SAJC P1 ans

1 Solutions 1 (i) y x O

2 1 (ii) Method 1: (Graphical Method) – Preferred and efficient method From the graph, for 247 4712xx x+ ≤− ++ , The intersection points have x-coordinates −4 and 2 respectively. 4 or 0 or 2xx x≤− = ≥ Method 2: (Algebraic Method) – Not preferred for this question 2 2 2 2 2 474 12 12 7 44 3 33 xx x xx x xxx xx xx x −+ + ≤+ +≤ ++ ≤≤ ≤ − + ( ) ( ) 2 2 3 30 4 0 4 or 0 xx x xx x xx xx −+ ≤ ++≥ +≥ ≤− ≥ and ( ) 2 2 3 20 20 0 or 2 xxx xx xx xx ≤+ −≥ −≥ ≤≥ Combining 4 or 0xx≤− ≥ and 0 or 2xx≤≥ on a number line, 4 or 0 or 2xx x≤− = ≥ .

3 2 (i) 2 (ii) ( a− , 0)  vertical asymptote xa=− (a , 0)  vertical asymptote x = a Minimum point ( ) 20, a−  Maximum point 2 10, a  − (a , 0) y x y = f(x) O

4 2 (iii) The tangents to the curve of 1 f( )y x= tends to a horizontal line / becomes parallel with the x – axis when x→±∞ . 3 (i) Let us consider an arithmetic progression with 12a= and 2d = , where n is the number of odd days. Then ( )( )12 1 2nun=+− . Let 42nu = , then ( )( )12 1 2 42n+− = gives 16n= . Hence it requires 16 odd numbered days to complete 42 km. Starting on 1st day, the last day is given by 1 + (16-1)(2) = 31. Therefore Mary completed 42 km on the 31st day of training. (shown) x = a y x O y = 0

5 (ii) Let us consider a GP, with a = 12, 1 100 xr = + and n = 31 for Leo to first run 42 km Using 1n nu ar −= 31 1 12 1 100 using GC, 4. 42 26 ( 2 d.p.) x x − + = = (iii) Total distance covered by Mary =2 distance covered during 1st 15 odd num bered days + distance covered on 31st day × ( ) ( )( ) ( )( ) ( ) ( ) 31 15Total distance covered = 2 2 12 15 1 2 422 12 1 1 0.042643 1 1 0.042643 822 745.516 1567.516 1570 3s.f.   + −× +  −+ + −+ = + = =

6 4 (i) Let T be the temperature of a heated body at t min. dd 0 , ( ) , > 0dd ddd ( ), where > 0dd d 1d d in out in out TT kT ktt TTT kT ktt t T kTt = = −θ = − =− −θ =−−θ Integrating with respect to both sides, 1 dd ln , where C is an arbitrary constant ln( ) , since . , where . C kt kt C t T ktT T kt C T kt C T T ee T Ae A e − − = −−θ −θ=− + −θ =− θ+> −θ= = +θ = ∫∫ Given that 25θ= , 25ktT Ae −= + . 0 When 0, 175, 175 25 150 d 150d dwhen = 0, 3,d 1150 3 50 kt tT Ae A T ket Tt t kk − = = = + ∴= =− =− ∴− =− ⇒ = 0.02150 25 tTe −∴= + For the apple pie to cool to 400C, 40T = ,

7 0.02 0.02 0.02 40 150 25 150 15 0.10 115.13 115min (3 sf ) t t t e e e t − − − = + = = = = 5 (i)  πππ 2 3 3ACB θθ− −=−= By Sine Rule,  sinsin BC AB ACBB AC = B A C 2

8   sin sin 2sin sin 3 2sin sin cos sin33 2sin π cos2 4si ππ cos 31 sin n 2 3 cos sin AB B ACBC ACB θ θ θ θθ θ θθ θ θθ = =  − = = − = − − (ii) Given that θ is a sufficiently small angle,

9 2 1 1 2 3 cos sin 31 2 3 1 33 3 34 4sin 4 4 4 41 13 3 4 ..33 . BC θ θθ θ θ θ θθ θθ θθ θθ − − = = − ≈ −−