JPJC 2024 J2 H2 FM Prelim P2 Solutions

1 JPJC J2 Preliminary Examination 2024 Further Mathematics Paper 2 Solutions Section A: Pure Mathematics [50 marks] Qn Suggested Solutions 1(a) 5 0 cos dxx   Let ( )f cosxx= , x 0 10  5  ( )f cosxx= 1 cos10  cos 5  Using Simpson’s rule with 2 strips, ( ) 5 0 2 2 2 22 1cos d 0 1 4 cos cos6 5 10 5 sin 2 1 4cos cos 210 30 10 10 2sin cos 1 4cos 2cos 110 10 30 10 10 1 cos 2 cos10 30 10 1 2 , let cos30 10 1 xx x x x x                      − + +           + +         + + − −  +  −  + = −  ( ) 2 22 2 2 2 2 4430 1 1 030 15 15 xx xx      + +         + + + −                     Therefore cos10  is approximately the root of 2 2 2 21 1 0.30 15 15 xx          + + + − =                    (b) Solving 2 2 2 21 1 030 15 15 xx          + + + − =                    using GC, 0.951051199 and 0.9944402928xx − Since 10  is acute, cos10  is positive. cos 0.951110  (4 dp)

2 Qn Suggested Solutions 2(i) C is a parabola, then 3k = C is an ellipse, then 3k  C is a hyperbola, then 03 k (ii) 22 22 2 2 2 22 2 2 2 2 22 4 2 3sin 2 3 sin 4 2 3 4 32 2 946 4 5 644 5 12 16 4 5 5 12 5 1 8 4 5 5 r rr x y y x y y x y y y x y y xy y x   = + += + + = + = − + = − + − + = − − =−  − −=       (iii) Eccentricity, 3 2e= and coordinates of the two foci are (0,0) and (0,4.8)

3 Qn Suggested Solutions 3(a)(i) The equation ( )f0 x = has no roots in the interval ( ),ab . (a)(ii) Let ( ) 2f cosec 3lnx x x=− By using linear interpolation in the interval  3, 4 , 1x = ( ) ( ) ( ) ( ) 3 f 4 4 f 3 f 4 f 3 + + = 3.95 (3 sf) ( ) 2 2 1f cosec 3ln 3ln sinx x x x x= − = − is undefined at 3.14 (3 sf)x == ( )f x has a discontinuity at x = and 34  , hence the method is not suitable. (b)(i) ( ) ( ) 1 10 1 21 1G sin 0.76629 0.7663ln 2 1G sin is undefined3ln (0.76629) xx xx − − = = =  == Since ln (0.76629) is negative. (b)(ii) ( ) ( ) 21 cosec 23 10 21 3 4 5 G e 1.49653 G 1.39819 1.40982 1.40793 1.408 1.40823 1.408 xx xx x x x = = = == = = = 1.408  The student can verify the correctness by evaluating ( )f 1.4075 and ( )f 1.4085 and check that ( ) ( )f 1.4075 f 1.4085 0  . x y O (a, f(a)) (b, f(b)) y = f(x)

4 Qn Suggested Solutions 4(a)(i) ( ) ( ) 111 cos isin cos isin22 cos (shown) zz      −+ = + + − = (a)(ii) ( ) 1 1 1 2 2 2 2 1 1 1cos 2 1 ...1 2 2 12 nn n n n n n n n n zz n n n nz z z z z z z z z z nn  − − − − − − + − + − =+         = + + + + + +         −−        ---(1) Flip the equation, 2 4 4 21cos ... 1 2 2 12 n n n n n n n n n n n nz z z z z z nn − − + − + − −        = + + + + + +         −−        --- (2) Since , where is an integer 0nn k k nk n k    =      −    (1)+(2), ( ) ( ) ( ) ( ) ( ) ( ) 2 2 4 4 4 4 2 2 ...1212cos 2 21 2 cos cos( 2) cos( 4) ... cos( )122 2 cos2 n n n n n n n n n n n n n n n n nnz z z z z z nn z z z z z znn nnn n n n n k      − − − + − − + − + − − + − −     + + + + + +       =      + + + + + +   −−         = + − + − + + −        = 0 ( 2 ) n k nk  = − Therefore, 0 1cos cos( 2 )2 n n n k n nkk = =−  (a)(iii) When n = 3,   ( ) 3 3 3 0 31cos cos(3 2 )2 1 cos3 3cos 3cos( ) cos( 3 )8 1 cos3 3cos4 k kk      = =−  = + + − + − =+ 

5 322 00 2 0 1cos d cos3 3cos d4 1 sin 3 3sin43 11 343 2 3          =+ =+  = − + =  (b) 2 (2 i) i (2 2 ) i 2 2 and 22 12 z xy xy xy xy    = + + + = + + = + = =+ =−

6 Qn Suggested Solutions 5(i) 22 22 22 2 2 1 2 2 d 0d d d xy ab x y y a b x y b x x a y += += =− Equation of tangent at P, 2 0 00 2 0 2 2 2 2 2 2 0 0 0 0 2 2 2 2 2 2 0 0 0 0 22 0 0 0 0 2 2 2 2 00 22 () 1 (shown) bxy y x x ay a y y a y b x x b x b x x a y y a y b x x x y y y x a b b a x x y y ab − =− − − =− + + = + + = + += (ii) Equation of tangent at P, 00 22 1x x y y ab += Intersect with directix ax e= , ( ) ( ) 00 22 00 2 2 0 0 2 0 0 1 1 since , x y ya a e b y y x b ae b c x cye cy a b c xaT e cy  += =− −==  −   ( ) ( ) ( ) 0 2 0 0 0 2 0 0 2 2 0 2 0 0Gradient of 0 Gradient of yPF x ae y xc b c x cyTF a aee b e c x cy a ae −= − = − − − = − −= −

7 ( ) ( ) 2 00 22 2 0 0 2 2 2 22 2 2 22 2 2 2 Gradient of Gradient of 1 1 1 since 1 1 1 b e c xyPF TF xc cy a ae be ac e bc ecaa a b ac c a b − =  − − =−  −   =− =   − =−  − =− = − Therefore, o 2 90PF T= (iii) 2 2 PF ePD = 2 2 cos cos PF ePT PF ePT   = = From part (ii), similarly, 1PF S is also o90 1 1 PF ePD = 1 12 cos cos PF ePS PF PFePS PT   =  = = (iv) ( ) ( ) ( ) ( ) 1 1 2 2 12 12 cos cos cos cos by (iii) (Since both angles are acute) PFSPF PS PFTPF PT SPF TPF SPF TPF = =  =   =

8 Section B: Statistics [50 marks] Qn Suggested Solutions 6(a) The sprint time of an athlete may not follow normal distribution and hence t-test is not appropriate. (b) Let K+ = number of ‘ + ’. m be the population median of difference H0: m = 0 H1: m > 0 A B C D E F G H I J Sign + + + + − + + − + Abs diff 6 4 10 11 0 5 1 7 2 3 Rank 6 4 8 9 5 1 7 2 3 P = sum of positive rank = 6 + 4 + 8 + 9 + 1 + 7 + 3 = 38 Q = sum of negative rank = 5 + 2 = 7 T = min(P, Q) = 7 At 5% level, we reject H0 if T  8 Since T  8, we reject null hypothesis, and conclude that there is sufficient evidence that the new training is effective at 5% level of significance.

9 Qn Suggested Solutions 7(a) Assumptions: (1) The delays occurring in a month is independent from the delays occuring in another month. (2) The average number of delays occurring in a month is a constant. (b) 2 4 6 46 2 () ( ) ( ) ( ) ( ) ( eee ~ Po      =e 2! 4! 6! e 221         P 0 2 4 6 P 0 P 2 P 4 P 6 6) 2 3 0 D DDD or or or D D       −−− − − = = = = + = + = + +++ = + + + 46 2 ( ) ( ) ( ) ( ) (P even P 0 P 2 P 4 P 6) e 22 12 360 + DD D D D   − = = = = + = + = ++ = + + + ( ) ( ) 2 2 3 4 5 6 2 3 4 5 6 2 4 6 46 2 1e1 e e e22 e 112 2! 3! 4! 5! 6! 2! 3! 4! 5! 6! e 2 2 2 22 2! 4! 6! e 22 +12 360                    − −− − − − + = +    = + + + + + + + − + − + − +             = + + +            = + + + ( ) 2P even 1( ) 1 e 2 D −== +

10 Qn Suggested Solutions 8(i) Let HT be the time spent in hours for customers during the holiday season. 341 3.41100 Ht == , ( ) 2 2 3411 40019156399 100 9900 Hs  = − =  Since 100 50n= is large, by Central Limit Theorem, 40019 99003.41, 100 HTN      approximately. For 95% confidence interval, ( )0.025z = 1.9600 95% CI for  = 1.9600 1.960 40019 40019 9900 99003.41 , 3.41 100 10 0 0    −+  = ( )3.02, 3.80 (ii) Let HT and NT be the time spent in hours for customers during the holiday and non-holiday seasons respectively. Let H and N be the population mean time spent in hours for customers during the holiday and non-holiday seasons respectively. 0H : 0 HN−= 1H : 0 HN− 257 2.57100 Nt == , ( ) 2 2 2571 37151103299 100 9900 Ns  = − =  Since 100 50n= is large, by Central Limit Theorem, 37151 99002.57, 100 NTN      approximately. Under 0H , test statistic