2024 CCHY A Math P1 Ans

* [Turn_over 2024 Preliminary Examination Secondary Four Express / Five Normal Academic CANDIDATE NAME MARKING SCHEME CLASS INDEX NUMBER ADDITIONAL MATHEMATICS 4049/01 Paper 1 22 August 2024 2 hours 15 minutes Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, class and index number on the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use paper clips, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an approved scientific calculator is expected, where appropriate. You are reminded of the need of clear presentation in your answers. Up to 2 marks may be deducted for improper presentation. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 16 printed pages. For Examiner’s Use Presentation Deduction – 1 / – 2 TOTAL 90 CHUNG CHENG HIGH SCHOOL (YISHUN)

2 [Turn_over Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 2 0bx cax + += , 2 4 2 b b acx a −± −= Binomial Expansion ( ) 1 22 12 n nn n n nr rnn nab a ab a b a b b r −− −  + =+ + ++ ++      , where n is a positive integer and ( ) ( ) ( )11! !! ! n nn n rn r rn r r − −+ = = −  2. TRIGNOMETRY Identities 22 1cs osin A A+ = 22sec 1 tanA A= + 22 coc sec 1 to AA += ( ) os sin sin c cos sinB A B ABA±± = ( ) oc sos c si sco ins nB A B ABA± =  ( ) atant tn taan n tan1 BB B AA A ±± =  sin 2 2sin cosA AA= 22 2 2cos2 cos sin 2cos 1 1 2sinA AA A A= − = −=− 2 2tantan2 1 tan AA A= − Formulae for ∆ABC sin sin sin abc ABC= = 2 22 2 cos Aa c bcb += − ∆ 1 sin2 bc A=

3 [Turn_over 1 (a) Given that the line 2y xk= − meets the curve 2 214 ky x kx= −+ , find the range of values of k. [4] 2 2 124 k x kx x k− += − 2 2 12 04 k x kx x k− +− + = 2 2 2 104 k x kx x k− − ++= , 2 2, 14 ka b k ck= = −− = + Line meets curve, 2 40b ac−≥ ( ) ( ) 2 12 2 4 10 4k kk −− − +≥  224 84 0k k kk+ +− −≥ 23 7 40kk + +≥ ( )( )3 4 10kk+ +≥ 11 or 13kk≤− ≥− (b) A student claimed that when 1 2k =− , the curve does not intersect the line. By showing your workings clearly, explain whether the statement is valid. [2] No intersection, 2 40b ac−< 23 7 40kk + +< ( )( )3 4 10kk+ +< 1113 k− < <− Since 1 2k =− does not lies within the range. The statement is not valid.

4 [Turn_over 2 A triangle has a base of 1 32 16 m22 48  −+ and a height of h m. Given that the area of the triangle is ( ) 22 2 3 m− , find, without using calculator, the value of h in the form 6 5 ab + where a and b are integers. [5] Simplifying base: 1 32 16 22 48 −+ = 14 21 6 22 43 −+ = 1 22 4 123 −+ = 3 43 42 66 6 −+ = 42 33 6 − = 42 33 6 66 − × = 83 92 6 − 1 83 92Area = 26 h−×× 83 922 2 3 = 12 h−−× ( )12 2 2 3 83 92 h − = − 24 2 12 3 8 3 9 2 83 92 83 92 h −+= × −+ 192 6 432 288 108 6 192 162h +−−= − 84 6 144 30h += ( )6 14 6 24 30h + = 14 6 24 5h +=

5 [Turn_over 3 Show that 23 5 21xx−+ is always positive for all real values of x. [3] Method 1: ( ) ( )( ) 2 2 4 5 4 3 21 227 0 b ac− = −− = −< Since discriminant < 0, there is no real roots Since the coefficient of 2 0x > , the graph lies entirely above the x- axis Hence, 23 5 21xx−+ is always positive for all real values of x. Method 2: 23 5 21xx−+ = 2 537 3xx −+ = 22 5537 66x   −− +     = 2 5 2537 6 36x  − −+ = 2 5 2273 6 12x −+ Since 2 5 06x −≥ for all real values of x 2 5 22730 6 12x −+> Hence, 23 5 21xx−+ is always positive for all real values of x. (ii) The curve ny ax= , where a and n are constants, passes through ( )2, 48 , ( )3, 108 and ( ), 192k . Find the values of a, n and k where 0k > . [3] 48 2 n a= 48 2na= ---- (1) 108 3 n a= 108 3na= ---- (2) 48 108 23nn= 3 108 482 n n = 39 24 n  = 2 33 22 n   =   Therefore n = 2 When n = 2, 2 48 2 a= a = 12 212yx= 2192 12 k= 216 k= 4 or 4 (rej)kk= =−

6 [Turn_over 4 (a) Given that ( ) ( )23f 43 xx x += − , find ( )f x′ . [3] Method 1: ( ) ( )23f 43 xx x += − ( ) ( ) ( ) ( ) ( ) ( ) 1 2143 2 23 43 4' 2f 243 x xx x x − − −+ − = − ( ) ( )( ) ( )( ) ( ) 1 2 43 2 23 43 2'f 43 x xx x x − − −+ − = − ( ) ( ) ( )( ) ( )( ) 1 2 4 3 4 32 2 32'f 43 xx xx x − − − −+= − ( ) [ ] ( ) ' 3 2 8 64 6f 43 xxx x −− −= − ( ) ( ) 3 2 4 12'f 43 xx x −= − Method 2: ( ) ( )( ) 1 2f 2 34 3xx x − = +− ( ) ( )( ) ( ) ( ) ( ) 13 22 1'f 2 43 23 43 4 2xx x x −− = − ++− −  ( ) ( ) ( )( ) ( )( ) 3 2 'f 43 2 43 232xx x x − = − −+ + − ( ) ( ) [ ] 3 2 'f 43 86 46xx x x − = − −− − ( ) ( ) 3 2 4 12'f 43 xx x −= − (b) Find the set of values of x for 38ln 85 xy x +=  − to be a decreasing function. [3] 38ln 85 xy x −=  − ( ) ( )ln 3 8 ln 8 5y xx= +− − d8 8 d 38 8 5 y x xx= −+− ( )( ) d 64 40 64 24 d 3 885 yx x x xx −− −= +− ( )( ) d 64 d 3 885 y x xx −= +− For decreasing function, d 0d y x < ( )( ) 64 03 885xx − <+− ( )( )3 8850xx+ −> 35 or 88xx<− >

7 [Turn_over 5 (a) Explain why 32 2 2 4 13 10 ( 4)( 2) xx x xx −+− +− is considered an improper fraction. [1] The degree of x in the numerator is the same as the degree of x in denominator thus, it is an improper fraction. (b) Express 32 2 2 4 13 10 ( 4)( 2) xx x xx −+− +− in the form of 2( 4)( 2) Bx CA xx ++ +− and hence, express it in partial fractions. [6] Using long division: 32 22 2 4 13 10 5 6 2( 4)( 2) ( 4)( 2) xx x x xx xx −+− + = ++− +− ( ) 22 56 ( 4)( 2) ( 4) 2 x Ax B C xx x x ++ = ++− + − ( )( ) 25 6 2 ( 4)x Ax B x C x+= + − + + When x = 2, 16 8 2 C C = = Comparing x2: 0 2 AC A = + =− Comparing constant: 624 628 1 BC B B = −+ = −+ = ( ) 32 22 2 4 13 10 1 2 2 2( 4)( 2) ( 4) 2 xx x x xx x x −+− − = +++− + −

8 [Turn_over 6 (a) Given that ( )( ) ( ) 26 91 8 21 1 2x x A x x Bx C− += + −+ −+ for all values of x, find the values of A, B and C. [3] ( )( ) ( ) 26 91 8 21 1 2x x A x x Bx C− += + −+ −+ Comparing x2: 6 = 2A A = 3 Comparing x: 9 AB−= −+ 6B=− Let 1x= 6 9 18 15 BC CB −+ = −+ = + 9C = (b) Given that ( ) 32f 2 11 3 36x x xx= − ++ , show that 23x+ is a factor of ( )f x and hence solve the equation ( )f0 x = . [5] ( ) 32f 2 11 3 36x x xx= − ++ ( ) ( )Let 2 3 be a factor of x fx+ ( ) ( ) ( ) ( ) 32 f 1.5 2 1.5 11 1.5 3 1.5 36 6.75 24.75 4.5 36 0 − =− − − +− + = − − −+ = Since ( )f 1.5 0−= , by factor theorem, 23x+ is a factor of ( )f x . ( )( ) 32 22 11 3 36 2 3x x x x ax bx c− ++= + ++ Comparing x3: A = 1 Comparing constant: 36 3 c= 12c= Comparing x2: 11 3 2ab−=+ 7b=− ( )( ) 32 22 11 3 36 2 3 7 12x xx x xx− ++= + −+ ( ) ( )( ) 2f 2 3 7 12x x xx= + −+ ( )f0 x = ( )( ) 22 3 7 12 0x xx+ −+ = ( )( )( )4 32 3 0xx x− − += 14 or 3 or 1 / 1.5 2xxx= = = −− or ( )( ) 32 22 11 3 36 4 2 9x x x x x bx− ++=− +−

9 [Turn_over 7 Without using a calculator, evaluate 555 25 log 9 2 log 6 4 log 3 log 4 +− . [4] 555 25 log 9 2log 6 4log 3 log 4 +− = 2 24 55 5 5log 9 log 6 log 3 log 4 +− = ( ) 2 5 5 log 9 36 81 log 4 ×÷ = 5 5 5 log 4 log 4 log 25 =