ACSI 2019 Y3EXP FYE AMath P2 Solutions

ACS(Independent)MathDept/Y3AddMathP2/2019/FinalExam 1 Answer all the questions. 1i) Sub ( )0, 1− into 20x py− −= ⇒ p = 2 1ii) Sub 22xy= + into 22 21x xy y− += 2 4 30yy+ += ( )( )1 30yy+ += 1( . ) 3y N A or y=−= − The other point is ( )4, 3−− 2. Let ( ) 32 7f x bx cx x c= + −+ ( ) ( ) ( ) ( ) 32 1 1 1 71fbc c−= − +− −−+ ( )71 7bc c−+− − += 2bc= ------------------------(1) ( ) ( ) ( ) ( ) 32 3 3 3 73fbc c= + −+ 27 10 21 43bc+ −= 27 10 64bc+= ---------------(2) Substitute (1) into (2): ( )27 2 10 64cc+= 1c= , 2b= 3. 2 1 2xx  = 3 1 2x = 0.794x= 1.26y= The point of intersection is ( )0.794,1.26

ACS(Independent)MathDept/Y3AddMathP2/2019/FinalExam 2 4. Base area of cuboid = ( ) 2 13+ = ( ) 24 23 cm+ Height of cuboid = 14 12 3 4 2 3 423 423 +− × +− = ( ) 56 28 3 48 3 72 16 4 3 −+− − = ( )4 5 3 cm−+ 5. Let ( ) 2 22 54 44 x x a bx c xxxx ++ + = + ++ , ( ) ( ) 22 54 4x x a x x bx c+ += + + + By comparing constants, 1a= By comparing x terms, 5c= By comparing x2 terms, 11 b= + 0b= ( ) 32 22 2 341 5 44 xx x xxxx −+− = + ++ x y

ACS(Independent)MathDept/Y3AddMathP2/2019/FinalExam 3 6 (i) Let ( ) 322 ( 2) ( 7) 3f y y my my= +− +− − ( ) ( ) ( ) ( ) 32 1 21 ( 2 )1 ( 7 )13f mm− =−+−−+−− − 2 2 73mm= −+ −− +− 0= (ii) 32 11 1 12 ( 2) ( 7) 3 022 2 2f mm   = +− +− − =       127 3044 2 mm−−++− = 1 2 2 14 12 0mm+−+ − − = 9m= (iii) ( ) 322 7 23fy y y y= + +− ( )( )( ) 322 7 2 3 12 1y y y y y ay b+ + −= + − + Comparing coefficient of x3, 1a= Comparing constants, 3b= ⇒ ( )( )( )12 1 3 0y yy+ − −= 11, 32y y or y= −= = 7a) 2 68mx m x+ + >− 2 8 60mx x m+ +−> 2 40b ac−< ( ) 284 6 0mm− −< 264 4 24 0mm−+< 2 6 16 0mm− −> ( )( )2 80mm+ −> 28m or m<− > and m >0 m > 8 7b) ( ) 22 21 2x kx k++= − ( ) 22 2 1 20x k xk+ + +−= ( ) ( )( ) 22 4 2 1 42 2b ac k k− = +− − 24 4 1 8 16kk k= + +− + 24 4 17kk= −+

ACS(Independent)MathDept/Y3AddMathP2/2019/FinalExam 4 22 2 114 1722kk   = −+ − +     2 14 16 2k=−+ Since discriminant > 0, the equation has real and distinct roots for all real values of k. Alternatively, ( ) 22 21 2x kx k++= − ( ) 22 2 1 20x k xk+ + +−= ( ) ( )( ) 22 4 2 1 42 2b ac k k− = +− − 24 4 1 8 16kk k= + +− + 24 4 17kk= −+ ( ) ( )( ) 22 4 4 4 4 17b ac− = −− < 0 and a > 0 ⇒ 24 4 17 0kk−+> Since discriminant > 0, the equation has real and distinct roots for all real values of k. 8 (a) 32 16 4 8 x x x − − = 34 32 2 22 2 2 x xx − − = ( ) 3432 222 x xx −−−