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−<⇒>