2022 A Level 9649 FM P1 (Solutions)

Paper 1 Remarks 1 (a) Ellipse (b) Given: 2PF PG a+= By cosine rule, 2 2 2 2( )( )cosPG FP FG FP FG = + − 2 2 2 22 22 2 (2 ) (2 ) 2 (2 ) cos 4 ( cos ) 4( ) cos ,1 cos1 cos a r r c r c r c a c a acr ac ca la c k a     − = + − − = − −= − − == −− where 2cla a=− and ck a= (c) k represents the eccentricity of the conic. 2 3d 2d y xy xx+= Integrating factor 22d ee xx x== ( ) 22 3d eed xx yxx = 22 22 22 2 2 2 1e 2 e d 2 1 e 2 e d2 1 ee2 xx xx xx y x x x x x x xc = =−  = − +   221 ( 1) e2 xy x c −= − + When 1, 1xy== : 11e ce c −=  = . 2211 ( 1) e2 xyx − = − + 3 311 62yx x=+ 2 2 d 1 1 d2 y xxx =−  22 42 42 d 1 1 1 11 1 2d 4 4 y xxx x x      + = + − + = +           Length of arc s=

2 2 2 3 33 3 3 3 d1d d 11 d [since , 0]2 11 23 1 1 1 2 3 3 1 1 1 1 1 6 2 2 3 1 1 1 23 t a t a t a y xx x x a tx x x ta ta at t t a au ta =+  = +  =−   = − − −  = + − − −   = − − −      Therefore, 1su t=− for all t when 3 41 0 3 (since 0)3 a aaa− =  =  . 4 1 2 3 1 2 3 3 1 2 1 2( 1) z z z z z z z z z z z + + =  − = + Case 1: 12 1zz = . Then 12arg( ) arg(1)zz = 12 12 arg( ) arg( ) 0 arg( ) arg( ) zz zz  + =  =− Then it cannot be that 12 and PP are either both above or both below the real axis. Case 2: 12 1zz  . Then 12 3 12 1 zzz zz += −    22 ( ) i( ) ( i )( i ) 1 ( ) i( ) ( 1) i( ) ( ) i( ) ( 1) i( ) ( 1) ( ) a c b d a b c d a c b d ac bd ad bc a c b d ac bd ad bc ac bd ad bc + + += + + − + + += − − + + + + + − − − += − − + + 3 22 ( )( 1) ( )( )Im( ) ( 1) ( ) b d ac bd a c ad bcz ac bd ad bc + − − − + += − − + + The denominator is non-negative. 22 22 22 Numerator ( )( 1) ( ) ( )( 1) ( ) ( ) ( )( 1) ( ) b d ac bd a d abc acd bc b d ac bd b d ac a d bc b d bd a d c b = + − − − + + + = + − − − + − + =− + + − +

We prove by contradiction. Case 1: Assume that imaginary parts of 1 2 3,,z z z are positive. Then 0bd+ and 0bd  3Im( ) 0z (contradiction) Case 2: Assume that imaginary parts of 1 2 3,,z z z are negative. Then 0bd+ and 0bd  3Im( ) 0z (contradiction) Therefore, 1 2 3,,P P P cannot be all above or all below the real axis of the Argand diagram. 5 22 22 4 4 2 2 d1 2e ed8 d 1 1 11 1 4e e 2e ed 2 64 8 xx x x x x y x y x − −− =−    + = + − + = +       ( ) ln 2 2 0 ln 2 2 2 2 2 0 ln 2 44 0 ln 2 44 0 4ln 2 4ln 2 d2 1 d d 112 e e 2e e d 16 8 112 2e e d 4 128 1 1 12 e e2 4 512 1 1 1 1 12 e ln 2 e2 4 512 2 512 1 1 12 16 ln 224 x x x x xx xx yA y x x x x x       −− − − − =+    = + +      = + +  = + −  = + − − −   = + −      1 255 512 16 512 61455 1 ln 24096 2   −  =+  6 (a) Let k be an eigenvalue of M. ( )det 0 det 0 k a k b b c k −= −=  − MI 2 22 22 22 ( )( ) 0 ( ) ( ) 0 ( ) ( ) 4( ) 2 ( ) ( ) 4 2 a k c k b k a c k ac b a c a c ac bk a c a c b − − − = − + + − = +  + − −= +  − += Since 22( ) 4 0a c b− +  for all real values of a, b, c, there are real solutions for k  the eigenvalues of M are real.

(b) For = , we require 22( ) 4 0 and 0a c b a c b− + =  = = . (c) Let 22( ) 4D a c b= − + . Then suppose 2 a c D ++= and 2 a c D +−= . To get the eigenvectors, • Solve () −=M I x 0 02 a c Da x by   ++ − + =   2 c a Dyx b −+= 21 2 2 x bx c a Dy b c a Dxb    = = =  −+ −+   x Eigenvectors corresponding to  are 2 , \{0} b c a D   −+ . • Solve () −=M I x 0 02 a c Da x by   +− − + =   2 c a Dyx b −−= Eigenvectors corresponding to  are 2 , \{0} b c a D   −− . Consider the dot product, ( ) ( ) 2 2 2 2 2 2 2 22 4 ( ) 4 ( ) ( ) 4 0 bb b c a D c a D c a D b c a a c b         = + − −   − + − −    = + − − − − = Therefore, every eigenvector corresponding to  is perpendicular to every eigenvector corresponding to  . 7 (a)(i) ( ) ( )1 0 0 2 1 20000 1 100M M b M b= − = − If the population remains constant, then 10 10000MM== and we have ( ) 11 2 1 100 200bb= −  = (ii) 1 0 1 100 0Mb  −  1 100b

(iii) 1As , , nnn M L M L+→ → → ( ) 112 1 1 22L L b L b L b L= −  − =  = 0 10000 0 100 0 100 given 0L L b L b b         Therefore 11100 2 200bb   From (ii), we gather that 1 100b if L is to be positive. Hence, for 110 10000, 200 100Lb    1110000 100 100 given 22L b L b b b L     = Therefore 10 200b (b) ( ) ( )2 1 1 1 9680 2420 1 2420M aM b M a b= −  = − ( )4 1 2420 (1)ab = − ( ) ( )3 2 2 1 2420 9680 1 9680M aM b M a b= −  = − ( ) 1 1 2 2420 (2)4 ab = − (1) 1 2420: 16 32 32 2420 1 2420(2) 1 2 2420 b bb b  −=  − = − − ( ) 31 2420 15 15 31 2420 15 22 515 2420 3 5 31(2420) 31(2420) 682 b b b = = = = = From (1), ( )( ) 35 682 44 1 2420 1 22 5 a b == − − 15 16 31 31 4 4 31 14a= = =− (ii) When 0 9679M = and 0 9681M = , the model predicts that the population oscillates between values approximately 2420 and 9680 for the first few values of n but it eventually becomes negative in value when 9 or 10.n= Hence the proposed model is not appropriate.

8 (a) The linear system can be written as =Ax b , where 1 1 1 2 1 1 413  =− A , x y z  = x , sin cos 1    = b . ( ) ( ) ( )det( ) 1 4 1 2 1 6 0= − − + =A . This implies that the linear system has either infinitely many or no solutions. (b) 2 2 1 4 3 1 2 4 1 1 1 sin 1 1 1 sin 2 1 1 cos 0 3 1 cos 2sin 4 1 3 1 0 3 1 1 4sin R R R R R R      →− →−        − ⎯⎯⎯⎯⎯ → − − −       − − −    3 3 2 1 1 1 sin 0 3 1 cos 2sin 0 0 0 1 cos 2sin R R R    →−  ⎯⎯⎯⎯ → − − − −− Solutions exist if and only if 1 cos 2sin 0− − = cos 2sin 1 + = ( )5 cos 1−= , where tan 2 = 1 1cos [since 2 ] 5       −− = −  −  − = 12 2 tan 2 − = = . (c) From GC, 0 1 2 1 1,,3 3 3 3S z z z= − − −  . The two solution sets represent distinct parallel lines with direction vector 2 1 3 − − . 9 (a) (b) (c)(i) 3 22 1 1 sin(2 1)2 r r y r x  − = =− 

i(2 1) 22 1 i 2 i2 1Im e 2 eIm (using sum of GP formula)1 2 e rx r r x x  − − = −  =   =  −  i 2 i2 2 i2 2 i2 i 2 i 24 2 e (1 2 e )Im (1 2 e )(1 2 e ) e 2 eIm 1 2 (2cos 2 ) 2 sin 2 ( sin ) 17 1 cos 216 2 20sin (shown)17 8cos 2 xx xx xx x xx x x x −− − − − −− −− −  −=  −−  −=  −+ −−= − = − (ii) 3 2 d (17 8cos 2 )(20cos ) (20sin )(16sin 2 ) d (17 8cos 2 ) y x x x x xx −−= − When 3 2 d 9(20) 200, 2.22d 9 9 yx x= = =  We expect a good approximation for the square-wave curve to have a very steep gradient at the origin. However, the gradient for this approximate curve is only around 2.22, which does not reflect the sharp edges of the original curve. Therefore, 3y is not a good approximation. 10 (a) Foci are ( 2, 0)− and ( 2, 0) . (b) Equation of light beam: tan ( 1)yx =− ( )1 cotxy  = + Since B is the point of intersection between the light beam and the hyperbola, ( ) ( ) ( ) ( ) 2 2 2 22 1 2 cot cot 1 cot 1 2 cot 0 y y y yy   + + − = − + = Since 0y , ( ) ( ) 2 22 cot 1 2 cot 2cot 2 tan cot 1 1 tan tan 2 y y     − =− −−== −− =− 2 2 21 tan 2 sec 2x  = + = Note that (1, 0)A lies on this curve, so