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 − −