SOTA 2023 Prelim MAA HL Paper 3 Solutions

Yr6/5/MATAA/HP3/Aug2023 6 pages © School of the Arts, Singapore Year 6 Mathematics: analysis and approaches - MARKSCHEME Higher Level Paper 3 Preliminary Examinations Friday 28 August 2023 1 hour Instructions to candidates • Do not open this examination paper until instructed to do so. • A graphic display calculator is required for this paper. • Answer all questions on the answer sheets provided. Write your name and class on each answer sheet, and attach them to this examination paper. • Unless otherwise stated in the question, all numerical answers should be given exactly or correct to three significant figures. • A clean copy of the mathematics: analysis and approaches formula booklet is required for this paper. • The maximum mark for this examination paper is [55 marks]. Question Marks 1 2 Name: ____________________________________ Class: ____________________________________ Index: ____________________________________ 55

Yr6/5/MATAA/HP3/Aug2023 - 2 - 1. (a)(i) At E So (a)(ii) Max height at (b)(i) (b)(ii) (c) (i) Or gradient . (c) (ii) Using the GDC to find the value. ( )01 c o s cos 1 2 ya q q qp == - = = ( )2 sin 2 2 OE x a OE a pp p == - = qp=( )1c o s 2ya a p=- = sin cos dyad dxaad qq qq = =- d d yd y d xd d x q q=´ d sin sin dc o s 1 c o s ya xa a qq qq==-- 1tan 30 3 = 1 3 = sin 1 1c o s 3 q q=- ( )1sin 1 cos 3 qq=- 2.0944 1802.09 120 q q p q = =´ =

Yr6/5/MATAA/HP3/Aug2023 Turn Over - 3 - (c)(iii) (c)(iv) (d) BC= their OE – 2 their OF (e) Can have FT. (f) (g) Part (f) + MP +QN 120q=( ) ( ) 1c o s 1 cos(2.09) 31.5 2 BF y a BF a BF a a q== - =- == ( )2.0944 sin(2.0944) 1.23 OF x a OF a == - = 22 ( 1 . 2 3 ) 3 . 8 2 3 2BC a a ap=- = 2.5981tan 30 BFAF a==  2 3.8232 2(2.5981 ) 20 9.0164 2.22 AD BC AF AD a a a a =+ =+ = = d 8sin 2d d 4 4 cosd y x qq qq = =- ( ) ( ) 22 3 2 22 2 dd ddd 4 4 cos 8sin 2 d 26.8163 26.8m(3sf ) b PQ a PQ PQ PQ xyL L L L p p qqq qq q æö æö=+ç÷ ç÷èø èø =- + = = ò ò 4 sin 2.283222MP ppæö=- =ç÷èø

Yr6/5/MATAA/HP3/Aug2023 - 4 - 3325 4 sin 2.1504422QN ppæö=- - =ç÷èø 26.8163 2.2832 2.15044 31.24994 31.2(3sf ) MPTQN MPTQN MPTQN =+ + = =

Yr6/5/MATAA/HP3/Aug2023 Turn Over - 5 - 2. (a)(i) (a)(ii) Replace one of the terms (b)(i) (b)(ii) u=(1−x)n⇒dudx=n(1−x)n−1(−1)dvdx=xn−1⇒v=xnn(1−x)nxnn⎡⎣⎢⎤⎦⎥01−xnnn(1−x)n−1(−1)dx01∫xn−1(1−x)ndx01∫=xn(1−x)n−1dx01∫xn(1−x)n−1dx01∫+xn−1(1−x)ndx01∫xn−1(1−x)n−1(x+1−x)dx01∫xn−1(1−x)n−1dx01∫=In−1u=xn⇒dudx=nxn−1dvdx=(1−x)n⇒v=(1−x)n+1(−1)(n+1)xn(1−x)n+1(−1)(n+1)⎡⎣⎢⎤⎦⎥01−(1−x)n+1(−1)(n+1)nxn−1dx01∫nn+1xn−1(1−x)n+1dx01∫nn+1xn−1(1−x)n(1−x)dx01∫nn+1(xn−1(1−x)ndx01∫−xn(1−x)ndx)01∫In=nn+1(12In−1−In)2n+1n+1In=n2(n+1)In−1In=n2(2n+1)In−1

Yr6/5/MATAA/HP3/Aug2023 - 6 - (c) n=1 Assume n=k is true n=k+1 Since n=1 is true and n=k+1 is true when n=k is true, hence by induction the statement is true for positive integers. (Note: must get at