2012 H2 Math Prelims Paper 2 Solution + Marking Scheme (final)

2 Sectio Qn Su 1 B r r 1(i) V d d d d  A d d d d d d d d 1(ii) d d r d d W d d 2012 Ye on A uggested S y Pythagora  22 15 1 5 30h h    2453V h   d 30d 20 30 d 0.050d V ht h t       Alternative   d 90d 3 d dd d d d d d 30 5 d 0.050d V hh hh tV h t h t        30 d 1 30d 2 h h r ht    d dd d dd When 5, d 0.0455d rr h th t h r t    ar 6 Pre olution as’ Theorem  2 2 5 (shown) h h   3h     2 2 d d 55 929 0.0 5 hh t          2 2 33 0 d 1 5 5 929 0.0 5 h h V t       2 1 2 2 30 h h    15d d 30 552 0.0 4 h t r t   eliminary m,  d d 509 cm/min h t   20 20 509 cm/min hh  2h     2 5 5 55 456 cm/min    15 y Exami n (3.s.f) n (3.s.f) 0.050929 n (3.s.f)  5 15 h ination P  P2 Markk Scheme Marking S AG1 – Cor formulation AG M1 – Corre differentiat A1 Alternati v M1 – Diffe & chain rul A1 M1 – Evalu M1– Corre differentiat A1 – Rate o = 0.0456 cm d 0.04d r t  Tota 1 e Scheme rrect n leading to ect tion e erentiation le uate r ect tion of decrease m/min or 456 cm/min al: 6 marks o s

2 Qn Suggested Solution Marking Scheme 2 G1 – Asymptotes G1 – Shape + exclude origin C1 is transformed to C2 by i) a reflection in the y-axis and ii) scaling by a factor of 1 2 along the x-axis. Note: Order of (i) and (ii) can be interchanged. B1 – (i) B1 – (ii) Note: Order of (i) and (ii) can be interchanged. G1 – Horizontal asymptote G1– Shape and x- intercept Total: 6 marks Qn Suggested Solution Mark Scheme 3 (i) (ii) G1 – Circle with centre at origin, radius 5 G1 – Perpendicular bisector of the line segment joining points representing (0,8) and (-8,0); passes through origin x y 1x 2y O x y 1 1 2y  O 0 P 5 (R) Re Q – 5 5 α |β| α locus (i) locus (ii) 8 – 8 β Im

3 From diagram, 3 4 4 33 55e 5 cos isin ( 1 i)44 2 55e 5 cos isin (1 i)44 2 i i p q                  Alternative arg 0p q        arg arg 0 arg argp qp q  Equation of circle: 222 5xy -------- (i) Equation of perpendicular bisector: y x -------- (ii) Solving (i) and (ii) : 5 (1 i ) 2 5 (1 i) 2 p q    From diagram arg(5 ) q  > 0 arg(5 ) | |p    (5 )arg (5 ) arg(5 ) arg(5 ) | | ( ;using corresponding angles) p q pq PRQ         ( at circumference, semi-circle)2   M1 – Use polar forms with correct radius and angle for p and q A1 A1 M1 – Solve simultaneous eqns (must see correct equations for both) A1 A1 B1 : 2  Total: 6 marks 3 4  4  4  P Q 5 5 0

4 Qn Suggested Solution Mark Scheme 4 (i)  50 1 11 1 99 49 99 1 0.15 0.075 (Shown)2 aa ad a a    M1 – Correct expression for 50a in terms of 1a AG1 4 (ii)  50 1 50 0.075 99 0.075 187.52n n a     M1 – Correct formula A1 4 (iii)      25 1 99 0.075 0.98 (0.075) 24 0.15 35.8 least 36 k k ba k k       M1 – For kb A1 4 (iv) Consider 1 1 10 . 9 8 0.991 0.98 1 0.98 0.98 0.01 227.9 least 228 h h b b h h      M1 – Correct inequality (must see inequality) A1 4 (v)  13 0 147 3 99 0.075 10 . 9 8 3.82 (3s.f.) m m b bb b          M1 – Correct formula A1 Total: 10 marks Qn Suggested Solution Marking Scheme 5 (i) Subst. 1 2 3   r into LHS of equation of 1p , we have LHS = 1 27 2 0 32            . = RHS.  A lies in 1p (Shown) Subst. 1 2 3   r into LHS of equation of 2p , we have B1

5 LHS = 13 257 2 3             . B1 5 (ii) Consider 32 4 75 2 6 22 5 2 1                12 4 :2 2 6 , 35 2 1 lm m                   r  Here, 24 4 26 2 52 1 1 6, 3 k k                   M1 – Consider cross product of the normals A1 – Correct equation formed AG1 – Correct method leading to AG 5 (iii) Acute angle = 1 33 7. 5 22cos 34.5 62 38                M1 – Correct formula used (condone w/o modulus) A1 5 (iv) Let foot of perpendicular be F. 14 14 22 62 35 14 4 62 . 2 0 51 (16 4 1) 4 12 5 1 kk OF k BF k kk k k k k k                                        Hence 5 0. 4 OF    Alternatively, 21444 42 222 483 111 22121 21 1 4 2 1 AF AB                                       M1 – Find either OF  or BF  M1 – Use dot product for appropriate pair of vectors and set to zero A1 M1 – Correct projection used

6 145 22 0 31 4 OF OA AF                Consider '2ABA BB F       where B’ is reflection of B in l. 13 7 '6 2 4 2 54 3 AB             Hence required line is 17 22 , . 33 mm               r  M1 – Correct method to find OF  from AF  A1 M1 – Correct vector equation formed using previous part A1 Total: 12 marks Section B Qn Solution Mark Scheme 6 (i) 810 2730k  List the applications in so me order e.g. by applicants’ names or loan amount, and randomly select the first application to process. Subsequently, select every 27th application until a sample of 30 applications is selected, going back to the front of the list if necessary. Alternatively (only applicable if k is an integer), List the applications in some order and divide the applications into 30 groups of 27 each. Randomly select a number from 1 to 27, e.g. 5, and pick the 5 th application in each of the 30 groups. B1 for every 2 points:  List applications in a certain order  Randomly select first application  Select every 27th application thereafter  Go back to the front of the list if necessary 6 (ii) A systematic random sample may not ensure that all categories of loan applicatio ns were processed within that day if the sampling interval coincides with a cyclic pattern in the list of applications. B1 – Disadvantage of a systematic random sample 6 (iii) Stratified random sampling ensures that each category of loan is proportionately represented in the sample. B1 – Advantage of stratified random sample (answer in context) Total : 4 marks

7 Qn Suggested Solution Mark Scheme 7(i) Let G and E be the number of Green Top taxis and EZCab taxis arriving in a randomly chosen 10-minute period respectively. G ~ Po(3) , E ~ Po(5) G + E ~ Po(3+5)  P( 7) 1 P( 6) 1 0.31337 0.68663 0.687 (shown) 3 s.f. GE GE      B1 - G + E ~ Po(3+5) AG1 7(ii) Required probability =P all taxis arrived were EZCab | at least 7 taxis arrived    P all taxis arrived were EZCab and at least 7 taxis arrived P at least 7 taxis arrived P( 7) P( = 0) P( 7) EG GE    1 P( 6) P( = 0) P( 7) EG GE      1 0.76218 0.049787 0.68663  = 0.0172 (3 s.f.) M1 – P( 7) P( =0) P( 7) EG GE   Note: Accept 0.687 for denominator (given) A1 Total : 4 marks

8 Qn Suggested Solution Marking Scheme 8 (i) Number of six-figure number that can be formed 7 6 5040P B1 – accept w/o working 8 (ii) Number of six-figure number that can be formed 67 117649 B1 – accept w/o working 8 (iii) Case 1: xx yy z