2019 A Level H2 FM 9649 P2 (Qns & Solutions))(30 Sep 2024)

2019 GCE A Level H2 Further Maths 9649 Paper 2 Solutions Section A: Pure Mathematics [50 marks] Question 1 For a given non-zero column vector x y = u , let M be the set of all 22 matrices X for which there exists a real scalar constant  such that =Xu u . (i) Show that M contains the zero 22 matrix and also that it is closed under the usual operations of matrix addition and multiplication by a scalar. [4] (ii) Write down a 22 matrix that is not in M for any non-zero u . Justify your answer. [2] [Solution] (i) Given M is the set of 22 matrices X for which there exists a real scalar constant  such that =Xu u , that is u is the eigenvector of X. 00 000 xx yy      =         , thus M contains the zero 22 matrix. Let X1 and X2  M, that is there exists real scalar constants 1 and 2 such that 11 =X u u and 22 =X u u Then 1 2 1 2 1 2 1 2( ) ( )    + = + = + = +X X u X u X u u u u , 12+ is a real scalar constant. 12+XX  M . Thus M is closed under matrix addition. Consider   , (X1)u = ( 1Xu ) =  1 u , 1 is a real scalar. Thus X1  M . Thus M is closed under matrix multiplication by a scalar. (ii) Consider X = 01 10  − then =Xu u  01 10 x x y x y y x y           =  =        −−         No solution for  for any non-zero u = x y   . Thus 01 10  −  M Or any other 2x2 matrix such that =Xu u has no real solution for  .

Question 2 A conic section has polar equation 1 , 0 2 .2 sinr =  − (i) By finding the equation of this curve in a standard Cartesian form, show that it is an ellipse [5] (ii) Determine the eccentricity of this curve and the coordinates of the two foci [4] (i) 1 2 sinr = − 2 sin 1rr −= 2221 x y y+ = + ( ) ( ) 22241x y y+ = + ( ) 2 2 24 1 2x y y y+ = + + 223 2 4 1y y x− + = 2 2113 4 1 33yx − − + = 2 21434 33yx − + = 2 2 19 3 314 y x  − += 2 2 22 1 3 1 21 33 yx  −+=       (ii) ( )1 ; 0,02e= and 20, 3  

Question 3 (i) On an Argand diagram, shade the region of the complex plane which represents the set of all complex numbers z for which 1 i 2z− +  . [3] (ii) For this set of complex numbers, determine the maximum value of ( )arg 2z+ , where ( )arg 2 ,z−  +  giving your answer in the form 1tan p− for some rational number p. [7] [Solution] (i) 1 i 2z− +   (1 i) 2z− −  The shaded region is inside the circle with centre at 1 – i and radius = 2 units (ii) Let max arg(z + 2) = arg(z – (−2)) =  AB is tangent to the circle at B, thus ABC = 90 0 AC = 1 ( 2) 10i− − − = and BC = 2 Thus AB = 10 2 2 2−= tanCAB = 21 222 = Let  be the acute angle AC made with the Real axis, thus tan  1 3= . Thus tan  = tan(CAB − ) = 11 23 11 23 tan tan 1 1 tan tan 1 ( ) 7 CAB CAB   −− ==+  + Thus max arg(z + 2) = 1 1tan 7 − Re(z) Im(z)  

Question 4 Let nP be the proposition that 2 4 2 3 5 2 1 11 n n a a a n na a a a − + + + +  + + + where 0, 1aa , and n is a positive integer. (i) Show that nP can be written in the form ( )f0 n  , where ( ) ( )( ) 2 2 2 2 1 1 1 f ( ) . 1 nnn a a n a n a + − − + − = − [3] (ii) By considering ( ) ( )f f 1n a n−− , prove by induction that nP is true for all positive integers n. [9] [Solution] (i) As a > 0, the given proposition 2 4 2 3 5 2 1 11 n n a a a n na a a a − + + + +  + + +  2 4 2 3 5 2 1(1 ) ( )( 1) nnn a a a a a a a n −+ + +  + + + +  2 1 2 22 (( ) 1) ( 1) ( 1) 11 nnn a n a a aa + − + −−−  2 2 2 2 ( 1) ( 1)( 1)f ( ) 0 1 nnn a a n an a + − − + −= − (ii) So Pn is the proposition 2 2 2 2 ( 1) ( 1)( 1)f ( ) 0 1 nnn a a n an a + − − + −= − , n is a positive integer. 4 2 2 2 2 22 ( 1) 2 ( 1) ( 1)( 1 2 )f (1) ( 1) 0 11 a a a a a a aaa − − − − + −= = = − −− as 0, 1aa Thus P1 is true Assume Pk – 1 is true that is f(k – 1) > 0 for some positive integer k Consider ( ) ( )f f 1k a k−− ( ) ( ) 2 2 2 2 2 2 2 2 ( 1) ( 1)( 1) ( 1)( 1) ( 1)f f 1 1 k k k kk a a k a a k a a k ak a k a +−− − + − − − − + −− − = − 2 2 2 2 2 2 ( 1 ) ( 1)( 1 1) 1 k k kk a a a a a k k a + − + − − − + + −= − 2 2 2 2 2 2 2 2 2 2 1 22 ( 1 ) 2 ( 1) ( 1 2 2 ) 11 k k k k k kk a a a ak a k a a a a a aa + + +− + − − − − + − − +== −− 2 2 2 2 2 22 ( ( 2 1) ( 2 1)) ( 1) ( 1) 11 kkk a a a a a k a a aa − + − − − − −== −− = k(a2 – 1)(1 + a2 + a4 + … + (a2)k – 1 ) > 0 as 0, 1aa f(k) = k(a2 – 1)(1 + a2 + a4 + … + (a2)k – 1 ) + a f(k – 1) > 0 as f(k – 1) > 0 by induction hypothesis. Since P1 is true and Pk – 1 is true  Pk is true. Thus by Mathematical Induction, Pn is true for n +.

Question 5 Use the substitution d 2d yu x y x=+ to solve the differential equation ( ) 2 2 dd 3 2 3 7dd yyx x y x xx + + + = + given that 2y= and d 1d y x = when x = 1. [13] d 2d yu x y x=+ ------ (1) Differentiate w.r.t. x: 22 22 d d d d d d 23d d d ddd u y y y y yxxx x x x xx = + + = + ----- (2) ( ) 2 2 dd 3 2 3 7dd yyx x y x xx + + + = + 2 2 d d d 3 2 3 7ddd y y yx x y x xxx  + + + = +   Sub eqn (1) and (2) into DE: d 37d u uxx + = + ----- (3) Integrating factor = dee x x = Multiply equation (3) throughout by integrating factor: ( )de e e 3 7d x x xu uxx + = + ( ) ( )d e e 3 7d xxuxx =+ ( )e e 3 7 dxxu x x=+ ( ) ( )e e 3 7 e 3 dx x xu x x= + −  ( ) 1e e 3 7 3ex x xu x C= + − + ----- (4) When 2y= and d 1d y x = when x = 1 From eqn (1): 5u= Sub x = 1, u = 5 , 15e 10e 3e 2e CC= − +  =− ( )e e 3 4 2exxux = + − ( )de 2 e 3 4 2ed xx yx y xx  + = + − ( )d 2e 2 3 4d ex yx y xx+ = + −

1d 2 4 2e 3d xyy x x x x −+ = + − Integrating factor: 2 2ln 2ee xx x  == Multiply throughout by integrating factor: ( ) 2 2 1d 2 3 4 2 ed xyx xy x x xx −+ = + − ( ) ( ) 2 2 1d 3 4 2 ed xx y x x xx −= + − ( ) 2 2 1 3 4 2 e d xx y x x x x −= + − ( ) ( ) 2 3 2 1 1 2 2 e 2e d xxx y x x x x −−= + + −  ( ) ( ) 2 3 2 1 1 22 2 e 2e xxx y x x x C −−= + + + + Sub 1, 2,xy== ( ) 222 3 2 2 5 CC= + + +  =− 1 22 1 1 52 2e xyx x xx − = + + + − 

Section B: Probability and Statistics [50 marks] Question 6 Hand grip strength is measured using a dynamometer. Measurements are given in kilograms. For males aged between 30 and 39 the mean grip strength for the dominant hand is known to be 46 kg. A researcher wished to investigate whether or not hand grip strength is affected by obesity. She took a random sample of obese males aged between 30 and 39, and measured their dominant hand grip strength, x kg. The data are summarised as follows. 22n= 1031x= 2 49410x = (i) Construct a 95% confidence interval for the mean grip strength for the dominant hand based on the data, giving the end -points of the interval correct to 2 decimal places. State an assumption required for your method to be valid. [5] (ii) State, with a reason, what conclusion the researcher should reach. [2] (i) 2 2 1 1031 34374941021 22 66s  = − =  95% confidence limits for the mean grip strength for the dominant hand = x (,  )n 12 −  t s n = 1031 34372.0796122 66 22    95% confidence interval for the mean grip strength for the dominant hand = (43.66, 50.06) Assumption: Assume that the grip strength for the dominant hand follows a normal distribution. (ii) Since 46 = lies within the 95% confidence interval, there is insufficient evidence at 5% level of significa