2025 Prelim 4E5N Presbyterian HS P1 Solutions

Name: Index No.: Class: PRESBYTERIAN HIGH SCHOOL ADDITIONAL MATHEMATICS 4049/01 Paper 1 25 August 2025 Monday 2 hours 15 min PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL PRESBYTERIAN HIGH SCHOOL 2025 SECONDARY FOUR EXPRESS / FIVE NORMAL (ACADEMIC) PRELIMINARY EXAMINATIONS SUGGESTED ANSWERS For Examiner’s Use Qn 1 2 3 4 5 6 7 8 9 10 11 12 13 Marks Deducted Marks Category Accuracy Units Notations Others Question No. Setter: Mr Tan Lip Sing Vetter: Ms Sabrina Tan This question paper consists of 23 printed pages and 1 blank page. TOTAL MARKS 90

2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 2 0,ax bx c+ += 2 4 2 b b acx a −± −= Binomial expansion 1 22( ) ... ... ,12 n n n n nr r nnn nab a a b a b a b b r −− −  + = + + ++ ++     where n is a positive integer and 2. TRIGONOMETRY Identities 22sin cos 1AA+= 22sec 1 tanAA= + 22cosec 1 cotAA= + sin( ) sin cos cos sinAB A B A B±= ± cos( ) cos cos sin sinAB A B A B±=  tan tantan( ) 1 tan tan ABAB AB ±±=  sin 2 2sin cosA AA= 22 2 2cos 2 cos sin 2cos 1 1 2sinA AA A A= − = −=− 2 2 tantan 2 1 tan AA A= − Formulae for sin sin sin abc ABC= = 2 22 2 cosa b c bc A=+− 1 sin2 bc A∆=

3 1 (a) Differentiate 2 3ln 1 x x   + with respect to x. [3] (b) Hence find 2 2 d1 x xx +∫ . [2] 1 (a) ( ) ( ) 2 2 2 3ln ln 3 ln 11 d12 d1 xLe t y x xx yx xxx = = −+ + = − + OR ( ) ( ) ( ) ( ) ( ) 2 22 2 2 22 22 2 2 3 132 1d 3d 1 3 36 1 31 1 1 x xx xy xx x x xx xx x xx +− + = + +− += × + −= + (b) 22 22 2 12 3 d ln11 21 3 d ln11 3ln ln 1 xx xcxx x xx dx x cx xx xxc x   −=+   ++   = −+ ++  = −+  + ∫ ∫∫ OR ( ) 2 22 d1 d1 1 y x A Bx C x xxxx −+= = + ++

4 Solve for A, B & C. 1, 2, 0AB C== −= ( ) 2 22 1 12 11 xx xxxx − = − ++ 22 22 2 12 3 d ln11 21 3 d ln11 3ln ln 1 xx xcxx x xx dx x cx xx xxc x   −=+   ++   = −+ ++  = −+  + ∫ ∫∫

5 2 It is given that 1cos 2A= and 1sin 2 B=− where 0 90 A°< < ° and 180 270B°< < ° . Find, without using a calculator, the exact value of ( )cos AB− , leaving your answer in the form 26pq + , where p and q are real numbers. [4]