2022 A Level H2 FM - Solutions

2022 GCE ‘A’ Level F Math Paper 1 1 y b P(r,θ) r θ -a F(-c,0) O G(c,0) a x Directrix: ax e= (a) L is an ellipse, centre O and length of major axis is 2a. (b) With ( ) ,0Fc− as the pole of polar coordinates, let P have polar coordinates ( ),Pr θ . Applying cosine rule in triangle FPG, ( ) ( ) ( ) 1 1 22 222 22 2 2 cos 4 4 cosPG r c r c r c cr θθ=+ − =+− . ( ) ( ) 1 22 2 222 22 2 4 4 cos 2 4 4 cos 2 1 cos FP PG a r r c cr a r c cr a r ac ar c a θ θ θ + = ⇒+ + − = ⇒+ − = − − ⇒= − Thus 22 ,ac clk aa−= = . Alternative Solution:

By DEF definition with ( ) ,0Fc− as the pole, cos cos r rc eaa ar cr c ce a θθ = ⇒= +− + − since c ae= which reduces to 22 1 cos ac ar c a θ − = − as above. (c) ck a= is the eccentricity of the ellipse. 2 Multiply DE by Integrating Factor = 22dee xx x∫ = gives ( ) 2 22 2 33d e e e edd x xx xy x y xxx = ⇒= ∫ ( ) 2 22 22 2 d d 2 2 ed 1 e e d (Integrating by parts)2 11 ee22 x u v x xx xx xx x x xx xC = = − = −+ ∫ ∫ Since 1y= when 1x= , 11eee e22 CC= − +⇒= . Thus ( ) 2 22 22 2111 1e e e e 1e22 2x xx xy x yx −= − +⇒= −+ . 3 ( ) ( )3444 22 433d1 1 6 d6 2 xx xxy xy xx x x  −++− = ⇒= = .

 2 d1d d t a ys AT xx= = + ∫ 24 2 4 2 3 33 11d 2 1 d2 11 23 1 1 11 23 2 3 t a t a t a x xx x xx x x ta ta  −= +  += = −   = −+ −     ∫ ∫ 4 3 436 3 1 311 1 6 23 111 23 tu t ttu t tt t asu ta += +− = −= −  ⇒=−+ −  1su t= − for all permissible values of t provided 13 4 411 0 3323 a a aa − =⇒ =⇒= . 4 12 1 2 3 123 3 12 1 zzz z z zzz z zz+++= ⇒= −

( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 22 ii ii i 1i i 1i 1 a bc d a bc d ac bd ac bd ad bc a c b d ac bd ad bc ac bd ad bc + ++= ++ ++ += − −+ + ++ + − −− +    = −−+ + ( ) ( )( ) ( )( )( ) ( )3 22 1Im 1a c ad bc b d ac bdz ac bd ad bc−+ + ++ −−= −−+ + Since the denominator is always positive, we consider the numerator of ( )3Im z . ( )( ) ( )( ) ( ) ( ) 2 22 2 22 2 2 1 11 a c ad bc b d ac bd a d bc b d b bd d a bd c db −+ + ++ −−= − − − − − − = −+ + −++ Since we wish to show not all points are either above or below the real axis, we need only examine 2 cases. Case 1: Both 12 and PP are above the real axis. In this case, ,0bd > . Hence ( ) ( ) ( )22 2 2 31 1 0 Im 0a bd c db z−+ + −+ + <⇒ < . So 3P is below the real axis. Case 2: Both 12 and PP are below the real axis. In this case, ,0bd < . Hence ( ) ( ) ( )22 2 2 31 1 0 Im 0a bd c db z−+ + −+ + >⇒ > . So 3P is above the real axis. Note: The case where 12 and PP are on opposite sides of the real axis automatically establishes the hypothesis to be proved and so need not be considered.

5 2 2 22 1d 1e e 2e e16 d 8x x xx yy x−−=+ ⇒= − 22 22d11 1 2e ed8 xxyx −  ⇒+ =+ −     44 2 22 22 111 4e e 2 64 14e e 16 12e e 16 xx xx xx − − − = + −+ = +  = + ln2 2 0 ln2 2 22 0 d2π1 d d 1 4 π e ed 16xx yAy x x x− = +  = +  ∫ ∫ ln2 44 0 ln2 44 0 4ln2 4 ln2 114π e ed 8 256 1114πe e4 8 1024 1 1 1 114π e ln 2 e4 8 1024 4 1024 1 1 114π 4 ln 28 16384 4 1024 61455 1π ln 24096 2 xx xx x x − − − = ++  = +− = + − −+ = + − −+ = + ∫

6 (a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 22 22 2 2 2 0 0 0 Method 1 11 024 11 1 24 4 1Since 0, eigenvalues are re l.4 : a ac ac ac a cb a ab bc ac b c a c ac b a c b ac b λ λ λ λλ λ λ −+ +  −+−+ + − =  − + = +− −= − =− += − + − +≥ −= M I ( ) ( ) ( ) 2 2 2 22 2 2 Method 2: 4 24 40 Eigenvalues (roots) are real. D a c ac b a ac c b ac b = +− − =− ++ = −+≥ ∴ (b) ( )2 2For , we need 4 0. Necessary and sufficient condition and 0. ac b ac b λµ= −+= = =

(c) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 11 24 From 1 if 0 Eigenvector associated 11When , eigenvector 1124 24 11Whe 0, using first r n , e2 ow: 4 ac ac b a x by y a xbb b a b a ab b c ac b ca ac b ac ac b c λ λλ λ λ λ λ λ λ = +± − + −= − = −− ≠   −  = ++ − +  −+ − + = +− − − − + −= =IM ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 22 22 2222 igenvector 11 24 11 11 24 24 11 044 Since the scalar product of the two eigenvectors are zero, they are perpendicular. b ca ac b bb ca ac b ca ac b b ca ac b    −− − +      •  −+ − + −− − +     =+ − − − −= 7 (a)(i) ( ) ( ) ( )( ) 1 10 0 1 21 When 0, 2 1 For population to remain constant 10,000 . 10000 2 10000 1 1000 1 0.00520 0 0 0.5 100 nn nM b M bM n M M bM M b b + ⇒= = = − = = − = = − =

(a)(ii) ( ) ( )( ) 10 0 21 0 2 10000 1 10000 0 1 0.01100 M M bM b b = −< −< >= (a)(iii) ( )21 0.50.5 1If 0 10000, but from (a)(ii) we need 0.01 , 200 0.005 0.01 1If 10000, and since 0, we have 200 0 0.0 5 0 L L bL bL b L Lb b b Lb b b = − = = << > < << >< << ⇒ ∴ ∴ > (b)(i) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 21 1 32 2 9680 2420 1 2420 2420 9680 1 9680 0.25 1 9680 2 1 24201 2 :16 16 704 5 1 22 5 1 9680 15 3 5682 5 15 682682 5 35U 1 When 1, sing 1 1 4 1 2420 1 : When 2, 1 4 61 8 nn n ab ab M aM b M n M aM b M ab n M aM b ab b bb b b M a b + ⇒= − ⇒= − = − −−−− −÷ = ⇒− = − − = ⇒ = − = = − = − −−−− = = = = − = − ( ) 312 7.7524 25 a ⇒= =

(b)(ii) 0 0 population dies out for 10.When 9679, the model reaches a max of 11 ,617 when 9 but When 9681, the model reaches a max of 11 ,646 when 8 but population dies out f . or 9 Mn Mn n n = = = = = = 8 (a) 111 Since 2 1 1 0, the system does not have uni que solution. 413 −=

(b) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 11 2 2 2 2sin 1 2 cos 2 4 31 3 1 2 : 4 3 2sin cos Comparing with 3 : 2sin cos 1 5 cos 1 where tan 2 1cos 5 tan 2 2tan 2 or 2 tan 2 Method 1: Checking n si xyz xyz xy z x yz z x y θθ θ θθ θθ θα α θα θα θπ θ −− ⇒ + + = −−−−− − + = −−−−− + + = −−−−− + ++ = + += −= = −= −= = + + + = = ( ) ( ) ( ) 1 1 1 2 tan 2 Using GC: 2sin cos 1 thus is rejected unique value of , 2 tan 2, for which solu tions to the system exists. 2 tan 2 sin 0.8 and cos 0.6 0.8 1 2 0.6 2 4 31 xyz xyz xy z π θθ θθ θ θθ − − − + += − ∃= = ⇒= = − −−−−− − + =− −−−−− ++ = − ++= −−−− ( )3 Solving with GC: 12 1 11 115 03 x y z λ        = +       −   