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     

2

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.