S4 A-Math SJI P2 2024 (with Answer) EXAM555

ST JOSEPH’S INSTITUTION PRELIMINARY EXAMINATION 2024 (YEAR 4) CANDIDATE NAME CLASS INDEX NUMBER ADDITIONAL MATHEMATICS Paper 2 Candidates answer on the Question Paper. 4049/02 20 August 2024 2 hours 15 minutes (0805 - 1020) READ THESE INSTRUCTIONS FIRST Write your class, index number and name on all the work you hand in. Write in dark blue or black pen in the space provided. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions. If working is needed for any question it must be shown with the answer. Omission of essential working will result in loss of marks. The use of an approved scientific calculator is expected, where appropriate. If the degree of accuracy is not specified in the question, and if the answer is not exact, give the answer to three significant figures. Give answers in degrees to one decimal place. For , use either your calculator value or 3.142, unless the question requires the answer in terms of . 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 document consists of 19 printed pages and 1 blank page.

2 [Turn over 1. ALGEBRA Quadratic Equation For the equation 02  cbxax , a acbbx 2 42  Binomial expansion   nrrnnnnn bbar nbanbanaba                  ......21 221 , where n is a positive integer and ! )1)...(1( )!(! ! r rnnn rnr n r n       2. TRIGONOMETRY Identities 1cossin 22  AA AA 22 tan1sec  cosec 2 A = 1 + cot 2 A 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  Cab sin2 1

3 [Turn over 1 Solve the inequality 282 2 4xx  and represent your solution on a number line. [6] 1 282 2 4xx      2 2 82 28 0 42 0 2 or 4 xx xx xx xx               2 2 22 4 22 4 0 64 0 46 xx xx xx x            Combining both inequalities: 42 o r 4 6xx     –4 –2 0 4 6

4 [Turn over 2 The expression 3cos2 6sin2  is defined for 0  . (a) Using cos 2R  , where 0R and 0 2  , solve the equation 3cos2 6sin2 5  . [6] 2(a) cos 2 3cos 2 + 6sin 2R      = cos 2 cos sin 2 sinRR    cos 3 sin 6 R R     22 236 45 R R    = 35 or 6.7082 (6.71) 16tan3     = 1.1071 (1.11)  35 c o s2 1 . 1 1 or  6.71cos 2 1.11  3cos2 6sin2 5    35 c o s2 1 . 1 1 5   5cos 2 1.11 35    15cos 35   21 . 1 0 7 1 0 . 7 2 9 7 2 , 0 . 7 2 9 7 2  0.189 , 0.918

5 [Turn over (b) Explain why P cannot be less than –7. [2] 2(b) Minimum value of P = 35 1  – 6.71 > –7 Since the minimum value is greater than –7, 3cos2 6sin2  cannot be less than –7.

6 [Turn over 3 The diagram shows a circle passing through the points A, B, C and D. The straight line MAN is a tangent to the circle. DB is parallel to MN and AD bisects angle MAC. (a) Show that triangle ACD is an isosceles triangle. [2] 3(a) Let x be MAD   alt. seg theoremACD x  MAD=   bisects CAD x AD MAC   Since ACD CAD x  , triangle ACD is an isosceles triangle, DA = DC. (b) By identifying another isosceles triangle, show that   2 AD AB CD  . [3] 3(b) Triangle DBA with AD = AB = DA AD DC AB DA AB AD DC   since AD = AB, DA AD AB DC     2 AD AB CD  (shown) A M N B C D

7 [Turn over 4 (a) Given that the coefficient of 2x is 100 in the expansion of   6 2 321 2 xmx     , find the values of m. [4] 4(a)     62 2 2233 32 1 4 4 1 6 15 ...22 2 xx xmx mx m x                              =     2 223341 5 4 6 . . .22 xx mm x                  2 23341 5 46 1 0 022 m              2135 36 100mm   2 36 35 0mm   35 1 0mm   1 or 35m (b) Determine if the term independent of x exists in the expansion of 100 2 1 2x x     . [3] 4(b) General Term =   100100 2 1 2 r r rCx x     =     100100 2 1 2 r r r rCx x     = 100 200 2 1 2 r rr rCx      200 3 0r  200 26633r   Since r is not a whole number, the term independent of x does not exist in the expansion.

8 [Turn over 5 (a) Solve the simultaneous equations   22 14xy  , 231xy . [5] 5(a)   22 14xy   --- (1) 231xy  --- (2) From (2), 13 2 yx  --- (2*) Sub (2*) into (1)   2 213 142 y y      2 216 9 21 44 yy yy     2216 9 4 8 41 6yy y y      213 2 11 0yy    13 11 1 0yy   11 13y or 1y Sub 11 13y into (2*) 111313 2x     = 10 13 Sub 1y into (2*) 13 1 2x   = 2

9 [Turn over (b) Find the set of values of m for which the line 1ym x  is a tangent to the curve 51 4 xy x . [4] 5(b) 1ym x  --- (1) 51 4 xy x  --- (2) Sub (1) into (2) 51 4 x mx mx   2254 4 4xm x m x   254 4 4 0mx m x    Line touches curve, discr = 0      2 44 5 4 4 0mm   216 64 80 0mm   2 45 0mm   51 0mm   5 or 1m

10 [Turn over 6 The volume V of the solid is given by the equation 112 ln 3 22Vx       , where x is the length of one of the sides of the solid, in cm. x increases from an initial value of 12 cm at a constant rate of 0.008 cm per second. (a) Find the rate of increase of V after 30 seconds. [4] 6(a) 112ln 3 22 d6 1d 32 12 6 Vx V x x x             After 30s, 12 30 0.008 12.24 cm x x    3 dd d dd d d 12 0.008d 12.24 6 d1 cm /sd 190 VV x tx t V t V t      (b) Explain why the volume of the solid never reach a maximum value. [2] 6(b) For , 0 : 12 06 d 0d xx x V x      Since d 0d V x, there are no stationary points. the volume of the solid will never reach a maximum value.