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] 2 13cos , sin22AA= = 11sin , cos 22 BB= −= − ( )cos cos cos sin sin 11 31 22 22 13 22 22 13 22 11 2644 AB A B A B−= +  = −+ −        = −− −−= = −−

6 3 Baking powder is poured onto a flat surface at a constant rate of 312 cm sπ − , forming a right circular cone. The radius of the cone is always 1 18 of its height. Find the rate of change of the radius of the cone after 3 seconds of pouring. 21Volume ofcone 3 rhπ = [5] Volume of cone, ( ) 231 18 63V rr rππ= = 218dV rdr π= After 3s, 2 36V ππ= ×= 366 1 r r ππ = = When 3, 1 ,t s r cm= = dV dV dr dt dr dt= × ( ) 2 2 18 1 1 cm/s9 dr dt dr dt ππ= × = Rate of change of radius of cone after 3 s is 1 cm/s9

7 4 A and B are the points of intersection of the line 4 21yx= + and the curve 34y x xy−= . (a) Find the coordinates of A and of B. [4] (b) Henry says that the line 245yx−= is perpendicular to the line AB. Is he correct? Justify your answer with workings. [3] 4 (a) ( ) 4 2 1 .........(1) 21 ........... 24 yx xy = + += 3 4 ...........(3)y x xy−= Substitute (2) into (3): ( )( ) 2 2 2 21 2134 44 33 224 6 34 8 4 8 2 30 2 14 3 0 13 24 11 28 xx xx x x xx x xx x xx xx x or x y or y ++  −=   +−= + +− = + + −= − += = =− = =− 11,22A = 31,48B = −− (b) Gradient of 245yx−= is 2 Gradient of AB 11 182 31 2 42 −− = = −− Since 121 2× ≠− , the 2 lines are not perpendicular to each other. Henry is not correct.

8 5 (a) Write down and simplify the first three terms in the expansion, in descending powers of x, of 8 32 x  − . [2] (b) Given that there is no x term in the expansion of ( ) 8 2 312 2x kx x −− −  , find the constant term in the expansion. [4] 5 (a) ( ) ( ) 82 87 6 2 883 332 2 2 2 ........12 3072 16128256 ......... x xx xx     − =+ −+ − +          = −+ + (b) ( ) ( ) ( ) ( ) 8 2 2 2 212 1 3072 161281 2 256 ......... 256 2 256 2 3072 3072 16128 ...... 256 512 6144 3072 16128 ...... x kx x x kx xx x k x k x k x k −− −  = −− − + +  = − + +−+ = −++ − + Since there is no x term in the expansion, 3072 512 0 1 6 k k −= = The constant term 256 6144 16128 6400 2688 3712 k= +− = − =

9 6 (a) Express 244 3y xx= −− in the form ( ) 2 px q r++ where p, q and r are constants. [2] (b) Hence, explain whether 24 4 30xx− −= has any real solutions. [2] 6 (a) ( ) 2 2 22 2 2 2 4 43 43 1143 22 14 13 2 142 2 xx xx xx x x −+− = − −−   = − −+− −− −    = − − +− = −−− (b) 24 4 30 0( ) xx y x axis − −= = − Since the maximum value of y is 2− , the graph of 244 3y xx= −− will not intersect the x -axis, 24 4 30xx− −= has no real solutions. OR 2 2 14 20 2 11 22 11 ()22 x x x no real solutions − − −=  −= −   − = ±−

10 7 Peter constructed an open fish tank with a rectangular base of length 4 l m, breadth l m, and height h m. He wanted the total outer surface area of the fish tank to be 5 2m . (a) Show that the volume of the tank, 3mV , is given by ( ) 32 54.5V ll= − [3] (b) Determine the area of the rectangular base for the tank to contain the maximum amount of water when filled to the brim. [4]