A Level 2015 P1 Solutions for 9780 CWY 2018

A Level 2015 P1 Solutions (edited for New H2 Math Syllabus 9758) Qn Suggested Solutions 1(i) 2: aC y bx cx   3 2 2 1 1.6 1 ( 0.7) 2.4 1.6 (1) 3.6 ( 0.7) (2) d2 2 2 (3)d a b c a b c ya b a bxx                      1(ii) When C crosses the x -axis, 2 0ay bx cx    By GC, 0.589x (3 d.p.) 1(iii) 2 3.593 5.187 7.303 As , 5.187 7.303 yx x x yx       The other asymptote is the line 5.187 7.303yx  . 2(i) (ii) By GC, 1.73Ax  , 0.414Bx  , 1.73Cx  (3 s.f.) From the graph, 1 21 x xx   when 1.73 0.414x   or 1.73x . Solving simultaneously, 3.59345 3.593a  5.18691 5.187b  7.30274 7.303c

3(i) 1 1 2f f ... f n n n n n                   is the sum of areas of the n rectangles in the sketch. As n , the space between the top of the rectangles and the curve  fyx decreases to 0, so that this sum approaches the area under the curve.   1 0 1 1 2lim f f ... f f d n n xxn n n n                      3(ii)   33 3 3 3 3 3 1 3 1 1 2 ...lim 1 1 2lim ... 1 1 2lim f f ... f where f x x x n n n n n n n n n xxn n n n                                      1 0 11 1 0 0 3 4 33 4 fd 3d 4 xx x x x        4 Total perimeter of shapes,      2 22 4 2d x y x xx y       4 (1)2 dxy   Total area of shapes, 21 2 (2)xA xy    Sub (1) into (2): x y x

  2 2 2 4 1 22 22 xdx xA d xx    Let d 40d2 Ad xx    . 8 dx 2 2 d 40d A x   when 8 dx .  The area is maximum when 8 dx . Maximum area 2 21 2 8 8 3 2 2 d d d d             (shown!)

5(i)     1 242 2 3 0 Scaling by factor parallel t T o the -axi ranslate s 133 4yy x y x y x            The graph of 2yx undergoes a translation of 3 units in the positive direction parallel to the x-axis, followed by a stretch of ¼ units parallel to the y-axis. 5(ii) 5(iii) 6(i) Using standard series from MF26,   23 23 8 3 (2 ) (2 )ln 1 2 (2 ) 23 22 xxxx x x x          6(ii)                  2 23 2 3 4 31 1 21 1 ... 2! 3! 1 1 2 ...26 c c c c c cax bx ax c bx bx bx ab c c ab c c cax abcx x x                Comparing coefficients of x : 2a 2x : 122 bbc c  3x :   2 8 31b c c 

    2 8 3 8 31 bc bc b b   or   8 3 8 3 8 3 1 1 bc bc b b        53 53 ,bc   Coefficient of 4x in  1 c ax bx is        35 3 8 13 5 5 532 104 6 27   7(i) 3 5OC a , 5 11OD b 7(ii) 3 5BC OC OB    ab Line BC :  3 5 ,OB BC      abrb  3 5 1 (1)   r a b (shown!) 5 11AD OD OA    ba Line AD :  5 11 ra ba   5 111 , (2)      r a b 7(iii) When the lines BC and AD meet, (1) = (2)         35 5 11 35 5 11 11 11                   a b a b ab As a and b are non-parallel, 3 5 1 0 (3)    and 5 11 1 0 (4)    Solving (3) and (4) simultaneously, 3 4  , 11 20  Sub into (1):    3 3 3 9 1 5 4 4 20 41OE    a b a b 11 1 20 4AE OE OA    ab 99 20 44ED OD OE    ab O A B C D 3 : 2 E

11 9AE ED  : 11: 9AE ED  8(i) For athlete A, the lap timings , 2, 4,T T T form an A.P. with first term T and common difference 2. Time taken for 50 laps  50 2 2 49(2) 25(2 98)TT   1 21 hours = 1 21 3600 5400 s ; 3 41 hours = 3 41 3600 6300 s . To meet the required time interval, 5400 25(2 98) 6300 216 2 98 252 118 2 154 59 77 T T T T         8(ii) For Athlete B, the lap timings 2,1.02 ,1.02 ,t t t form a G.P. with first term t and common ratio 1.02. Time taken for 50 laps     50 50(1.02) 1.021.02 1 50 11 t t    To meet the required time interval,     50 50 5400 50 1.02 1 6300 108 1.02 1 126 63.845 74.486 63.9 (3 74. s.f.)4 t t t t          Set of values of t = { : 63.9 74.4}.tt   8(iii) If each athlete takes 1 21 hours, then 63.845t  and T = 59 [from parts (i) & (ii)]. On their 50th laps, Athlete A takes 49(2) 157Ts . Athlete B takes 491.02 168.47ts . Time difference 157 168.47 11 (nearest second)s   9(a) Given 2 * w w is purely imaginary,

2 2 i for some * * i kkw w ww k   2 22 ( i ) ( i ) i 2 i i a b a b k a b ab bk ak        Comparing real and imaginary parts, 22 (1)a b bk   2 2 (2)ab ak k b    Sub (2) into (1): 2 2 2 2 2 2 2 1 3 323 aa b b a b b a b        3i awa   10(i)   12 2 0 2 0 2 Area of Area of cos d sin sin sin 0 1 (1) AA xx x                    1 4 2 4 2 24 2 2 Area of 2 cos d (by symmetry) 2 sin 2 sin 2 (2) sin 21 2 A xx x                (1) – (2): 2 1 (2 2) 2 1 (3)A      (2)  (3):  1 2 222 2 21 1 2 2 1 A A    10(ii) The shaded region is the region described. Rotating it 2 radians about the y-axis gives the volume  0 2 22 0 2 1 11 22 sd d (show n n)ixy yy  . The working of someone who forgot how to integrate sin x.

10 (iii) 1sin siny u u y   When y = 0, 1 00sinu   ; 2 2y , 1 2 2 4sinu   . 4 00 4 0 2 1 2 22 2 (sin d ( d d (shown d)) d co )s yu u y u u uy u            Exact volume,   4 0 4 2 4 0 0 2 d si cos 2 ) s( innd V u u u u uu uu             Now 2 2 2 4 0 16 4 2 32 snsn 0i iu u       and           0 4 0 4 0 4 4 0 2 42 22 82 ( d c) sin co s sin soduuu u u u u u                 2 32 3 2 2 2 32 8 2 22 32 4 2 ( 2) units V           11(i) 32sin 3sin, cos , 0 2xy      23si c sd d nox   ,   26sin cos cos 3sind d ( sin )y         223sin 2cos sin    22 2 22 d d d d d d 2cos 3sin 2cos sin 3sin cos co sin sin si s cos 2cot tan (shown) (1 n ) y y x x                   2 d d d sind cos 2 vwu u w u u u vu     Integration by Parts: d d d cosd sin 1 vwu u u u w vu    

11(ii) 0 2cotd tnd a 0 (2)y x      (2) tan : 22 tan 0  2 2 2 tan 2 tan 0 as tan (0 )            C has a turning point when tan 2  . From the diagram,   3 3 2 3 22 33 sinx        2 2 21 3 3 2 3 cos3s 3 iny      The turning point is   326 2 93 , . Differentiating (1) w.r.t x,   2 22 2 22 2 d2cosec sed d c d 2cosec sec 3sin cos y xx        As  is acute, 2cosec , 2sec  , 2sin  and cos are all positive 2 2 6 d 3 2 d 0y x   , The turning point is a maximum point. 11(iii) 23sin si cos 0 0 orns 0co y        0  or 2   (as 20  ) 0x or 1x ddd d xy x y       222 0 2 0 2 1 0 4 Bounded area d 3sin 3sin d 9si cos cos cond s 0.884 (3 d. (shown!) p.) yx                 x y C 0 P y = ax 1 Diagram for parts (iii) & (iv):

11(iv) To find P, we can substitute the parametric equations of C into y ax :   23 2 2 3 3sin or 3cos cos sin sin 3cos sin 0 sin 0 sin 0 (rejected as 0 0) o r tan a a a a x                     3 (tan shown)a  For P to be the maximum point of C, 3 3 2 2, using result from 11(itan 32 2 i) 2a a       