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N14/5/MATHL/HP1/ENG/TZ0/XX/MMARKSCHEMENovember 2014MATHEMATICSHigher LevelPaper 120 pages

–2–N14/5/MATHL/HP1/ENG/TZ0/XX/MThis markscheme is the property of the International Baccalaureateand must not be reproduced or distributed to any otherperson without the authorization of the IB Assessment Centre.

–6–N14/5/MATHL/HP1/ENG/TZ0/XX/MSECTION A1.(a)g ( x) 1 1x 3A1A1Note: Award A1 for x 3 in the denominator and A1 for the “ 1 ”.[2 marks](b)x 3y 1A1A1[2 marks]Total [4 marks]2.(a)using the formulae for the sum and product of roots:(i) 4(ii) 12A1A1Note: Award A0A0 if the above results are obtained by solving the originalequation (except for the purpose of checking).[2 marks](b)METHOD 1 2 2 2 2 required quadratic is of the form x 2 x 4q (M1) q 8A1 2 2 p 2( ) M12 412p 16A1 Note: Accept the use of exact rootscontinued

–7–N14/5/MATHL/HP1/ENG/TZ0/XX/MQuestion 2 continuedMETHOD 2replacing x with2xM12 2 2 2 8 1 0 x x 8 16 1 0x2 xx 2 16 x 8 0p 16 and q 8(A1)A1A1Note: Award A1A0 for x 2 16 x 8 0 ie, if p 16 and q 8are not explicitly stated.[4 marks]Total [6 marks]

–8–3.N14/5/MATHL/HP1/ENG/TZ0/XX/MMETHOD 1 OP (1 s ) 2 (3 2 s ) 2 (1 s ) 2 ( 6s 2 12s 11 )A1Note: Award A1 if the square of the distance is found.EITHERattempt to differentiate:attempting to solved OPdsd OPds2 12 s 12 M12 0 for s(M1)s 1(A1)ORattempt to differentiate:attempting to solve d OPds 6 s 2 12 s 11 6s 6d OP 0 for sdsM1(M1)s 1(A1)OR 2attempt at completing the square: OP 6 ( s 1) 2 minimum valueoccurs at s 1 5 M1(M1)(A1)THEN the minimum length of OP is5A1METHOD 2 1 the length of OP is a minimum when OP is perpendicular to 2 1 1 s 1 3 2s 2 0 1 s 1 attempting to solve 1 s 6 4 s 1 s 0 (6s 6 0) for ss 1(R1)A1(M1)(A1) OP 5A1Total [5 marks]

–9–4.(a)(i)(ii)N14/5/MATHL/HP1/ENG/TZ0/XX/Muse of P ( A B ) P ( A) P ( B )P ( A B ) 0.2 0.5 0.7(M1)use of P ( A B ) P ( A) P ( B ) P ( A)P ( B )P ( A B ) 0.2 0.5 0.1 0.6(M1)A1A1[4 marks](b)P ( A B)P ( B)P ( A B ) is a maximum when P ( A B ) P ( A)P ( A B ) is a minimum when P ( A B ) 00 P ( A B ) 0.4P ( A B) A1A1A1Note: A1 for each endpoint and A1 for the correct inequalities.[3 marks]Total [7 marks]5.use of the quotient rule or the product rule3 t 2 2 2t 2t 6 2t 2 24t 2 or C (t ) 2 3 t 2 2 3 t 2 3 t 2 2 3 t 2 M1A1A1Note: Award A1 for a correct numerator and A1 for a correct denominatorin the quotient rule, and A1 for each correct term in the product rule.attempting to solve C (t ) 0 for tt 3 (minutes)C 3 33 mg l or equivalent. 1(M1)A1A1Total [6 marks]

– 10 –6.N14/5/MATHL/HP1/ENG/TZ0/XX/Mdu1 dx 2 xA1dx 2 (u 1) duNote: Award the A1 for any correct relationship between dx and du. 1 xxdx 2 (u 1) 2duu(M1)A1Note: Award the M1 for an attempt at substitution resulting in an integral onlyinvolving u.1 2 u 2 duu2 u 4u 2 ln u ( C ) (A1) x 2 x 3 2 ln 1 x ( C )A1A1Note: Award the A1 for a correct expression in x, but not necessarily fullyexpanded/simplified.Total [6 marks]7.(a)p (3) f (3) g (3) g (3) f (3)(M1)Note: Award M1 if the derivative is in terms of x or 3. 2 4 3 1 11A1[2 marks](b)h ( x) g f ( x) f ( x)h (2) g (1) f (2) 4 4 16(M1)(A1)A1A1[4 marks]Total [6 marks]

– 11 –8.N14/5/MATHL/HP1/ENG/TZ0/XX/Mlet P ( n) be the proposition that (2n)! 2n (n !) 2 , n consider P (1) :2! 2 and 21 1! 2 so P (1) is trueR1assume P ( k ) is true ie (2k )! 2k (k !) 2 , k M12Note: Do not award M1 for statements such as “let n k ”.consider P (k 1) : 2(k 1) ! (2k 2) (2k 1) (2k )! 2(k 1) ! (2k 2) (2k 1) (k !)2 2kM1A1 2(k 1) (2k 1) (k !) 2 2k 2k 1 (k 1) (k 1) (k !) 2 since 2k 1 k 1R1 2k 1 (k 1)! A12P (k 1) is true whenever P (k ) is true and P (1) is true, so P ( n) is true for n R1Note: To obtain the final R1, four of the previous marks must have been awarded.Total [7 marks]

– 12 –9.N14/5/MATHL/HP1/ENG/TZ0/XX/M(a)2 t correct for [1, 2]A12 t correct for [2, 3]A1[2 marks](b)EITHERlet q1 be the lower quartile and let q3 be the upper quartilelet d 2 q1 q3 2 and so IQR 2d by symmetryuse of area formulae to obtain1 2 1d 24(or equivalent)1d or the value of at least one q.2ORM1A1A1let q1 be the lower quartileq1consider1 (2 t ) dt 4M1A11obtain q1 2 12A1THENIQR 2A1Note: Only accept this final answer for the A1.[4 marks]Total [6 marks]

– 13 –10.(a)N14/5/MATHL/HP1/ENG/TZ0/XX/Muse of the addition principle with 3 termsto obtain 4C3 5C3 6C3 ( 4 10 20)number of possible selections is 34(M1)A1A1[3 marks](b)EITHERrecognition of three cases: (2 odd and 2 even or 1 odd and 3 even or 0odd and 4 even)(M1)545454(M1)A1 C2 C2 C1 C3 C0 C4 ( 60 20 1)ORrecognition to subtract the sum of 4 odd and 3 odd and 1 even fromthe totalC4 5C4 5C3 4C1 ( 126 5 40)9(M1)(M1)A1THENnumber of possible selections is 81A1[4 marks]Total [7 marks]

– 14 –N14/5/MATHL/HP1/ENG/TZ0/XX/MSECTION B11.(a)x e3 y 1(i)M1Note: The M1 is for switching variables and can be awarded at any stage.Further marks do not rely on this mark being awarded.taking the natural logarithm of both sides and attempting to transpose f 1 ( x) 13 (ln x 1)x or equivalent, for example x 0 .(ii)M1A1A1[4 marks](b)111ln x (ln x 1) ln x ln x (or equivalent)3331ln x (or equivalent)2x e 12M1A1A1A1 1 coordinates of P are e , 2 1 2A1[5 marks](c)coordinates of Q are (1, 0) seen anywheredy 1 dx xdyat Q , 1dxy x –1A1M1A1AG[3 marks]continued

– 15 –N14/5/MATHL/HP1/ENG/TZ0/XX/MQuestion 11 continued(d)let the required area be Aee11A x 1dx ln xdxM1Note: The M1 is for a difference of integrals. Condone absence of limits here.attempting to use integration by parts to find ln xdx(M1)e x2 x [ x ln x x]1e 2 1A1A1x2Note: Award A1 for x and A1 for x ln x x .2Note: The second M1 and second A1 are independent of the first M1 and the first A1.e21 e2 2e 1 e 222 A1[5 marks](e)(i)METHOD 1consider for example h ( x) x 1 ln x1h (1) 0 and h ( x) 1 xas h ( x) 0 for x 1 , then h ( x) 0 for x 1as h ( x) 0 for 0 x 1 , then h ( x) 0 for 0 x 1so g ( x) x 1 , x (A1)R1R1AGMETHOD 21x2g ( x) 0 (concave down) for x the graph of y g ( x) is below its tangent ( y x 1 at x 1 )so g ( x) x 1 , x g ( x) A1R1R1AGNote: The reasoning may be supported by drawn graphical arguments.continued

– 16 –N14/5/MATHL/HP1/ENG/TZ0/XX/MQuestion 11