2020 A Level H2 FM - Solutions

2020 Paper 1 Solutions 1 ( ) 44 cos isin cos isin cos 4 isin 4 (by De Moivre's theorem) z z θ θ θθ = + = + = + 4 3 22 3 4cos 4icos sin 6cos sin 4icos sin sin cos4 isin4θ θθ θ θ θ θ θ θ θ+ − − +=+ Comparing real and imaginary parts, 4 22 4 33 cos4 cos 6cos sin sin sin4 4cos sin 4cos sin θ θθ θ θ θ θ θθ = −+ = − ∴ ( ) ( ) 33 4 22 4 33 4 4 22 4 4 4 24 4cos sin 4cos sintan4 cos 6cos sin sin 4cos sin 4cos sin cos cos 6cos sin sin cos 4tan 4tan (shown)1 6tan tan θ θ θθθ θθ θ θ θ θθ θ θθ θ θ θθ θθ −= −+ −÷ = − +÷ −= −+ 2 (i) Put 0θ = in 11 2sin 2r θ= + gives 1r = . So polar coordinates ( )1, 0≡ . Put 3 π2θ = in 11 2sin 2r θ= + gives 212 1 2 2r = += + . So polar coordinates 31 2, π2≡+ . (ii) 1d 11 2sin cos2d 2 rr θθ θ= + ⇒= . 22 2 22d 1 1 111 2sin cos 2 4sin 3sind 2 2 22rr θ θ θθθ += + + = + +   . ππ 2 22 00 d 112 d 2 2 4sin 3sin d 15.135d 22rLr θ θ θθθ=+ =++ ≈ ∫∫ .

3 12 12 12 12 2 1 2 2 2 Let ( 1) ! ( 1) ! ( 2) 2/1 2 , 7/2 ! 7/2 2 10 1 2 72 2 31,22 1! ! )!2 ! 2 3(2 nn n nn n n n n nn n nn nX YY Y mm m Y An B Y AB Y AB AB X nY n X nn X XX X nn n XY n n Y Y −− −− −− = = − − = = = = − += = = + =+= = −− −− = = += = = + = = 4 (i) ( ) ( ) f 3 27 27 14 14 0 f 4 64 36 14 14 0 =−−= −< =−−=> (ii) ( ) ( ) ( ) ( ) 1 3f 4 4f 3 f3 f4x += + Since ( ) ( )f3 f4 = , 1x is the midpoint of 3 and 4. 3.5β ≈

(iii)(a) ( ) ( ) ( ) 3 2 f 9 14 f' 3 9 f '' 6 xx x xx xx =−− = − = When 34 x≤≤ , ( )f '' 0x > . Since f is increasing and concave upwards in [ ]3, 4 , β is an underestimate of α . (iii)(b) Since β is an underestimate, [ ]3.5,4γ ∈ ( )f 3.5 2.625=− ( ) ( ) ( ) ( ) 3.5f 4 4 f 3.5 f 3.5 f 4 3.5789 3.58 (3sf) γ += + ≈ = 5 (i) ( ) ( )3 12 3 i 2 i 5 5iz zz= =+ +=+ . P3 P2 P1 O ( ) ( ) ( ) ( )3 12 1 2 11 arg arg arg arg arg 5 5i arg 3 i arg 2 i 1 11 π tan tan4 32 z zz z z −− = = + ⇒ + = ++ +  ⇒= +    (ii) Set 212 1i, i n nkz nk z k++=++ = + . Then

( ) ( ) ( ) 2 22 3 12 22 22 111ii 1 i 21 21 i n nk n nk n nkz z z nk nk nkkkk n n nk k n nk k kk     ++ ++ ++= = ++ + = + −+ ++         + ++ + ++= + ( ) 22 21 in nk k nk+ ++= + ( ) ( ) ( ) ( ) ( ) 22 2 3 12 1 2 2 22 21 1arg arg arg arg arg i arg i arg i 1 21 arg i arg i arg i since 0 n nk k n nkz zz z z n n kkk n nk n nk kn nk kk   + ++ + += = + ⇒ + = +++ +      + + + ++⇒ += +++ + > 11 1 2 11 tan tan tan 1kn n k n nk−− −    ⇒= +      + ++     since 2 1, , 0 n nknn k k+++> . (iii) 111111tan tan tan237−−−  = +     (from (ii), using 2, 1nk= = ) 11 1111tan tan tan7 12 17−− −    = +         (from (ii), using 7, 5nk= = ) 111 111tan tan tan17 46 27−−−  = +     (from (ii), using 17, 29nk= = ) 111 111tan tan tan27 32 173−−−   = +       (from (ii), using 27, 5nk= = ) 11111π tan tan4 32 −− ∴= +    (from (i) )

11 11 1 11 1 1 11 1 1 1 112 tan tan37 11 12 tan tan tan3 12 17 11 1 12 tan tan tan tan3 12 46 27 11 1 12 tan tan tan tan tan3 12 46 32 −− −− − −− − − −− − − −  = +       =++            =+++              =++++          1173 6 (i) (ii) (iii) 12 3 2, 34, 196418EE E= = = 9 2734, 196418FF= = Conjecture: 3nnE F= Let nP be the proposition that 3nnE F= for 0n≥ When n = 0, 0 0 13 1 1RHS F LHS E F LHS = = = = = = 0P is true. Assume kP is true for some 0k ≥ i.e. 3kkE F= When n=k +1,

3 1 22 2 3 3 2 23 33 33 5( ) where 3 ( () 5 [( )] 5 [ )] 5 3 (5 3) (5 3) 11 )(5 3)5 1 ) 3( 1 3 3 3( 1 3 ( 3( 1 3( 1)( 3 ( ) )) 5 ( )) 5 ( kk k k k rr rr rr rr rr r rr r r r r r r r r r rr rr r r rr r k r E r EE E E FF αβ αβ αβ αβ α α β αβ β α β α β α β αβ αβ α β αβ αβ + − = −= − = − = − − + −− −− − − −− − = = −− −− = − = −− = −− 1 3 33 3 3(3 ) 3 ) as 1, ( )( ( 1) ( 5 1) 1 ) k kk r rr rF F F αβ αβ αβ + = −= − = = =− −= = Thus 1kP+ is true. Since 0P is true and if kP is true implies kP is true, thus by Principle of Mathematical Induction, 1kP+ is true for 0n≥ 7 22 22 2 2 cos dd cos sindd d dd d sin cos sin cosd dd d ddcos 2sin cosdd uy x uy x yxxx u y yy x x x yxx xx x yyx x yxxx = = − = −+ −− =−−

Since 2 2 dd cos 2 sin cos 0ddyy x x ay xxx − += ( ) 2 2 dd cos 2 sin cos 1 cos 0ddyy x x yx a yxxx − − ++ = ( ) 2 2 d 10d u aux ++ = When 1a=− , ( ) 2 2 d 0d d d sec u x u Ax u Ax B y Ax B y Ax B x = = = + = + = + When 3a= , 2 2 d 40d u ux += Auxiliary equation: 2 40 2i k k += =± cos2 sin2u A xB x∴= + ( ) cos cos 2 sin 2 sec cos 2 sin 2 y xA xB x y xA x B x = + = + 8 (i) When 0t = , 0NN= When 1t = , 02NN=

d d 1 d ad ln ee e , where e at C at C N aNt NtN N at C N AA = = = + = = = ⌠⌡ ∫ 0 02e a NA NA = = Solving, 002e e2 ln2 a a NN a = = = (ii) 2d d N aN kNt = − When d1000, 0dNN t= = 21000ln2 1000 0 ln2 1000 k k −= =

(iii) ( ) ( ) ( ) 2 1 d 1dln 2ln 2 1000 1 d ln2d11 1000 1 1 1000+ d ln2d11 1000 ln ln 1000 ln 2 ln ln 21000 Nt NN Nt NN NtN N N N tC N tCN = − =− = − − −= + = +− ⌠ ⌡ ⌠  ⌡ ⌠   ⌡ ∫ ∫ ∫ ( ) ( ) ln2 ln2 1 1 ee1000 e ( e) 1000 2 1000 2 1000 1 ()12 21 t t C C t t tt N N AA N NA ANA AA A − = ⋅− = = = −⋅ = = =++ When 0, 50tN= = , 1 1 100050 1 19 A A = + = 1000 19 2 1tN −∴= ⋅+

(iv) Let 950N = , 1000950 19 2 1t−= ⋅+ 2 1000ln 1 /19950 ln2 1ln 19 ln2 2ln19 ln2 t t  −−= =− = There could be other factors that could affect the growth rate. For example, disease outbreaks can increase the death rate of the rodents. 9 (i)(a)(b) 111 11 11 111 333 1 3 1 2 2 3 1 2 nn n n nn n nn n AA B C BC AB B C −−− −− −− = = + + = + + Let n nn n x A B C  = Thus 1 1/3 1/3 1/3 0 1/3 2/3 1/2 1/2 0 nnx x −  = 0 700 200 400 x  = Thus 1 700 26 50200 20 3400 27A x    = =       so 111 1300/3, 1000/3, 450BCA = = =