RI 9740/02 2016

1 | Page Section A: Pure Mathematics Question 1 No. Suggested Solution Remarks for Student d 0.1d V t = 1tan tan where depth of water2 r rh h hhαα=⇒= = = 2 231 11 1 3 3 2 12V rh h h hππ π = = = 2 1 33 2 3 d1d d4 d 1 36When 3, 12 1 36 d0.1 4d d 0.0251d Vh htt V V hh h t h t π π π π π = = = ⇒=  =   = Rate of increase of depth of water is 0.0251 m per minute. Raffles Institution H2 Mathematics Solution for 2016 A-Level Paper 2

2 | Page Question 2 No. Suggested Solution Rem arks for Stud ent (a)(i) 22 2 2 2 23 dcos d , cosd 12 d1sin sin d 2 , sind 1 21 1 dsin cos cos d , sind 1 22 d1sin cos sin 1, cosd vx nx x u x nx x ux nx x nx x x v nxn n xn vx nx x nx nx x u x nxn nn n x ux nx x nx nx c v nxn n n xn = = = −= = = −− + = = = + −+ = = − ∫ ∫ ∫ (ii) ( ) ( ) ( ) ( ) ( ) 2 2 22 23 22 2 2 2 22 2 2 22 1 22cos d sin cos sin 22 2 cos 2 cos 4cos 2 2cos 2If is even, cos d 4 2 2 6If is odd, cos d 4 2 6 x nx x x nx x nx nxnn n nnnn nnn n x nx x ann n x nx x ann ππ π π π π π π π πππ π ππ ππ ππ = +− = − = − = −= = = += = ∫ ∫ ∫

3 | Page (b) ( ) ( ) 2 22 0 2 2 20 22 20 2 5 29 9 25 9 5 d d1Volume d 9 2d d2 d 099 d 25 9 19 d2 1 91 d2 19 ln2 19 1 ln 9 ln 525 14 5 ln25 9 uxyx u x x x ux xx x xux xx x xu x u uu uuu uu π π π π π π π π = = − ⇒= − ⇒= − = =⇒= − = =⇒= − −=− = −  = −− = −− + + = +  ∫ ∫ ∫ ∫ ∫ Question 3 No. Suggested Solution Remarks for Student cos , 1 cos , for 0 2xt t y t t π=− =− ≤≤ (i)

4 | Page ( ) ( ) 0 1 cos 0 0 or 2 0 cos 0 1 or 2 cos 2 2 1 1, 0 and 2 1, 0 y t t xx π π ππ π = ⇒− = ⇒= ⇒=− = − = − = − −− ( ) cos , 1 cos , for 0 2 dd 1 sin , sindd d sin 0 sin 0d 1 sin 0, , 2 From above points where meets the -axis, max point is when cos 1, 1 cos 2 1, 2 xt t y t t xy tttt yt txt t Dx t xy π ππ π π ππ π π =− =− ≤≤ = += = = ⇒=+ ⇒= = ∴ = −= + = −= + (ii) ( )( ) ( ) cos 1 0 0 0 0 Area d 1c o s 1s i n d 1 cos sin sin cos d 11 cos sin sin 2 d2 1sin cos cos 2 4 13sin cos cos 2 44 aa a a a a yx t tt t t t tt t t tt tt t t aa a a − − = = −+ =−+− =−+− = −−+ = −−+ + ∫ ∫ ∫ ∫ (iii) 1 2 d1 d2 t y x π= = ( ) ( ) ( ) 2 11 , 122 Thus , equation of normal is 12 2 21 1 is ,0 2 is 0, 1 1 11area of triangle 1 12 24 t xy yx yx E F OEF ππ π π π π πππ = ⇒= = −= − − = −++ +  + +∴ = += +  

5 | Page Question 4 No. Suggested Solution Remark s for Student (a) ( )3i 1 3i 1zz−−=⇒ − + = 2arg where tan 5z αα= = (a)(i) (ii) Equation of circle: ( ) ( ) 22 3 1 1 ...(1)xy− +−= 3 1 1 Im Re 0 2 2 Z1 Z2 α 3 1 1 Im Re 0 2 2 Z1 Z2 α

6 | Page Equation of line: 5 2y xc= −+ Sub (3, 1) into line, we have 15 171 22 cc= − +⇒= 5 17 ...(2)22yx= −+ Solving (1) and (2), the two values are 3.37 + 0.0715i, 2.62 + 1.93i (b)(i) i 42 2i 8ew π−= −= 1 3 8133i i2 i3 44 4 22 811 i 122 37i ii 4 12 12 8 e 2e 2e 2 e , 1, 0,1 2e , 2e , 2e kk k zw zw zk z zz ππ ππ π πππ −    − −+         −     −−        = = = = = = =− = = = (ii) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 21 arg * 2 arg * 2 arg * arg 2 arg since * is a positive real number2 1 arg 2 371 or or ...42 2 2 7 15or or ...44 4 4 Required 7 n n n n ww w ww ww w w ww w nw n n n π π π π π ππ π π ππ π π − − − = = += = = −= −−= − − −= − − =

7 | Page Section B: Statistics Question 5 No. Suggested Solution Remarks for Student (i) ( ) ( ) ( ) 1 1 2 1 4 1 11PPP + 7 2 7 3 7 6 42RBY    ∗+ ∗+ ∗= + =       (ii) ( ) ( ) ( ) 21 P 473P| 11P 11 42 BBW W     ∗   = = = (iii) ( ) ( ) ( ) 112141 4P P P 3! 3!7 2 7 3 7 6 1029RBY       ∗× ∗× ∗ = =            

8 | Page Question 6 No. Suggested Solution Remarks for Student P D A F Total Male 2345 1013 237 344 3939 Female 867 679 591 523 2660 Total 3212 1692 828 867 6599 (i)(a) 3939 59.69 606599 = ≈ (b) 679 10.29 106599 = ≈ (ii) Part (i) only takes into account gender and dept NOT age. Age group may not be evenly spread across all dept and gender. (iii) Null hypothesis, 0H : 37µ = Alternative hypothesis, 1H : 37µ < Perform an one-tailed test at 5% significance level. Under H0, 140~ N 37, approximately by Central Limit The orem since 80 is large80Xn  = Managing director’s belief should be accepted 0H is rejected⇒ ( )-value P 0.05 0 34.824 p Xx x = <≤ ⇒<≤ ∴Set of values of x is ( ]0,34.8 (iv) Now, perform an one-tailed test at α % significance level. Managing director’s belief should not be accepted 0H is not rejected⇒

9 | Page ( )-value P 35.2 100 0.086809 100 0 8.6809 0 8.68 pX α α α α =<> ⇒> ⇒<< ⇒<≤ ∴Set of values of α is ( ]0,8.68 Question 7 No. Suggested Solution Remarks for Student (i) Number of ways is 4 3 24P = (ii) Number of ways is 10 6 4 333 720 120 24 576PPP−− = − −= Applying complement principle (iii) ( ) ( ) 8 1 !3! 1Required probability 10 1 ! 12 −= = − Grouping method (iv) ( ) ( ) 7 37 1! 5Required probability 10 1 ! 12 P−= = − Slotting method

10 | Page Question 8 No. Suggested Solution Remarks for Student (a)(i) (3, 87.4) should be excluded. Read question carefully. Question requires students to use GRAPH paper, with given scale. Is (1,72.5) acceptable? (ii) Scatter diagram does not display linear relationship, since as x increases, y increases at a decreasing rate. Thus it should not be modelled as an equation of the form y ax b= + (iii) cyd x= + All x and y values are positive, thus d > 0 to account for large values of x when 0.c x → Values of y increases as x increase, so c < 0 (iv)