2023 CCHM Prelim AMath P2 - MS

2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/02 [Turn over Name: Class: Class Register Number: PRELIMINARY EXAMINATION 2023 SECONDARY 4 ADDITIONAL MATHEMATICS 4049/02 Paper 2 Tuesday 29 August 2023 Candidates answer on the Question Paper. 2 hours 15 minutes MARKS SCHEME This document consists of 20 printed pages and 2 blank pages.

2 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/02 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 2 0ax bx c+ + = , 2 4 2 b b acx a −  −= Binomial expansion nrrnnnnn bbar nbanbanaba ++     ++     +     +=+ −−− ......21)( 221 , where n is a positive integer and ! ( 1) ... ( 1) !( )! ! n n n n n r r r n r r  − − +== − . 2. TRIGONOMETRY Identities 22sin cos 1AA+= 22sec 1 tanAA=+ 22cosec 1 cotAA=+ sin( ) sin cos cos sinA B A B A B =  cos( ) cos cos sin sinA B A B A B= tan tantan( ) 1 tan tan ABAB AB = sin 2 2sin cosA A A= 2 2 2 2cos 2 cos sin 2cos 1 1 2sinA A A A A= − = − = − 2 2 tantan 2 1 tan AA A= − Formulae for ABC sin sin sin a b c A B C== 2 2 2 2 cosa b c bc A= + − 1 sin2 ab C=

3 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/02 [Turn over 1 (a) The line 4 3 2xy=+ intersects the curve 2 50x xy− + = at the points A and B. Find the midpoint of AB. [5] 2 4 3 2 ...(1) 5 0 ...(2) xy x xy =+ − + = From (1): 3 4 2 42 ...(3)3 yx xy =− −= Sub. (3) into (2): ( )( ) 2 22 2 2 42 503 3 4 2 15 0 2 15 0 2 15 0 5 3 0 5 or 3 sub. into (3): 146 or 3 xxx x x x xx xx xx x y −− + = − + + = − + + = − − = − + = =− =−  143, 3A−− and B(5, 6) A1 M1 f.t. – substitution M1 f.t. – solving quadratic M1 – either correct x or y values 14 635 3Midpoint of , 22 21, 3 AB  −+−+=   = M1 f.t. – midpoint formula

4 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/02 (b) Find the least value of the integer h for which 2 5hx x h++ is positive for all real values of x. [3] (c) Given that the line 3y x p=+ is tangent to the curve 2 5y x x q= + + , where p and q are integers, prove that p and q are consecutive numbers. [4] 2 3 ... (1) 5 ... (2) y x p y x x q =+ = + + Sub. (1) into (2): 2 2 2 35 5 3 0 20 x p x x q x x q x p x x q p + = + + + + − − = + + − = a = 1, b = 2, c = k – c Line is tangent to curve → 2 40b ac−= ( )( ) 22 4 1 0 4 4 4 0 4 4 4 1 qp qp qp qp − − = − + = =+ =+ Since q = 1 + p, q will always be the next number after p. Hence, p and q are consecutive numbers (proved). M1 – substitution M1 – forming quadratic M1 f.t. – any use of discriminant A1 – with explanation 2For 5 0,hx x h+ +  ( ) ( )( ) 2 discriminant 0 25 4 0 25 4 0 5 2 5 2 0 55 or 22 hh h hh hh  − − − +  −  Since h > 0, 5 2h .  Least integer value of h = 3. B1 – discriminant A1 M1 f.t. – factorising quadratic

5 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/02 [Turn over Alternative: 2 5 d 25d y x x q y xx = + + =+ Since line is tangent to curve and gradient of line = 3, 2 5 3 22 1 x x x += =− =− 2 3 ... (1) 5 ... (2) y x p y x x q =+ = + + Sub. (1) into (2) and x = −1: ( ) ( ) ( ) 2 2 35 3 1 1 5 1 34 1 x p x x q pq pq qp + = + + − + = − + − + − + =− + −= Since the difference between q and p is 1, q will always be the next number after p. Hence, p and q are consecutive numbers (proved). M1 – equate d d y x to 3 M1 f.t. – finding x M1 – substitution A1 – with explanation

6 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/02 2 (a) By considering the general term in the binomial expansion of 7 3 2x x  − , explain why there are only odd powers of x in this expansion. [3] (b) Find the term independent of x in the expansion of 7 32 25 2xx xx    −−       . [3] ( ) 73 1 7 2 r r rTx r x − +  =−  ( )( ) ( ) ( ) 21 3 21 4 7 2 7 2 rrr r r xxr xr −− − =− =− Power of x = 21 – 4r Since 4r is an even number for all non-negative integer values of r and 21 is an odd number, then 21 – 4r is always an odd number. Therefore, there are only odd powers of x in this expansion. A1 – conclusion B1 – general formula M1 f.t. – finding powers of x Consider 21 4 1r−= , 4 20 5 r r = = ( ) ( ) ( ) 7 5 21 4 53 2 2 2 72 5 5 2 ... 2 ... 2 5 5... 672 ... 2 x x x xx x x xx x −      − − = + − + −             = − + −  Term independent of 672 5 3360 x=−  =− A1 M1 f.t. – identifying x term M1 f.t. – expansion

7 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/02 [Turn over 3 The expression 7sin 3cos+ is defined for 0° ≤ θ ≤ 360°. (a) Using ( )sinR + , where R > 0 and 0° < α < 90°, solve the equation 7sin 5 3cos=− . [5] ( ) ( ) 7sin 5 3cos 7sin 3cos 5 58 sin 23.1985... 5 5sin 23.1985... 58     =− += +  = +  = 1 5basic angle sin 58 41.0359... − =  = 23.1985... 41.0359... or 180 41.0359. .. 17.8 or 115.8 (1 dec. pl. )   + =  −  =   ( )7sin 3cos sin sin cos cos sin R RR         + = + =+ Comparing, 7 cos and 3 sinRR == 22 1 73 58 3tan 7 3tan 7 23.1985... R   − =+ = = =  = ( )7sin 3cos 58 sin 23.1985...   + = +  M1 – attempt to find α [only accept 1 3tan 7 −   or 1 7tan 3 −   ] M1 f.t. – substitute R-form M1 f.t. – basic angle A1 M1 – attempt to find R

8 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/02 (b) State the largest and smallest values of ( ) 2 7sin 3cos 12+− and find the corresponding values of θ. [4] ( ) ( ) ( ) 22 2 largest value of 7sin 3cos 12 58 12 or 46 58 12+− −−=− = occurs when or 2723.1985... 90 0 + =   66.8 or 246.8 (1 dec. pl.) =   ( ) ( ) 22 smallest value of 7sin 3cos 12 0 12 12 + − = − =− occurs when 23.1985... 180 or 360 + =   156.8 or 336.8 (1 dec. pl.) =   B1 B1 B1 B1

9 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/02 [Turn over 4 The diagram shows a circle passing through the points A, B, C, D and E. The straight line FDG is tangent to the circle at D while FAB and FEC are secant lines. Given that angle FDA = angle ADB, (a) show that triangle ABD is an isosceles triangle, [2] (b) prove that AF BF EF CF =  . [3] A D C B E G F ABD FDA = (alternate segment theorem) = ADB Since ABD ADB = , they form base angles of isosceles triangle. Thus, triangle ABD is an isosceles triangle. (shown) B1 B1 FAE FCB = (exterior  of cyclic quadrilateral) AFE CFB = (common ) Thus, triangle AFE is similar to triangle CFB. AF EF CF BF= (ratio of corresponding sides are equal) AF BF EF CF =  (proved) B1 B1 B1

10 2023 Preliminary Exam/CCHMS/Secondary 4/Additional Mathematics/4049/02 Alternative: FBE FCA = (s in same segment) EFB AFC = (common ) Thus, triangle EFB is similar to triangle AFC. AF CF EF BF= (ratio of corresponding sides are equal) AF BF EF CF =  (proved) B1 B1 B1