2017 SNGS Sec 4 AM Prelim P1 Solutions

1 The mass of a radioactive substance reduces by half every 4 hours. It is given that m0 is the mass of the substance at a particular time and that m is the mass of the substance t hours later. Calculate the value of the constant k in the relationship ktemm  0 . [3] ktemm  0 i.e when 4t sub 0 2 1 mm kemm 4 00 2 1  ek ln42 1ln  2 1ln4 1k or 2ln4 1 or 0.173 (3.s.f.) Note: m0 is a constant When t = 0, initial 0mm 2 Solve the equation    xxx 212 3227363  . [4]    xxx 22 32273633  Let xu 3 , 22 22763 uuu  02765 2  uu    0953  uu 3u , 5 9u 33 x , 5 93 x (no solution) 1x 3 The function f is defined, for all values of x, by xexx 2)(f  . Find the range of values of x for which f is a decreasing function. [6] xx exxexf 22)('  )2( 2xxex  0)(' xf Since 0xe for all real values of x, 02 2  xx 0)2(  xx 02  x 0 –2 + + – x

2 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/01 4 (i) On the same diagram, sketch the curves 3 1 2xy and 3 1 8   xy . [2] (ii) Find the coordinates of the intersection points of the two curves. [3] (i) (ii) 3 1 2x 3 1 8   x 4 3 1 3 1   x x 43 2 x Take cube 64432 x 8x When x = 8, 4y When x = –8, 4y Coordinates: (8, 4) and (–8, –4) Note: 2 3 2 3 3 2 4)( x 2 3 4x =8 Only 1 answer 5 The equation of a curve is x xy 25 8   . A particle moves along the curve in such a way that the y-coordinate of the particle is decreasing at a constant rate of 2 units per second. Find the possible y-coordinates of the particle at the instant when the x-coordinate of the particle is increasing at 11 71 units per second. [6]        2 25 28)1(25 x xx dx dy     2 25 16225 x xx     2 25 11 x  dt dx dx dy dt dy    11 18 25 112 2    x   2 25 11 9 11 x  Note: mistake in expansion

3 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/01   2 259 x 325  x 1x , 4x 3 12y , 3 11y 6 A curve has equation xy 42  and a line has equation )1(  xmy . Find the range of values of m for which the line )1(  xmy intersects the curve xy 42  at two distinct points. [5]   2 4 mmxx  2222 24 mxmxmx  0)24( 2222  mmxxm 042  acb      0424 2222  mmm 0441616 442  mmm 01 2  m    011  mm 11  m 7 The diagram shows part of the graph of cb xay    sin . (i) State the amplitude and period. [2] (ii) Find the value of each of the constants a, b and c. [3] (i) Amplitude = 3 Period = 8 Period 4 1 (ii) 3a       b 1 28  4 11 b Note: a is negative based on the inverted sine curve y x 4 –2 –4π –2π 0 2π 4π 6π 8π 10π 12π 1 –1 + – – m

4 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/01 4b 1c 8 (i) Sketch the graph of xy  4 . [2] A line of gradient m passes through the point (0, 1). (ii) In the case where m = –2, find the coordinates of any point of intersection of the line and the graph of xy  4 . [3] (iii) Determine the set of values of m for which the line intersects the graph of xy  4 at one point. [2] (i) Steps to sketch 1) When x = 0, y = 4 2) When y = 0, 4x , x = 4 or x = - 4 Need to show correct x and y- intercepts (0, 4), (4, 0), (–4, 0) (ii) 12  xy …(1) xy  4 …(2) or xy  4 …(3) (1) = (2), xx  412 3x and 7y (n.a.) (1) = (3), xx  412 33 x 1x and 3y Intersection point is (–1, 3) Note: (–3, 7) is (n.a.) because the y-coordinate is greater than the maximum y-value of the modulus graph. (iii) 1m or 1m Read from graph y = –2x +1 m = 1 m = –1

5 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/01 9 (i) Prove that AA AA secsin1 costan  . [3] (ii) Hence, find the exact solutions of xcx xx osec2sin1 2cos2tan  for 20  x . [5] (i) LHS A A A A sin1 cos cos sin   AA AAA sin1cos cossinsin 22    AA A sin1cos 1sin   Acos 1 Asec = RHS (ii) xA 2 xx cosec2sec  xx sin 1 2cos 1  xx sin2cos  xx sinsin21 2  01sinsin2 2  xx    01sin21sin  xx 1sin x , 2 1sin x 2 3x ref 6  6 5,6 x Sub part (i) into LHS, RHS remains Must be exact values

6 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/01 10 A particle, moving in a straight line, passes through a fixed point O with a velocity of 3 m/s . The acceleration, a m/s2, of the particle, t seconds after passing through O is given by tea 2.06.0 . (i) Find the value of t when the particle is at instantaneous rest. [4] (ii) Find the distance travelled by the particle in the first 5 seconds. [4] (i) tea 2.06.0  dtev t2.06.0 cev t  2.03 Sub v =3, t=0 6c 63 2.0  tev Sub v = 0, 063 2.0  te 63 2.0 te 2lnln 2.0 te  465.32.0 2lnt 3.47s (ii)   dtes t 63 2.0 ctes t  615 2.0 Sub s = 0, t = 0 15c 15615 2.0  tes t At 2.0 2lnt , 7944.5s At 5t , 226.4s Distance travelled =  3628.7)226.47944.5(7944.5 7.36 m Alternative method Distance travelled =      5 2ln5 2.02.02ln5 0 6363 dtedte tt =     5 2ln5 2.02ln5 0 2.0 615615 tete tt  = 2ln301530150152ln3015 2ln2ln  eee = 2ln3015602ln3015  e =5.7944 +1.5686 =7.36 m

7 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/01  P Q R S O 4 cm 11 The diagram shows a trapezium PQRS inscribed in a semicircle with centre O. The radius of the semicircle is 4 cm. Angle POQ = angle SOR = θ radians. (i) Show that the area, A cm2, of the trapezium PQRS is given by  2sin8sin16 A . [2] (ii) Given that θ can vary, find the value of θ for which the area of the trapezium is a maximum. [6] (i) Horizontal = cos4 Vertical height = sin4 Area =    cos428sin42 1   cossin16sin16   2sin8sin16  Area = 2       2sin42 1sin42 1 22  Using supplementary s, Area = 2       2sin42 1sin42 1 22  (ii)  2cos16cos16 d dA 02cos16cos16   02coscos   01cos2cos 2   01coscos2 2      01cos21cos   1cos  , 5.0cos    (NA) ref 3  3   or 1.05 (3s.f.) Check for max  2sin32sin162 2 d Ad At 6   , 06.412 2 d Ad , max value Note: different angles, do not use R formula 4 cm  4 cm

8 CHIJ SNGS Preliminary Examinations 2017 - Additional Mathematics 4047/01 12 In the diagram, AB is a diameter of the circle with centre O. CS and BT are the tangents to the circle at C and B respectively. UCS, ACT and BST are straight lines. Prove that (i) triangle ABC and triangle ATB are similar, [2] (ii) angle SCT = angle CTS, [3] (iii) S is the midpoint of BT. [2] (i) TABBAC  (common angle)  90ACB (right angle in semi-circle)  90ABT (tangent perpendicular to radius)  90ABTACB triangle ABC and triangle ATB are similar. (ii) ABC = ATB = CTS (ABC similar ATB, proven in i) ABC = ACU (tangent chord theorem) ACU = SCT (vertically opposite s) CTS = SCT (iii) BS = CS (tangents from external point) Since CTS = SCT (proven in ii) CS = ST (base s of isosceles triangle) BS = ST S is the midpoint of BT. O A C B S