5*. 10 units2 (example 5 page 480) 6*. -cos4x + c (example 19a page 467) 1. (a) (1) = 2 (A1) f = 4x A1 f ′(x) evidence of finding the gradient of at =1 M1 f x substituting =1 into e.g. x f ′(x) finding gradient of at =1 A1 f x = 4 e.g. f ′(1) evidence of finding equation of the line M1 2 = 4(x 2 = 4(1) + e.g. y b – –1), = 4x 2 AG N0 5 y – (b) appropriate approach (M1) 4x 2 = 0 e.g. – A1 N2 2 x= (c) (i) bottom limit = 0 (seen anywhere) (A1) x approach involving subtraction of integrals/areas (M1) (x) area of triangle, e.g. f f ∫ – ∫ – ∫l correct expression A2 N4 e.g. (ii) METHOD 1 (using only integrals) correct integration (A1)(A1)(A1) substitution of limits (M1) e.g. area = A1 N4 METHOD 2 integral and triangle) area of triangle= (A1) correct integration (A1) substitution of limits (M1) e.g. correct simplification (A1) e.g. area = A1 N4 9 [16]
5*. 10 units2 (example 5 page 480) 6*. -cos4x + c (example 19a page 467) 1. (a) f (1) = 2 (A1) f ′(x) = 4x A1 evidence of finding the gradient of f at x =1 M1 e.g. substituting x =1 into f ′(x) finding gradient of f at x =1 A1 e.g. f ′(1) = 4 evidence of finding equation of the line M1 e.g. y – 2 = 4(x –1), 2 = 4(1) + b y = 4x – 2 AG N0 5 (b) appropriate approach (M1) e.g. 4x – 2 = 0 x= A1 N2 2 (c) (i) bottom limit x = 0 (seen anywhere) (A1) approach involving subtraction of integrals/areas (M1) e.g. ∫ f (x) – area of triangle, ∫ f – ∫l correct expression A2 N4 e.g. (ii) METHOD 1 (using only integrals) correct integration (A1)(A1)(A1) substitution of limits (M1) e.g. area = A1 N4 METHOD 2 integral and triangle) area of triangle= (A1) correct integration (A1) substitution of limits (M1) e.g. correct simplification (A1) e.g. area = A1 N4 9 [16]
semester_1_review.docx