2022 ZHSS AMath Prelims P2 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 4049/02 PAPER 2 13 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 21 printed pages (including this cover page) For Examiner’s Use 90

2 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 1 (a) By using suitable substitution or otherwise, solve the equation ( )e 3 e 5ex x x− −= , giving your answer correct to 2 decimal places. [4] ( )( ) 2 2 2 2 3 15 35 5 3 0 Let 5 3 0 1 1 4 5 3 10 1 61 10 1 61 (rej. negative)10 1 61ln 10 0.384156 0.38 (2 d.p) x x xx xx x x ee ee ee ue uu u e x −= −= + − = = + − = −  − −= −= −+= −+=  =− =− B1- Quadratic equation M1- Quadratic Formula M1 A1

4 (b) Solve 75xx+ = − . [3] ( ) ( )( ) 2 2 2 7 10 25 11 18 0 9 2 0 9 or 2 (rejected) 9 75 75 x x x x x x x xx x xx x x + = − + − + = − − = == + = − −  = = + 2 Given that tan p = and that  is acute, find in terms of p, (a) ( )sin − [1] ( ) 2 sin sin 1 p p − =− + =− (b) tan 2   − [1] 1tan 2 p   −= M1- Square both sides M1- Solving quadratic equation A1 B1 B1- o.e

5 3 (a) Express 23 36 106xx− − − in the form 2()a x b c++ . [2] ( ) ( ) ( ) 2 2 22 2 3 36 106 1063 12 3 1063 6 6 3 3 6 2 xx xx x x − − − =− + + =− + − + =− + + (b) Hence, sketch the graph of 23 36 106y x x=− − − on the axes below. Indicate clearly the coordinates of the points where the graph crosses the y-axis and the turning point on the curve. [3] y x M1- Taking out common factor A1 B1- Correct shape B1- Correct coordinates for maximum point B1- Correct coordinates for y-intercept (-6, 2) (0, -106) x x

6 4 (a) Given that 2 sin cos xy x += , show that 2 d 1 2sin d cos yx xx += . [3] ( ) ( )( ) 2 22 2 2 cos cos sin 2 sin2 sin cos cos cos 2sin sin cos 1 2sin (shown)cos x x x xdx dx x x x x x x x x − − ++ = ++= += (b) Hence, evaluate 4 0 2 2 2sin dcos x xx  +  in exact form. [4] ( ) 4 0 0 2 4 1 1 2sin 2 sin cos c 2 os 1 2 22 2 xxdxxx  ++ = = + − = −  ( ) ( )   ( )   22 0 2 2 4 4 4 00 4 0 4 0 21 21 21 2 2 2sin 1 2sin 1 cos cos cos 2 sec 2 tan 2 1 0 2 xxdx dx dxx x x xdx x      ++ =+ =+ =+ =+ − − − − =     M1- Quotient/Product rule A1- Correct simplification leading to answer M1- Showing anti-derivative B1- Splitting 2 2 2sin cos x x + A1- Integrate 2sec x to get tan x A1 M1- Differentiating cos x to get sin x−

7 (c ) Find the y-coordinate of the maximum turning point of 2 sin cos xy x += when 02 x  . [6] 2 6 1 2sin 0cos 1 2sin 0 1sin 2 Basic angle 6 7 6 11, x x x x x   + = += =− = = 7 6  − 7 6  7 6  + d d y x positive 0 negative Shape It is a maximum point at 7 6x = . 11 6  − 11 6  11 6  + d d y x positive 0 negative Shape It is a minimum point at 1 6 1x = . 72 sin 6-coordinate of m 3 ax. turning point 7cos 6 y   + = =− M1- 0dy dx = M1- Finding basic angle A1 M1- Values of x M1- Use of 1st or 2nd derivative test to determine the maximum point M1- Use of 1st or 2nd derivative test to determine the min point

8 5 The diagram shows a straight coastline ABC. A turtle, at point X, at sea, swam 35 m in a straight line to point B, where John was throwing food to feed the turtles. The turtle then turned 90 and swam 50 m in a straight line to point Y at sea. Angle 90XBY =  , angle XBA = and XA and YC are perpendicular to ABC. (a) Show that 35cos 50sinAC =+ . [2] ) c 90 sin 50sin 35cos 50si os 3 5co n (sh s own AB XB YBC BYC BC BY BC AC A A B B BC        = = = = −  = = =+ =+ Y A B C X B1- trigonometric ratios BYC  = must be seen B1- Finding AB and BC AG 35 m 50 m

9 (b) Express AC in the form of ( )cosR − , where 0R and 0 90   . [4] 1 22 35cos 35 50 50 3 9 5 55. 7 50sin 3725 5 14 tan 5 0079 5.0 A R C   − + = = = =  = = +   = ( )5 149 cos 55.0AC  = −  (c) John’s father claims that the coastline AC is 70 m. Without measuring, John said that it was incorrect. Explain, with clear workings, how John arrived at this conclusion. [1] m M 6 ax . . v 3 alue 149 1 0 27 61.0 m < 70 f5 o AC = = = cannot be 70 m.AC M1 M1 B1 A1- Accept 3725 Do not accept 61.0 or 55.007 B1- Comparison needs to be seen Accept equating and showing that there is no solution

10 6 (i) Write down the general term in the binomial expansion of 12 3 px x  − , where p is a constant. [1] ( ) ( ) ( ) ( ) ( ) ( ) 12 1 3 312 12 4 12 12 1 12 1 r r n r r rr rrr pTx r x x p xr xpr − + −− −  =−  =− =− (ii) Write down the power of x in this general term. [1] Power of 12 4xr=− (iii) The constant term in the binomial expansion of 12 3 px x  − is 27500− . Find the value of p. [3] 12 4 0 3 r r −= = ( ) ( ) ( ) ( ) 3 3 3 3 3 12Constant term 3 12 275003 220 27500 125 125 5 p p p p p p =−  − =− − =− − =− = = B1 B1 B1 M1- equate to -27500 A1