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