TJC JC2 H2 Maths 2012 Prelim Paper 1 Solutions

TJC_JC2_H2_Maths_2012_Prelim_Paper_1_Solution 1 An ellipse E has equation 229 16 144xy . (i) Find d d y x in term of x and y. [1] (ii) Show that the point P with coordinates ( 4cos, 3sin ) lies on E. [1] (iii) Find the equation of the normal to the ellipse E at the poin t P, in the form of y mx c where m and c are single trigonometric expressions of . [3] Solution: (i) 229 16 144xy  18 32 0 dyxy dx  9 16 dy x dx y (ii) Since 2 2 2 29 16 9(4cos ) 16(3sin )xy    22144(cos sin ) 144   Hence, point P lies on C. (iii) Equation of normal at point P: 16(3sin )3sin ( 4cos )9(4cos )yx     3(cos ) 9cos sin 4(sin ) 16cos sinyx         74 tan sin33yx   

TJC_JC2_H2_Maths_2012_Prelim_Paper_1_Solution 2 Consider the equation     322 1 2i i 2 2i 0z z a b z       , where a and b are real. Given that 2 is a root of the equation, find the values of a and b. [3] Given also that 1i is another root, find the third root of the equation. [3] Solution: Since 2 is a root of the equation,         32 2 2 1 2i 2 i 2 2 2i 0 ab           16 4 8i 2 2 i 2 2i 0ab         10 6i 2 2 iab    by comparing coeff., 5, 3ab   Let z  be the third root.         322 1 2i 5 3i 2 2i 2 2 1 iz z z z z z             By comparing constant term, we have   2 2 1 i 2 2i       2 2 1 i 2 2i 1= 2       

TJC_JC2_H2_Maths_2012_Prelim_Paper_1_Solution 3 HIMHEYS' Confectionery recently created three types of chocolate: Organic white, Organic milk and Organic dark, available in 250g bars. Cocoa butter, an essential ingredient in chocolate bars, makes up 25%, 20% and 15% of the mass o f an Organic white, Organic milk and Organic dark chocolate bar respectively. To prepare for the official launch of the ir chocolate bars at an upcoming Food Expo, HIMHEYS' decides to manufacture a total of 300 bars for the event, with more than 70 bars of each type. The confectionary intends to use 14kg of cocoa butter in the production of t he above batch of chocolate bars. I f the number of milk chocolate bars is to be smaller than the number of white chocolat e bars, d etermine how many organic chocolate bars of each type can be produced. [6] Solution: Let D, M and W be the number of dark, white and milk chocolate bars (250g organic) respectively. 300D M W   -------- (1) 15 20 25250 250 250 14000100 100 100D M W                        ie. 3D + 4M +5W =1120 -------- (2) Solving (1) and (2): 80 , 220 2D W M W    Given 70, 70, 70D W M W    , 70 220 2 WW    73.3 75W   Since D, M and W are integers, we get 74, 72, 154W M D  

TJC_JC2_H2_Maths_2012_Prelim_Paper_1_Solution 4 Referred to an origin O, the p osition vectors of two points A and B are a and b respectively. The points P on OA and Q on AB are such that OP = 2 PA and 5AQ = 4 QB. Show that the equation of the line l passing through P and Q can be written as r =  2 43 a b a , where  . [4] The point X on l is such that AX is perpendicular to l. If 2, 1ab and a is perpendicular to b, show that the position vector of X is  1 11 415 ab . [4] Solution: Since OP = 2PA, 2 3OP   a Since 5AQ = 4QB, therefore AQ:QB=4:5 Using Ratio Theorem, 54 9OQ   ab 54 99 a + b 14 99PQ OQ OP        ab  1 49 ba Since the line passes through P and is // 4 ba therefore an equation of l is r =  2 43 a b a where l Î Since X lies on l, we have  2 43OX t    a b a for a particular value of t Since AX is perpendicular to l,  40AX    ba    2 4 4 03 t       a b a a b a  1 4 4 03 tt      a b b a Since a and b are perpendicular, 0  ab 221 16 03 tt    ab 1 4 16 03 tt    1 15t     2 1 1 4 11 43 15 15OX       a b a a b

TJC_JC2_H2_Maths_2012_Prelim_Paper_1_Solution 5 The functions f and g are defined by 2 1f : 6 , 2 2x x x x      ,   2g : ln 9 , 3 3x x x     . Determine whether each of the following function s exists and give a definition (including the domain) of the function if it exists. (a) 1f , (b) gf . [9] Solution: (a) Any horizontal line ,y k k cuts the graph of ()y f x at most once, so by the horizontal line test, f is 1-1. Therefore 1f exists. Now Domain of f1= Range of f = 50, 2   Let 2 2 266y x x y x x         22 60x x y       21 1 4 6 2 y x       21 25 4 2 y Since 1 2x ,   2 21 1 4 6 1 25 4 22 y yx     So 2 1 1 25 4 5f : , 0 22 yxx   (b) Range of f = 50, 2   , Domain of g = (3,3) Since Range of f  Domain of g, gf exists.   2gf ( ) g 6x x x     2 2ln 9 6 xx      2ln 3 xx   So gf:x   2ln 3 xx , 12 2x  

TJC_JC2_H2_Maths_2012_Prelim_Paper_1_Solution 6 Given that  ln 1 sinyx , show that (i) d cosd y yex x  , [2] (ii) 332 32 d d d d d30d d d d d y y y y y x x x x x                . [3] Find the Maclaurin’s series for y up to and including the term in x3. [3] Hence, or otherwise, show that 2cos 11 sin 2 xx xx    . [2] Solution: (i)  ln 1 sin 1 sin yy x e x     Diff wrt x, we have d cosd y yex x  (ii) Diff wrt x again, we have 2 2 2 dd sindd yy yye e x xx    Diff wrt x again, we have 3 2 2 3 2 2 3 d d d d d d2 cosd d d d d d y y y yy y y y y ye e e e xx x x x x x                              3 2 2 3 2 2 3 d d d d d d d2d d d d d d d y y y y yy y y y y y ye e e e ex x x x x x x                              332 32 d d d d d30d d d d d y y y y y x x x x x                 When x = 0, 23 23 d d d0, 1, 1, 1d d d y y yy x x x    231101 2! 3!y x x x      2311 26y x x x     i.e. 2311ln(1 sin ) 26x x x x     [Hence] Diff wrt to x, we have 2cos 1 11 sin 2 x xxx     [Otherwise]     22 1 22cos 1 1 1 1 1 11 sin 2 2 2 x x x x x x x xx                      

TJC_JC2_H2_Maths_2012_Prelim_Paper_1_Solution 7 (a) A finite arithmetic progression has n terms and common difference d. The first term is 1 and the sum of the last 5 terms exceeds the sum of the first 4 terms by 193. (i) Show that 5nd 21d192 = 0. [3] (ii) Given also that the 6th term of the progression is 16, find n. [2] (b) A sequence U is formed in which the nth term is given by nte where nt is the nth term of an arithmetic progression with first term 1t =1. (i) Show that U is a geometric progression. [2] (ii) Given that the sum to infinity of even -numbered terms of U is 8 63 e , find the common ratio of U. [3] Solution: Method 1 (a)(i)     54 193 542 ( 1) 2 ( 6) (2 3 ) 193 2 2 2 nnS S S nn n d n d d            22 2 ( 1) 2( 5) ( 5)( 6) 8 12 386 10 11 30 12 394 10 42 384 0 5 21 192 0 n nd n n n n d d n d nd n d nd d d nd d nd d                        (a)(ii) 1 5 16 3 d d   Method 2 1 2 3 4 4 193n n n n nT T T T T S         4 1 4 54( ) ( ) 19322 nnT T T T     54( ( 5) ( 1) ) (2 3 ) 19322a n d a n d a d        5 (2 2 6 )