AJC H2 MATH P1 SOLUTIONS

Anderson Junior College Preliminary Examination 2011 H2 Mathematics Paper 1 (Solutions) 1 12cos dx xx  22 12 4 23 12 4 2 12 4 2cos d22 1 14cos d24 1 1cos 122 xx x x x x xx xx x x xx C               2 Let the price of a X-box console be $ x, a Kinect sensor be $y and a Game DVD be $z. 499 0.9 0.85 0.9 439.15 0.95 0.75 0.8 426.30 xyz xy z xy z      Aug matrix = 11 1 4 9 9 0.9 0.85 0.9 439.15 0.95 0.75 0.8 426.30     rref = 1002 4 7 0101 9 9 001 5 3    x = 247 , y = 199 , z = 53 Employees of Company P will pay $[0.9(247)+0.8(199)+0.85(53)] = $426.55 > $426.30. No, it will not be more attractive for employees to purchase all the 3 items from own company. 3      2 0xaxb xxc    , x c , 0x Since  2 0xc as x c ,   0xaxb x     0xx a x b  x a or 0 x b and x c Replace x by ln x . ln x a or 0l n x b  and ln x c 0 ax e or 1 bx e  and cx e

4 (i) tantx  22sec 1dt x tdx   = 2 2 2 11 11 1 dtt t t      = 2 1 12 dtt = 11 tan 2 2 tc  = 11 tan 2 tan 2 x c  (ii) volume is same as Exact volume = 2 4 20 1 d1s i n xx      41 0 1 1 tan 2 tan 2 tan 2 2 x         5 (a) Method 1: sin 2 cos dx xx 1 sin 3 sin d2 1c o s 3 cos23 1c o s 3 cos23 xx x x x C x x C         Method 2: sin 2 cos dx xx y =2 2 1 1s i ny x  2 12 1s i ny x 

1 2 3 [f ( )]2sin cos d [use f ( )[f ( )] d ] 1 2 cos3 n n xx x x xxx c n xC        (b) (i) dd cos 1, 2cos 2dd xy tttt  dd d2 c o s 2 dd dc o s 1 yy t t x tx t  When d 0,d y x  33 1 1 3cos 2 0 2 , , , 22 44 4 4 22 tt t x         At point A, 1 ,142 x y  1y  is the equation of the tangent to the curve at point A. Or Since 0 t  , the maximum and minimum values of y (i.e. sin 2y t ) is 1 and -1. The y- coordinate of point A is 1 and since the tangent to this max pt is a horizontal line ( d 0,d y x  ), therefore the equation of the tangent to the curve at point A is y = 1. (ii) y x A

 1 42 4 4 44 Area 1 d 31 sin 2 cos 1 d4 2 31 sin 2 cos sin 2 d4 2 3 1 1 cos3 cos 2 cos42 3 22 31 2 1 1 43 2 23 2 31 2 2 46 3 yx tt t tt t t x tx                                          6  1 2s i n 2y x    2 2 2cos2 2c o s 2 2s i n 2 dy x y xdx x    Differentiating wrt x:  2 2 2 2 2 2 4c o s 2 22 s i n 2 2( 2 cos 2 ) 4 sin 2 dy d y y x y xdx dx dy d y y x y xdx dx      2 2 2 22 2 2 124 s i n 2 2 4s i n 2 ( ) d y dy dy yxdx y dx dx dy d y y x showndx y dx        (i) 332 2 32 2 24 8s i n 2 8 c o s 2d y dy dy d y dy yx y xdx y dx y dx dx dx             At x =0 , y = 1 2 , dy dx = 1 2 , 2 2 dy dx = 1 , 3 3 dy dx = -1 By Maclaurin’s theorem , y = 1 2 1 2 x+ 21 2! x + 3(1 ) 3! x +… = 2311 1 1 22 2 6x xx  (up to term in x3) (ii)  1 2s i n 2 x =  13 1 11 82s i n 2 1 2 . . .22 6 xxx             23311 8 1 11 2 2. . . 2. . . . . .22 6 2 2 xxxx              

=   3 2311 8 1 11 2 2 .. 2 ... ...22 64 8 xxx x      = 3 23121. . .23 xxx x    2311 1 1 ..22 2 6xx x    which is the same as the series obtained by using Maclaurin’s theorem. 7 (i) 224 8yz , 21 8xz  Expressing z and x in terms of y, 24zy , 23 0xy (2 30) (24 )V xyz y y y  3227 87 2 0yy y   (ii) 0dV dy  26 156 720 0yy  2 26 120 0yy Using G.C, 6 or 20yy y = 6 is not a feasible solution as x will be negative. 2 2 12 156dV ydy   When 20y  , 2 2 84 0dV dy   Hence, when 20y  , Maximum volume = 20(2 20 30)(24 20) 800   (iii) Let t be the time in seconds when robot A starts to move. m = 2t and n = t-1 Distance between A and B = l,  222 21 3 10lt  Differentiating wrt t,   d22 2 1 3 3d llt t   At n = 4, t = 5

   22 63d9 d 3461 0 l t    cm/s Method 2:  222 20 10lm n  Since m = 2n + 2,  222 18 3 10ln  Differentiating wrt n, 26 1 8 3dllndn   At n = 4, l 2= 102 + 62. 22 d1 8 d 10 6 l n   22 22 dd d 1 8 1 8 9 1dd d 3410 6 10 6 ll n tn t       cm/s 8 (i)    2 2 22 22 2 2 axy xa x a ax axdy ax a x dx xa xa      Set dy dx  0    22 20 2 0 0 2ax a x ax x a x or x a       For all negative values of a, there will be two distinct values of x thus two stationary pts. (shown) Or 22 44 0BA C a  for all negative values of a. 2cm/s 1cm/s 2t t-1 A B 21-3t 10

(ii) (iii)   242 1xk x x   a = -1  22 22 2 1 x kx y kxx       insert a circle, centre (0,0) , radius k to cut curve twice .  0 < k < 20 9 (i) ---(1) 2 2 ---(2) 2 2 2 1 ---(3)                           The first and second equation has only 1 solution i.e. =0 and = 0 and it is obvious that equation (3) will be inconsistent for this solution; this implies that l1 and l2 are non- intersecting lines. Since l1 and l2 are non-parallel lines as 11 2 2 where is a scalar 22 kk             Since l1 and l2 are non-parallel and non-intersecting lines, l1 and l2 are skew lines. (ii) Let 2 and 2 22 2 OX OY                        11 0 2211 2+ 2 0 1 122 22 2 1 1 1 OZ OX OY                                                 Since  and  can be any real number, the locus of Z is a plane that passes through (0, 0, - 23 2ax aya x ax ax a    Max pt is (-2a, -4a2) Min pt is (0,0)

1) and parallel to both –½ i + j + k and ½ i + j + k , Therefore 11 022 111 11 1                    is a normal to the plane p. The equation in scalar product form is 00 0 :10 1 1 11 1 p         r (iii) Let 2 and ' 2 22 2 OS OS                        Method 1: '2 2 2 2 22 2 22 2 SS                                     This vector will be parallel to the normal of p. 00 0 '2 2 1 1 2 2 22 2 1 1 22 2 SS k k k k                                         Solving, 1 4  => 1 4 1 2 3 2 OS       Coordinates of S is 11 3,,42 2  Method 2: Let F be the midpoint between S and S’,  11 '2 222 22 2 OF OS OS                 

and 0 21 2 22 1 22 OF OS k n k k k                                    Equating the position vector of point F, 0 1 22 2 222 22 2 22 22 2 kk kk                                   Solving, 1 4  => 1 4 1 2 3 2 OS       Coordinates of S is 11 3,,42 2  10 (i) 1 2yf x   y B(- 1 2 , 2) 1x x   '' 3 2 ,0A  