RVHS 2020 H2 P2 Math Soln V1

Suggested Solution A lvl H2 Math P2 1i Method 1: Let 2y ax bx c= + + be the required quadratic equation. d 2d y ax bx =+ Since the quadratic curve have a minimum point at ( )1, 2− , 0 2 ----- (1)ab=+ ( ) ( ) 2 2 1 1 2 ------- (2) a b c abc − = + + + + =− When d2, 5d yx x== . ( )5 2 2 4 5 ------ (3) ab ab =+ += Solving (1), (2) and (3) via GC, 51, 5,22a b c= =− = . Hence, 251 522y x x= − + . Method 2: Since the quadratic curve have a minimum point at ( )1, 2− , it is has the equation of the form ( ) 2 12y k x= − − . ( )d 21d y kxx =− . When d2, 5d yx x== . ( )5 2 2 1 5 2 k k =− = Hence, ( ) 2 2 5 122 51 5.22 yx y x x = − − = − +

Suggested Solution A lvl H2 Math P2 2ai (A) The sequence is increases by adding terms of a geometric progression (powers of 2). In particular, 1 2n nnuu+ =+ . (B) The sequence is a constant sequence, where the value remains constant at 5. 2aii Method 1: ( ) ( ) ( ) 21 3 4 5 2 5 2 5 2 2 5 5 4 15 2 4 15 5 8 35 2 8 35 5 16 75 u u p u p p u p p u p p = − = − = − − = − = − − = − = − − = − Hence, 101 16 75 11 p p =− = Method 2: 1 1 25 5 2 nn n n uu uu + + =− += Hence, 5 4 4 3 3 2 2 1 5 106 5322 5 58 2922 5 34 1722 5 22 11.22 uu uu uu upu += = = += = = += = = += = = =

Suggested Solution A lvl H2 Math P2 2bi ( ) 3 4 27 2 2 7 7 2 5 21 v a b v b a b a b = + − = + + − − = + − Hence, ( ) 43 2 2 5 21 2 2 7 14 21 7. vv a b a b b b = + − = + − =− + = 2bii ( ) ( ) ( ) 3 4 5 2 7 7 7 2 14 7 2 2 14 7 5 28. v a a va v a a a = + − = + =+ = + + + − = + 2ci Let the sequence be 1 2 3, , ,...x x x and 12 ...nnS x x x= + + + . ( ) ( ) ( )( ) ( ) 1 3232 2 2 11 4 1 11 1 4 1 3 3 1 11 2 1 4 3 25 16 n n nx S S n n n n n n n n n nn −=− = − + − − − − + − = − + − − + = − + 2cii ( ) ( ) 3 2 3 2 32 11 4 3 11 3 4 3 11 4 60 0 m m m m m m − + = − + − + + = By GC, 2,3,10 10 (as 3.) m mm =− =

Suggested Solution A lvl H2 Math P2 3(i) d 6d y t = and d 6d x tt = . d 6 1 d6 y x t t== . At ( )14,11 11 6 1 2 tt = −  = . Hence the gradient of the tangent is d1 d2 y x = . Therefore, the gradient of the normal is 1 21/ 2 − =− . Equation of the normal: ( )11 2 14 2 39. yx xy − =− − += 3(ii) When 25 1,12 6xt== and when 14, 2xt== . Required area ( ) ( )( ) ( ) ( ) 25 12 1 6 1 6 14 2 232 32 32 2 1 39 d 11 1422 1216 1 6 d 4 12112 3 4 1 1 12112 2 3 2 12 3 6 6 4 2057 units .18 yx t t t tt = + −  = − + = − +     = − − + +         =   3(iiia) Required area Scale by factor of 2 parallel to -axis Scale by factor of 3 parallel to -axis 2 2057 2 3 18 2057 units .3 x y =   = (14,11)

Suggested Solution A lvl H2 Math P2 3(iiib) Scale by a factor of 2 parallel to x-axis: Replace x with 2 x . Scale by a factor of 3 parallel to y-axis: Replace y with 3 y . Curve D; 2 232 642 18 3613 x t xt y ytt  =+  =+  =− =− Method 1: Considering x, 4 6 xt −= (reject – ve as 1 6