SAJC_9758_2022_Prelim_P2_Solutions

1 [Turn Over St Andrew’s Junior College 2022 Preliminary Examination H2 Mathematics Paper 2 (9758/02) Section A: Pure Mathematics Q Solutions Mark Scheme 1(i) ( )1 22 21 32 43 12 1 1 1 ! ... 11 ! 1! 1 1! NN nn nn NN NN N n uun uu uu uu uu uu uu N N − == −− − − =− −  +−  +−=  +−+− =− =− =− 

2 [Turn Over Q Solutions Mark Scheme (ii) As 1,0 !N N→ → 2 1 ! N n n n= −  → –1 which is finite, hence 2 1 ! N n n n= −  converges and the sum to infinity is –1. (iii) ( ) ( ) ( ) ( ) 54 87 46 22 21 1 ! ! 11= !! 11 1 1 4 ! 6! 11= 4! Replace by 2 + 70 1 NN nn N nn nn nn nn n nn n N N ++ == + == −− =− −− − = − − − +  −+  

3 [Turn Over Q Solutions Mark Scheme 2(i) 1l is parallel to 1 , 11 1 0 4 31 pp        − − =  =−          Let point P be the point (9, 0, 0) which lies on 1 and let point Q be (2, −4, 3). Distance between 1l and 1 1 7 1 4 4 4 1 3 1 10 2 31 18 4 1 PQ −           −               = = =    units (ii) Let the required Cartesian equation by 4x y z d+ + =

4 [Turn Over Q Solutions Mark Scheme Distance between the two planes = 9 1 4 1 d−    9 10 2 29 or 11332 d d− =  = − Since (2, −4, 3) lies on 4 11x y z+ + =− , the required Cartesian equation of plane is 4 29x y z+ + = (iii) Let y = 0, 3 2 10 0, 55 xz xzxz −=  = =−−=  . So a common point between the planes is (0, 0, −5).

5 [Turn Over Q Solutions Mark Scheme A vector parallel to the l3 = 3 1 1 2 11 2 1 1 3 a a a −           −  − =          − − −      3 0 1 2 : 0 1 , 5 1 3 a l a  −       = +       −−    r Alternative Solution 1 Let x = 0, 2 10 0, 55 yz yzay z − − =  = =−− − =  3 0 1 2 : 0 1 , 5 1 3 a l a  −       = +       −−    r Alternative Solution 2 Let z = 0, 3 10 5 10 5 ,5 1 3 1 3 xy axyx ay aa −= − ==−= −− 3 1 2 1 2 5: 1 1 ,13 0 1 3 aa l a a  −−       = +     −     −    r Alternative Solution 3

6 [Turn Over Q Solutions Mark Scheme 3 2 10 3 10 2 --- (1)x y z x y z− − =  − = + 5 5 --- (2)x ay z x ay z− − =  − = + Solving (1) and (2), 3 2 2 (1 2 )x y x ay x a y− = −  = − From (2): ( ) ( )1 1 25 , 51 3 1 3 ay z x z aa −= + = +−− ( ) 5 10 1 513 13 x a z yz az a z −+  =+  −  −  3 1 2 1 2 5: 1 1 ,13 0 1 3 aa l a a  −−       = +     −     −    r (iv)

7 [Turn Over Q Solutions Mark Scheme Given  be the acute angle between l3 and 1 , ( ) 2 2 1 2 1 14 1 3 1 sin 1 2 1 (1 3 ) 18 a a aa  −              −    = − + + − ( ) 2 2 1 2 1 14 1 3 1 3 181 2 1 (1 3 ) 18 a a aa −              −    = − + + − (given) ( ) ( ) ( ) 2 2 1 2 4 1 3 3 181 2 1 (1 3 ) 18 aa aa − + + − = − + + − ( ) 2 2 65 54 181 2 1 (1 3 ) a aa − = − + +