Math HL Nov 97 p2 MARK

Math HL Nov 97 p2 MARK - 1. (a) (b) (i) (ii) (iii) N97/5...

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Unformatted text preview: 1. (a) (b) (i) (ii) (iii) N97/5 10/H(2) Let =‘p3 +p2 -5p—2, then = 23 +22 -10—2 = 0 (M1)(C1) Therefore p = 2 is a solution of f ( p) = 0. (C1) Now f(p)=(p—2)(p2 +3p+1)=0 (Students may use other approaches, use your discretion) The rest of the solutions can be found by using the quadratic formula onp2+3p+1=0 (MI) Hence these solutions are p = _3 :‘5 (A1) Arithmetic sequence: 1, 1+ p, 1+ 2p, 1+ 3p (A1) Geometric sequence: 1, p, p2 , p3 (A1) (1+2p)+(1+3p)=p2+p3:>p3+pZ—5p—2=0 (C2) Therefore, from part (i), p = 2, p = -3 (C1) (a) The sum to infinity of a geometric series exists iff lpl <1, (C1) Hence, p = -3 is the only such number. MD (b) The sum to infinity of the geometric series is _“_ =___1__ (c2) — r 1_ J3 — 3 2 . 2 l 1 Which converges to 5 _ J5- = '2- + 1-043 (A1) (0) The sum of the first 20 terms of the arithmetic series can be found by applying the sum formula S20 =10(2a+19d)=10(2+19p) (C1) So, 520 = t0[2 +19[‘/—§§'—3-D = -265 + 95J§ . (A1) N97/5 10/H(2) 3 2 (i) (a) The vectors 71 = —2 , and 72 = 1 are parallel to L and L2 4 —1 re ectivel . 5P y (CI) Also, if we let :7 = —?.z7 +11] + 7.1;, then u'll=-6—22+28=0andu-12=-4+ll—7=0 (CI) Therefore the required vector is perpendicular to both lines. ( C1) (b) L and L2 have the following parametric equations: L,:x =3t+2, y=1—2t, z=4t; and L,:x=2u-2, y=a+1, z=2—u (CI) For an intersection to happen, the following system must be consistent 3t+2=2u—2 (1) (R1) 1 - 21‘ = u + l (2) 4t = 2—u (3) “311 fi'om (2) and (3) we have t = 1. Hence u = -2. However these values (A1) do not satisfy (1) since 5 ¢ -6. Therefore the lines cannot meet. (R1) - (0) Since :7 = —2? +1 1f+ 71?, is perpendicular to both L1 and L2, and since the plane contains one of the lines and is parallel to the other, therefore 17 is normal to the plane. The equation of this plane is of the form (R1) —2x+11y+7z= d. The point (2, l, 0) lies on L, and on P, therefore (R1) —4+11+0=d,henced=7andtheequationis (A1) —2x+11y+7z= 7. (ii) (a) The required system is —l k +1 —1 x O 1 1 k -2 y = k + 2 2 2 'k z kz-Zk—S (C2) ('1 f°r each error) (b) det(M)=02>2(k—2)—k—(k+l)(k—2(k-2))-0=0 =k2—2k—8=0 (M1) :>k=4ork=—2 (A1) (A1) a ‘v-«v a -~ — 2:Jt‘"Tlf'sfifv‘vugfifivqum warms; , \‘ 1‘. er" N97/5 l 0/H(2) (c) (i) Using Gaussian elimination on the following matrix will lead us to the solution: —1 k+l -1 0 1 1 k—2 k+2 2 2 k kz-Zk-S «1) ‘ with 2 elementary row operations, the system will be -1 k +l -l O 0 k + 2 k —- 3 k + 2 2 — 0 0 —k+4 k 4k 12 (CI) Therefore the value of 2 can be read fiom the last row z_E—M—u MD (6 4—k (ii) When k = 4, the value of z is undefined, and hence the system has no solution. (R1) When k = —2, the system is a homogenous system, and hence it in ; admits at least one solution. It is reduced to x + y = 0 (R1) z = 0 1% Hence it has an infinite number of solutions. (A1) For all other values of k, the system has a unique solution. (A1) (9 ,. ‘ F VWV—vwmr v—g—nzg-vgn—v-t “ha-ng. ",Vr; ~~ m- v-_-L-,1-<r~_-~-;»4--r-,v~ v -- -» ‘-A.‘Y\:." {73 “ 'JV—vwwu ~r '.".v.,-\ n u“ . .\ w Us», n -» arr-va'ut 'r' .r-wv-zm n.~.~'~f\«~ 4-.- ~y w, ..~-=-‘~]~3an'_’“ N97/5 lO/H(2) 4. (i) (a) y = 2xsinx + cost fay —25ian+4xeost-25in2x=4xc052x ' WZXAG) . 1:/2 2 x/2 2 00 Required volume V = 1: Jo cos x) dx — nj; (x cos x) dx (CI) 1:]2 = ifs/2111+ cost) dx = E] (x + xcost) dx 2 o 2 0 (CI) «l2 and now using integration by parts, or otherwise, on Jox COS 2x dx will yield 1:/2 V = g(%x2 +éxsin2x + 1cost] 4 ° (MINA!) 2 =£[“__l_l]=1(nz _4) 2 8 4 4 16 (A1)(AG') (ii) (a) g(x)=x+l—e‘=>g’(x)=l—e‘:Owhenx=0, also, g’(x) < 0 when x > 0 and g’(x) > 0 when x < 0; (MIX/fl) g(x) will increase for x < 0 and decrease for xi> 0, hence the function has a maximum value at x = 0, g(0) = 0. .‘.g(x)SOVxeR,=>x+l—e‘$0,=>x+13e‘ (10%“) (R1) Cb)(i) From the previous result, we have 6‘ -1 2 x, and since it is given that x > O, and we know that ex > 0 dividing both sides of this inequality by xez" will yield cx _ 1 1 (R1) ,x > 2: , that is f(x)>e'2". xe e CI) 6“ -1 e“ l—e“ ( (‘9 f(x)=T':= x xe e xe (R1) . (iii) also from (ii)(a) and since g(x) _<_ 0 for all x ER, and since replacing x by —x will result in reflecting the function about the y—axis, therefore its maximum will not change, and hence g(—x) S 0. (R2) g(-x) = —-x + l - e“ S 0 => 1 - e" s x, and dividing both sides of this inequality by the positive expression JceJr will yield f (x) < e“ . . (M) l (iv) The two results above give e’“ s f (x) s e" , and since lim e‘z" =1, and lim e“ =1 X-eo’ 390’ f then as x approaches zero from the right 1 g f (x) s l , therefore (C1) lim f (x) = l x—>O* "(Room ..e«.~.~.~ (mg—v u; vn v-A~“." -—‘ - " I' ' ‘I r ‘ 3 1- hum-m.” A w;~;~-~-tetra-mfg:vafi'sjlharsrrefi‘lr"q'wj§wW_ N97/510/H(2) 5. (i) (a) To prove that a set is a group under an operation, we need to prove that: (i) The set is closed under multiplication: b consider any two elements x = [a d] and y = [ e f) in S. 0 (C1) Now x.y=(ae+bg af+bh} ce + dg cf + dh Also (CI)(A1) (ae + bg)(cf + dh) — (ce+ dg)(af + bh) = aecf + aea'h + bgq“ + bgdh - aecf - bceh — adgf — bgdh = ad(eh—gf)+ bc(g’—eh) = (eh —g‘)(ad —bc) = 1. (M1) Therefore x-y ES. (CI) 4% (ii) The operation is associative. This is true for multiplication of matrices, hence true for this case. (R1) (CI) 1 0 (iii) There is an identity element. Since 0 1 is such that ad — bc = 1, hence it belongs to S. (R1)(C1) (iv) Every element has an inverse. Since for every x E 5, ad - bc = det(x) ¢ 0, therefore it is invertible. (R1) ( CI) (b)(i) To prove that G is a subgroup of S, it is enough to prove that it is non-empty, closed, and that it contains the inverse of each of its elements. (R1) 1 0 . . . . Clearly 0 1 e G, hence the first condition 13 satisfied. " v b (CI) The elements of G are of the form [:2 a) , also, [ ‘1: 3: f]:( 22-1); af+gfj eG, and the second — a — e - — a ae — _ _1 (R2) a b a —b condition is satisfied. Also _b a = b a 5 G: (R I” C 1) Hence G is a subgroup of S. . 0 —l —l O l 0 (i1) k=[1 o],andsincek2=[ o 1J,andk“=[o I), therefore the order of k is 4. (M2) (A1) 10 .-..—‘,-..._V.«_..._.!...V“,I _ ‘,3_,$‘_.‘._,‘.§...“eta.- ,7 .- ._. .7. .,. .,,. . a" .l y. l .a . 0., .mwr. .. _...'. ‘ NV... .. _. ., . - .t we a.“ . T. ~_..-—. 7. yr. - --«-- g-v— « "."vge.-;v‘_ ...1 (iii) The elements of H are k, k2, and k‘ which were found in part (b) 0 1 above. The fourth element is k3 =[ 1 0). (iv) The multiplication table for H is (c) (i) By setting up a multiplication table, we can verify that the set is a group. The set is closed under the operation since the product of any two elements is another element of the set. Also there is an identity element 1, and every element has an inverse since 1 appears in every row once. The operation is associative. Since multiplication of complex numbers is a commutative operation, the group is Abelian. (ii) Two groups G and H are isomorphic if there is a one to one correspondence fi'orn G to H which preServe the group operation. The correspondence, 1<—>k4, —1<—>k2,i<—>k3,-i <-> kis sucha 1-1 correspondence. Hence H and K are isomorphic. Since the two groups are isomorphic, and K is Abelian, then H has to be Abelian. (ii) A cyclic group is a group G with at least one element a, such that every element b e G can be expressed as b = ai for some i < n, and a" = e where n is the order of the group. Clearly a° =1, and a generates the group elements. Also an“ -_— e2"i =1, hence the order is n + 1. (iii) For the group to be Abelian, ab = ba for all a and b in G. NOW, (ab)2 = (ab)(ab), and (ab)2 = azbz, therefore, abab : aabb, multiplying on the left by a“ and on the right by b“ will yield ab = ba. Hence, the group is Abelian. 11 N97/510/H(2) (C2) (C2) (-1 for each error) (R3) (C1) (C1) (C3) (R1) . (C2) (R2) (CI) (M2) (R1) . -~v~-g~.cvwrw 5‘99???“ at: 6. (i) (a) Any layout is acceptable. 1%; 4/4: (b) An isomorphism is a 1-—l correspondence so that images of any two (0) adjacent vertices are also adjacent and conversely. By looking at the adjacency matrix of G, we observe that F has an out degree of l and so does X in H, we also observe that B has an out degree of 4 and so does Z. Therefore these should be images of each other in the required isomorphism. We also see that C has an out degree of 3 and so does U in H. D has a 2 out and 3 in degrees, so does W. By similar reasoning, we have A and Vcorrespond as well as E and Y. Hence the required isomorphism is: A<—>V,B<—>Z,C<-)U,D<—>W,E<—>Y, andFHX. For a simple connected graph G, with at least three edges, to be planar, then according to Euler’s formula e s 3v — 6 but 14 >18 — 6, therefore G cannot be planar. (ii) (a) Prim’s algorithm requires that we start‘at any vertex at random and consider it as a tree, then look for the shortest path that joins a vertex on this tree to any of the remaining vertices and add it, then repeat the last step until all the vertices are on the tree. Starting at F, (other forms can be used) 1. The shortest path fi'om F to ABCDEGH is to E, add FE => weight added = 90 2. The shortest path from FE to ABCDGH is to G, add F G with weight 120, 3. The shortest path fiom GFE to ABCDH is to H, add GH with weight 220, 4. The shortest path fiom GFEH to ABCD is to C, add F C with weight 230, 5. (ABD) Add CB with weight 240, 6. (AD) Add BA with weight 180, 7. (D) Add AD with weight 200. Therefore a minimal tree has a weight of 1280. 12 N97/5 1 0/H(2) (C4) (-1 for each error) (C1) (R1) (R2) T (C1) (R1)(C1) (R1)(CI) (R5) (-1 for each error) fl"! a. -'-,|1\"<("V'1“q" v»v~,~-. mq_—.,.., N97/5 10/H(2) 6R, A (b) To find the minimum path fi'om A to H using Dijksu'a’s algorithm we (A1) assign a weight of zero to the first vertex and a weight of co to all vertices that are not in the path. Then we start adding vertices adjacent to the present ones that have a minimum weight. At each iteration, the length of the path to a given vertex is evaluated and the minimum is assi ed. g“ (R2) A (0) AB = 180, => B (130) AD = 200, z: D (200) AC = 270, => 0 (270) ADF = 480 is the minimum path to F, a F (480) ADE = 480 is the minimum path to E, => E (480) (R5) (-1 for ACG = 620 is the minimum path to G, => G (620) eaCh error) ADFG = 600 is the'minimum to G, = New G (600) @ ADEH = 840 is the minimum to H, :9 H (840) i ' ADFGH = 820 is the minimum to H, => NEWH (820). This is a minimum path fi'om A to H. (A1) (iii) (a) A » B C D (b) There is no such walk since the graph has 4 vertices with odd degree. Therefore there is no Euler’s cycle nor a path. (RI) ' (C1) If the door between A and B is locked then there will be a Euler path through the whole graph. There will be only two vertices, C and F with (121) an odd degree. 13 .w ,_ . an. . .. h... N, ., a”: -. x FM”: "A. -~. .w i.“ ....< . .w . w. -,-- I~iq'\'7§5\')‘31"¢"¥w wvwwflf? N97/5 l Oil-1(2) Therefore our walk should start at one end at the other. Such a walk (RI) could be the following: FBFEFDBEDCAFACF C. T (A2) (G) Since the degree of each vertex is larger than 11- = 3 , there is a . . . . . . 2 (R1) Harmltoman c1rcmt. Such a c1rcu1t could be FABEDCF. (A1) 14 ‘ ~VJ~MKMW...¥._V,W...\.Vuh,v.07,” . ... ,. r . _. .. « - w r M w. .. V’ r. .. 7 , . _ .. .-.fi ».:«,y..;i.3.\.‘$.¢wnmfll 7. (a) The confidence interval gives a range of values which has a 95% probability of containing the mean of the population. 0’ 0‘ That is the interval 1? “196—, I? — 196—] if calculated r eatedl , ( J? J; ep y then 95% of such intervals will contain the population mean. 1.96 is the number of standard deviations away from the mean and is read from the standard normal table. According to the scheme above, the confidence intervals are: Male: 1741-1.96=(172.824, 175.176) Female: 173 i 1.96 = (170.844, 175.156) Not enough evidence of any difference between the performance of males and females since each of the confidence intervals contain the mean of the opposite sample. (b) Finding the confidence interval for the difference between the means is similar to the process above, however the standard error is different. The confidence interval is 2 2 (flu‘flr)=EM‘iFi—196 i+i£ nM n; 2 2 =174—173~_~196.’1—2—+l = (-1.456, 3.456) 400 100 The result here confirms our previous findings, since this interval contains zero. Which indicates that the situation where the two means are equal is also a member of the confidence interval. 1. Form the null hypothesis: How” ~,u,- = 0 The alternate hypothesis: H a : y M — M; > O 2. The test statistic is _ 5M —EF —(pM —y,) = 174—173—0 z,_ =o.79s 5;, s; (144+121 __+_. —" —- M 400 100 "M 3. Critical value for Zc for a one-tail test is 1.645. Since Z. = 0.798 < 26 = 1.645 - it is non-rejection region - we do not have enough evidence that males out perform females on mathematics exams. 15 9-71f‘, "raw-aw [gnu \f‘flf"fl"?‘f7"""“?1a-v<fi~ .-_...-y-—- 1...... .i , 4 "a. e .4.V...-...,~.4.‘.».\... v.,...»..,i...., .7- iv N97/5 10/H(2) (R2) ( C1)(A1) (CD041) (R2) (RI)(C1) (C1)(A1) (R1)( C1) (C2) (C2)(AI) (C1) (R1) (A1) ' ‘ 'x”-"'“5""~" 7“" "WT-"W 3W???“ (d) This is a matched-pairs difference test since the samples are not (6) M. \.-..»...-:.~...T_,.;, ‘1“. “mm”, 1 . independent. We study the differences between pairs and we test the differences with a t-distribution since the sample size is small and we do not know the standard deviation of the population. ' The number of degrees of freedom is 9. The differences are: l, 2,—2,—1, 0,-2,—1,1,1,—2. Hence the average difference is E = —03 and standard deviation sd = 1.494. 1 Hozd = 0 H1217 e o 2. Test statistic T = L031 = —0.635 1.494/Jfi 3. The critical value with 9 d.f. is £262. Since —2.262 < T < 2.262 , there is not enough evidence of any difference in speed between the two groups. This is a goodness of fit test. If the distribution were normal, then the expected frequencies should not be very different from the observed ones. Using the normal distribution with mean 174 and standard deviation 12, the probabilities corresponding to the different classes are 0.0228, 0.0989, 0.248, 0.322, 0.217 and 0.0912 respectively. Therefore the corresponding expected frequencies are 9, 40, 99, 129, 87 and 36 respectively. H 0: distribution is normal with mean 174 and standard deviation 12. H1: distribution is not normal with mean 174 and standard deviation 12. 2 2 (fa—f.) 1 o 9 81 81 16 = ——_=_.+._+_.+__+.__+_, N°W 1‘ 2 f, 9 4o 99 129 87 36 that is z} = 221. But Z3 =11.07 (5 d.f., 5% upper tail). And since the test statistic does not lie in the rejection region, we do not have enough evidence to reject the hypothesis that the distribution is normal with mean 174 and standard deviation 12. Students using the continuity correction will get: 12 = 2.75. Conclusion does not change. 16 N97/5 lO/H(2) (R1) (R1) (C1) (A2) (C1) (C1) ( C1) (Al) (R1) (R1) (A1) ' (RI) (C2) (A1) (R2) (i) (a) The ratio test is appropriate here. The ratio of consecutive terms in the 36 ,. 36 series is r(n) = $11 = M. = ’ and an "3641114 4 n hence lim r(n) = i x 1 = i < 1 . Thus the series converges by the ratio test. (b) The ratio of consecutive terms in the series is 31 " ! Whig: (3n+ )“1100 n a,l (3n)!100 (n+1)! 1 1 _fixmx(3n+3)(3n+2)(3n+l) =I3_Ox(3n+2)(sn+1) and as n -> so, r(n) —> co, and so the series diverges. (ii) (a) Using the trapezium rule, the area under the curve is‘approximated by b — a L = 2n [f(xo) + f(x,) + 2{f(xx )+.. .+f(xu-1 M 1 0100+O.199+0296+0297+0.479+ ' ‘[0 + 0'8“ + 2{ 0565 + 0.644 + 0.717 + 0.783 7;, = 0.450 2 (b) The volume in question is V = 1t£[ f (x)] dx , hence the function values in the table have to be squared. b—a Vz{ 3n }[y0 +4y1+2y2+...+4yn_l+yn]x 1t 02 +4x 01002 +2 x 01992 +4 x 02962 + lief—0 2x02972 +4xo.4792 +2x05652 + 4x 0.6442 + 2 x 0.7172 +4 x 0.7832 + 0.8412 V 2: 02681: = 0.843 cubic units. 17 .. .VJ‘HWr i7 ..:.. ,V—‘n ._..—‘-_~.., --q~1g':~hv ‘I— r;......~.w,——m»_ -,- 3..., , v. .7 .,...\r.,‘ w “A”. - w .- «a»: -..V. “NW-«w . ~f‘wr-‘1v_".‘v“"-1“fl"fWlfi1'w’S'FW’53‘" N97/510/H(2) (C1) (M2)(A1) (R1)(A1) ‘ (CI) J (M2) (A1) (R1)(C1) (R2) 012) (A1) (R2) (C2) (M2) (A1) N97/5 I 0/H(2) (iii) (a) The Maclaurin series for a function f (x) is x2 xn xn+l x= 0+ ’0+— "O+...+— (“0+ ("we f() f()xf() 2!f() nlf()(n+1)!f () (R2) where 0 < < |x| . Thus, .763 XS sinx=0+x+0xx2-———-+0xx" +—cosc, O<|c|<|x|. 3! 5! (C2) 2 ’53 (b) In the equationx = smx , replace smx byx ~33 to get (R1) 3 sinx z x—% ,or x3 + 6x2 —6x = 0, yields the non- zero roots as “L... «WP—Mg (CZ) Obviously x = —3 — Jl‘S— cannot approximate x2 = sinx , (CI) thus x0 = —3 + Jig is the only approximate solution. (AG) Now lsinx —le= x -§-+£g-COSC—X2 (C1) 0 0 0 0 x3 x5 = x0 mfg—x3 +5—zcos x5 = 3—00st (C1) 5! where 0 < c < Jig — 3. Since {cosxl $1 for all x , it follows that s 5 J3 —3 s lmflzis xo <(__l<£<_1_ (x2) 5! 5! 120 200 x5 Since sinx0 —x3 = geese, where 0 <c < 5—3, and inthis interval the cosine function is positive, it follows that Sin Io ‘xg is positive. (122) - 2 smx -x . . (c) I1=xo-&;roi_2—x°o, andbysettmgx0=—3+s/l_5_weobtam (RI) 2:, =O.8767. (A1) 18 ...
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Math HL Nov 97 p2 MARK - 1. (a) (b) (i) (ii) (iii) N97/5...

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