SAJC H2 Math P2 ANS

1 H2 Math Prelim Paper 2 Solutions and Markers’ Comments 1(i) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 1 2 1 3 f d 6 d 3 9 6 1 2 6 + d 3 3 9 1 2 6 2 + + d 3 9 3 62 ln 3 ln 9 tan 3 3 2 ln 3 ln 9 2 tan 3 x x x x x x x xx x x xx x x xx x C xx x C − − − = − + +  =   − +    =   − + +       = − + + + +             = − + + + +         ∫ ∫ ∫ ∫ 1 (ii) Area ( ) ( ) ( ) ( ) ( ) 0 3 2 0 23 0 2 1 3 f d 6 3 d 3 9 2 ln 3 ln 9 2 tan 3 x x x x x x xx x − − − − = − = − +     = − + + +         ∫ ∫ ( ) ( ) 2 2 ln 3 ln 9 ln 3 3 ln12 2 6 92 ln units 34 3 3 π π     = + − + − − −                 = +     +      2 (a) 1 st Lesson - 40 minutes 2 nd Lesson - 45 minutes 3 rd Lesson - 50 minutes + 5 + 5

2 ⋮ ⋮ This is an arithmetic progression: 1 40 u = and common difference, d = 5 60 hours = (60 × 60) minutes = 3600 minutes Total time (in mins) after n lessons, nS = [ ]2(40) ( 1)(5) 2 n n+ − = [ ]75 5 2 n n+ ∴ For Adrian to attend at least 60 hours of lessons, [ ]75 5 3600 2 n n+ ≥ Method 1: ∴ For Adrian to attend at least 60 hours of lessons,

3 [ ]75 5 3600 2 n n+ ≥ ⇒ 2 15 1440 0 n n + − ≥ ⇒ 46.181 n ≤ − or 31.181 n ≥ (rejected 0n >∵ ) ∴ Adrian has to attend a minimum of 32 lessons before he is qualified to take the test. Method 2: Using GC to set up the table of values Using GC to tabulate a table of values of nS for various values of n, n [ ]75 5 2 n nS n = + Comparison with 3600 31 3565 < 3600 32 3760 > 3600 ∴ Adrian has to attend a minimum of 32 lessons before he is qualified to take the test. 46.181 − 31.181

4 2 (b) 1 1 1 1 , For 2, 2 n n n n n n n n n n n n n n u S n n u S u u u S u u u u S u + + + − − = ∈ ≥ = += += = = ℤ ∵ 1 constant, n n u u + =∵ { }2 3 4 , , , u u u … follows a geometric progression with common ratio 2. ( ) ( ) 1 1 1 1 2 2 1 1 1 1 2 1 2 1 2 1 2 1 2 N N r r r r N N N u u u u u u u u u u + + = = = + − = + − = + − = = ∑ ∑ ∵ 3(i) Since N lies on 1l ,

5 1 3 1 4 2 ON λ λ λ −    = − +       for some λ ∈ ℝ . 1 3 4 1 4 5 2 3 3 3 6 4 3 2 AN ON OA λ λ λ λ λ λ = − −        = − + −             − −     = − +     − +   Since AN is perpendicular to 1l ,

6 3 4 0 2 3 3 3 6 4 4 0 3 2 2 9 9 24 16 6 4 0 29 21 21 29 AN λ λ λ λ λ λ λ λ −    • =       − − −         − + • =         − +     + − + − + = = = Hence, 1 3 34 / 29 55 / 29 29 42 / 29 21 0 2 29 21 29 21 1 4 ON   −    −       = =               +            −   +       The coordinates of N are 34 , , 55 42 29 29 29       − 3(ii) 2 3 0 : 3 5 , 6 1 l µ µ         = + −         −   ∈  r ℝ

7 Since the plane 1Π is parallel to both lines, the normal to 1Π can be obtained by taking the cross-product of the two director vectors. 3 0 4 5 2 1 6 2 3 3 1 15 5 −        = × −         −            = − = −             1n Hence, 2 1 5     = −       1n .

8 Since 1Π contains 1l , 1Π contains the point with p.v. 1 1 0     −      . Hence, Cartesian equation of 1Π : 2 1 2 1 1 1 5 0 5 2 5 2 1 2 5 3 x y z x y z x y z                 • − = − • −                         − + = + − + =

9 3(iii) 1Π : 2 5 3 x y z − + = --(1) 2Π : 3 4 x y z − + = --(2) Using G.C., 3 41 : 1 13 , 0 1 s s l     = − +     ∈          r ℝ Alternative solution: Direction vector of 3l

10 2 3 1 1 5 1 1 5 (2 15) 2 3 4 13 1         = − × −             − +     = − −     − +       =       Since 1 3 1 1 0 1         − • −             =4, and 1 1 0     −      lies on 1∏ , 1 1 0     −      lies on both 1∏ and 2∏ , resulting it to be a common point on both planes and must line on 3l . Hence, 3 41 : 1 13 , 0 1 s s l     = − +     ∈          r ℝ (iv) If the three planes intersect at one point, it implies that 3Π cuts the line 3l at exactly one point.