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 xx y x x x y x x x c c c c      −− = − − =− − − + − =− − − +    − =− − − +           =  11180cos 18sin 4 4 63y x x x =− − − + 6

5 PHSS 2021 Prelim Additional Math Paper 1 Solutions 7 (a) 20.1 4 1.8h d d=− + + h = −0.1𝑑2 + 4𝑑 + 1.8 = −0.1[𝑑 − 40𝑑 − 18] = −0.1[𝑑2 − 40𝑑 + (−20)2 − (−20)2 − 18] = −0.1[(𝑑 − 20)2 − 418] = − 1 10 (𝑑 − 20)2 +41.8 (b) The coordinates of the turning point are (20, 41.8). The greatest height reached by the object is 41.8 m and its corresponding horizontal distance is 20 m. (c) When h = 0, −0.1(𝑑 − 20)2 + 41.8 = 0 −0.1(𝑑 − 20)2 = –41.8 (d – 20)2 = 418 d– 20 = ±√418 d = 20 ± √418 = 40.445 or –0.44504 (5 fig) Since the ball travelled 40.4 m (> 40m) when it hits the ground. Therefore the canon is not precise. 8 1cos 2A= ; 3sin 2A= 1sin 2 B=− ; 1cos 2 B=− (a) cos (A – B) = cos A cos B + sin A sin B = ( 1 2) (− 1 √2) + (√3 2 ) (− 1 √2) = − 𝟏 𝟐√𝟐 − √𝟑 𝟐√𝟐 = −𝟏−√𝟑 𝟐√𝟐 = −√𝟐−√𝟔 𝟒 = − 𝟏 𝟒 √𝟐 − 𝟏 𝟒 √𝟔

6 PHSS 2021 Prelim Additional Math Paper 1 Solutions (b) cos A = 2 cos2 𝐴 2 − 1 2 cos2 𝐴 2 − 1= 1 2 cos 𝐴 2 = ±√3 4 cos 𝐴 2 = √3 2 9 (a) Volume of gift box, V = 2.5x2h When V = 27000, 2.5x2h = 27000 h = 10800 𝑥2 (b) ( ) ( ) ( ) ( ) 2 2 2 2 2 112 2.5 2 2 2.5 2 2 2.5 44 5 8.75 108005 8.75 945005 (shown) A x hx hx hx hx A x hx A x x x Ax x    = + + + +        =+ =+  =+ (c) d𝐴 d𝑥 = 10𝑥 − 94500 𝑥2 For stationary points, d𝐴 d𝑥 = 0 10𝑥 − 94500 𝑥2 = 0 x3 = 94500 10 x = 21.1 (to 3 fig) d2𝐴 d𝑥2 = 10 + 189000 𝑥3 When x = 21.141, d2𝐴 d𝑥2 > 0. Thus when x = 21.1, A is a minimum.

7 PHSS 2021 Prelim Additional Math Paper 1 Solutions 10 (a) Remainder = f(–5) = 0 (–5)3 + a(–5) + b = 0 b – 5a = 125 (1) Remainder = f(3) = 24 (3)3 + a(3) + b = 24 3a + b = –3 (2) (2) – (1): 8a = –128 a = –16 Subst. a = –16 into (1): b = 45 (b) f(x) = x3 – 16x + 45 = 0 (x + 5) is a factor of f(x). 𝑥3 – 16𝑥 + 45 𝑥+5 f(x) = (x + 5)(x2 – 5x + 9)=0 (𝑥 + 5)[(𝑥 − 2.5)2 + 2.75]=0 Since (𝑥 − 5 2) 2 + 2.75 > 0 for all real values of x (or discriminant = -20 < 0), f(x) has only 1 real root where x = –5. 11 82(a) 31 3 1 3 since 1, 90 and parallel to , is a right-angled trapezium. AB AD AB AD m m m m BAD AB CD ABCD −= − = =−  =−  = 

8 PHSS 2021 Prelim Additional Math Paper 1 Solutions ( ) (b) Let be midpoint of 3 5 8 4, 4,622 is the midpoint of 4 5 6 4 9, ,52 2 2 M AC M E MC E ++== ++   ==       ( ) ( ) 2 (c) eqn. of side : 4 35 3 11 (1) 3 27 (2) subs. (1) into (2) 6, 7 6,7 3 4.5 6 31Area = 8 5 7 82 1 94.5 872 3.75 units DC y x yx yx xy D − =− = − −−−−− + = −−−− == =− = (a) 1 tan tan 1 2sec 2tan 1 tan 1 xx xxx −+ −=+− LHS = 1 tan tan 1 tan 1 tan 1 xx xx −+ −+− = (𝑡𝑎𝑛𝑥−1)2+(tan 𝑥+1)2 1−tan2 𝑥 = 2 tan2 𝑥+2 1−tan2 𝑥 = 2(tan2 𝑥+1) 1−tan2 𝑥 = 2(sin2 𝑥+𝑐𝑜𝑠2𝑥 𝑐𝑜𝑠2𝑥 ) 𝑐𝑜𝑠2𝑥−sin2 𝑥 𝑐𝑜𝑠2𝑥 or 2 2 2sec 2 1sec x x− = 2 𝑐𝑜𝑠2𝑥 𝑐𝑜𝑠2𝑥−sin2 𝑥 𝑐𝑜𝑠2𝑥 or 2 22 2 2cos 1 cos cos x xx − 12

9 PHSS 2021 Prelim Additional Math Paper 1 Solutions = 2 𝑐𝑜𝑠2𝑥−sin2 𝑥 = 2 𝑐𝑜𝑠2𝑥 = 2 sec 2x = RHS (proven) (b)2𝑠𝑒𝑐2𝑥 = 5 𝑐𝑜𝑠2𝑥 = 2 5 ∝= 𝑐𝑜𝑠−1 2 5 = 66.421° 2𝑥 = 66.421,293.579,426.421,653.579 𝑥 = 33.2,146.8,213.2,326.8(𝑟𝑒𝑗) (c) Principal value of cos−1 (− 1 2) = 2π 3 . 13 (a) ( )30 18 6ve=− = 12 ms-1 (b) 30 18 6 te=− 318 6 te− =− 3 3 3 ln 3 ln 3 0.366203 0.366 s A1 te t t t = = == = (c) 318 6 tve=− 318 tdvae dt= =− 0,a Decreasing velocity (d) 318 6 tve=−

10 PHSS 2021 Prelim Additional Math Paper 1 Solutions 3 3 3(0) 3 d 18 6 d 18 2 0; 0 0 0 2 2 18 2 2 t t t s v t e t s t e c st ec c s t e = = − = − + == = − + = = − +  (e) ( ) ( )3 0.36620 0.36620 18 0.36620 2 2 2.5916se = − + = m ( ) ( )32 2 18 2 2 2 768.85se= − + =− m Total distance = 2(2.5916) + 768.85 = 774.03 m Average speed = 774.03 2 = 387.01 = 387 m𝑠−1