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1. u n 1 Consider 10 n 1 un 1 1 1 10 n 10 n 1 10 n M1A1 n10 10 A1 1 as n 10 A1 1 1 10 R1 So by the Ratio Test the series is convergent. R1 [6] 2. (a) lim x x 1 lim x x x e e M1A1 =0 (b) AG Using integration by parts a 0 e e x e x dx xe x ae a = 1 ae (c) M1 a a 0 0 x dx x a 0 a a e Since ea and aea are both convergent (to zero), the integral is convergent. Its value is 1. A1A1 A1 A1 R1 A1 [9] IB Questionbank Mathematics Higher Level 3rd edition 1 3. (a) Rewrite the equation in the form dy 2 x2 y 2 dx x x 1 M1A1 2 x dx M1 = e 2ln x A1 1 x2 A1 Integrating factor = e = Note: Accept (b) 1 as applied to the original equation. x3 Multiplying the equation, 1 dy 2 1 3 y 2 2 x dx x x 1 (M1) d y 1 2 2 dx x x 1 (M1)(A1) y dx 2 2 x x 1 M1 = arctan x + C A1 Substitute x = 1, y = 1. M1 1 4 C C 1 A1 4 y x 2 arctan x 1 4 A1 [13] 4. (a) The area under the curve is sandwiched between the sum of the areas of the lower rectangles and the upper rectangles. Therefore dx 1 1 1 1 1 1 1 3 1 3 1 3 ... 1 3 1 3 1 3 ... 3 3 4 5 6 x 3 4 5 which leads to the printed result. IB Questionbank Mathematics Higher Level 3rd edition M2 A1 2 (b) We note first that 3 dx 1 1 2 3 x 2 x 3 18 M1A1 Consider first n 1 n 1 3 1 1 1 1 1 1 3 3 3 3 ... 3 2 3 4 5 6 1 1 1 1 8 27 18 n n 1 1 3 1.22 which is an upper bound 263 216 = 1 1 1 1 1 3 3 3 ... 3 2 3 4 5 1 1 1 8 18 85 255 1.18 (which is a lower bound) 72 216 = M1A1 M1A1 A1 M1A1 M1A1 A1 [15] 5. (a) (b) Constant term = 0 f ( x) 1 1 x f ( x) f ′′′ (x) = A1 A1 1 1 x 2 2 1 x 3 f (0) = 1; f (0) = 1; f (0) = 2 A1 A1 A1 Note: Allow FT on their derivatives. f ( x) 0 1 x 1 x 2 2 x 3 ... 1! 2! 3! = x x2 x3 2 2 IB Questionbank Mathematics Higher Level 3rd edition M1A1 AG 3 (c) 1 1 2 x 1 x 2 (A1) 1 1 1 ln 2 2 8 24 M1 = (d) 2 0.667 3 Lagrange error = A1 f n 1 (c) 1 (n 1)! 2 n 1 1 1 = 4 1 c 24 2 6 6 1 1 2 4 (M1) 4 1 1 24 16 giving an upper bound of 0.25. (e) Actual error = ln 2 2 0.0265 3 The upper bound calculated is much larger that the actual error therefore cannot be considered a good estimate. A1 A2 A1 A1 R1 [17] IB Questionbank Mathematics Higher Level 3rd edition 4