TJC 2026 Prelim P2 Ans

Q1 [Solution] (a) 1 1 2 33 22x x a = + = + 2 2 12 3 3 3 3 15 2 2 2 4x x a a      = + = + + = +           3 3 1 2 3 4 3 3 3 3 57 2 2 2 2 8x x a a        = + = + + + = +               (b) By observing the pattern, 12 1 3 3 3 ...2 2 2n n xa −      = + + + +           1 33 122 3 12 n a − −=+ − ( ) 11 333 1 3 322 nn aa −−   = + − = − +       (c) As n→ , 1 3 2 n−  → , i.e., nx → Thus the sequence  nx diverges.

Q2 Solution (a) Using ratio theorem, OC  = 41 27 32 2 3 −+                   = 3 1 2    (b) OD  = 4 1 22 2− = 2 1 1  − C, D and E are collinear  CE  = t CD  for some t  CD  = 1 2 1 − −− and CE  = 2 1 2 k −  −− = 1 12 2 1 k −  −  − We have 1 22k − =−  k = − 3 (c) Length of projection = OE AB AB    = 11 33 00 3 1 9 k        −     +  = 13 10 k− Given: 13 10 k− = 10  1 – 3k = 10 or 1 – 3k = −10 Hence k = − 3 or k = 11 3

Q3 Solution (a) Subst z = p + pi into ( ) 2 2 i + 4i 0,z z q− + + = ( ) ( ) ( ) ( )( ) 2 2 2 2 22 22 0 i 2 i 4i i 0 2 i 2 i 4 i 4 0 4 i 2 4 i 4 0 2 4 4 i i 4 p p p p q p p p p p q p p p p q p p q p p + − + + + + = + − − + + − + = − − + − + = − − + − + = Comparing real and imaginary parts: 2 2 2 4 0 --- (1) 4 4 0 --- (2) p p q pp − − + = − + = From (2), ( )4 1 0 pp− + = 0 (rej 0) or 1p p p =  =  Subst into (1): ( ) ( ) 2 2 1 4 1 0 6 qq− − + =  = (b) Let r be the other root of the equation. ( ) ( ) ( )( )( ) 2 2 i + 4i 6 2 i 1 iz z z z r− + + = − + − + − Comparing coefficient of z0: ( ) ( )( )( ) ( )( ) ( ) ( ) 6 2 i 1 i 6 2 i 1 i 6 2 2i i 1 6 3 i 3 i 3 i 6 3 i 93 i10 5 5 r r = − + − + − =− −+ =− + − + −=− +− −=− =− + OR Let ( ) ( )( ) ( )( ) 2 2 i + 4i 6 1 i 2 iz z z z r− + + = − + − + − Find r by comparing coefficient of z0: Then the other root is z = 2i r −+ = 2i 2 i 2 i r −−− + − − =…

Q4 Solution (a) 2 1xt=− , 32 31y t t= − + , t > 0 d 2dx tt = , 2d 36dy ttt =− 2d 3 6 3 ( 2)d 2 2y t t txt −= = − At stationary points, d 0dyx =  t = 2 Coordinates of the stationary point of C is ( ) 2 3 22 1, 2 3(2) 1− − + = (3, −3) (b) (c) When 4t = , ( )d3 4 2 3d2yx = − = , 15x= , 17y= Equation of the tangent to C at P is ( )17 3 15yx− = − , i.e., 3 28yx=− (d) Tangent: 3 28yx=− C: 2 1xt=− , 32 31y t t= − + Tangent at P meets curve C: ( ) 3 2 23 1 3 1 28t t t− + = − −  32 6 32 0tt − + =  (t – 4)2(t + 2) = 0  t = 4 (which is point P) or t = − 2 Since we are given that t > 0 , the tangent at P does not meet curve C again. y x

Q5 Solution (a) d2 2sin 2cos dxx  − =  = When 0, sin = 1 2 When 2, sin = 0 0 x x   = −  =− =  = ( ) ( ) ( ) ( ) 20 22 0 2 0 2 2 0 2 2 0 2 0 2 4 2 d 4 2sin 2cos d 2cos 2 1 sin d 4 cos d 2 1 cos 2 d sin 22 2 2 0 0 2 xx                 − − − − − − − = − =− = =+ =+   = − − −  =     (b) ( ) ( ) ( ) 2 2 2 2 2 4 2 4 2x y y x− + − =  =  − − Required volume = Volume of hemisphere − 2 2 0 dyx ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) 232 0 2 22 0 2 22 0 23 0 2 2 214 2 2 4 2 d23 16 4 4 4 2 4 2 d3 16 8 2 4 4 2 d3 216 8 433 16 8 4 16 0 033 4 8 or xx x x x x x x xx            = − − − − = − − − − + − − = − − − − − −  −= − − +    = + − − − +   =−    ( ) 4 2 −

Q6 Solution (a) Method 1: (Using Binomial Distribution) Let X be the number of tokens (out of 3) that land in a red cup. ( )~ B 3, 0.7X The probability distribution of S is s 0 1 3 ( )P Ss= ( ) 3 P0 0.3 0.027 X = = = P(X = 1) + P(X = 2) 1 0.027 0.343 0.63 = − − = ( ) 3 P3 0.7 0.343 X = = = ( ) ( ) ( ) ( )E 0 0.027 1 0.63 3 0.343 1.659 (shown) S = + + = Method 2: (Without using Binomial Distribution) Let R be the event that a token lands in a red cup. 3P( = 0) = P( ) 0.3 0.027S R R R  == !PP 3! 3!2! 2( = 1) = P( ) ( )S RR R RRR    +  = ( ) ( ) 220.7 (0.3) 3 0.3 (0.7) 3 0.63 +  = 3P( 3) 0.7 0.343S = = = ( ) ( ) ( ) ( )E 0 0.027 1 0.63 3 0.343 1.659 (shown) S = + + = (b) ( ) ( ) ( ) 2 2 2E 1 0.63 3 0.343 3.717S = + = ( ) ( ) ( ) 22 2 Var E E 3.717 1.659 0.964719 (exact) S S S=−  =− = (c) To have a chance of winning the stuff toy (total score of 100 points), he needs to play at least 100 3  34 games.

Q6 Solution (d) Let the total score of n games, 12 ... nT S S S= + + + For a total score of 100 points, need at least 34 games. Since n > 30 is sufficiently large, by Central Limit Theorem, ( )~ N 1.659 , 0.964719T n n approximately Given: ( )P 100 0.9T  Method 1: Using GC Thus least n is 67 Method 2: By Standardisation ( )P 100 0.9T  100 1.659P 0.9 0.964719 nZ n −  100 1.659 0.964719 n n − ≤ −1.2816 100 1.659 1.2816 0.964719nn− − ( ) ( ) 2 100 1.659 1.2816 0.964719 0nn− +  ( ) ( ) 2 1.659 1.2816 0.964719 100 0nn − −  Since 0,n  , 8.1525n  66.5n Thus least n is 67 n ( )P 100T  66 0.8829 < 0.9 67 0.9173 > 0.9 −7.3937 8.1525

Q7 Solution (a) Required number of ways 7! 5040== (b) Required number of ways = total no. of ways – no. of ways all in blue or silver – no. of ways exactly one person is in blue or silver ( ) 88 12 2 2 238 C= − −  = Alternative method No. of ways = 8C2 + 8C3 + 8C4 + 8C5 + 8C6 = 238 (c) Case 1: Beng and Charlie seated on round table of 3 No. of ways ( ) ( ) 6 1 3 1 ! 5 1 ! 6 2! 4! 288C=  −  − =   = Case 2: Beng and Charlie seated on round table of 5 No. of ways ( ) ( ) 6 3 5 1 ! 3 1 ! 20 4! 2! 960C=  −  − =   = Total required number of ways 288 960 1248=+= Alternative method Required number of ways = total no. of ways with no restrictions – no. of ways Beng and Charlie are seated at different tables = ( ) ( ) 8 5 3 1 ! 5 1 !C  −  − ( ) ( ) 6 4 3 1 ! 5 1 ! 2C−  −  −  = 2688 − 1440 = 1248

Q8 Solution (a) Let X and Y be the time taken, in minutes, to complete task A and B by a contestant respectively. X ~ N(m, 1.25) Y ~ N(7, 1) Let T X Y=+ , total time taken (in mins) by a contestant to complete both tasks A and B. ~ N( 7, 2.25)Tm + Given: ( )P 22.8 0.031XY+  = ( )22.8 7P 0.031 2.25 15.8 1.8662961.5 13.0 mZ m m −+ =  − = = (b) ~ N(20, 2.25)T X Y=+ Given: ( )P 0.1Tk= From GC, ( )P 18.0777 0.1T = Thus 18.1k = (to 3 s.f.) (c) ( ) ( ) ( )1 2 3E 3 13 3 7 60T T T+ + = + = ( ) ( ) ( )1 2 3Var 3 1.25 3 1 6.75T T T+ + = + = 1 2 3 ~ N(60, 6.75)T T T++ Required probability ( )1 2 3P 62T T T= + +  0.779 (3 s.f.)= (d) ( )E2XY− = ( )13 2 7 1− =− ( )Var 2XY− = ( ) 21.25 2 1 5.25+= 2 ~ N( 1, 5.25)XY−− Required probability = P(X − 2Y > 0) = 0.331 (3 s.f.)

Q9 [Solution] (a) Since the correlation coefficient of 0.93244 is close to 1, it suggests that rice crop yield and amount of rainfall has a strong positive linear correlation. (b) The scatter diagram shows that the data points increase at a decreasing rate and then later decreases and not increasing at a constant rate . Thus the diagram does not support the linear correlation suggested in (a). (c) As x, the amount of rainfall increases, the rice crop yield is unlikely to increase because too much water can severely damage crops.