g′(x) = 0, 1– 2ln x = 0 A1 A1 N2 3 [7] 5. (a) substituting into the second derivative M1 e.g. 3 × f″ = –5 A1 since the second derivative is negative, B is a maximum R1 N0 (b) setting f′(x) equal to zero (M1) evidence of substituting x = 2 (M1) e.g. f′(2) correct substitution A1 e.g. correct simplification e.g. 6 – 2 + p = 0, + p = 0, 4 + p = 0 A1
semester_1_review.docx