PJC P2 MS

2012 J2 FYE P2 1 (i)     291 2 1 11 2 1 cos3 2 13Since is small, h 2 sin 2 2                (ii) (iii) 22    (iv)           1 ' 1 13Let ln ln 1 3 ln 22 1 d 3 d 1 3 2 1 d 1 7 when 0, 3 2 d 2 2 2 170 0 Verified 22 n n n n nn y y n n y n n y y n nyn ny f f                                          11 13 1 3 22 1 1 .... 1 3 ...22 1 1322 1 7 1 7 ,2 2 2 2 n nn n n n n n n n n nn nn ab                          

2 (i)  4 1 4 22 zi z i zz              13 13arg 2 arg 2 arg 24 3 4 4z z z           ` (ii) z nearest to origin is complex represented by A. Method 1 Using triangle ADE, 1cos 2 24 2 DE DEAD       , 1sin 2 24 2 AE AEAD       Therefore  2 i 2 2z   Method 2 DC = 222 2 2 2 =>AC= 2 2 2 Using triangle ABC,   1sin 2 2 2 2 24 2 AB ABAC          1cos 2 2 2 2 2 24 2 BC BC OB BC OCAC            Therefore  2 i 2 2z  

3 Total volume of revolution of four rectangles = 222 2 1 1 1 1 5674 1 2111 444                           2 2 2 2 4 4 4 4 4 9 10 11 12                             4 2 1 4 8r r     Total volume of revolution of n rectangles =       22 2 2 2 2 2 1 1 1 1 12 1 2 2 1 2 2 2 21 1 1 1 n r n n n n n n n n n n nrnn                                                               (i) Exact volume of solid formed     22 1 21 1 1 d 1 61 xx x       Since estimated volume < exact volume of solid formed     22 11 11 6622 nn rr n nn r n r        (ii)     2211 1lim lim 6622 nn nn rr nn n r n r         4(a) By long division, 2 222 411144 12 x xx x       Hence, 1A , 1B , 1 2C  . The sequence of transformations is: (i) Scaling parallel to x-axis by a factor of 2 (ii) Reflection about the x-axis (iii) Translation of 1 unit in the direction of the y-axis

(b) (i) (ii) 5   1 1 (i) f 1 From graph, there exists no horizontal line that cuts the graph at 2 more points f is 1-1 f exists for any , (ii) Since 1, Let , 1 1 1 f ( ) 1 xx xx xyx x xy y x yx y xx x                                f x               1 (2) (2012) (2013) (2013) f ( ) f f f f f f f 0 xx x x x x x x             x y 0 f ( ) yx  1 ,9 2 4y x y 0 f ( 1) yx 1 1 4y From sketch, any line, y = k will cut the graph at most once.

      22(iii)g( ) 3 6 2 3 1 1 min. value = 1 R = 1, Since \ and fg does NOT exists g f x x x x D           