PH 2021 AMath Prelim P1 Anskey

1 PHSS 2021 Prelim Additional Math Paper 1 Solutions 1 4y = 2x + 1 (1) 31 4xy−= (2) From (1), y = 2𝑥+1 4 (3) Subst. (3) into (2): 2 34 421 6 3 4 8 4 xx x x x x −= + + − = + ( )( ) 28 2 3 0 2 1 4 3 0 0.5 / 0.75 xx xx x + − = − + = =− Subst. x = 0.5 into (3): y = 2(0.5)+1 4 = 0.5 Subst. x = −0.75 into (3): y = 2(−0.75)+1 4 = −0.125 𝑀𝑖𝑑𝑝𝑜𝑖𝑛𝑡𝐴𝐵 ( 0.5−0.75 2 , 0.5−0.125 2 ) = (−0.125,0.1875) 2 (a) Amplitude, b = ( )62 42 −− = Maximum value = a + 4(1) = 6 a = 2 (b) a shifts the graph vertically from the x-axis by a units

2 PHSS 2021 Prelim Additional Math Paper 1 Solutions (c) 1.5 cycle x-intercepts between 60 and 120; y-intercept at 2 3 (a) ( ) ( ) ( ) 522 5 2 2 51 1 5 2x x x x x x − + = + − + + − + +  ( ) 2 2 3 4 2 1 5 5 10 2 1 5 15 A1 x x x x x xx = − + + − + + = − + + (b) ( ) ( ) ( ) 11 2 11 11 3 1 11 11 1 nnn nn nT mx x m xnn − − − − +    = − = −       𝑇4 = (11 3 )𝑚8𝑥2 = −165𝑚8𝑥2 𝑇8 = (11 7 )𝑚4𝑥−10 = −330𝑚4𝑥−10 165𝑚8 = 128(330𝑚4) 𝑚4(𝑚4 − 256) = 0 m = 0 or 4 𝑚 = ±4 (a) f(x) = 0.05[11+(3x + 1)3] f′(𝑥) = 0.05(3)(3x + 1)2(3) = 0.45(3x + 1)2 Since f′(𝑥)  0 and 1 3x− , f is an increasing function. 4

3 PHSS 2021 Prelim Additional Math Paper 1 Solutions (b) When V = 0.95, 0.95 = 0.05[11+(3x + 1)3] (3x + 1)3 = 8 3x + 1 = 2 x = 1 3 When x = 1 3, 𝑑𝑉 𝑑𝑥 = 1.8 d𝑥 d𝑡 = 1 d𝑉 d𝑥 × d𝑉 d𝑡 = 1 1.8 × 0.081 = 0.045 m/s (c) When the volume of liquid in the container = 0.95 m 3, the height of liquid in the container is increasing at 0.045 m/s. 5 (a) When t = 0, 50 = 𝑝 𝑞+27 p – 50q = 6400 (1) When t = 2, 250 = 𝑝 𝑞+24 p – 250q = 4000 (2) p – 50q = 6400 (1) p – 250q = 4000 (2) (1) – (2): 200q = 2400 q = 12 Subst. q = 12 into (1): p = 7000 (b) 550 = 7000 12+27−1.5𝑡 27–1.5t = 8 11 t = 7−log2 8 11 1.5

4 PHSS 2021 Prelim Additional Math Paper 1 Solutions = 4.9729 (5 fig) = 5 years (round up) (c) t = 10 N = 7000 12+27−1.5(10) = 583.14 (5 fig) = 583 clients (round down) No it will have 583 clients after 10 years. 2 2 d 1 1 5cos 2sind 6 3 y xxx =+ d 1 1 5cos 2sin dd 6 3 d 1 130sin 6cos d 6 3 118 30sin 6cos63 118 30 6 22 8 15 3 4 d1 30sind6 y x x xx y x x cx c c c c y x  =+ = − + = − +    = − +       = − + =− =  16cos 4 3 1130sin 6cos 4 d63 11180cos 18sin 4 63 1199 3 180cos 18sin 4 63 3399 3 180 18 4 22 4