IJC H2 MATH P2 Solution

1    2015 IJC H2 Maths Prelim 2 Paper 2 (Suggested Solution) 1 The parametric equations of a curve are 21e , ettxy  . (i) Find the equation of the normal to the curve at the point   21e, ep pP  . [3] (ii) This normal meets the x-axis at the point Q. Find the cartesian equation of the locus of the mid-point of PQ as p varies. [4] Q1 Solutions (i) d ed tx t  and 2d 2ed ty t  d 2ed ty x  OR 1 e e 1 ...... (1)ttxx   2e .........(2)ty Sub (1) to (2), 2(1 )yx d 2( 1) 2ed ty xx   At point P, tp . Hence of gradient normal at P 1 2ep Equation of the normal:  2 1e( 1 e ) 2e p p pyx    211 1 ee 2e 2 2 p p pyx     OR Let the equation of normal be ym x C 

2    2 2 1e( 1 e ) 2e 11e 2e 2 pp p p p C C       Equation of the normal 211 1 ee 2e 2 2 p p pyx     (ii) At point Q, where 0y , 31e 2 ep px  . Mid-point of PQ: 32 11e e ,e 2 p pp . Let 31e e p px  and 21 e 2 py . Eliminating p: Since 1 2e( 2 )p y , 13 221( 2 ) ( 2 )x yy  .   1 21( 2 )12x yy      22 1( 2 ) 1 2x yy 

3    2 (i) Find the general solution of the differential equation 2 3 2 d 2 d yx x x   . [3] (ii) It is given that 1y when 1x . On a single diagram, sketch three members of the family of solution curves for 0x  . [5] Q2 Solutions (i) 2 23 2 d2 1 d y x xx  2 d1 1 d y Cxx x    1 lnyx C x Dx   (ii) When 1x and 1y  , 110 CD D C     Hence, 1 ln 1yx C xx   y x 0C 0C (1,1) 0C

4    3 (i) Given that sin 2f( ) cos 2 3 xx x   , where x is sufficiently small, find the series expansion of f( )x in ascending powers of x, up to and including the term in 2x . [4] (ii) Use your answer to part (i) to give an approximation for 0 f( ) d n x x in terms of n. Evaluate this approximation in the case where 0.5n , leaving your answer in 6 decimal places. [3] (iii) Use your calculator to fi nd an accurate value for 0.5 0 f( ) dx x , correct to 6 decimal places. Explain why this value is more than the approximation obtained in part (ii). [3] Q3 Solutions (i) Since x is small, sin x x and  2 22cos 2 1 1 2 2 xx x  .  22 sin 2 2 2f( ) cos 2 3 4212 3 xx xx x xx     f( )x    12242xx     121 2142 xx      21 2 1 ...42 xx     211 1 24 4 x x  (ii) 0 f( ) d n x x 2 0 11 1 d24 4 n x xx  23 0 11 1 28 1 2 n xx x    2311 1 28 1 2nn n  When 0.5n ,

5    0.5 23 0 11 1f ( ) d (0.5) (0.5) (0.5)28 1 2 0.29166667 0.291667 (6 dec pl) xx      (iii) For an accurate value, 0.5 0 f ( ) d 0.2932183197 0.293218 (6 dec pl) xx    Method 1: From the above sketch, it can be observed that the graph of f( )yx is higher than the graph of 211 1 24 4y xx  for 00 . 5x  . Hence for 00 . 5x , the area under the curve f( )yx , given by 0.293218 , is more than the area under the curve for 211 1 24 4y xx  , which is given by the value of 0.291667 obtained in part (ii). Method 2: The full expansion of f( x) includes terms with powers of x higher than 2, all with positive coefficients. The integration of the terms with limits 0 to 0.5 will result in the sum of positive values. T hus the answer in part (iii) is larger than the approximation obtained in part (ii). y x O 0.5 f( )yx   211 1 24 4yx x 

6    4 (a) The complex number z satisfies the relations 33z  and 33 izz  . (i) Illustrate both of these relations on a single Argand diagram. [3] (ii) Find exactly the maximum and minimum possible values of 2 z . [4] (b) The complex number w is given by 2 3i 2i      . Without using a calculator, find (i) w and the exact value of arg w , [4] (ii) the set of values of n, where n is a positive integer, for which *nww is a real number. [4] Q4 Solutions (a) (i)   (ii) Min value of 2 z = OP 2 = 2 93cos 42   Max value of 2 z = OQ 2 = 22 33 3 2(3)(3) cos 4  = 18 9 2 33z    33 izz 

7    (b) (i) Method 1: 252 i 136 ii66 i4 3i 2 e ee 22 i 2e w             1w  and arg 6w  Method 2: 22 2 3i3i 4 1422 i 22 i w       2 3iarg arg 22 i 2a r g 3 i a r g 2 i 2 522 64 6 w                  Method 3: 2 3i 22 i 32 3 i1 24 i2 22 3 i 4i 31 i22 w        1w  and arg 6w  (ii)  *arg arg arg 1 6 nww n w w n    Since *nww is a real number, and n is a positive integer,

8     1 , {0}6 6 1, {0} nk k nk k         

Or 6 5, nk k     : , 6 5, nn n k k 

9    5 A manufacturing company has three factories that produce packets of instant noodles. The manager wants to test whether the lead co ntent in its latest batch of instant noodles produced exceeds the legally permitted leve ls. The number of packets of instant noodles produced from each factory for the latest batch is shown in the table below. Factory Number of packets produced A 4000 B 2000 C 1000 To carry out the test, a sample of 100 packet s will be chosen from this batch of instant noodles produced. (i) Describe how the sample could be c hosen using stratified sampling. [2] (ii) State one advantage of using stratified sampling in this context. [1] Q5 Solutions (i) Factory Number of packets to be selected for the sample A 4000 100 577000  B 2000 100 297000  C 1000 100 147000  From each factory, select randomly the number of packets of instant noodles as according to the table a bove. The total number of packets selected form the stratified sample required. (ii) Stratified sampling ensures that the sample chosen is representative of the batch of instant noodles produced by a ll the three factorie s, as it allows each factory to be represented proportionally.

10    6 A jackpot game machine at an arcade cont ains 4 slots where each of the first 2 slots displays any of the twelve zodiac signs and each of the next 2 slots display any of the twenty-six letters of the alphabets A ‒Z. The jackpot is won if the 4 slots display two identical zodiac signs and two identical letters. Find the probability that a random game played at the machine results in (i) two different zodiac signs and two different vowels, [2] (ii) winning the jackpot, [2] (iii) exactly two identical zodi ac signs or exactly two identical letters or both. [3] Q6 Solutions (i) Prob required 12 11 5 4 12 12 26 26 55 2028   (ii) P(Winning jackpot) 12 1 26 1 12 12 26 26 1 312   (iii) Let A be the event for 2 identical Zodiac Signs. Let B be the event for 2 identical letters. P( ) P( ) P( ) P( ) 12 1 26 1 1 12 12 26 26 312 37 312 AB A B AB      (Method 2) Prob required P(same zodiac signs and different letters) P(different zodiac signs and same letters) +P(same zodiac signs and same letters) 12 1 26 25 12 11 26 1 1 12 12 26 26 12 12 26 26 312 37 312           