RVHS 9758 2024 Promo Solutions

River Valley High School 1 2024 JC1 H2 MATH 1 Solution [5] Solving system of linear equations 2 32 11Substitute 2, 2 11 4 2 2 2 4 22 ---(1) d2 d dWhen 1, 0 d 20 2 0 ---(2) d3When 2, d2 3 4 4 2 6 ---(3) abyc xx ab c a b c y a b x x x yx x ab ab yx x ab ab = + + −−  − + =−  − + =− =− − == − − =  + = = =− − − =−  + = Solving (1), (2) and (3), from GC, 6, 12, 2a b c=− = = Equation of C is 2 6 12 2y xx=− + +

River Valley High School 2 2 Solution [5] Transformation of graphs (i) (b) y O 1x=− 2y= 4( ,0)3 (0, 8)− f (3 2)yx=+ x

River Valley High School 3 3 Solution [5] Solving inequalities (i) 2 2 2 2 9 15 9 105 4 05 x x x x −− +− + − Since 22 4 0 as 0xx+   , hence 2 50x − ( )( )5 5 0xx+ −  55 + − + − 5 or 5xx−  (b) Replacing x with e x , 2 9 1e5x −− Using result in (i), ( ) e 5 or e5 (Reject as e 0 for all real ) ln 5 1 ln 5 2 xx x xx x −    Therefore 1 ln5 2x .

River Valley High School 4 4 Solution [6] Sigma Notation (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 3 3 1 21 33 11 2 2 2 2 2 2 2 2 22 2 1 112 1 2 2 144 11 2 1 1 4 1 1 4 2 14 nn r n k n nn kk rk kk n n n n n n n n n n n + = = + + == += =− = + + − + = + + − + = + + −    Therefore 4a= (b) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 3 3 3 3 11 2 33 11 2 2 2 2 2 2 2 2 2 2 42 1 3 ... (2 1) (2 ) 8 112 2 1 8 144 2 1 2 1 21 2 kk rr kk rr k r r rr k k k k k k k k kk kk == == + + + − = − =− = + − + = + − + =− =−  

River Valley High School 5 5 Solution [8] Graphing techniques (i) Since C has asymptote 44xb=  =− Method 1 ( )( )2 3 423 44 x x kkyx xx − − += − + = −− , where k is a constant By comparing ( ) 22 2 11 122 20 44 x x kx axy xx − + +++== −− , we obtain 11a=− . Method 2 By long division, 22 20 4 52 2844 x ax ay x a xx + + += = + + +−− By comparing 2 3 2 8y x x a= − = + + , we obtain 11a=− . (b) (i) When 22 11 2011, 4, 4 xxa b y x −+=− =− = − (iii) ( ) 322 20x ax x b+ + =− + ( ) 2 22 20 (1)x ax xbxb ++ =− + −−−+

River Valley High School 6 There is 1 intersect between the graph of C and ( ) 2 4yx=− − , implying 1 real root for equation (1).

River Valley High School 7 6 Solution [8] Integration Techniques (a) (i) ( ) 33 44 4 14dd1 4 1 1 ln 14 xx xxxx xc =++ = + +  (a) (ii) ( ) 24 2 12 12dd12 1 1 tan2 xx xxx x xc− =+ + =+  (b) ( ) ( ) ( ) -1 2 1 2 2 2 4 3 2 1 2 4 3 2 1 2 4 2 1 2 4 tan d 211tan d 2 2 1 1 tan d21 1 1 4 tan d2 4 1 11 tan ln 124 x x x xx x x x x xx x x x xx x x x x x x c − − − −    =−     +     =− + =− + = − + +     

River Valley High School 8 7 Solution [9] Functions (a) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 Let 6 2, 1 62 62 3 9 2 73 37 37 37 3 7 (reject 3 7 as 1) y x x x y x x y x x yx yx xy xy xy x y x y x = − − − =− + − =− − + =− − − + = − − − = − − = − =  − = − − = + − − 1 ff ( , 9]DR− = = −− 1 7 9f: ,3 ,xx xx− −− −  (b) (i) ( ) 2 2 2 2 2 2 g ( ) axa xax ax axa ax xa ax a x a xa ax ax ax a ax a x  −=  − − −= −− − = −+ = = (ii)   2 -1 2 -1 -1 -1 -1 g ( ) g g ( ) g ( ) g g(g( )) g ( ) g( ) g ( ) (Shown) xx xx xx xx =  = = = (iii) For 1fg− to exist, -1 fgRD . Since -1 ggR D ( , ) a= = − and fD ( , 1]= − − ,

River Valley High School 9 1a − Alternatively, from (ii), since 1g( ) g ( )xx −= , Since -1 ggR R ( , ) a= = − and fD ( , 1]= − − 1a −

River Valley High School 10 8 Solution [10] Integration (a) y = 1 y x