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