PHS 2024 A Math Prelim P2 MS

Name: Index No.: Class: PRESBYTERIAN HIGH SCHOOL ADDITIONAL MATHEMATICS 4049/02 Paper 2 20 August 2024 Tuesday 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 2024 SECONDARY FOUR EXPRESS / FIVE NORMAL (ACADEMIC) PRELIMINARY EXAMINATIONS MARK SCHEME Q1 – 8 Mr Gregory Quek Q9 – 10 Mr Tan Lip Sing

2 1 (a) Write down, and simplify, the first three terms in the expansion of 5 23 x  − in descending powers of x. [2] (b) Given that there is no term independent of x in the expansion of ( ) 5 2 25 3 ,ax x +−  hence find the value of the constant a. [3] ( ) ( ) 52 435 52 2 23 3 5 3 3 ... 2x x x      − = + − + − +             5 2 .23 810 1080243 .. xx x− −= +  + B2: Three correct terms (B1: Two correct terms) ( ) ( ) 2 2 5 2 425 810 108235 03 ...ax ax x xx + − = +   −+  + Term independent of x = ( )( ) ( ) 2 2 85 10 0243 x ax +  M1: Derive terms indep. of x 1215 1080 0a + = M1: Equate terms to zero 1215 1.1251080a =− =− A1: Accept 118− or 9 8−

3 2 In the figure, ABCD is a rectangle inscribed within a semicircle of radius 4 cm and centre O. It is given that AB = x cm and BC = y cm. (a) Show that the area of the rectangle, A cm, is given by 21 64 .2A x x=− [2] (b) Find the exact value of x for which A has a stationary value. Give your answer in the form 2,k where k is an integer. [4] O A B C D x cm y cm 2 22 14 2yx =−  2116 4yx=− M1: Correct application of Pythagoras Theorem 2116 4A x x=− 21 644A x x=− M1: Factorise and simplify surd 21 642A x x=− (a.g.) ( ) ( ) 1 22 2d 1 1 1 64 2 64d 2 2 2 A x x x xx − = − − + −  M1, M1: Product rule ( ) ( ) 11 2 2 2 22d 1 1 64 64d 2 2 A x x xx − = − − + − ( ) ( ) 1 2 2 2 22 2 d 1 32 64 64d2 64 Ax x x xx x − −= − − + − =  − For stationary value, 2 2 d 32 0d 64 Ax x x −== − M1: Equate dA/dx to zero 232 0 x−= 32 4 2x== A1

4 3 The diagram shows a triangle ABC is inscribed in the circle with centre O. BD is a tangent to the circle at B and AB is parallel to CD. Point M is the midpoint of BC. (a) Prove that triangles ABC and BCD are similar. [3] (b) Prove that ABMO is a trapezium. [2] (c) Prove that 2 2 BCOM CD= . [3] *Penalise 1m per question for any missing or incorrect reasons. A B C D O M ( . , // )ABC BCD alt s AB CD =  M1 ()BAC CBD alternate segment theorem=  . ( )Triangles ABC and BCD are similar AA similarity A1 Since and are the midpoints of and re spectively,O M AC BC // ( )OM AB midpoint theorem M1 . ( )ABMO is a trapezium one pair of parallel sides A1 ( . )AB BC corr sides of similar sBC CD= M1 Since )2 (midpoA int tM heoBO rem= M1 2OM BC BC CD= 2 2 BCOM CD= A1

5 4 Milk is poured into an empty cup and heated. The temperature, C,mT  of the milk in the cup, t minutes after it is heated, is modelled by the formula, ( )5 2 20. t mT =+ (a) State the initial temperature of the milk. [1] Coffee is poured into another empty cup. The temperature, C,cT  of the coffee in the cup, t minutes after it is poured, is modelled by the formula, ( )60 2 25. t cT − =+ (b) Find the time taken for the temperature of the coffee to drop to 35 C. [3] (c) Find the time taken for the milk and the coffee to reach the same temperature. [4] Initial temperature of milk = ( ) 0 5 2 20 25 C+ =  B1 ( )60 2 25 35 t− += ( ) 35 25 12 60 6 t− −== M1: Isolate ( )2 t− ( ) 1lg 2 lg 6 t− =  M1: Take lg on both sides ( ) 1lg 2 lg 6t −=  ( )l 2.58491lg g 26t = −  = 2.58 min (3 )t sf A1 ( ) ( )5 2 20 60 2 25 tt − + = + M1: Equate mT to cT ( ) ( ) ( ) 2 5 2 20 2 60 25 2 t t t + = + M1: Multiply 2t throughout / obtain quad. eqn. ( ) ( ) 2 5 2 5 2 60 0 tt − − = ( ) ( ) 2 2 2 12 0 tt − − = Let u = ( )2 t , 2 12 0uu− − = ( )( )4 3 0uu− + = M1: Solve quadratic equation 43u or u= =− (rejected) ( )24 t = 2t = min A1 Let u = ( )2 t , 605 20 25u u+ = + 25 5 60 0uu− − = 2 12 0uu− − =

6 5 It is given that 3 2 2 3f ( ) 2 13 6x x x y xy y= − − − . (a) Show that 3xy− is a factor of f ( ).x [2] (b) If y = 1, find an expression in fully factorised form for f ( )x . [3] (c) Hence solve the equation 6 4 22e e 13e 6 0z z z− − − = and show that the solution may be written in the form ln p , where p is an integer. [3] 3 2 2f (3 ) 2(3 ) (3 ) 13(3 ) 6y y y y y y= − − − 3 3 3 3f (3 ) 54 9 39 6 0y y y y y= − − − = M1: Sub. into f(x) & simplify Since f (3 ) 0y = , by Factor Theorem, 3xy− is a factor of f ( ).x AG1 ( ) 32 2 Let f ( ) 2 13 6 3 2 2 x x x x x x bx = − − − = − + + Comparing 2x term: ( )( )1 3 2b− = + − 5b = ( ) 2f ( ) 3 2 5 2x x x x  = − + +  A1 ( )( )( )f ( ) 3 2 1 2x x x x = − + + A1 M1: Comparing coefficient (or long division) 2 6 4 2 Let e , we get 2e e 13e 6 0 z z z z x = − − − = ( )( )( ) 2 2 2e 3 2e 1 e 2 0z z z − + + = M1: Sub. 2e zx= into (b) ( ) ( )2 2 2e 3 2e 1 e 2z z z or rejected or rejected = = − = − A1: Seen 2e3z = 2lne ln 3z = 2 ln 3z = 1 ln 32z = ln 3z= A1

7 6 (a) Given that tan 2cosec= , show that 2cos 2cos 1 0.+ − = [3] (b) Using part (a), find the exact value of cos in simplest form, given that 0 90 .   [3] (c) Hence find the value of 2sec 