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