CJC P2 Ans

1 2020 JC2 H2 PRELIMS Paper 2’s Suggested Solutions 1: Let y = ax3 + bx2 + cx + d, where a, b, c and d are real constants. When x = 0, y = 0, ∴ d = 0 Given x = 2, y = 0, a(23) + b(22) + c(2) = 0 8a + 4b + 2c = 0 ------ (1) Given x = 2.55, y = – 0.0631, a(2.553) + b(2.552) + c(2.55) = – 0.0631 ------ (2) Differentiating y with respect to x, 2d 32d y ax bx cx = ++ Given x = 0.785, d 0d y x = , 3(0.7852)a + 2(0.785)b + c = 0 ------ (3) Solving (1), (2) and (3) using GC, a = 0.0993, b = – 0.497, c = 0.596 (3 s.f.) Hence, y = 0.1x 3 – 0.5x2 + 0.6x.

2 2: dcos ec cos ec cotd xx θ θθθ= ⇔= − ( ) 2 6 32 3 22 33 6 22 3 6 2 3 6 2 3 6 2 3 3 2 6 11 d cos ec cot d 1 cos ec cos ec 1 1 cot d cos ec cos ec 1 1 cot d cos ec cot 1 dcos ec sin d sin d x xx π π π π π π π π π π π π θ θθ θθ θθ θθ θθθθ θθ θθ θ = − −− =− − =− =− =− = ∫∫ ∫ ∫ ∫ ∫ 6 3 6 3 (shown) cos 2 1 d 2 1 sin 2 22 π π π π θ θ θ θ θ −= = − ∫ ∫ 22sin sin1 63 2 26 23 = 12 ππ ππ π    = −− −    

3 3: (i) ( )( ) 3 31 34 31 34 AB rr r r ≡+++ ++ When 13,13 14rA= −= = −+ When 43,13 41rB= −= = −−+ ( )( ) 3 11 31 34 31 34rr r r∴ ≡−++ ++ ( )( )11 3 11 31 34 31 34 11 47 11 7 10 11 3 23 1 11 3 13 4 11 43 4 nn rr rr r r nn nn n = = = − ++ ++  =−+  −+  −+ −+  − ++ = − + ∑∑ 

4 (ii) ( )( )7 3 13 13 1 3 22 25 25 28 28 31 3 1 3 4 r rr ∞ = ×+×+×+= ++∑ ( )( ) ( )( ) 6 11 33 3 13 4 3 13 4 11 1 1lim 43 4 43 ( 6 )4 111 4 4 22 1 22 rr n rr rr n ∞ = = →∞ = −++ ++ = − −−  ++  = −− = ∑∑

5 4: (i) d 3d x kxt = − Since the population remains constant when xp= , 03 kp= − 3k p= Thus d3 3d x xtp= − (ii) d3 3d x xtp= − ( )d 33 3 d x xp xptpp −= = − 13 ddxtxp p =−∫∫ 3ln xp tc p−= + 3 e tcpxp + −= 3 e tpxpA−= where ecA=± 3 e tpxA p= + ( )3 0 55 pAe p A p= +⇒=− Thus ( ) 3 5e tpxpp= −+ (shown) (iii) Since the population of trees decreases over time, 50 5pp−<⇒> Let x = 0, ( ) 3 05 e tppp= −+ 3 e5 tpp p−= − ln35 pp tp−= − or ln35 ppt p= − or ln35 ppt p =  − since p > 5

6 Thus, Starting from the initial year t = 0, it takes ln35 pp p   − years for the tree population to be fully depleted. x t (0, 5) ln , 035 pp p     −

7 5: (ai) 1 2AD AO OD a b= + = −+        1 2BE BO OE b a= + = −+        Method 1 (Using Triangular Law of Vectors) 1 2 1 2 ( ) for some Also, ( ) for some OF OA AF a a b OF OB BF b b a λλ µµ = + =+ −+ ∈ = + =+ −+ ∈                 Equating we get, 11 22(1 ) (1 )ab a bλλµ µ− + = +−   Since andab are non-parallel and non-zero, we can equate coefficients. 2 2 1 1 µ λ λ µ  −= = − ∴  Solving, 2 3λµ= = 2 11 33 1 32( ) aaOF b ab∴= ++ +−=       (shown) Method 2 (Using vector equation of lines) 1:, 2ADl r a ba λλ= + −∈    1:, 2BEl r b ab µµ= + −∈    At the point of intersection of the two lines, 11 22a ba b abλµ+ −= + −    ( ) ( )1111 22ab baλλ µµ−+ = −+  Since and ab are non-zero and not equal, we can equate coefficients, 2 2 1 1 µ λ λ µ  −= = − ∴  Solving, 2 3λµ= = 2 11 33 1 32( ) aaOF b ab∴= ++ +−=       (shown) Method 3 (Using property of medians) By property of medians, ( )1 2 2 2 12 3 3 33BF BE b a a b= = −+ = −       

8 So, ( )121 333OF OB BF b a b a b=+= +−= +           (Shown) Method 3 (Using Ratio Theorem and property of medians) By property of medians, : 1:2DF FA = By ratio theorem, ( ) ( ) 1 2221 3 33 abOA ODOF a b ++= = = +        (Shown) (aii) 1 2 ( ) using the midpoint theoremOC a b= +    . 1 3 21 2 32 3 () () OF a b a b OC = + = +=       Since for some scalar where 0 1OF kOC k k= <<     , point F lies on OC. (bi) PQ q p= −    . Since ( ) (since . . ) 0 PQ n q p n qn pn pn pn q n p n ⋅= − ⋅ =⋅−⋅ =⋅−⋅ = =          PQ n∴⊥    (shown). So, line // plane PQ Π Since P is a point on line PQ, but P is not on plane Π ( 0OP n p n⋅=⋅≠      ), line PQ is parallel to Π but non-intersecting. (b)(ii) Point P with position vector p  lies on the plane. PQ q p= −    is parallel to the plane. Since n is perpendicular to plane Π, and plane Π is perpendicular to the new plane, another vector parallel to this plane is n . ∴ New plane : ( ) , ,rp qp nλ µ λµ= + −+ ∈    . Equivalent answers: New plane : ( ) , ,rq qp nλ µ λµ= + −+ ∈    or ( ) , ,r q pq nλ µ λµ= + −+ ∈    New plane : () ()r qp n pqp n −×= −×     or () ()r pq n p pq n −×= −×     New plane : () ()r qp n qqp n −×= −×     or () ()r pq n q pq n −×= −×    

9 6: (ai) P (all counters blue) 6 4 10 4 1 14 C C= = (aii) Case 1 : P(3 blue,1 red) 64 31 10 4 8 21 CC C ×= = Case 2 : P (all blue) 1 14= P(at least 3 blue) = P(3 blue, 1 red) + P(all blue) 8 1 19 21 14 42=+= (aiii) P (all counters red) 4 4 10 4 1 210 C C= = P(at least 1 counter of each colour is drawn) = 1 – P(all blue) – P(all red) 1 1 971 14 210 105= −− = (aiv) P(at least 3 blue | at least 1 of each colour) P(at least 3 blue at least 1 of each colour) P(at least 1 of each colour) ∩= P(3 blue,1 red) P(at least 1 of each colour) = 8 4021 97 97 105 = = (b) Since P(at least 3 blue | at least 1 of each colour) = 40 19 97 42≠= P (at least 3 blue), the two events are not independent.

10 7: (i) Any two of the following: • the event whether a JC2 student sign up for the GP tuition is independent of one another. • The probability of a JC2 student sign up for the GP tuition is a fixed constant 0.08. • There are a fixed number of JC2 students received the flyers. (ii) Let G be the r.v. denoting the number of students sign up for GP tuition out of 120 students. ( )~ 120,0.08GB (a) ( ) ( )15 1 15PG PG>= − ≤ ( ) 0.030115PG => (b) ( ) ( )1Var G np p= − ( ) ( )120(0.08) 0.98 11048.83 or 125 Var G = = (iii) ( )30 or 31 0.037986PM = = ( ) ( )30 31 0.037986PM PM= += = From GC, Since p > 0.08, p = 0.125 0.0703 0.125 P(M = 30) + P(M = 31) = 0.037986 P(M = 30) + P(M = 31) p