2026 EJC Prelim P2 (Soln)

1 Error! Reference source not found. EJC 2026 H2 Math Prelim P2 (Solutions) Section A: Pure Mathematics [40 marks] 1 Do not use a calculator in answering this question. Find the complex numbers z and w which satisfy the simultaneous equations ( )3 2i 3 3 11izw+ −= − and ( )1 i 6i 10 8izw−−= + . [4] Solution ( ) ( ) ( ) ( ) ( )( ) 3 2i 3 3 11i (1) 1 i 6i 10 8i (2) (1) 2i : 4 6i 6i 22 6i (3) (2) (3) 5 7i 12 2i 12 2i 5 7i 74 74i 1 i (4)5 7i 5 7i 74 Sub (4) into (1) 3 2i 1 i 3 3 11i 3 3i 2i 2 3 3 11i 1 5i 3 3 11i zw zw zw z z w w w + − = − −−−−− − − += + −−− × − + − = + −−− − − = −+ −+ + −−= × = =− − −−−−+ + −− − =− −− − +− =− −− − =− −3 4 6i 4 2i3 w w = − = −+

2 Error! Reference source not found. 2 The curve C has equation cos 1 sin xy x = + for π0 2x≤≤ . The region R is bounded by the curve C, the x -axis, and the lines 0x= and π6x= . (a) Use the substitution 1 sinux= + to find the exact area of R. [4] The region S is bounded by the curve C, the line 1 2 y= , and the y-axis. (b) Verify that 1 2 y= when π6x= . Hence find the volume of the solid formed when S is rotated through 2π radians about the x-axis. [3] Solution (a) ( )dd1 sin cos OR by implicit diff, 1 0 cosdduxux x xxu= + ⇒= = + d1 d cos x ux⇒= When 0x= , 1u= . When π6x= , 32u= . 6 0 3 2 1 3 2 1 3/2 1 cosArea of d 1 sin cos 1 dcos d 2 322 2 62 1 xRx x x uxu u u u π = + = = = = − = − ∫ ∫ ∫ y x O R

3 Error! Reference source not found. (b) When π6x= π 3cos 16 2 π321 sin 62 y= = = + (verified) Method 1 [using 2221π dyy x −∫ ] 2 2π 6 0 3 cos 1V ) olume required πd 1 sin 2 0.40157 n (3 s.0.402 u its f. x x x  = − +  ≈ = ∫ Method 2 [using 21π d Vol. of cylinderyx −∫ ] 2 2π 6 0 3 cos 1 πV ) olume require 4 d = π dπ 61 si (3 s.f. 5 n2 0.401 7 0. 02 units x x x   −  +  ≈ = ∫ S y x O

4 Error! Reference source not found. 3 A curve C has equation 2 2 5 16 18 4 xxy xx ++= − . (a) State the equations of the asymptotes. [2] (b) Using an algebraic method, show that 2 2 5 16 18 4 xx xx ++ − cannot lie between two certain values (to be determined). [4] (c) Hence sketch the curve C, indicating clearly any turning point(s). [3] Solution (a) ( ) ( ) 22 22 5 4 20 16 185 16 18 44 36 185 4 xx xxxxy xx xx x xx −+++++= = −− += + − The asymptotes are 5, 0, 4yxx= = = (b) ( ) ( ) 2 2 22 2 5 16 18Let 4 5 16 18 4 5 16 4 18 0 xx k xx x x kx kx kx kx ++ = − + += − − + + += We want values of k for which this equation has no solution, i.e. Discrimant 0< ( ) ( )( ) ( )( ) 2 2 16 4 4 5 18 0 2 25 13 0 2 1 13 0 kk kk kk + −− < + −< − +< 113 2k−<<∴ (c) x=0

5 Error! Reference source not found. 4 Referred to the origin O, the points A and B have position vectors a and b respectively. O, A and B are not collinear. Point C has position vector 2 +ab and point D lies on AB extended such that AB BD= . (a) Show that the position vector of D is 2 −ba . [2] (b) Given that OC OD= , use suitable scalar products to show that 22 833⋅= −ab b a . [2] (c) Given also that 2=ba , show that 9cos 16θ = , where θ is the angle between a and b . [2] (d) Given further that a is a unit vector, find the exact area of triangle OAC . [4] Solution (a) By ratio theorem, 2OD+ =a b   2OD∴= − ba   (shown) Alternative ( )22AD AB= = − ba     ( )2 2 (shown)OD OA AD= + = + −=−a ba ba       (b) Since OC OD= , 22+= −ab ba . Then 22 22+=−ab ba Then ( ) ( ) ( ) ( )2 .2 2 .2+ += − −ab ab ba ba ( ) ( ) ( ) ( ) 22 22 LHS 2 2 4 4 ...(1) RHS 2 2 4 4 ...(2) = + ⋅ + = + ⋅+ = − ⋅ − = − ⋅+ ab ab a a b b ba ba b a b a 22 (1) (2) : 8 3 3 (shown)= ⋅= − ab b a (c) 22 22 833 8 cos 3 3θ ⋅= − = − ab b a ab b a Given 2=ba , ( ) ( ) 2 2 22 8 2 cos 3 2 3 916 cos 9 cos (shown) 16 θ θθ = − =⇒= aa a a aa (d) ( ) ( ) 11 area of 2 2 1 1 2 22 2 sin2 AOAC θ = × + = × +× = × ×== a ab a aab a b a a 0 ab  22= =ba and 2 9 175 5sin 1 16 697cs 16 1o 1 6θθ ⇒== −== ( )( ) 61 57 5712 area of 62 11AOAC  ==  

6 Error! Reference source not found. 5 (a) Given that x is sufficiently small, find the series expansion of πtan 3x − up to and including the term in 2x . [4] (b) Hence find an approximation for the value of 7πtan 24− , giving your answer in the form 2ππab c+ + , where a, b and c are exact real constants to be determined. [3] (c) Find the series expansion of πln sec 3x − up to and including the term in 2x . [3] Solution (a) Method 1 (use small angle approx. then binomial expansion) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 2 22 2 πtan tanπ 3tan 3 1 3π3 1 tan tan3 (1 ) (2 )3 1 ( 1) 3 3 ...2! 3 1 3 3 ... 3 3 3 3 3 ... 3 4 4 3 ... x x xx x x xx x xx xx x x xx −  − −= − +  + −−= − +− + + = − −++ = − −+ + = +− + ≈ − − Method 2 (repeated differentiation) f( ) tan 3xx π= −  , 2f ( ) sec 3xx π′ = −  , 2f ( ) 2sec tan 33xxx ππ′′  = −−    f(0) tan 3 3π= −= − , 2f (0) sec 4 3π′ = −=  , 2f (0) 2sec tan 8 333ππ′′ = − −= − 2f( ) tan 3 4 4 33x x xx π= − =− + − +… (b) 7 24 24 ππ π 3x x⇒=−= − 2 2 7π ππ π πtan tan 3 4 4 324 24 3 24 24 π 3π3 6 144      − = − ≈− + −           = − +− i.e. 133, , 6 144a c b= =−=−

7 Error! Reference source not found. (c) Method 1 (integrate both sides) 2 2 ππln sec t an d33 3 4 4 3 .... d 3 2 ... x xx xxx xx c  −= −   = −+− + = − + ++ ∫ ∫ When 0, ln2xc= = 2πln sec ln 2 3 23x xx −≈−+ Method 2 repeated differentiation g( ) ln sec 3xx π= −  g ( ) tan 3xx π′ = −  2g ( ) sec 3xx π′′ = −  g(0) ln sec ln2 3π= −=  , g (0) tan 3 3π′ = −= − 2g (0) sec 4 3π′′ = −=  2g( ) ln sec ln2 3 23x x xxπ= − = − + +… Method 3 (use ln for expansion) 1 1 ln sec ln lncos cos cos sin sin3 3 33x x xxπ π ππ −= =  − +  1 ln 13cos sin22xx =  +   2 2 ln 13 2x x= −+ 2 ln2 ln 1 3 2xx=−+−  22 2 3 2ln2 3 22 xxxx  − = − − − +… 2 23ln2 3 22xxx= − − − +… 2ln2 3 2 xx= − + +…

8 Error! Reference source not found. Section B: Probability and Statistics [60 marks] 6 A fair tetrahedral die has its four faces numbered 1, 2, 3 and 4. A game is played by rolling the die twice, and the number on the face touching the table is recorded each time. Let X be the score obtained from the two rolls, defined as follows: • If the two numbers recorded are different, X is the absolute difference between them. • If the two numbers recorded are the same, X is their common value. (a) Find the probability distribution of X. [2] (b) Find the expectation and variance of X . [2] To encourage citizens to adopt an active lifestyle, a town council awards fitness cash vouchers to participants. Each participant plays the game three times and is awarded a $2 cash voucher for every point scored. The participant cashes out the vouchers only after playing the game three times, where a one-time administration fee of $0.50 is deducted. Let W be the random variable denoting the value, in dollars, of the cash vouchers received by a participant after the administration fee has been deducted. (c) Find E(W) and Var(W). [3] Solution (a) Method 1: table of outcomes Roll 2 Roll 1 1 2 3 4 1 1 1 2 3 2 1 2 1 2 3 2 1 3 1 4 3 2 1 4 7P( 1) 16X = = , 5P( 2) 16X = = , 3P( 3) 16X = = , 1P( 4) 16X = = Method 2: consider individual cases 1 17P( 1) P{(1,2),(2,3),(3,4),(1,1)} 6 16 16 16X = = = ×+ = 1 15P( 2) P{(1,3),(2,4),(2,2)} 4 16 16 16X = = = ×+ = 1 13P( 3) P{(1,4),(3,3)} 2 16 16 16X = = = ×+ = 11P( 4) P(4,4) 16 16X = = = = (b) From GC, E( ) 1.875 (Exact) Var( ) 0.859375 (Exact) X X = = Or 2Var( ) (0.92702) 0.859(3s.f)X = = (c) ( )1232 0.5W XXX= ++ − ( ) ( )E 2 3 1.875 0.5 10.75W =× −= ( ) 2Var( ) 2 3 0.8593