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 − = − + + − Squaring both sides, ( ) 2 2 2 65 54 1 324 61 2 1 (1 3 ) a aa − == − + + −

8 [Turn Over Q Solutions Mark Scheme ( ) 2 22 65 1 1 4 4 1 1 6 9 6 a a a a a − =− + + + − + ( ) ( ) ( ) 2 2 2 2 2 2 65 1 13 10 3 6 6 6 5 13 10 3 6 36 60 25 13 10 3 0 a aa a a a a a a a − =−+ − = − + − + − + − = ( )( ) 2137 350 213 0 1 137 213 0 2131 or 137 aa aa aa − − = − − =  = = 3(i) ( ) ( ) 24 24 33cos3 1 ... 2! 4! 9 271 ...28 xxx xx = − + + = − + +

9 [Turn Over Q Solutions Mark Scheme 24 24 24 2 24 24 2 4 4 24 9 27ln(1 cos3 ) ln 1 1 ... 28 9 27ln 2 1 ...4 16 9 27ln 2 ln 1 ... 4 16 9 27 ...9 27 4 16ln 2 ... ...4 16 2 9 27 81ln 2 ...4 16 32 9 27ln 2 4 32 x x x xx xx xx xx x x x xx  + = + − + +   = − + + = + − + + − + + = + − + + − + = − + − + = − − ...+ (ii) 0.5 0.5 24 00 0.5 35 0 0.52 46 0 9 27ln(1 cos3 )d ln 2 ... d 4 32 9 27(ln 2) ... d4 32 99(ln 2) +...2 16 64 0.04929 (5 d.p.) x x x x x x x x x x x x xx + = − − +  = − − + = − −     (iii) Using GC, 0.5 0 ln(1 cos3 )d 0.04900 (5 d.p.)x x x +=

10 [Turn Over Q Solutions Mark Scheme (iv) From the diagram, it can be seen that the graphs of ( )ln 1 cos3y x x=+ and 2 49 27ln 2 4 32y x x x= − − are close to each other mostly from x = 0 to x = 0.5. Hence, the approximated value of 0.5 24 0 9 27ln 2 ... d4 32x x x x − − + from (ii) is approximately equal to the actual value of 0.5 0 ln(1 cos3 )d 0.04900 (5 d.p.)x x x += in (iii). y x O x = 0.5