HCI 2023 JC1 Promos 9649 (Solutions)

1 2023 H2 Further Math Promotional Examination Suggested Solutions 1. With respect to the origin as the pole, a curve C has polar equation of the form sin 2ra b  for   , where a and b are constants, and b is non-zero. (a) Find the range of values of a b so that there are tangents to C at the pole. [2] (b) For the case where a = 1 and b = 2, sketch C, stating clearly the equations of the tangents to C at the pole, and the equations of the line of symmetry of C. [4] Solution 1.(a) Let sin 2 0 sin 2 ab a b     For there to be solution, 11 11 a b a b      1.(b) Let 1 2sin 2 0 1sin 2 2 57 1 12, , , 66 6 6 57 1 1,, ,12 12 12 12              d 4cos2 0d 332, , , 22 2 2 r        Lines of symmetry are 33 ,, ,44 4 4     

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3 2. Consider the equation f( ) 0x  , where f( ) c o s 1xx x    and  is very close to zero. (a) By sketching the curve cos 1yx x  , show that f( ) 0x  has a single root , and that  is very close to 0. [3] (b) Use two iterations of the Newton-Raphson method, with initial approximation 0 0x  , to show that 21 21     . [You may assume for  small, sin , 2 cos 1 . 2   ] [5] Solution 2.(a) f( ) c o s 1 0 cos 1 xx x xx          Graph of cos 1yx x  passes through the origin, hence the horizontal line y  , where  is very close to zero, will cut the graph of cos 1yx x  near the origin. Hence the root  to the equation f( ) cos 1 0xx x     , which is the x- coordinate of the intersection point between cos 1yx x  and y  , is close to 0. Since 1s i n 1x  , d sin 1 0d y xx    Hence graph of cos 1yx x  is decreasing, therefore the horizontal line y  will cut the graph of cos 1yx x  exactly once. Therefore f( ) c o s 1 0xx x     has a single root  which is close to 0. cos 1yx x 

4 2.(b) f( ) c o s 1xx x    f' ( ) s i n 1x x   Newton-Raphson method 1 f( ) f' ( ) n nn n xxx x   0 0x  ,  0f x  ,  0f' 1x  0 10 0 f( ) f' ( ) xxx x    1x  ,    1f cos 1 cos 1x     ,   1f ' sin 1 sin 1x      1 21 1 f( ) cos 1 f' ( ) s i n 1 xxx x        Since  is very small, 2 cos 1 2  and sin   , and 2 2 2 cos 1 sin 1 2 1 1 (shown)21 x                 

5 3. Points F and G lie on the x-axis at the points (, 0 )c and (, 0 )c respectively, with c > 0. The curve L is defined as the locus of points P for which the total distance 2FP PG a , where ac . It is given that FP r and GFP  . (a) Show that 1c o s lr k   , giving the constants l and k in terms of a and c. [4] A, B, C, D are points on L such that the chords AB and CD both pass through F, and AB is perpendicular to CD. (b) Find 11 FA FB in terms of a and c. [2] (c) Find 11 AB CD in terms of a and c. [3] Solution 3.(a) For triangle PFG, 222 2( )( ) cosPG PF FG PF FG  22 2( 2 ) ( 2) 2( 2) c o sar r c rc    22 22 2 4c o s 4 4 4 cos 1c o s cr ar c a acr ac ca ar c a          Therefore 2cla a ck a   L r F O G P

6 3.(b) Assuming Point A forms angle  with axis. 11 FA FB 22 1c o s 1c o s ( ) 22 kk ll a lac       3.(c) 22 1c o s 1c o s ( ) 1 cos 1 cos 2 1c o s AB FA FB ll kk ll kk l k           22 31c o s 1c o s 22 1s i n 1s i n 2 1s i n CD FC FD ll kk ll kk l k                 22 2 2 2 22 22 11 1 s i n 1 c o s 22 2 2 2 2( ) kk AB CD l l k l ac aa c      F B D C A

7 4. A sequence 123, , , uuu  is such that 1 2 3u  and 21 1 4 1 nn n uu   , for all 2n . (a) Find the values of 23 4, and uu u , giving each answer as a fraction in its lowest term. Make a conjecture for nu in terms of n. [2] (b) Use Mathematical Induction to prove your conjecture in (i) for all positive integers n. [ 4 ] (c) Deduce the value of 2 2 1 41n n    . [ 3 ] Solution 4.(a)    21 2 32 2 43 2 12 1 3 31 5 542 1 13 1 4 53 5 743 1 14 1 5 76 3944 1 uu uu uu           Conjecture: 1 21n nu n   4.(b) Let P(n) be the statement: 1 21n nu n   for all n  ℤ+ . When n = 1, LHS = 1 2 3u  RHS 11 2 LHS2(1) 1 3   P(1) is true. Assume P(k) is true for some k  ℤ+, i.e. 1 21k ku k   To prove P(k+1) is also true, i.e. 1 2 23k ku k   . LHS = 1ku  2 1 4( 1) 1 ku k   (given) 2 11 21 48 3 k k kk    (by assumption)

8 11 21 ( 23 ) ( 21 ) k kk k             2 (1 ) ( 23 ) 1 23 21 25 2 23 21 21 2 23 21 2 RHS23 kk kk kk kk kk kk k k          Since P(1) is true, and P(k) is true  P(k+1) is true, by Mathematical Induction, P(n) is true for all n  ℤ+. 4.(c)  2 2 1 2 1 41 N n N nn n n uu        12 23 1 1 : : NN N uu uu uu uu      21 32 1 21 ( 2 1 ) 1 32 2 1 N N N N     21 1 322 ( 2 1 ) 11 62 ( 2 1 ) N N     As 1,02(2 1)N N   . Hence 2 1 11 41 6n n    .

9 5. A function f is defined as 31f( )x xx  , x  , 1x . (a) Using the trapezium rule with 3 ordinates, the value for 2 f( )d a x x , where a  , 2a , is approximately 5.56. Find the value of a correct to 1 decimal place. [3] An approximate value for 2.5 2 f( )dx x is to be found using Simpson’s rule. (b) Explain why it is not possible to use Simpson’s rule with 3 strips to find an approximate value for 2.5 2 f( )dx x . [ 2 ] (c) Estimate 2.5 2 f( )dx x using Simpson’s rule with 4 strips. Leave your answer correct to 1 decimal place. [2] (d) A student made the following claim: ‘Since Simpson’s rule will give an exact value for 2.5 2 f( )dx x if f( )x is of degree 3 or less, the value for 2.5 2 f( )dx x calculated in part (c) should be taken instead of the value for 2.5 2 f( )dx x given in part (a).’ Comment on the student’s claim. [2] Solution 5.(a) Using trapezium rule with 3 ordinates, 31 3 3 3 2 3 3 2 2 12 2 1d( 2 ) 2 [ ( ) ] ( )222 2 (2 )21 5 4 1 42 4 2 a a axxx a aa aa aaa                Given 3 3(2 )21 5 4 1 ( ) 5.5642 4 2 aa aaa        3 3(2 )15 4 1 22.24 () 024 2 2 a aaa a       Using GC,

10 2.654242a (rejected since 2a ) or 1.606305a (rejected since 2a ) or 0.0522726a (rejected since 2a ) or 2.4988872 2.5 (1 d.p.)a 5.(b) Simpson’s rule requires number of strips (intervals) to be in pairs (3 ordinates form a pair) to form a parabola so that an estimate using the parabola can be carried out. If there are 3 strips, then 1 nor 2 parabolas cannot be formed and thus Simpson’s rule cannot be carried out. 5.(c) 33 3 2.5 2 33 2.5 2 4 11 78 1 9 824 [ ( ) ( ) ]28 1 7 8 1 9f( )d 18 8 13 2[( ) ] (2.5)81 8 2 . 5 5.542481152 5.5 (1 d.p.) xx             5.(d) f( )x is not a polynomial of degree 3 or less, hence student’s claim that Simpson’s rule will give an exact value for 2.5 2 f( )dxx is not valid. Since both 5.6 given in part (a) and 5.5 found in part (c) are estimates, it is not meaningful to make comparisons and choose one over another. Note: A polynomial is an expression consisting of variables and coefficients where all the powers of the variables are positive integers or zero.