KSS_2023_Prelims_4NA AMaths P1 (Solutions)-1

1 | Page 2023 Sec 4NA Preliminary Examinations Paper 1 1 18 7 ( 1)(4 1) 1 4 1 18 7 (4 1) ( 1) 1 18 7 5 5 1 4 1118( ) 7 ( 1)44 55 24 2 18 7 5 2 ( 1)(4 1) 1 4 1 x AB xx x x xA xB x Let x A A Let x B B B x xx x x + = +−+ − + += ++ − = += = =− − += −− =− =− + = −−+ − + 2 2 2 2 3 512 3 3 40 (3) 4(3)(4) 39 xx x xx D ++=− + += = −= − Since D < 0, the line will not intersect the curve. 3(i) 2 2 (5 3)( 2) 6 5 10 3 6 6 5 7 12 0 ( 1)(5 12) 0 121(reject) or 5 xx x xx xx xx xx + −> − + −> −−> + −> <− > (ii) x = 3 4(i) ( ) ( ) ( ) 2 32 32 2 2 2 21 32 5 43 26 4 3 3 3 0 xx x xxx xx xx xx x xx −− + + −− −+ −− −− − −− −− − 5 3 2 2 2 80 ( 2)( 2 4) 0 20 2 40 2 2 4(1)(4) 12 0 x x xx x or x x x or D −= − ++= −= + += = =− = −< Since 2 2 40xx+ += has no solutions, x = 2 is the only real root of the equation 3 80x −= 4 (ii) x + 3 is a factor of f(x). 6(i) 2 2 2 22 22 2 43 32( 2 ) 2 32( 2 ( 1) ( 1) ) 2 52[( 1) ] 2( 1) 52 xx xx xx xx −−+ = − +− = − ++ −− −− = − +−= −++ 4 (iii ) 2( 3)(2 1) 0 ( 3)(2 1)( 1) 0 13 or or 12 x xx x xx x + −−= + + −= = −− 6(ii ) Maximum value is 5.

2 | Page 7 3 2 2 4 2 3, 4 23 At 5, 4(5) 2 98 52 98 26 0.531 cm/s (to 3 sf)49 x dVVx x dx dVx dx dV dV dx dt dx dt dx dt dx dt = −+ = − = = −= = × = × = = Rate of change of x = 0.531 cm/s 8 2 2 2 2 2 2 32 32 32 d 12 2d 12 2 6 2 (2,15) , 27 6(2) 2(2) 27 7 6 27 6 27 2 7 2, 15 15 2(2) 2 7(2) 11 2 7 11 y xx dy x dx x x cdx dyAt dx cc dy xxdx y x x dx x x x c When x y c c y xx x = − = − = −+ = − + =  → = = −+ = −+ = −++ = = ∴= −+ + =− = −+− ∫ ∫ 9a (i) 1 3sin ( ) 602 − = ° 9b (ii) 9a (ii) 1cos ( sin 45 ) 135− − °= ° 9b (i) Period = 180°, Amplitude = 3 10 (i) 2 22 ( 7 2) ( 7 2) 7 47 4 7 47 4 22 22 22 AC AC cm reject AC = ++ − = + ++− + = = =− 10 (ii) 2 22 72tan 72 ( 7 2) ( 7) 2 7 47 4 1 1 47 33 A −= + −= − −+ −= = 11 (i) 2 2 sin 2 1 cos 2 2sin cos 1 2cos 1 2sin cos 2cos sin cos tan xLHS x xx x xx x x x x RHS = + = +− = = = = 11 (ii) sin 2 21 cos 2 tan 2 1.1071 1.1071, 1.1071 1.11, 4.25 x x x BA x π =+ = = = + =

3 | Page 12 (a) 4 1 34 2 1 34 2 3 3 2 3 3 2 12 14 12 ( ) ( 12 )( 2 )2 12 4 12 ( 12 ) 12 [4(1 2 ) ] (1 2 ) (4 7 ) (1 2 ) xy x x xx xdy dx x x xx x x x xx x xx x − − = + +− + = + +− += + +−= + += + b (i) 25 52 4 4 4 ( 1) 2 ( 1) (5)( 1) ( 1) [2( 1) 5 ] ( 1) (7 2) y xx dy xx x xdx xx x x xx x = − = −+ − = − −+ = −− b (ii) Since y is decreasing, 0dy dx < 4 4 ( 1) (7 2) 0 Sin ( 1) 0 , (7 2) 0 20 7 xx x ce x for all x xx x − −< −≥ −< << 13 2 28yx x=++ 22dy xdx = + At x = 1: 4, 11dy ydx = = Equation of tangent: 11 41 y x − =− 47yx= + At x = 4: 2(4) 2 10, 32dy ydx = += = 1Gradient of normal is 10− Equation of normal is 32 1 4 10 1432 10 10 12 3210 5 y x yx yx − =−− −= − + = −+ Solve the 2 equations simultaneously: 124 7 32 10 5 124 2510 5 8 326 ; 3141 41 xx x xy += − + = = = 8 32(6 ,31 )41 41P