KSS_2023_Prelims_4NA AMaths P2 (Solutions)-1

1 | Page 2023 Sec 4NA Preliminary Examinations Paper 2 1(a) 2 2 2 3 32 2( 3) 3(3 2) 2( 3) 3(3 2) 2 2 24 8 22 2 22 26 96 11 0 0 or 11 x x x x x x x xx x xx xx xx − + − + − −+ = = = −= − − −= = = 1(b ) Midpoint of PQ = (−2, 7.5) Gradient of PQ = 10 5 5 2 ( 6) 8 − =−− Gradient of ⊥ bisector = 8 5− Equation of ⊥ bisector 7.5 8 25 8 43 5 10 y x yx − =−+ = −+ 2 Area of square base ( ) 2 10 2 10 2 20 2 12 4 5= + = + += + Height 52 28 5 52 28 5 12 4 5 12 4 5 12 4 5 12 4 5 + +−= = × + +− 64 128 5 1 2 5 cm64 += = + 3(a) 5sin 7A= Using Pythagoras Theorem: 24x= 24 2 6cos 77A= = sin2 2sin cos 5262( )( )77 20 6 49 A AA= = = 3b (i) sec(2 15 ) 2 cos(2 15 ) 0.5 x x − °= − °= Basic angle = 60° 2 15 60 ,300 37.5 ,157.5 x x − °= ° ° = °° 3b (ii) 2 2 2 7sec 6 tan 23 0 7(1 tan ) 6 tan 23 0 7 tan 6 tan 16 0 (7 tan 8)(tan 2) 0 AA AA AA AA + −= + + −= + −= − += 8tan 7A=  0.852A= (3 sf) or tan 2A=−  2.03A= (3 sf) 4a (i) 32 2 41 60 5002 3 41 60 Nt t t dN ttdt = −+ − + = −+− To find max, let 0dN dt = 23 41 60 0 ( 3 5)( 12) 0 5(rej) or 123 tt tt tt −+−= −+ − = = = 4a (ii) 32 41(12) (12) 60(12) 5002 1004 N = −+ − + = 4(b ) ( ) ( ) 2 222 2 22 3 4 3(2 ) 6 ( 4) 4 When 0, 4 0 Therefore, 0 for all 0. y x dy x x dx x x xx dy xdx = + −−= = + + > +> <> Hence y is a decreasing function.

2 | Page 5(i) 52 411(1 ) 5 (1 )d xdx x x −−= − 5(ii ) From (i), rewrite as 52 4 6 411(1 ) 5 (1 ) 5 ( 1)d x xxdx x x −−−= −= − 2 64 1 2 64 1 2 55 1 ( 1) 1 5 ( 1)5 1 1 11 1(1 ) (1 )5 5 2 160 x x dx x x dx x − − − = − = − =−= ∫ ∫ 6(i) 22 22 22 4 6 12 0 4 4 6 91 249 ( 2) ( 3) 25 xy xy xx yy xy +−+−= − ++ + += ++ − ++ = Centre = (2, −3) & Radius = 5 units 6 (iii) 3 19 44yx= −+ When y = 0, 19 3x= When x = 0, 19 4y= Area of triangle 1902 01 3 19 192 3044 1 361 19= ( 19)2 12 2 475 1919 or19.8(to3sf)24 24 ABC = − −+ = = 6(ii ) 1 ( 3) 4Gradient 52 3 3Gradient of tangent at 4 CP P −−= = − =− Equation of tangent at P: 13 54 3 151 44 3 19 44 y x yx yx − =−− −= − + = −+ 7(i) 32 3 3 12 12xxx x− −+= + 32 2 3 40 ( 3 4) 0 ( 1)( 4) 0 0 or 1 or 4 xx x xx x xx x x − −= −−= + −= = − When x = 4, y = 16 A (4, 16) and B (2, 14) 7 (ii) [ ] 4 32 2 442 3 2 2 Area of 1 (14 16)(2) 3 3 122 330 1242 30 24 14 30 10 20 unit ABC x x x dx xx xx = + − − −+ =− −− + =− − =−= ∫

3 | Page 8(i) cos4 4cos sin10 10sin 10sin 4cos x x y y AF θ θ θ θ θθ = = = = = + (ii) 22 1 10sin 4cos 10 4 116 2 29 4tan 10 21.8 2 29 sin( 21.8 ) AF R AF θθ α θ − = + = + = = = = ° = +° 8 (iii) 116 sin( 21.8 ) max length of 116 when 21.8 90 68.2 AF AF θ θ θ = +° = + °= ° = ° 8 (iv) 10cos 68.1986 4sin 68.1986 7.4278 6 tan 68.1986 2.4 Max possible height of train 7.4278 2.40 5.03m (to 3 sf) >3.6 m CF h h = °+ °= = ° = = − = Yes, a train of height 3.6 m can pass through the tunnel.