VJC 2026 Prelim P2 Ans

VJC 2026 H2 Math Prelim Paper 2 (Solution) 1 Solution (a) Method 1: y ax b= + Subt (0, 2v): ( )2 0 2 2 va b b v bv = +⇒= ⇒= ± Subt (-v, 0): ( )0 0 a v b av b ba v = −+ ⇒− += ⇒= Hence, a = 2 and b = 2v or a = –2 and b = –2v The expressions are: ( )f 22x xv= + and ( )f 22x xv= −− Method 2: ax by ax b ax b+= += −− Consider y ax b= + : Gradient 02 20va v−= =−− , y-intercept b 2v= Hence 2222xvy ax b xv+= += −− The expressions are: ( )f 22x xv= + and ( )f 22x xv= −− (b) (i) f (2 2 )y xv= − Method 1: 1. Translate f( )yx= by 2v units in the positive x-direction 2. Stretch resultant f( 2 )y xv= − parallel to the x-axis by factor 12 . Method 2 Since a < 0, f( ) 2 2x xv= −− ( ) f(2 2 ) 22 2 2 4 2 y xv y xv v xv = − = − − − = −+ y x O 0.5v 2v f (2 2 )y xv= −

(b) (ii) Since a < 0, f( ) 2 2x xv= −− 2 Solutions (a) If w is a root of ( ) 0Pz = , then 642 0w aw bw c+ + += Consider ( ) ( ) ( ) ( ) 642 642 0 0 Pw w aw bw c w aw bw c − = −+ −+ −+ = = +++ = Hence −w must also be a root. (b) Since all coefficients are real and 1i+ is a root, by conjugate root theorem, 1i− is a root. From (a) which states that if w is a root, then w− is also a root. ( )1i−+ and ( )1i−− are also roots. Hence 3 other non-real roots are 1i ,1i a n d1i−−− −+ . (c) Method 1: ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) 642 1i 1i 1i 1i z az bz c z z z z zkzk +++ = − + − − −−− −−+ − + Compare constant term: ( ) ( ) ( ) ( ) ( )( ) ( ) 2 1i 1i 1i 1i 4 4 where 0 kk c kc ck c − + − − − − − − − +  − =     −= = ±− < Hence, the remaining roots are 42cc −±− = ± . y x O –2v ( )fyx= ( )f 22y x xv= = −−

Method 2: [expand ( )( ) 22zwzw z w− +=− ] ( )( ) ( )( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( ) ( )( )( ) ( )( ) 642 2 22222 222 42 1i 1i 1i 1i 1i 1i 2i 2i 4 z az bz c zzzzz m z n z z z mz n z z z mz n z z mz n +++ =−+ ++ −− +− + + = −+ −− + + =− + ++ =+ ++ Compare z term: 4m = 0 0m⇒= Compare constant: 4n = c 4 cn⇒= To get the remaining 2 roots 2 0 4 42c cczz −+ = ⇒= ±− = ± 3 Solution (a) Using ratio theorem, 22 OA OC a cOM ++= =     Method 1: Given ab bc×=×  ( ) 0 0 0 abbc abcb ac b ×−×= ×+×= + ×=         Since a and c are non-parallel and b is non-zero, // ac b+  // 2acOM OB +=    Since O is a common point and // OM OB     , O, B and M are collinear. Method 2: ( ) ( ) ( ) 2 1 2 1 2 0 given acOM OB b abcb abbc ab bc +×= ×  = ×+× = ×−× = ×=×              Hence // OM OB     . Since // OM OB     and O is a common point, O, B and M are collinear.

(b) Given angle between p  and q  is 1 1cos 6−  ( ) 1 1cos cos 6 1 given is a unit vector6 pq pq pq p q − ∴⋅=   ⋅=      ( ) ( ) ( ) ( ) ( ) 22 2 22 2 2 2 2 10 2 5 10 4 5 10 10 4 5 5 20 100 4 40 100 3 60 0 13 60 0 6 3 10 0 10 03 p q pq p q pq p qp q pqpq p pq q p pq q p pq pp pp pp +=− += − + ⋅+ = − ⋅− + ⋅+ = − ⋅+ − ⋅= −=  −= = >                         4 Solution (a) (i) ( )ln cos 1y xy x+ = −−− Differentiate w.r.t. x, ( )1d d sin 2ddyy yx xyx x + + =− −−− Differentiate w.r.t. x, ( ) 222 22 2 1d 1 d d d d cos 3d d dd dy y yy y xxyx y x x x x− + + + =− −−− When x = 0, ( )From 1 : ln 1 y= ⇒ ey= ( ) 1dFrom 2 : e 0 ed yx += ⇒ 2d edyx =− ( ) ( ) 2 222 22 1d 1From 3 : e 2e 1ed eyx −−−= − ⇒ ( ) 2 2 2 d e 3e 1d yx = − ( ) 2 22 e 3e 1e e ...2yx x −= −+ +

ii) To find the first term of the series, we need to sub stitute x = 0, however cosec 0 is undefined. Hence, no Maclaurin’s series for y exist. (b) 3πcos 4 22 3π 3πcos cos sin sin44 22 x x xx x  − − + = − ( )( ) ( ) ( ) ( ) ( ) ( ) 1 2 2 11 22 2 2 22 1 cos sin 2 2 2 1 1 2 1 using small angle approx22 13 11 22= 1 1 ...2 2 2 2! 13 1 1 ...2 2 28 13 122 xx x x xx xx xx x xxx xx − −− =−+ −  ≈ −− + −     −−    −+ + +− − + − +      = −+ + + + +      = −− − ( ) 2 22 2 2 ...8 22 15 1 ...2 28 1 5 4816 xxx xx xx  + +++  = −+ + +  ≈ +− 5 Solution (a) 225 2 x xay x− −+= − ( ) ( ) ( ) 2 2 225 2 52 0 yx x x a x y x ya −= − −+ + + +− − = For x to not have real roots, discriminant < 0 ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 2 2 2 2 2 5 42 2 0 10 25 16 8 0 26 25 8 0 13 169 25 8 0 13 144 8 0 13 144 8 13 144 8 0 13 144 8 13 144 8 0 y ya y y ya yy a ya ya y ay a y ay a + − −−< ++ ++ < + ++< + − ++< + − −< ++ − +− − <    −− − − −− + − <    13 144 8 13 144 8ay a−− − << −+ − The two values are: 13 2 36 2 a−− − and 13 2 36 2 a−+ −

(b) 22 5 182922xxyx xx−−= = − −−−− Equations of asymptotes are 29yx= −− and 2x= (c) Turning points (by GC): (-1, -1) and (5, -25) Axial intercepts (by GC): (0, 0) and (-2.5, 0) (d) ( ) ( ) 43 43 22 2 2 2 5 2 40 242 5 22 25 225 2 xxx x xx x x xx xx xx + + −= −= − − −= − − −−= − Add 2 2y x= into part (c). From the graph, the equation has 2 real roots. 6 Solution (a) Let X and Y be the number of defective microchips in the first and second sample of n respectively. ( )B , 0.2Xn and ( )B , 0.2Yn 2 Batch rejected 1 Batch rejected2 0 Batch accepted 1 Batch accepted X YX Y X >⇒ ≥⇒= ⇒  = ⇒ ≤⇒ (5, -25) (0, 0) (-1, -1) (-2.5, 0) y x y = -2x – 9 x = 2 x = 0 2 2y x= y = 0 × ×

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) [ ] ( ) ( ) ( )( ) ( )( ) ( )( ) 1 22 1 22 1 22 P 1 P 2P 0 P 0 P 1 P 2P 0 0.8 0.2 0.8 0.2 0.8 0.812 10.8 0.2 0.8 0.04 0.82 0.8 0.2 0.8 0.02 1 0.8 n n nn nn nn PX X Y X X XY nn nnn n nn −− −− −− = ≤+ = = == += += =  = ++    −= ++ = + +− (b) ( )( ) ( )( ) 1 220.8 0.2 0.8 0.02 1 0.8 0.1nnn nn−−+ +− < From GC, n P 18 0.102 19 0.0851 Least n = 19 (c) Let W be the number of batches (out of 100) accepted ( )B 100, 0.398W  ( ) ( ) ( )P 40 45 P 45 P 39 0.399 WWW≤≤ = ≤ − ≤ = 7 Solution (a) ( ) ( )( ) ( ) PP P 7 320P 7 5 12 ABBA A A ∩= = = ( ) ( ) ( ) 371P 'P P 5 20 4AB A AB∩= − ∩= −= (b) ( ) ( )( ) ( ) 7 P 920P P 7P 20 9 ABAB BB∩= ⇒= = ( ) ( ) ( ) 917P PP ' 20 4 10AB B AB∪ = + ∩ = += OR ( ) ( ) ( ) ( ) 39 7 7P PPP 5 20 20 10AB A B AB∪= + − ∩= +−= (c) ( ) ( ) 3P ' ' 1P 10AB AB∩= − ∪=

(d) ( ) ( )P ' ' ' 1 P 0.335ABC ABC x∩∩ = − ∪∪ = − 0.215 0 0.215xx− ≥⇒≥ 0.315 0 0.315xx−≥ ⇒ ≤ Greatest and least possible values of ( )P ABC∩∩ are 0.315 and 0.215 respectively. 8 Solution (a) Method 1 ( ) ( ) ( ) ( )P 3 P , ', 2! P , ', 2! P , ', 2! 6 53 5 5 63 4 3 65 22! 2! 2!14 13 12 14 13 12 14 13 12 972 2184 81 (Shown)182 X RR R WW W BB B= = ×+ ×+ × +++           = ×+ ×+ ×                       = = Method 2 ( ) ( ) ( ) ( )P 3 P _ 2! P _ 2! P _ 2! 6 5 54 3 5 24 6 3 522! 2! 2!14 13 12 14 13 12 14 13 12 972 2184 81 (Shown)182 X RW BW RB= = ×+ ×+ × +++           = × + ×+ ×                       = = ( ) ( ) ( ) ( ) 653 222 14 14 14 222 P 2P P P 65 54 32 OR 14 13 14 13 14 13 4 13 X RR WW BB CCC CCC = = ++         = + + ++                 = ( ) 81 4 45P 41 182 13 182X == − −= x A B C 0.6 0.7 0.42 x x ×− = − 0.35 x− 0.45 0.7 0.315xx× −= − 0.17x− 0.215x− 0.035x−

OR ( ) ( ) 6 5 3 45P 4 P _ 3!14 13 12 182X RWB    == = ×=       x 2 3 4 P(X = x) 4 13 81182 45182 (b) ( ) ( ) E 2.9396 Var 0.55129 X X = = ( )E 200 50 200 50E( ) 53.02XX−= − = ( )Var 200 50 2500Var( ) 1378.23XX−= = Expectation and va