2022 SCGS AMath Prelims P2 w Ans

SINGAPORE CHINESE GIRLS’ SCHOOL PRELIMINARY EXAMINATION 2022 SECONDARY FOUR O-LEVEL PROGRAMME CANDIDATE NAME Solution CLASS 4 REGISTER NUMBER CENTRE NUMBER INDEX NUMBER ADDITIONAL MATHEMATICS 4049/02 Paper 2 Wednesday 31 August 2022 2 hours 15 minutes Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, class, register number, centre number and index 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 staples, 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 the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of an approved electronic scientific calculator is expected, where appropriate. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. 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. FOR EXAMINERS USE Q1 Q5 Q9 Q2 Q6 Q10 Q3 Q7 Q11 90 Q4 Q8 Q12 The Question Paper consists of 17 printed pages and 1 blank page. [Turn over

2 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation 2 0ax bx c   , a acbbx 2 42 Binomial Theorem   nrrnnnnn bbar nbanbanaba                  221 21 , where n is a positive integer and ! )1()1( )!(! ! r rnnn rnr n r n        2. TRIGONOMETRY Identities 22 22 22 2 2 2 2 2 sin cos 1 sec 1 tan cosec 1 cot sin( ) sin cos cos sin cos( ) cos cos sin sin tan tantan( ) 1 tan tan sin 2 2sin cos cos 2 cos sin 2cos 1 1 2sin 2 tantan 2 1 tan AA AA AA A B A B A B A B A B A B ABAB AB A A A A A A A A AA A                  Formulae for ABC 2 2 2 sin sin sin 2 cos 1 sin2 a b c A B C a b c bc A bc A     

3 [Turn over 1. The equation of a curve is 2 8y x kx k    , where k is a constant. (a) Find the range of values of k for which the curve intersects the x-axis. [3] (b) In the case where k = 2, show that the line 43yx is a tangent to the curve. [2] (a) (b) 2 80x kx k          2 4 1 8 0kk     2 4 32 0kk     8 4 0kk   8 or 4kk  2 2 6 4 3x x x     2 6 9 0xx   Discriminant =      2 6 4 1 9    = 0  line is a tangent to the curve OR 2 6 9 0xx     2 30x line touches the x-axis at only one point where x = –3  line is a tangent to the curve

4 2. [The volume of a cone of height h and base radius r is 21 π3 rh .] An empty, inverted cone has a height of 60 cm and base radius 20 cm. The circular base is held horizontal and uppermost. Water is poured into the cone at a constant rate. (a) When the depth of the water in the cone is h cm, show that the volume of the water in the cone is 3π 27 h . [2] The water level is rising at a rate of 3 cm per minute when the depth of the water is 12 cm. (b) Find the rate at which water is being poured into the cone, leaving your answer in terms of . [3] (a) 20 60 rh 3 hr  21 π33 hVh   3π 27 h (shown) (b) 2d π d9 vh h  d d d d d d v v h t h t  2π 12d 3d9 v t  48π cm3/min 3. Solve the equation ln 1 2log yye , giving your answers in terms of e. [4] ln 1 2log yye 2ln 1 lny y   2 ln ln 2 0yy      ln 1 ln 2 0yy   ln 1 or ln 2yy  2 1 or ye e

5 [Turn over 4. It is given that cos siny px qx , where p and q are constants. Given that 6pq and 2 2 2 d 12d y qyx  when x = 0. Calculate the values of p and q. [8] d sin cosd y p px q qxx   2 22 2 d cos sin d y p px q qx x   When x = 0, 2 2 2 d d y p x  1y 22 12pq      12q p q p   2qp ---- (1) 6pq ---- (2) (1) + (2) q = 4, p = 2

6 5. (a) By considering the general term of the binomial expansion 22 n xk x   , where n is a positive integer, show that n is a multiple of 3 when the binomial expansion has a constant term. [3] (b) Given that n = 9 and the constant term is 2625 2 , find the value of k. [2] (c) Hence, find the term independent of x in the expansion 3 2452 n x x k x    . [3] (a) (b) (c) General Term 22 n r r n xk r x              30nr 3nr  n is a multiple of 3 r = 3   93 39 1 2625 3 22 k       k = 10 9 3 3r  4r Coefficient of 3x =   94 49 1 104 2      = 39375 Term independent of x =  1 262539375 452  = 2625

7 [Turn over 6. A curve is such that 2 2 d 2d y kxx  , where k is a constant. The curve has a minimum gradient at 1 3x . (a) Show that k = 6. [1] The normal to the curve at (1, 4) is 2 9 0yx . (b) Find the equation of the curve. [6] (a) (b) 2 2 d 2d y kxx  1 203 k k = 6 2d 32d y x x cx        2 3 1 2 1 2 c   1c 32y x x x d    4 1 1 1 d    3d  Equation of the curve: 32 3y x x x   

8 7. Given that and . (a) Show that f (x) + g(x) = . [3] (b) By expressing f (x) + g(x) in the form R cos (2x + ) + q where R > 0, q > 0 and , find the minimum value of and the corresponding values of x for . [6] (a) (b) = = = f (x) + g(x) = = Min = f ( ) 2 24sin cosx x x 2g( ) 10(1 cos )xx 5cos 2 12sin 2 17xx 0 2  2 f ( ) g( )xx 0 x    22 24sin cos 10 1 cosx x x   210cos 24sin cos 12x x x  5 cos 2 1 12sin 2 12xx   5cos 2 12sin 2 17xx 2 2 1 125 12 cos 2 tan 17 5x        13cos 2 1.18 17x 2 f ( ) g( )xx 21 13 17 15  cos 2 1.1760 1x 2 1.1760 2x  2.55x

9 [Turn over 8. Given that ln 2y x x for x > 0. (a) Find d d y x . [3] (b) Show that the y-coordinate of the turning point is 1 4e . [3] (c) By considering the sign of , determine the nature of the turning point. [2] (a) 1 ln 22y x x d 1 1 ln 2d 2 2 yx xxx         11ln 222 x or 1ln 2 2x (b)  1 ln 2 1 02 x ln 2 1x 12x e 1 2x e 11ln4y ee   1 4e (c) x 0.18394 0.18394 0.18394+ d d y x  0 + slope The stationary point is a minimum point. x y d d

10 9. (a) Given that 1 23 xy x   , show that   3 d2 d 23 yx x x   . [3] (b) Hence find   3 3 d 23 x x x . [4] (a)        11 22 12 3 2 3 2 1d 2 d 2 3 x x xy xx            1 22 3 2 3 1 23 x x x x         3 2 23 x x   (shown) (b)     3 21 d 2323 xx xc xx          33 21dd 232 3 2 3 xx x x c xxx           3 2 3 1d 2 2 3 d 2323 xx x x x c xx            1 22 2 31 123 22 xx c x         3 23 x c x     3 3 d 23 x x x   39 23 x d x  