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 equations: c1 = −18 , c2 = 11 and c3 = 1

5 Eccentricity, e, is the ratio between the distance of a point, P, on the conic and a focus, F, to the distance between the point and a directrix d That is PF PFe Pd PQ== . e = 1 , S is a parabola e = ½, S is an ellipse e = 2 , S is a hyperbola (b) (i) e = 2 and F(0, 0) and x = -1 x = −1

6 Let P(x, y) any point on the hyperbola. 22 2( 1) xy x + =−−  x2 + y2 = 4(x + 1)2  x2 + y2 = 4x2 + 8x + 4  3x2 – y2 + 8x + 4 = 0 (ii) Equation of conic S is 222 4 2 34 3 224 33 ()4( ) 1 3 9 ( ) xyyx ++ − =  − = (hyperbola) Consider ( ) 22 222 23 3 1() xy −= or 3x2 – y2 = 4 3 Using c = ae  c = 2 3 (2) = 4 3 So the other focus point for conic S is ( 8 ,03− ) (iii) Method 1 S is a hyperbola with 2 2 4, , and 233 3 a b c e= = = = . Let P be a point on the hyperbola and line PQ be the tangent to S at P. WLOG, let OP > O’P = x, such that OP – O’P = 2a = 4 3 By the reflective property of hyperbolae, 'OPQ O PQ = . Denote this angle by  . Let O’P = x, OP = 4 3x+ and let the distances from O to T and O’ to T be d1 and d2 respectively. Then d1d2 = (OP sin )(O’P sin ) = 24( ) sin 3xx  + = 2 4 1 cos 2 32xx −  +     = 221 4 1 4 cos 22 3 2 3x x x x    + − +       --- (*) Consider triangle OPO’: OO’ = 2ae = 8 3 . Using the cosine rule, 22 228 4 4 2 cos 23 3 3 x x x x      = + + − +           = 22 8 16 42 2 cos 23 9 3x x x x + + − +  Hence 2 228 16 4 4 2 2 cos 23 9 3 3 x x x x      − = + − +           221 4 1 4 1 64 16 4 cos 22 3 2 3 4 9 9 3x x x x       + − + = − =           Substituting into (*), we have d1d2 = 4 3 , a constant value.

7 Method 2 [Coordinate geometry – straight forward but more tedious] Consider the equation of tangent to the hyperbola 3x2 – y2 = 4 3 (before S is translated) so its foci are ( 4 ,03 ) and a point P(x0, yo) on the hyperbola 6x – 2y dy dx = 0  3dy x dx y= Equation of tangent to the hyperbola at P is: 0 0 3 ( ) ( ) 3 ( )o o o o o o xy y x x y y y x x xy− = −  − = − 22 003 3 0ooyy x x x y− + − = ------ (*) Distance between O( 4 ,03 ) to the tangent line (*) is : OM (refer to the diagram) = 2 2 2 24 3 2 2 2 2 00 0 3 ( ) 3 4 3 99 o o o o o o oo x x y x x y y x y x − + − − + − = ++ Since P(x0, yo) lies on the hyperbola 3x2 – y2 = 4 3 (before S is translated), 22 4 33 ooxy−= , thus OM = 224 4 3 3 2 2 2 2 00 0 3 ( ) 3 4 99 o o o o oo x x y x y x y x − + − −= ++ Similarly, distance between F2(− 4 ,03 ) and the tangent line (*) is O’L (refer to the diagram) = 2 2 2 24 4 3 3 2 2 2 2 2 2 0 0 0 0 3 ( ) 3 4 3 4 9 9 9 o o o o o o o o o o x x y x x y x y x y x y x − − + − + − +== + + + The required product is not affected by translation along the x-axis, thus The product is = OM  O’L = 4 3 22 0 4 9 o o x yx − +  4 3 22 0 4 9 o o x yx + + = 2216 16 0099 2 2 2 2 4 0 0 0 3 16 16 9 3 9o xx y x x x −− =+ − + = 21 09 2 1 0 9 16 4 12( ) 3 x x − =− is a constant Graph of S. The required product is not affected by translation

8 Section B: Probability and Statistics [50 marks] [Solution] (a) The probability distribution of the difference in level of anxiety in adults before and after treatment may not follow normal distribution. Hence t-test may not be appropriate. (b) Let X = Anxiety level before treatment and Y = Anxiety level after treatment. D X Y=− Let md = median of D Let di = xi − yi for i = 1, 2, ..., 9, then Adult A B C D E F G H I di 19 10 -2 -7 13 12 1 -4 -6 Rank of |di| 9 6 2 5 8 7 1 3 4 H0 : md = 0 (no difference after treatment) H1 : md > 0 (treatment is effective) Level of significance: 5% P = sum of the ranks corresponding to the positive di = 1 + 6 + 7 + 8 + 9 = 31 Q = sum of the ranks corresponding to the negative di = 2 + 3 + 4 + 5 = 14 Tcal = min{P,Q} = Q = 14 Since n = 9, at 5% level of significance, critical region = {t: t  8} Since Tcal = 14 > 8 falls outside the critical region, we do not reject H0. Hence, there is insufficient evidence at 5% level of significance to conclude that the treatment is effective.

9 (a) Method 1 Let  be the population mean mass in grams of a filled packet of rice. 0H : 1030= (1005g + 25g) (machine operating correctly) 1H : 1030 (machine not operating correctly) Level of significance: 5% Using GC, unbiased estimate of population variance, 22 2.63894s = Under H0, Test Statistic: 92.63894 10 1030X t− Using GC, p-value = 0.178 > 0