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 6t .) Hence, 418 3 6 xy −=− . Method 2: Considering y, 3 18 yt += Hence, 2 2 2 364 18 69 454 54 6 225. yx yyx x y y +=+ ++=+ = + +

Suggested Solution A lvl H2 Math P2 4(i) Let O be the centre of the square, M be a midpoint of one of the sides of the square and P be the top vertex of the pyramid. We have , , , 90 .2 aOM PM h OP H MOP= = =  =  Also, 30 2 15 2 aa h h= +  = − . By Pythagoras theorem, ( )( ) 2 22 2 2 22 15 15 (sub 15 ) 2 225 15 (Shown) ahH aaH h h aah a =+   = − +     = − = − =− 4(ii) Let V denote the volume of the pyramid. ( ) ( ) 2 2 4 4 5 1 225 153 15 225 15 2593 V a a V a a a a =− = − = − 34d 252 100d3 VV a aa =− . At max V, d 0d V a = ; 34 3 25100 0 3 25100 0 3 0 or 12 aa aa aa −=  −= == Clearly, 0, 0aV== , which is not the maximum. If ( ) ( ) 2 3112, 12 225 15 12 144 5 cm3aV= = − = . This is the maximum volume. 4(iiia) Let A be the required area of the 4 triangular faces.

Suggested Solution A lvl H2 Math P2 ( ) ( ) 2 2 2 2 4 2 2 15 2 30 225 15 225 0 (as 15 0 for all .) 225 cm . ahA aa aa a aa =  =−  =− = − −  − −  = The largest area is 225 cm2 when 15 0 15aa− =  = . 4(iiib) When ( ) 215, 225 15 15 0 0a H H= = − =  = . As the pyramid’s height is 0 cm, this implies the pyramid is flat (2- dimensional).

Suggested Solution A lvl H2 Math P2 5(i) Her possible points are 0, 4, 10 and 25. 5(ii) Let X denote Tina’s score. ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( )( )( ) ( )( ) ( ) ( )( ) ( ) ( ) E 0 P 0 4P 4 10P 10 25P 25 10 2 24 2 2 1 25 10 3 1 3 3 1 3 3 1 3 16 8 40 25 25 3 3 1 81 33 3 3 1 27 11 .31 X X X X X rrr r r r r r r r r r r r r r r r r r = = + = + = + = −−= + + ++ + + − + + −= + −= + −= + ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( )( )( ) ( )( ) ( ) ( )( ) ( ) ( ) 2 2 2 2E 0 P 0 4 P 4 10 P 10 25 P 25 100 2 216 2 2 1 625 10 3 1 3 3 1 3 3 1 3 64 32 400 625 625 3 3 1 1089 657 3 3 1 363 219 .31 X X X X X rrr r r r r r r r r r r r r r r r r r = = + = + = + = −−= + + + + + + − + + −= + −= + −= + Hence, ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) 22 2 2 2 22 2 2 2 Var E E 363 219 27 11 3 1 3 1 3 1 363 219 729 594 121 31 1089 294 219 729 594 121 31 360 300 340 . 31 X X X rr rr r r r r r r r r r r rr r =− −− =− ++  + − − − + = + − − − + −= + +−= +

Suggested Solution A lvl H2 Math P2 5(iii) ( ) 2 2 22 2 2 360 300 340 38 31 360 300 340 342 228 38 18 72 378 0 4 21 0 3 ( 7 is rejected as 1.) rr r r r r r rr rr r r r +− = + + − = + + + − = +−= = =− 

Suggested Solution A lvl H2 Math P2 6(i) Note: s.d. is 1.2 min, so 5 3.6 3 1.2 =  . This means, by empirical rule, that 0 to 10 should cover almost all the distribution. 6(ii) ( ) 2~ N 5,1.2T ( )P 6 0.202328 0.202 (3 s.f.) T = = 6(iii) We need to consider he time he takes to leave home, then the amount of time he takes to walk to work. ( ) ( ) 22E 21 5 26, Var 1.2 3 10.44T W T W+ = + = + = + = ( )~ N 26,10.44TW+ ( )P 30 0.107864 0.108 (3 s.f.) TW+  = = 6(iv) ( ) ( ) 22E 19 5 24, Var 1.2 6 37.44T D T D+ = + = + = + = ( )~ N 24,37.44TD+ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) ( )( ) P weather is fine lateP weather is fine | late P late P weather is fine P 30 P weather is fine P 30 P weather is not fine P 30 0.7 0.107864 0.7 0.107864 0.3 0.163400 0.6063435 0.606 (3 s.f.) TW TW TD = += + + +  = + = = T/min 0 (8am) 5 (8.05am) 10 (8.10am)