2022 AHS AMath Prelims P1 Ans

1 Secondary Four Additional Mathematics Preliminary Examinations 2022 AHS 2022 S4 Preliminary Examinations AM P1 Answer Key 1(a)  P 2 3x x  2 29 121 . 3 9x 1(b)    3 2 2 2 5 6 27 0 4 9 2 3 0 x x x x x x         For 2 4 9 0x x   , Discriminant 20 0  . Hence there is no solution for 2 4 9 0x x   . 3For 2 3 0, 2x x   . Hence the equation P( ) 0x  has only one real root 4(i) 0.000121 4(ii) 10.3% 4(iii) 0.0001210.0121 percent/yearte 5    2 22 10 2 1 3 2 133 2 1 x x xxx x       6(i) Period of f π 6(iii) Number of solutions is 2 6(i) 7(i) A B 7(ii) (a)   4tan 3A B  7(ii) (b)   5tan 12A B  7b (iii)    tan 2 tanA A B A B      8(a)(i) 2k   8(a) (ii) Coordinates of S are  5, 3  8(b) Gradient of tanQW  8(c) Coordinates of W are 11 5,2 2     

2 Secondary Four Additional Mathematics Preliminary Examinations 2022 8(d) Show that , ,MQ MP MS and MR are equal, these are radii, so a circle can be drawn to pass through the vertices of the quadrilateral PQRS. Centre of the circle is M. 8(d) Show that QR SR , angle 90QRS   (Right angle in a semicircle). Show that QP PS , angle 90SPQ  (Right angle in a semicircle) QS is the diameter of a circle, with P and R lying on its circumference. Hence a circle can be drawn to pass through the vertices of the quadrilateral PQRS . 9(a) 7.1202p 10.2x 9(b) 59 3k  10(i)  22 2 ln 1x x x  10(ii)     2 2 21 1 1 ln 12 2x x x c    11(a) B  2, 4 11(b) 215 units3 12(a) 30.4sin 2 45.6cosM    12(b) 66.0 g , maximum. 13(a) 90GBD   (Right angle in a semi-circle) 90ECD   (Right angle in a semi-circle) GBD ECD   BDG CDE  (Common angle) Therefore triangle DBG is similar to triangle DCE (AA) 13(b) 90FGE   (Tangent perpendicular to radius) FGE ECD   GEF DEC  (Vertically opposite angles) GE EC (Given) Therefore triangle GEF is congruent to triangle CED. (ASA) Hence GF CD 13(c)  AlternateSegment TheoremBGA BDG CDE    90GBD   (Right angle in a semi-circle)  180 90 sum of angles on a straight line 90 ABG      90ECD   (Tangent perpendicular to radius) ABG ECD  So triangle ABG is similar to triangle ECD. (AA) From (b), triangle EGF is congruent to triangle ECD. Therefore triangle ABG is similar to triangle EGF.