TJC H2 Maths P1 Solutions

1 (i) Find the derivative of 24 x with respect to x. [ 1 ] (ii) Given the differential equation 2 2 2 d41 d yx x  , find y in terms of x. [4] [Solution] (i)   1 2 2 2 d1 42d2 4 ux xxx x      (ii) 2 2 2 d41 d yx x 2 2 2 2 d1 d 4 d1 dd 4 y x x y xx x     1sin 2 x C 1sin d2 xyC x  1 2 1sin d22 1 2 xxx xC x x     1 2 sin d2 4 xxx xC x x    12sin 4 2 xx xC x D    from part (i) 2 (a) The point A has coordinates (3, a, b) where ,ab  . Given that A lies on the xy-plane and the magnitude of the position vector of A is 5, find the values of a and b. [ 3 ] (b) The real numbers c and d are such that the vectors dc mi j k and cdnij k are perpendicular to each other. Show that  2 1cmn . [3]

[Solution] (a) Since (3, a, b) lies on the x-y plane, b = 0. 5AB   22230 5a   2 16a 4 or 4a  (b) Since 1 and 1 c dd c         are perpendicular, 2 1 02 1 c dd d c c           sin90 o m n mn mn 2 2 22 2211 1dc cd cd      2212 1cc c    3 Given  2f zp z q z r  where p, q and r are complex numbers such that f1 2 i . The equation  f0 z  has roots 1i and 12 i . Find p, q and r. [6] [Solution] Since  f0 z  has roots 1i and 12 i ,     f1 i 1 2 i ,zp z z k      Since f1 2 i ,  11 i11 2 i 2 ip      i2 i 2 ip 22 ip  ip      fi 1 i 1 2 izz z          2i1 i + 1 2 i 1 i 1 2 izz        2i2 3 i 1 3 izz       2i3 2 i 3 izz     Therefore, ip , 32 iq  and 3ir

4 Without the use of a graphic ca lculator, solve the inequality 1025 2x x  . [3] Hence find the solution to the inequality 102cos 5 2c o s   , where 02   . [3] [Solution]   25 2 1 01025 0 22 xxx xx     2 2 2 02 2 02 xx x xx x     1 2x  or 02 x For 102cos 5 2c o s   Replace by cosx  , we have 1cos 2  or 0c o s 2 For 12 4cos 23 3      For 0c o s 2 0c o s 1    30 o r 2 22      24 30 or or 223 3 2         5 A souvenir company received an order to pr oduce a souvenir that must satisfy all of the following conditions: (1) The souvenir is a solid cuboid with a square base. (2) The souvenir is made using 1m3 of superior clay. (3) The external surface of the souvenir must be coated with a special-mixed glow paint. Find the dimensions of the souvenir, in m, such that the amount of special paint n e e d e d i s t h e m i n i m u m . [ 7 ]

[Solution] Let the length of the square base be x m and the height of the cuboid be y m. Volume of cuboid = 1 m3 2 2 11xy y x  ---(1) To minimise use of paint, surface area S, has to be kept at a minimum. 224Sx x y ---(2) Sub (1) into (2): 22 2 1424 2Sx x x x x    2 d4 4d S xx x 2 23 d8 4d S x x For S to be minimum, Set d 0d S x  3 2 440 1 0xx x     1x When x = 1, 2 1 1y x , 2 23 d8 41 2 0d S xx    Therefore, the required dimension is 1m by 1m by 1m. 6 A contagious disease was found to infect a village with a population of 10000 people. Let P, in thousands, be the number of infected people t days after the start of the outbreak. The disease spread at a rate that is proportional to the product of the number of infected people and the number of non-infected people. It was found that when P reaches half the initial population of the village, the disease is spreading at a rate of 1000 people per day. Show that the spread of the disease can be modelled by the differential equation  2 25 5d d2 5 PP t  . [ 2 ] Given that 100 people are infected by the disease initially, find P in terms of t. [3] Explain what will happen to the village population in the long term. [2] [Solution] d 10d P kP Pt  When P = 5,

  d 15 1 0 5 1d 1 25 P kt k       22 25 5d1 0 d2 5 2 5 PPP P t   22 11 dd 2555 Pt P   15 51ln where < 10 and is arbitrary constant10 5 5 25 2ln ' 10 5 P tC P CP P tCP       2 5 10 tP AeP  where 'CA e  2 5 2 5 2 5 10 10 1 t t t PA e P AeP Ae    When t = 0, 100 1 1000 10P , 11 0 1 10 1 99 A AA  2 5 22 55 10 e 1099 11e 9 9 199 t tt P e     As 2 5,e 0 t t    , 10P Therefore, since number of infected people will eventually become 10 000, the whole village people will eventually be infected by the disease.

7 A convergent geometric sequence of positive terms, G has first term a and common ratio r. Write down, in terms of a and r, an expression for the nth odd-numbered term of G. [ 1 ] If the sum of first n odd-numbered terms of G is equal to the sum of all terms of G after the nth odd-numbered term, show that 22 121 0nnrr   . (i) Hence find the value of r when 5n . [ 3 ] (ii) In another sequence H, each term is the reciprocal of the corresponding term of G. If the nth term of G and H is denoted by nu and nv respectively, show that a new sequence whose nth term is ln n n u v    , is an arithmetic progression. [ 4 ] [Solution] The nth odd numbered term would be 22 21 n nUa r    . Given that nn nUU U U U    13 2 12 2 1  ie nn n naa r a r a r a r a r       22 2 2 1 2 2 1   n nar ar rr    2 21 2 1 11   nnrr rr r    22 11 11 1   nnrr r   22 111 nnrr   22 121 0 [Shown] When n 5 , we have rr  10 921 0 So r  1 (NA convergent series) or .0 892

Now    1 11 n nV ar . Letting the nth term of the new sequence be nT , we have  ln ln lnn nn n n UTU V V     ln ln ln ln          111an r an r ln ln 22 1an r   ln ln ln ln   1 22 1 22 2nnTT a n r a n r ln r 2 , a constant Hence the new sequence is an arithmetic progression. 8 The functions f and g are given by f : x  2 81 3xx  , x  , x 4 , g : x  a e x , x  . (i) Show that 1f exists and express 1f in a similar form, stating the domain c l e a r l y . [ 3 ] (ii) Determine the largest integer value of a such that fg exists. [2] Alternative:         ln ln ln ln ln ln ln ln ln ln constant                   n nn n n n n n n nn n n n UTU V U U UV arTT U U r ar 1 11 2 2 22 2 2

(iii) For the largest value of a obtained in (ii), find fg(x) and state the domain and t h e r a n g e o f f g . [ 4 ] [Solution] (i) Any horizontal line y = k, k , cuts the graph at most once, therefore f is a one-one function. Thus, 1f exists. Let y = 2 81 3 , 4xx x  =  2 43x 43xy   43xy  or 43xy  (rej 4x )  1f : x  43 x  , 3x (ii) For fg to exist, gfRD =  ,4 Largest integer value of a = 4 (iii)  fg f 4 e xx  =   2 4e 8 4e 1 3xx   = 216 8e e 32 8e 13xx x   = 2e3 x  fg gDD fgR3 , OR   2 2 4e 4 3 e3 x x      Provided you are confident with your Complete the Square.

9 Given that ln e xy , show that  2 2 dd e1dd xyy xx   . [ 2 ] (i) Find the Maclaurin’s series for ee x y , up to and including the term in 3x . [ 4 ] (ii) Find the first three non-zero terms of the Maclaurin series for ee xxy  . Hence find in terms of e, the approximate