MI_8865_2022_Prelim_Solutions_Vetted

1 © Millennia Institute 8865/01/PU2/EOY/22 Solution PU2 MATHEMATICS Paper 8865/01 Section A: Pure Mathematics Qn Solution 1(i) [3] Let x, y and z be the cost of one serving of beans, vegetables and nuts respectively. Amy's : 2 3 9.1 (1) Betty's : 3 4 2 14 (2) Cathy's : 3 3 8.1 (3) Using GC, 1.2, 1.5, 2.2 x y z x y z xy x y z + + = −−−−− + + = −−−−− + = −−−−− = = = The cost of one serving of beans, vegetables and nuts is $1.20, $1.50 and $2.20 respectively. 1(ii) [1] Method 1 Original Price of the salad = 6.32 7.90.8 = 1.2 + 1.5a + 2.2 = 7.9 a = 3 Method 2 0.8(1.2) + 0.8(1.5a) + 0.8(2.2) = 6.32 a = 3 Qn Solution 2(a)(i) [2] Method 1 Substitute x = 3 and y = p into 2 2 2 58 3 5 8 3 5 8 0 81 or (rejected)3 p x y pp pp pp −= −= − − = =− = Method 2 Substitute x = 3 and y = p into 22 2 2 2 9 3 30 9 9 9 30 3 9 3 12 0 3 4 0 41 or (rejected)3 x y x xy pp pp pp pp + + = + + + = + − − = − − = =− =

2 © Millennia Institute 8865/01/PU2/EOY/22 Solution Qn Solution 2(ii) [3] Substitute p = –1 into ( ) ( ) ( ) ( ) 2 22 2 2 2 2 2 2 58 58 8 5 (1) Subst (1) into 9 3 30 8 5 9 3 8 5 30 8 5 64 80 25 9 24 15 30 8 5 29 87 58 0 1(1st set of solution) or 2 8 5 2 2 p x y xy xy x y x xy y y y y y y y y y y y yy yy x −= −= = + −−−−− + + = + + + + + = + + + + + + + = + + + + = =− =− = + − =− The other set of solution is x = –2 and y = –2. 2(iii) [2] ( )( ) 2 2 5 50 5 5 0 5 or 5 x x xx xx  − + −  −  Qn Solution 3(i) [3] ( ) ( ) ( ) ( ) ( ) ( ) 2 2 5 32 11 d 7 10d d 5 7 5 10 0d 1 7 31When 5, 5 5 10 5 1 3 2 6 Equation of tangent at : 31 056 31 x y xxx y x x y y P y y m x x yx y = = − + = − + = = = = − + + = − = − − = − = 6 5 5−

3 © Millennia Institute 8865/01/PU2/EOY/22 Solution Qn Solution 3(ii) [3] 3(iii) [2] Method 1 (Using GC) 5 32 1 2 1 7 31Required area 10 1 d 3 2 6 11.4 (3 s.f.) x x x x= − + + − =  Method 2 (Algebraic) ( ) ( ) ( ) ( ) ( ) 5 32 1 2 5 32 1 2 54 3 2 1 2 34 2 43 211 22 1 7 31Required area 10 1 d 3 2 6 1 7 25 10 d3 2 6 1 7 25 103 4 2 3 2 6 755 25 5 5 512 6 6 7 1 25 1 5 12 6 2 6 2 x x x x x x x x x x x x = − + + − = − + −       = − + −             = − + − −   − + −        ( )10.416 0.973 11.4 (3 s.f.)     = − − = y x 31 6y= 3217 10 132y x x x= − + + (0, 1) (–0.0967, 0) P 315, 6   (2, 9.67)

4 © Millennia Institute 8865/01/PU2/EOY/22 Solution Qn Solution 4(a)(i) [2] ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 22 2 3 3 3 d 1 d 1 1 d d 33 2 1 2 1 1d 213d 1 2 2 1 23 44213 3 2 1 xx xx xx x x x − − −     =    −−       =  −  = − − =− − =− − 4(a)(ii) [2] ( ) ( ) ( ) 33 3 3 2 2 dd 4e 4 edd 4 3 e 12 e xx x x xx x x = = = 4(b)(i) [3] ( ) 22 2 22 31 3 2 2 2d 2 d 4 4 d 4431 4