2022 ZHSS AMath Prelims P1 Ans

[Turn over ZHONGHUA SECONDARY SCHOOL PRELIMINARY EXAMINATION 2022 SECONDARY 4 EXPRESS/ 5 NORMAL (ACADEMIC) Candidate’s Name Class Register Number MARKING SCHEME ADDITIONAL MATHEMATICS 4049/01 PAPER 1 12 September 2022 2 hour 15 minutes Candidates answer on the Question Paper READ THESE INSTRUCTIONS FIRST Write your name, class and register number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use paper clips, glue or correction fluid. Answer all questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the 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 for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 90. ___________________________________________________________________ This question paper consists of 22 printed pages (including this cover page) For Examiner’s Use 90

2 [Turn over Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 02 =++ cbxax a acbbx 2 42 −−= Binomial expansion nrrnnnnn bbar nbanbanaba ++     ++     +     +=+ −−− 221 21)( , where n is a positive integer and ! )1()1( )!(! ! r rnnn rnr n r n +−−=−=       2. TRIGONOMETRY Identities 1cossin 22 =+ AA AA 22 tan1sec += 22cosec 1 cotAA=+ BABABA sincoscossin)sin( = BABABA sinsincoscos)cos( = BA BABA tantan1 tantan)tan(  = AAA cossin22sin = AAAAA 2222 sin211cos2sincos2cos −=−=−= A AA 2tan1 tan22tan −= Formulae for ABC C c B b A a sinsinsin == Abccba cos2222 −+= Abcsin2 1=

3 [Turn over 1 Express ( )( ) 32 2 2 4 18 24 x x x xx − + − −+ in partial fractions. [6] 3 2 3 2 32 2 2 4 8 2 4 18 M1 attempt at l ong division (2 4 8 16) x x x x x x x x x − + − − + − − − + − 7 2 x−− ( )( ) ( )( ) ( )( ) 32 22 22 2 2 4 18 7 2 2 2 4 2 2 4 72 B1( 2) ( 4)24 7 2 ( 4) ( )( 2) Let = 2, 7 2+2 (8) M1 attempts to find , , 2 x x x x x x x x x x A Bx C xxxx x A x Bx C x x A A B C A − + − + =− − + − + + ++ =+ −+−+ + = + + + − = = 2Comparing the coefficients of and const ant, 0 , 2 2 4 2 2 8 2 x A B B A B AC C = + =− =− =− =− ( ) 32 2 A2 all correct A,B,C 3 (A1 any 2 correct values) 2 4 18 2 C x x x xx = − + − −+ ( ) 2 2 2 32 A1( 2) ( 4)4 x xx −+= − − −+

4 [Turn over 2 Without using a calculator, (i) show that 31tan 285 13 += − , [3] tan 285 tan 75 B1 = tan(45 30 ) tan 45 tan 30 = M1 using add ition formula1 tan 45 tan 30 11 33 = 1 31 3 =−  − +  + − −   + − − 31= 31 31 = 13 +− − + − (ii) express 2sec 285 in the form 3ab+ , where a and b are integers. [4] ( ) 22 2 2 sec 285 1 tan 285 B1 31 = 1 13 3 2 3 1 = 1 1 2 3 3 4 2 3 = 1 B1 correct expansion of 4 2 3 = = +   ++ − +++ −+ ++ − 2(2 3) 2 3 1 M1 multiplying by conjugate sur ds 2(2 3) 2 3 4 4 3 3 = 1 1 = 8 4 3 A1 +++ −+ +++ +

5 [Turn over 3 Find the range of values of k for which the curve 2 8y kx x=+ lies entirely above the line y x k=− . [4] ( )( ) ( )( ) 22 2 For 8 , 7 0 0 B1 and 7 4 0 M1 0 and 7 2 7 2 0 kx x x k kx x k k k k k k k +  − + +   −   − +  770 and or B122 7Ans: A12 k k k k  −  

6 [Turn over 4 A curve is such that ( ) 2 223 d 4 15 d 32 y xx x =− − − and the point P(1, −6) lies on the curve. The gradient of the curve at P is 3. Find the equation of the curve. [6] ( ) ( ) ( ) 23 12 2 d 4 15 3 2 d M1 attempt to integra te d 15 3 24 = A1 (or equivalent)2 ( 1)(3) 25 = 32 dAt P, 1, 3 d 3 2 5 M1 y x x xx xx C Cxx yx x C −− −− = − − − −− − +−− ++ − == = + +  ( ) ( ) 12 1 1 substituting to find 4 d 2 5 3 2 4d 2 5ln(3 2) ln(3 2) 4 3 2 d B11 3 3 Substituting 1, 6 5ln1 6 2 4 3 0 C C y xxx x x xy x D x x C xy D D −− − − −= = + − − −−= + − + − = +− = =− − =− + − + =  A1 2 5ln(3 2) 4 A13 xyx x −=− + −

7 [Turn over 5 The equation of a polynomial is given by 32( ) 3 5 3 2p x x x x= − − + . (a) Show that ( 2)x− is a factor of ()px . [1] ( ) ( ) ( ) 32 (2) 3 2 5 2 3 2 2 24 20 6 2 0 By factor theorem, ( 2) is a factor of ( ) . p x p x = − − + = − − + = − Alternatively by long division, need to state remainder = 0 (b) Hence, solve the equation ( ) 0px = , expressing non-integer roots in surd form. [3] ( ) ( )( ) ( )( ) 2 2 2 2 3 1 M1 comparing coefficient of 6 5 1 2 3 1 0 B1 correct quadrat ic factor 1 1 4(3)( 1) 2 or 2(3) p x x x bx x b b x x x xx = − + − − + =− = − + − = −  − −== 1 13 2 or A1 6 xx −==

8 [Turn over 6 The diagram shows a circle passing through the points P, Q and R. The points X and Y lie on QP and RP respectively. The tangent ST to the circle at P is parallel to YX. (a) Prove that Q, R, Y and X lie on a circle. Let (angles in alternate segment) B1 (alternate angles, parallel to ) B1 = 180 (adjacent angles on a straight line) B1 SPY x PQR SPY x PYX SPY ST YX x XYR PYX  =   = =  =   = − 180 since 180 =180 B1 and are angles in opposite segments B1 hence x XQR PQR x XYR XQR x x XYR XQR = −   = =   + = − +    , , and lie on a circle.Q R Y X [5]

9 [Turn over 6 (b) The line RP is extended to Z such that angle PZT = 90  . Explain why a circle passing through the points P, T and Z has its centre at the midpoint of PT. [2] Since 90°, it is an angle in semicircle , B1 is the diameter of the circle passing through , and B1 with centre at the midpoint of . PZT PT P T Z PT =

10 [Turn over 7 (a) The equation of a curve is ( ) 3 25y x x=− . Find the range of values of x for which y is decreasing. [4] ( )