2021 A level H2 FM Paper 2 (Solutions )(updated 03 Oct 2024)

1 2021 Further Mathematics Paper 2 (9649/2) Section A: Pure Mathematics [50 marks] (a) Let ( ) 1f x x= Using Simpson’s rule, ( ) ( ) e 1 1 e 1 e 1d f 1 4f f 162xx −  +   + +     e 1 e1 2 e 1 4 1ln 1 + 6ex + −+  e 1 8 11 1 +6 e 1 e − +  + ( ) ( ) ( ) e e 1 8e e 1e11 6 e e 1  + + + +−  + ( ) ( ) ( ) 3 2 26e e 1 e e 8e 8e e 1+  − + − + − 32e 3e 15e 1 0+ − −  Hence e is approximately a root of 32 3 15 1 0x x x+ − − = . (shown) (b) 32 3 15 1 0x x x+ − − = Using GC, 0.06582, 5.6319, 2.6977x x x=− =− = Since e > 0, e  2.6977 (to 4 d.p)

2 [Solution] (a) Note that the centre of circle lies above the half-line. (b) Let Q represents zo In the right-angle triangle, OCQ, where C is the centre of the circle 2 2 2 25 6 7 25 60oz OQ OC= = − = + − = = 2 15 In  OCQ , tan COQ = 5 2 15  COQ = 0.5732033 Also tan−1( 7 6 ) = 0.862170 Thus arg zo = 0.862170 – 0.5732033  0.289 (correct to 3sf) Use above geometric method is faster! S arg(z) = 4  6 7 5zi− − = =  (6, 7) 5

3 dcos sin cos sin cos sind xx t t t t t t t t t t= −  = − − =− dsin cos sin cos sin cosd yy t t t t t t t t t t= +  = + − = 22 2 2 2 2dd sin cosdd xy t t t ttt    + = +       2tt== (since 30 4t  ) A ( ) 3 4 0 2 sin cos dt t t t t  =+  ( ) 3 4 2 0 2 sin cos dt t t t t  =+  ( ) ( )  333 3 444 42 00 00 2 cos 2 cos d sin sin dt t t t t t t t t    = − + + −       333 444 2 00 0 9 2 3 22 2 sin 2sin d cos16 2 4 2 t t t t t      = + − + +                 3 4 2 0 9 2 3 2 3 2 22 2cos 132 4 8 2 t     = + + + + − −    ( ) 29 2 9 2 22 2 2 132 8 2  = + + − − − −  29 2 9 2 3 223 32 8 2  = + − −  29 2 9 2 3 2 616 4 = + − − units2

4 [Solution] (a) Rank of a matrix is the dimension of the column space represented by the matrix. Wrong: number of non-zero rows in reduced form of the matrix; dimension of the matrix 1 3 1 2 2 5 1 5 1 2 7 10 5 3 0 9     −  1 3 1 2 0 1 1 1 0 1 6 12 0 12 5 1   −−  −−  − − − 1 3 1 2 0 1 1 1 0 0 7 13 0 0 7 13   −  −  − 1 3 1 2 0 1 1 1 0 0 7 13 0 0 0 0   −  −  There are 3 pivot elements (or 3 non-zero row in the REF form of M), so the rank of A is 3 (b) The four row vectors of A are not linearly independent as indicated by an empty row of the REF form of A. (c) Let 1 2 3 5 1 2 1 3 3 5 2 0 1 1 7 9 2 5 10 c c c                        = + +                       −        Solving the equat