ACSI 2019 Prelim Exam Y6 HL Paper 1 Solutions (for students)

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6 / 2019 Prelim Exam / Paper 1 / Solutions 1 PRELIMINARY EXAMINATION 2019 YEAR 6 IB DIPLOMA PROGRAMME MATHEMATICS HIGHER LEVEL P a p e r 1 S O L U T I O N S SECTION A Qn Solution 1. (a) Since the total probability is 1, 23 22 155kk      this takes the form of the sum to infinity of an geometric progression and so 2 2 5 121 5 k     Therefore 43 25 5k  and so 15 4k . (note that sum to infinity does exist because 22 155r   )

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6 / 2019 Prelim Exam / Paper 1 / Solutions 2 Qn Solution (b)     23 2122 55 2 5 21 1 15 2 2 2 45 5 5 115 41 15 5 2 2143 5 5 21. 5 x x x x PX x                              Comments : For (b), many students used integration when it is a discrete random variable. For (b), some students incorrectly concluded that there are x terms in the sum when there are actually 1x terms. 2. (a) The composite function exists since ,222 g fRD    sin 2 , 0,12fg x x x   Also accept   cos 2 , 0,1fg x x x since  sin 2 sin 2 cos cos 2 cos sin 2 0 cos 2 1 cos 222 2x xxx x x       (b) Comment : Generally well done.

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6 / 2019 Prelim Exam / Paper 1 / Solutions 3 Qn Solution 3. (a) 02 2 25 3 11 2 PQ                 321 752 21 3 PR                    303 725 211 QR         A vector perpendicular to the plane PQR is  21 9 4 1 3 32 6 2 8 23 4 3 1 PQ PR                           Therefore the vector equation of the plane is 13 0 13 82 8 1 5 11 1                     r The cartesian equation is 13 8 15xy z   (Accept 13 8 15xy z   ) NB : can also take PRQ R    or PQQ R    (b) The distance of the plane from the origin is 22 2 15 15 15 169 64 1 23413 8 1   The area of the plane PQR is 1 2 13 12 3 4822 1 PQ PR        So the volume of the pyramid formed is 1 234 15 5 15 32 2 6 234     Comment : Some students did not give the proper Cartesian equation answer to (a) 4. 2 2xye   Differentiate with respect to ,x 212 0 xydy edx   When 2 10, 2, ln 2 2 yxe y   So  12 12 0dy dx   1 2 dy dx  The equation of the tangent at 10, ln 22   is

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6 / 2019 Prelim Exam / Paper 1 / Solutions 4 Qn Solution 11ln 2 022 2l n 2 2l n 2 0 yx yx yx        (accept 11ln 222yx ) Comment : Students are not confident with implicit differentiation. Some students found the equation of the normal instead of the equation of the tangent. 5. (a) Put the system of linear equations into a matrix and use row operations to solve 21 31 3 2 2 31 3 111 0 11 1 0 1 1 1 0 233 2 01 1 2 0 1 1 2 31 0 1 01 3 31 0 0 1 0 2 5 RR RR R R kk k                                For no unique solution, 10 0,k so 10k (b) If 9,k  the last equation is 25z Then 2 2 25 23yz   and   23 25 2xy z     The coordinates of the point of intersection is  2, 23, 25 Comment : For (a), a common mistake is to equate 10 25k  . This is conceptually incorrect as it assumes there is a unique solution. 6. Let ,ux then 1 2 du dx x  and so 2dx udu Replacing dx with 2ud u and x with ,u   2 2 2 2 cos cos 2 2c o s 2 sin 4 sin 2s i n 4 c o s 4 c o s 2 sin 4 cos 4sin 24 s i n 4c o s xx d x u u u d u uu d u uu u u d u uu u u u d u uu u u u c xx x x c                  Comment : Students made careless mistakes doing the integration by parts. 7. (a) Let nP be the proposition that   2 1 1111 1 !! n rn r rr n rn     for any .n   To prove 1 :P

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6 / 2019 Prelim Exam / Paper 1 / Solutions 5 Qn Solution LHS of 1P :  21 11 113 1!   RHS of  1 1 11:1 1 2 1 31!P       . Thus 1P is true. Assume that kP is true for some k   . That is,   2 1 1111 1 !! k rk r rr k rk     LHS of 1 :kP                       21 1 22 1 1 2 1 22 1 1 1 11 ! 11 1111 !1 ! 11 1111 1!1 ! 1 11 1 1 11! 11 1 21 1 1 R H S o f 1! 1! k r r k rk r kk k k k k rr r kkrr rk kkk kk kk kk kkPkk                                    Since 1P is true and kP true 1kP  is true, by mathematical induction, nP is true for all .n   (b)        20 20 2 222 311 20 2 111111 !!! 20 1 2 111 1 120 ! 2 ! 21 31120! 2! 21 3 .20! 2 rr r rr r rr rr rr rrr                       Comment : Generally alright. For (b), many students made the fatal mistake of computing   20 3 22 11 1111 !! rr rr rr rr rr    instead. 8. (a)  12 01 1 2 10 1 102 2 12 211 22 11 2 x x P Xe d x d x e ex e e ee e               

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6 / 2019 Prelim Exam / Paper 1 / Solutions 6 Qn Solution (b)  111 1 1 1 5 102 1 1 1 22 2 . 7 1 2 3 6 2PX e                So 02 0 . 5PX  Also,  11 101 1 1 22PX e     So the median lies in 01 x (NB : Do not accept “otherwise” method) (c) Let m be the median of .X Then 0 1 2 m xed x  11 2 1 2 ln 2 m m e e m       Comment : For (b), many students did not refer to the answer in (a) when justifying why the median lies in 01 x , as required by the question. SECTION B Qn Solution 9. (a)        33 23 3 23 3 33 3 11 11 2 11 xx x x x x xx x ex edy dx e ex e x e x ee ey              METHOD 2 :   3 33 3 3 1 Differentiate w.r.t. , 1 1 x xx x x ye x x dy ey edx dy ydx e dy eydx        

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6 / 2019 Prelim Exam / Paper 1 / Solutions 7 Qn Solution (b) From earlier, 3 2 ,x dy x dx e   ALTERNATIVELY : 3 3 3 33 3 1 11 2 x x x xx x dy eydx xe e x x ee e          THEN : when 0, 2dy xdx  . When 23 12,x ye e   . So a turning point is at .ye Since we are finding the maximum and minimum points in a bounded interval  1, 4 , there is a chance that the endpoints might take a y value that is higher or lower than .ye Checking the endpoints, when 1, 0xy when 34,xy e Since 302 ee The maximum value is e and the minimum value is 0 Comments : Students knew that they had to compute 0,dy dx  but incorrectly deduced that 3 0xe   , and further concluding that 3.x For those who did correctly find that 2,x they forgot to check the endpoints of the bounded interval  1, 4 to check the values of y at the endpoints. Some students found the maximum and minimum values of dy dx instead of y (c) From 3 2 ,x dy x dx e        332 22 3 3 3 12 12 3 xx x x x ex edy dx e x e x e          When 2 2 0, 3dy x dx  So 0 31 2y e 

ACS (Independent) / Mathematics Department / Mathematics HL / Year 6 / 2019 Prelim Exam / Paper 1 / Solutions 8