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 443 x x x x xx x x x x x xx xc x xc x − −      − = − +           = − + = − + + − = − − +   4(b)(ii) [2] 11 1 1 1 1 44 4 4 4 4 ee d e d 4e 1 4 xxx x x x c c +++ + = = + = + Qn Solution 5(i) [2] Using GC, ( ) 5 d 15ln 1 5 10 2.5d t ttt = + − + =− The population of the moth is decreasing at a rate of 2.5 thousand moths per year at the 5th year after the natural bird predator is introduced. 5(ii) [5] ( )15ln 1 5 10P t t= + − + d 15 5d1 P tt=−+ To find stationary point,

5 © Millennia Institute 8865/01/PU2/EOY/22 Solution Qn Solution ( ) d 0d 15 501 15 51 15 5 1 5 10 2 P t t t t t t =  − =+ =+  = + = = Performing first derivative test: t 2− (1.99) 2 2+ (2.01) d d P t 0 (0.0167) 0 0 ( 0.0166− ) Slope Hence P is a maximum at 2t= The maximum value of P ( )15ln 2 1 5(2) 10 16.5= + − + = The maximum population is 16 500 (3 s.f.). 5(iii) [2] 5(iv) [2] The rate of change of the population of the natural bird predators is 0 initially, i.e. 0Q= when 0t= . ( )( ) 00 3 0 1 2 0 ba ba ab = + −   = − = 5(v) [1] Using GC, ( ) 4 0 5 3 1 5 2 dt=31.798=31.8 (3s.f.)tt+ −   Population is 31.8 hundred (or 3180) ( )8.87, 0 t P O

6 © Millennia Institute 8865/01/PU2/EOY/22 Solution Section B: Probability and Statistics Qn Solution 6(i) [2] Vowels: A, O, U, I, E Consonants: F, V, R, T There can only be one arrangement for vowels (V) and consonants(C) to alternate: V, C, V, C, V, C, V, C, V, No. of ways 5! 4! 2880=  = 6(ii) [3] Complement: all the 4 consonants are together V V V V V C C C C No. of ways for all 4 consonants together 6! 4!= No. of ways 9! 6! 4!=− 345600= Qn Solution 7(i) [2] Let X be the number of faulty mechanical pencils, out of 15. Then ( )~ B 15, 0.07X ( ) ( ) 15P P 3.754 P3 0.98247 0.982 (3s.f.) XX X   =  = == 7(ii) [3] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 15P 1| P 1| 3 4 P 1 3 P3 P 1 3 P3 P 3 P 0 P3 0.65729 0.657 (3s.f.) X X X X XX X X X XX X    =     =  =   − ==  ==

7 © Millennia Institute 8865/01/PU2/EOY/22 Solution Qn Solution 8(i) [2] ( ) ( )P( ) 1 P P 1 0.2 0.5 0.3 B A B A B   = −  −  = − − = 8(ii) [3] Method 1: Given A and B are independent, A and B are independent. ( ) ( ) ( )P P P 0.5A B A B =  = ( ) ( ) 0.5P P 0.5 1 0.3 5 7 A B=  = − = ----------------------------------------------------------------- Method 2: Given A and B are independent, A and B are independent. ( ) ( ) ( )P P P 0.2A B A B    =  = ( ) ( ) 0.2P P 0.2 1 0.3 2 7 A B  =  = − = ( ) ( )P 1 P 21 7 5 7 AA =− =− = ----------------------------------------------------------------- A B

8 © Millennia Institute 8865/01/PU2/EOY/22 Solution Qn Solution Method 3: Given A and B are independent, ( ) ( ) ( )P P PA B A B =  ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) P P P P P P P P A B A B A B A B A B  = + −  = + −  ( ) ( ) ( ) ( ) 0.8 P 0.3 P 0.3 0.5 0.7P 5P 0.714 (3s.f.)7 AA A A = + −  = == 8(iii) [1] ( ) ( )P | P (since and and independent) 51 7 2 0.286 (3s.f.)7 A B A A B = =− == Alternatively, ( ) ( ) ( ) ( ) ( ) ( ) ( ) PP | P PP (since and and independent)P P 2 0.286 (3s.f.)7 ABAB B AB ABB A  =  = = == Qn Solution 9(i) [2] 9(ii) [2] 0.94517 0.945 (3 s.f.)r== Since r = 0.945 is close to 1 , there is a strong positive linear correlation between the maximum ambient temperature and the mean ambient temperature. 9(iii) [2] Using GC, 0.42 14.326 0.42 14.33 (2 dp) yx yx =+ =+ x y (29.4, 26.7) (34.0, 28.8) (34.0, 28.8)

9 © Millennia Institute 8865/01/PU2/EOY/22 Solution Qn Solution Note: It is recommended to sketch the regression line at the same time the scatter diagram is sketched in part (i). 9(iv) [2] When x = 31, ( )0.42 31 14.326 27.346 27.3 (3 s.f.) y y =+ == Since r = 0.945 is close to 1 and x = 31 is within the data range, the estimate is reliable. Qn Solution 10(i) [3] 10(ii) [2] P(all three offspring are of same gender) = P(MMM) + P(FFF) 0.5 0.65 0.65 0.5 0.65 0.65 0.4225 =   +   = y x (29.4, 26.7) y = 0.42x + 14.33 Male Female 0.5 0.5 Female Female Female Male Male Female Male Male Female Male Female Male 0.65 0.65 0.35 0.35 0.65 0.65 0.65 0.65 0.35 0.35 0.35 0.35

10 © Millennia Institute 8865/01/PU2/EOY/22 Solution Qn Solution 10(iii) [3] P(first offspring is a male| at least two female offspring) P(first offspring is a male at least tw o female offspring)= P(at least two female offspring) P(MFF) P(MFF)+P(FMF)+P(FFM)+P(FFF) 0.5 0.35 0.65  = = 0.5 0.35 0.65 0.5 0.35 0.35 0.5 0.65 0.35 0.5 0.65 0.65 0.2275   +   +   +   = 10(iv) [2] Method 1 Required probability ( ) 20.4225 1 0.4225 3 0.309 (3 s.f.) = 