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