HL Test 2018 Calculus Option [50 marks]

Transcription

1 HL Test 208 Calculus Option [50 marks] a. Given that n > lnn for n > 0, use the comparison test to show that the series n=0 is divergent. [ marks] METHOD ln(n + 2) < n + 2 (A) > n+2 (for n 0) A Note: Award A0 for statements such as n=0 > n=0. However condone such a statement if the above A has alrea been n+2 awarded. n=0 (is a harmonic series which) diverges n+2 Note: The R is independent of the As. Award R0 for statements such as " diverges". n+2 R so n=0 diverges by the comparison test AG METHOD 2 > (for ) A ln n n n 2 Note: Award A0 for statements such as n=2 > n=2. However condone such a statement if the above A has alrea been ln n n awarded. a correct statement linking n and n + 2 eg, n=0 = n=2 or n=0 = n=2 ln n n+2 n Note: Award A0 for n=0 n n=2 n (is a harmonic series which) diverges (which implies that n=2 diverges by the comparison test) R ln n Note: The R is independent of the As. Award R0 for statements such as n=0 n deiverges and " n diverges". A Award AA0R for arguments based on n= n. so n=0 diverges by the comparison test AG [ marks]

2 b. () n Find the interval of convergence for n=0. [7 marks] applying the ratio test () n n+ ln(n+) () n = (as n = ln(n+) A M Note: Condone the absence of its and modulus signs. Note: Award MA0 for n. Subsequent marks can be awarded. series converges for considering = and = M Note: Award M to candidates who consider one endpoint. when =, series is n=0 which is divergent (from (a)) A Note: Award this A if n=0 is not stated but reference to part (a) is. when =, series is n=0 ( )n A n=0 ( )n converges (conditionally) by the alternating series test (strictly alternating, u n > u n+ for n 0 and n (u n ) = 0) so the interval of convergence of S is A Note: The final A is dependent on previous As ie, considering correct series when = and = and on the final R. Award as above to candidates who firstly consider = and then state conditional convergence implies divergence at =. [7 marks] < < < R 2. The function f is defined by 2 + < 2 f () = { a 2 + b 2 [6 marks] where a and b are real constants Given that both f and its derivative are continuous at = 2, find the value of a and the value of b.

3 considering continuity at = 2 2 f () = and 2 + f () = 4a + 2b (M) 4a + 2b = A considering differentiability at = 2 f () = { < 2 2a + b 2 (M) 2 f () = and 2 + f () = 4a + b (M) Note: The above M is for attempting to find the left and right it of their derived piecewise function at = 2. 4a + b = A a = and b = 2 4 A [6 marks] a. Find the value of 4. [ marks] R 4 = R 4 (A) Note: The above A for using a it can be awarded at any stage. Condone the use of. Do not award this mark to candidates who use as the upper it throughout. = R [ 2 ] R (= [ 2 ] ) M = R ( (R )) 2 = 2 A [ marks] b. Illustrate graphically the inequality n=5 < 4 < n=4. n n

4 AAAA A for the curve A for rectangles starting at = 4 A for at least three upper rectangles A for at least three lower rectangles Note: Award A0A for two upper rectangles and two lower rectangles. sum of areas of the lower rectangles < the area under the curve < the sum of the areas of the upper rectangles so n=5 < 4 < n=4 n n AG c. Hence write down a lower bound for n=4. n [ mark] a lower bound is A Note: Allow FT from part (a). [ mark] 2 d. Find an upper bound for n=4. n [ marks]

5 METHOD n=5 < n 2 64 (M) + n=5 = + n 2 64 (M) n=4 <, an upper bound A n 64 Note: Allow FT from part (a). METHOD 2 changing the lower it in the inequality in part (b) gives n=4 < (< n= ) n n n=4 < R [ 2 ] R n 2 (M) (A) n=4 <, an upper bound A n 8 Note: Condone candidates who do not use a it. [ marks] The function f is defined by f () = (arcsin ),. 4a. Show that f (0) = 0. f () = 2 arcsin () 2 MA Note: Award M for an attempt at chain rule differentiation. Award M0A0 for f () = 2 arcsin (). f (0) = 0 AG The function f satisfies the equation ( 2 )f () f () 2 = 0. 4b. By differentiating the above equation twice, show that ( 2 )f (4) () 5f () () 4f () = 0 where f () () and f (4) () denote the rd and 4th derivative of f () respectively.

6 differentiating gives ( 2 )f () () 2f () f () f ()(= 0) MA differentiating again gives ( 2 )f (4) () 2f () () f () f () () f ()(= 0) MA Note: Award M for an attempt at product rule differentiation of at least one product in each of the above two lines. Do not penalise candidates who use poor notation. ( 2 )f (4) () 5f () () 4f () = 0 AG 4c. Hence show that the Maclaurin series for f () up to and including the term in 4 is [ marks] attempting to find one of f (0), f () (0) or f (4) (0) by substituting = 0 into relevant differential equation(s) (M) Note: Condone f d 2 arcsin () (0) found by calculating ( ) at = 0. 2 (f (0) = 0, f (0) = 0) f (0) = 2 and f (4) (0) 4f (0) = 0 f (4) (0) = 8 A f () 2 (0) = 0 and so A 2! 4! Note: Only award the above A, for correct first differentiation in part (b) leading to f () (0) = 0 stated or f () (0) = 0 seen from use of the general Maclaurin series. Special Case: Award (M)A0A if f (4) (0) = 8 is stated without justification or found by working backwards from the general Maclaurin series. so the Maclaurin series for f () up to and including the term in 4 is 2 + AG 4 [ marks] 4d. Use this series approimation for f () with = to find an approimate value for π 2. 2 substituting = into M 2 the series approimation gives a value of so π ( ) 4 Note: Accept A 48

7 Consider the differential equation y = p + where R, 0 and p is a positive integer, p >. 5a. Solve the differential equation given that y = when =. Give your answer in the form y = f (). [8 marks] METHOD y = = p + integrating factor (M) = e M = e ln (A) = A d y y = p 2 + y 2 ( ) = p = p + C p (M) A Note: Condone the absence of C. y = p + C p substituting =, y = C = M p Note: Award M for attempting to find their value of C. y = ( p ) p A [8 marks] METHOD 2 dv put y = v so that = v + M(A) substituting, M dv (v + ) v = p + (A) dv = p + dv = p M v = p + C p A Note: Condone the absence of C. y = p + C p substituting =, y = C = M p Note: Award M for attempting to find their value of C. y = ( p ) p A [8 marks] 5b. Show that the -coordinate(s) of the points on the curve y = f () where = 0 satisfy the equation p =. p

8 METHOD find and solve = 0 for p = (p p ) = 0 p p = 0 p p = M A Note: Award a maimum of MA0 if a candidate s answer to part (a) is incorrect. p = p AG METHOD 2 substitute = 0 and their y into the differential equation and solve for p p = 0 ( ) + = p + M p = p p p A p p = Note: Award a maimum of MA0 if a candidate s answer to part (a) is incorrect. p = p AG 5c. Deduce the set of values for p such that there are two points on the curve y = f () where = 0. Give a reason for your answer. there are two solutions for when p is odd (and p > A if p is even there are two solutions (to p = ) p and if p is odd there is only one solution (to p = ) R p Note: Only award the R if both cases are considered. International Baccalaureate Organization 208 International Baccalaureate - Baccalauréat International - Bachillerato Internacional Printed for San Jose High School