NJC H2 MATHS P1 Solutions

Suggested Solution to 2015 SH2 H2 Mathematics Preliminary Examination Paper 1 1 Let PN be the statement 11 0 3 2 1 1( 1)!3 ( 1)!3 N nN n n nN     for all N ≥ 0. When 0,N  0 1 0 1 0 01 3 2 3(0) 2 2LHS ( 1)!3 (0 1)!3 3 12RHS 1 LHS(0 1)!3 3 n n n n             Thus, 0P is true. Assume Pk is true for some 0k , i.e. 11 0 3 2 1 1( 1)!3 ( 1)!3 k nk n n nk     . Consider 1Pk : To show 1 12 0 3 2 1 1( 1)!3 ( 2)!3 k nk n n nk      1 1 1 0 12 0 12 12 22 32LHS of P ( 1)!3 3 2 3( 1) 2 ( 1)!3 ( 2)!3 1 3 5 1 ( 1)!3 ( 2)!3 ( 2)(3) 3 5 1 ( 1)!( 2)3 3 ( 2)!3 3 6 3 51 ( 2)!3 ( 2)!3 1 k k n n k nk n kk kk kk n n nk nk k kk kk k k k kk kk                                  2 12 3 5 3 6 ( 2)!3 1 1 RHS of P( 2)!3 k kk kk k k           1P is true P is truekk  . Since 0P is true, and 1P is true P is truekk  , by mathematical induction, PN is true for all N ≥ 0.

2 (i) 2 (ii)

3 (i) Method 1:                  12 2 24 24 4 2 2 4 4 2 4 2 4 4 2 2 4 4 4 2 cos 2 cos 2 11 221 ... 1 ...2 4! 21 2 ... 1 ... 3 21 2 2 ... 3 2or 1 2 2 ... 3 1 3 x xxx xx xx xx x x xx x x x xx x x x xx                                   3 (ii) 24 2 1cos 2 3 1 1 1 1 1 21511 3 3 3 9 243 24311 3 2 215 8cos 3 243 9 1720 2187                     

4 (i) Either a = 0 OR b = 0 OR a and b are perpendicular to each other. 4 (ii) Projection vector of a – b onto a     2 2 ˆˆ() ˆ ˆ ˆ ˆ0 ( 0 since 0) (since )                     a b a a a a b a a aaa b a b aaa aa a a a a a a Alternatively, From the diagram, since ,ab the projection vector of vector of a – b onto a is a itself. 4 (iii) Method 1: Geometrical Definition of Cross Product Since ,ab the angle between a and b is 90°. Hence,     ˆsin 90 ˆsin 90 ( 1) (1) (shown)       ab a b n a b n ab ab ab Method 2: Area of Triangle or Rectangle

Area of triangle with sides and 1 2 11 22 (shown)    ab a b a b a b a b ab Method 3: Length of Perpendicular Component Since ,ab the component of a perpendicular to b is a itself. Hence, ˆ (shown)      a b a baa b ab ab a b a b

5 (i)                      11 1 1 1 2 1 2 3 4 2 1 4 2 3 1 1 1 4 1 5 11 37 11 59 ... 11 2 2 1 2 2 3 11 2 1 1 2 1 3 11 2 1 2 3 1 1 1 114 3 2 1 2 3 1 4 2 3 2 1 4 3 2 1 2 3 11 3 2 1 2 3 nn rr r r r r nn nn nn nn nn nn n nn                                        5 (ii) (a)               5 4 11 1 2 1 2 3 11 2 1 2 3 2 1 2 3 1 1 1 5lim 3 2 1 2 3 3 (9) 11 1 1 503 3 99 5 99 r rr n rr r r r r n nn                                 

5 (ii) (b) Let k = r + 1. Then                    00 11 11 11 11 11 2 1 2 5 2 1 2 5 1 2 1 1 2 1 3 11 2 1 2 3 2 1 2 3 12 3 2 3 2 5 n r n rr rn r k n n kk r r r r rr k k k k n nn                                  Alternatively,                0 1 1 1 2 1 2 5 1 1 1 1 1 5 3 7 5 9 2 1 2 5 11 (2 1 1)(2 1 3) (2 2 1)(2 2 3) 1 2( 1) 1 2( 1) 3 1 2 1 2 3 12 3 2 3 2 5 n r n r rr nn nn rr n nn                                   For students who ignored the “Hence” condition            00 1 1 1 2 1 2 5 4 2 1 4 2 5 1 1 1 4 1 5 11 37 11 59 ... 11 2 1 1 2 1 3 11 2 1 2 3 11 2( 1) 1 2 3 1 1 1 1 1 4 1 114 3 2 3 2 5 4 3 2 3 2 5 nn rr r r r r nn nn nn n n n n                                             

6 (a) Using GC, 2.2631a 0.67529b . For 112 1 x x  , 2.26 1 x   or 0.676x OR 2.26 1 x   or 0.675x 6 (b) Let the distances for the swimming, cycling and running stage be s km, c km and r km respectively. Kandy: 11.73 26 12 Landy: 12.43 28 9 Mandy: 13.92 30 7 s c r s c r scr          By GC, 3.37, 181, 43.1 (3s.f)s c r   . Therefore, the distances for the swimming, cycling and running stage are 3.37 km, 181 km and 43.1 km respectively. s = 29133/8635 c = 313404/1727 r = 372582/8635 y O x

7 (a) 2 2 ( 1)( 2) , , 33 ( 3) ( 1)( 2) 3 3 2 ( 3) (3 2) 0 xxy x x x x y x x xy y x x x y x y                 R For the values of y for which the graph exists, the equation 2 ( 3) (3 2) 0x y x y     has solutions for x. Hence, discriminant (with respect to x) ≥ 0 2 2 2 ( 3) 4(3 2) 0 6 9 12 8 0 6 1 0 yy y y y yy             Solving 2 6 1 0,yy   26 6 4(1)(1) 2 6 32 2 y   6 4 2 2 3 2 2   Hence, 2 6 1 0 3 2 2 or 3 2 2y y y y        7 (b) Equation of translated curve is ( 1)( 2) 2 33 xxy x       At the intersection between C and the line 2 ,3y x = 0 or x = 2.33333 7or 3   . Hence volume of revolution 2.33333 2 0 3 ( 1)( 2) 2 d33 2.6434 2.643 units (to 3 decimal places) xx xx              

8 (i) From the graph, the range of f is [0, 1] 8 (ii) Largest possible value of k is .2  8 (iii) 8 (iv) From the symmetry of the sketch, required area 2 2 2 2 1 1 00 2 0 0 2 0 f ( ) d f ( ) d cos d 1 cos 2 d2 1 sin 2 units2 4 4 x x x x xx x x x x               8 (iv) Alteratively, let f ( ).yx Then