All Calculus Sl Marks
101
1. y = x 3 + 1 x y d d = 3 x 2 = Slope of tangent at any point Therefore at point where x = 1, slope = 3 (M1) ⇒Slope of normal = – 3 1 (M1)(A1) ⇒Equation of normal: y – 2 = – 3 1 ( x – 1) 3 y – 6 = – x + 1 x + 3 y – 7 = 0 (A1) (C4) Note: Accept equivalent forms eg y = – 3 1 x + 2 3 1 [4] 2. (a) p = 3 (A1) (C1) (b) Area = ∫ π 2 0 d cos 3 x x (M1) = 2 0 ] sin 3 [ π x (A1) = 3 square units (A1) (C3) [4] 3. (a) y = 2 1 ) 4 3 ( 4 3 x x − = − 2 1 ) 4 3 ( 2 1 d d − − = x x y (–4) (A1)(A1) (C2) Note: Award (A1) for each element, to a maximum of [2 marks]. (b) y = e sin x x y d d = (cos x)(e sin x ) (A1)(A1) (C2) Note: Award (A1) for each element. [4] 1
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