DHS ans

DHS 2019 Year 6 H2 Math Prelim P1 Mark Scheme 2019 Year 6 H2 Math Prelim P1 Mark Scheme Qn Suggested Solution 1 Number sold before 7 pm Number left after 7 pm Banana 10 10 Chocolate 45 5 Durian 20 10 Let the selling price of banana cake, chocolate cake, durian cake before discount be $b, $c, $d respectively.

29.50 1 10 45 20 730 2 9 4 146 2 0.6 10 5 10 880 730 10 5 10 250 2 2 50 3 abc bcd bcd bc d bc d bc d " " " Solving (1), (2), (3) using GC, a = 8.50, b = 9, c = 12 The selling price of banana cake, chocolate cake and durian cake is $8.50, $9 and $12 respectively. Qn Suggested Solution 2(a) 2 2 2 2 2 0 3 0 01 1 30 11 2( 21 1 1 2 (2 3 9 )( 4) (3 ) ( 29 9) 0 9 3) x x xx x xx x xx xx d

d

d

d 3 23o r 3 4xx? d d 4 3 –3 ++ – – + x 00000000000000000000000 4))4))4))4)4)4))))4)4)4 3)33))))))))))))33))

ddd ++ ++ – – www.KiasuExamPaper.com 218

2 DHS 2019 Year 6 H2 Math Prelim P1 Mark Scheme (b) 3 3 2 2 2 n (3) e 0 30 e 0 3 3 si ce for all or 3 ax bx ax bx ab x ab x a b aa bb x x xx

! ! ! Qn Suggested Solution 3(i) Since A, B and C are collinear and AB ba JJJG 5P? 55 65 OC OB BC

=b b a ba JJJ G JJJ G JJJ G (ii)

1 65 ( 1)2 731 2 OE k OE ON OAOO O O OO §· ¨¸©¹ b b a a b a JJJ G JJJ G JJJG JJJ G 72 610 27 7 kOO Ÿ Ÿ 6 7OE P? bb JJJ G Qn Suggested Solution A C B O N E B B E E 6 7777777777777 b bbbbbbb 6 www.KiasuExamPaper.com 219

3 DHS 2019 Year 6 H2 Math Prelim P1 Mark Scheme 4 (i) Since the shape of the curve is For f to be 1-1, the largest b can take is the x-coordinate of the turning point. 2 2 1f' ( ) 1 () 110 () 1 x xa xa xa r x-coordinate of turning point is a+1, since ba! For graph to be 1-1, 1bad , (ii) Let f( )yx 2 2 2 2 2 2 2 1 1 (1 ) (1 ) 1 1 (1 ) 1 0 (1 ) (1 ) 1024 (1 ) ( 1) 124 (1 ) ( 1) 124 yx x xy x x xy y x x xy x y yyxy yyx yyx §· ¨¸©¹ §· ¨¸ ©¹ r Since 37,22 §· ¨¸©¹ is a point on the curve of f( )yx , 2(1 ) ( 1) 124 yyx 2 1 (1 ) ( 1)f( ) 1 24 xxx The domain of 1f is the range of f = [3, )f . x = a y x 2222 ( 111111111111) 111 4444444 yyy ( ((((((( ( 22222222222 1 )))) (((( 1111)))) 11 )) (( ) ( )) ((

www.KiasuExamPaper.com 220

4 DHS 2019 Year 6 H2 Math Prelim P1 Mark Scheme Qn Suggested Solution 5(a) Series of transformations: 2 2 2 22 ln 1 1ln ln 1 21 21ln ln(2 ) 4 xy x xxy xx xxy xx p p (b) (i) (ii) Qn Suggested Solution 6 >@ >@ iarg( ) 1 ( 1) 2aarg ( ii rg 1 arg 1 1)nwn n n 1. Reflect in the x-axis (replace y with y) 2. Scale by factor 1 2 parallel to the x-axis (replace x with 2x) y x O y x O www.KiasuExamPaper.com 221

5 DHS 2019 Year 6 H2 Math Prelim P1 Mark Scheme (i) (ii) >@ >@

>@ >@ >@ >@ 12 12 1 2 1 1 a i arg( ) arg iiarg i arg i i i arg(1) 2arg(1 i) arg(1 2i) a 2 rg(1 i) 2 arg(1 2i) arg(1 3i) arg(1 rg arg( ) arg( ) arg( ) 1 ( 1) 1 ( 1) 1 1( 1 ) a r g 1 2 arg 1 ( 1) n n n n k k n k n k z ww ww kk k kk k w w w } } §

· ¨¸¨¸©¹

¬

ªº ¼ ¦ ¦ ¦ > >@ > @ 1 4 2arg(1 3i) arg(1 4i)) arg[1 ( 2)i] 2arg[1 ( 1)i] arg(1 i) arg[1 ( 1)i] 2arg(1 i) arg[1 ( 1)i] arg(1) arg(1 i) arg(1 i) arg[1 ( 1)i] ʌ DUJ L DUJ> L@ nn n nnn nn nn ­ ° ° ° °® ° ° ° °¯ # (iii) As ,nof 11 22ʌDQG ʌarg(1 i) arg[1 ( 1)i]nno o Hence 1 4rg ʌa. nz o (argand diagram with y = -x line to show argument) Thus )I m ( )Re( nnzz Qn Suggested Solution 7 (i) 22 4sin2 2 16sin 2 ( 2) x x T T ---- (1) 22 4cos2 3 16cos 2 (3 ) y y T T ---- (2) (1) + (2) gives 22(2 ) (3 )1 6xy 222222222 222222222 (((( 222222222222222222) ( ( ------ (1(11111)))))))))))))))))))))))))))))))))))))) 3333 y

3 www.KiasuExamPaper.com 222

6 DHS 2019 Year 6 H2 Math Prelim P1 Mark Scheme Hence C is a circle with centre (2 , 3 ) and radius 4 units. (ii) d 8cos 2d x TT and d 8sin2d y TT gives d tan 2d y x T For 3ʌ8T , 22 2x 32 2y d 1d y x Equation of tangent: 32 2 1 ( 2 2 2 )yx Equation of normal: 32 2 2 2 2yx So T (0,1 4 2) and N(0,5) Hence the area of triangle NPT 1 (4 2 4)(2 2 2)2 (2 2 2)(2 2 2) 12 8 2 units2 Alternatively, Let E be the point closest to P along the y-axis. Since d 1d y x at P, the triangle TPE is such that ET EP and 90oTEP ( . P T N E y x PPPPPPPPPPPPPPPPPPPPPP NN NN EEE www.KiasuExamPaper.com 223

7 DHS 2019 Year 6 H2 Math Prelim P1 Mark Scheme The normal at P i.e. d 1d y x . the triangle NPE is such that EN EP and 90oNEP ( . Therefore the two triangles are congruent, and the area of triangle NPT

2 122 2 2 2 2 22 22 2 12 8 2 ªº «»¬¼ Qn Suggested Solution 8 (i) x –2 y + 3z = 4 ---- (1) 3x + 2y – z = 4 ---- (2) Solving (1) and (2) using GC gives 2 0.5xz 1 1.25yz zz Hence L : 22 15 , 04 OO §· §· ¨¸ ¨¸ ¨¸ ¨¸¨¸ ¨¸©¹ ©¹ r \ (ii) P3 : 5 1 6 k §· ¨¸ ¨¸¨¸©¹ r < If the three planes have no point in common, 25 5 0 46 k §·§· ¨¸¨¸ ¨¸¨¸¨¸¨¸©¹©¹ < 10 5 24 0 2.8 k k Ÿ ? ¹ hrrrrrrrrrreeeeeeeeeeeeeeeeeeee ppppppppppppplallaaaaaaaaanenenenenenenenenenenenenenenenenn sssssssssssss hahahahahahahahahhahahahhahaahhahaaah vveeeeeeeveeveeeeeeeeveee nnnnnnnoooooo poppppp ······ 5 ¸¸¸¸¸¸ ···· ···· www.KiasuExamPaper.com 224

8 DHS 2019 Year 6 H2 Math Prelim P1 Mark Scheme (iii) 2 1 0 OQ o §· ¨¸¨¸ ¹ © Distance required 5 2.8 6 5 2.8 6 1.42 units (3 s.f.) 68. 2 11 0 1 12.8 84 §· ¨¸˜¨¸©¹ ¨ §· ¨¸¨¸

§· ¸¨¸© ©¹ ¹ Alternative 2 1 0 OQ o §· ¨¸¨¸ ¹ © and let 0 0 1/6 OY o §· ¨¸¨¸©¹ where Y is a point on P3 Shortest distance from Q to P3 22 2 52 5 2.8 1 2.8 61 / 6 6 1.42 units 68.845 ( 2.8) 6 YZ o §·§ · §· ¨¸¨ ¸ ¨¸ ¨¸¨ ¸ ¨¸ ©¹© ¹ ©¹ << (iv) Plane containing Q and parallel to 3 :P 5 2.8 6 25 where 1 2.8 5(2) 2.8( 1) 6(0) 12.8 06 5 2.8 6 12.8 xy z d d xy z §· § · ¨¸ ¨ ¸ ˜ ¨¸ ¨ ¸© ? ¹© ¹ Since 12.8 1 0,!! 3P is in between the above plane and the origin. Thus O and Q are on the opposite sides of 3.P nddddddd QQQQQQQQQQQQQ aaaaaaareeeeeeeeeeeeeee oooooooooon n nn n nn n nnn nn n ththththththththththththththhhe eee oppppopppppppppppppppppppppppppppppppppp siisiisiitetetete ssssssidididididide www.KiasuExamPaper.com 225

9 DHS 2019 Year 6 H2 Math Prelim P1 Mark Scheme Qn Suggested Solution 9(i) (a) d1 e1d2 yy x

2 2 32 2 2 22 2 2 d1 d ed2 d dd1 dd dd d d d d d11 2dd d d d d d yyy xx yy xx yy y y y y y xx x x x x x §· ¨¸©¹ ªº§· § · ¨¸ ¨ ¸«»©¹ © ¹¬¼ (b) 243 43 dd d d 12 2dd d d yy y y xx x x ªº§·§ · «»¨¸¨ ¸ ©¹© ¹«»¬¼ 23 4 23 4 0, 0 (given) d1 d 1 d d 1, When , 0 , d2 d 4 d d 8 xy yy y y xx x x 11 23 84 4 1 2 21 2 02! 4! 11 8 192 yx x x xx x } } (ii) 1 2 1 2 d 1 d e1d d 1 1 e y y y x y x 1 2 1 2 1 2 1 2 1 2 d de e de ln | e | ee e ln( 1 11 e) y y y y xCyx x yx y A y xC xC A

r ³³ ³ When 0, 0xy 1 2 1 2 11 0 22 e) e) 0l n ( ln( x A y A

? ) 0000000 000000000000 yyyyyyyyyyy 0,0000,000,0,, 000 000000 e)e)e)e)e)e)e) 0 AAAAA www.KiasuExamPaper.com 226

10 DHS 2019 Year 6 H2 Math Prelim P1 Mark Scheme Alternative (for integration) 1 2 11 22 1 2 1 2 1 2 1 2 de ee de e 1 1 1 1 (1 1 l ) de ne|1 | y yy y y y y y y y xC xC xC xC y

³ ³ ³ 1 2 1 2 1 2 1 2 l 1 ln | 1| ln l ne e e e 1 ln | | n e e yy y y y y x xC x C xC C

# (iii) 5|f( ) g ( ) 00|.xx From GC, { : 0 2.43}xxd \ Qn Suggested Solution 10(i)

2 3 d d 1 (2 1) 1 2 2 3 (2 1)3 3 12 1 21 21 1 1 21 1 xx x x xx C xx C x xx x xx C x ªº ¬¼ §·

§· ¨ ¨ ¸ © ©¹ ¹

¸ ³³

2222222222222222222 11 2 xxxxxxxxxxxxxxx (2(2(2((2(2(2(2((((2(2(2(2((2(2((2((2((2(2(2(2(2(2(2(222(2(22(222(22222(2((2(2((2(2 x 1111111 xx 1111 C

1

§§§·§§· ··§§§§§§§§§§§§§§ 11 (((((((2(((((( 1))) 11 (2(((((((2(((2(2(2(2(2(((2((2(((2(2(2((22(2(((2(2(((2(2( 1)))) xxxxxxxxxxxxxxx 22222222222222 xx 11111111111 1)1)1)1) 1)) ©©©¹©© ¹¹¹©©©©©©©©©©©©©©© ( )))) 33333333333333333333333333 xxxx (2((2 1)1)1)1) 1)1) ¨¨¨¨¨ ¨¨¨¨¨ ¸¸ ¸¸

www.KiasuExamPaper.com 227