2019 A Level H2 FM - Solutions

2019 Paper 1 1 1 12 3 1 45 111 1 11 1 111 11 1 11( 1 11 1,, (1 ) ( 1) 1 1 11, 11 1 1) 11 uu uuu uu α α αααα αα α αα α α α αααα ααα α α αα +++ +− − +−− −− − ++ − ++= = = = = =−−+ −− − + + −+= = = = =−+ + − +−− + Thus n= 5 gives the first value of n where nu α= α cannot be -1, 0 or 1 2 (i) ( ) 3 4 44 11d 44 b b a a xx x b a= = −∫ (ii) 2 bah −= ( ) ( ) ( ) ( ) ( ) 3 33 3 3 3 2 2 33 32 23 32 23 44 1d4 32 2 1 13 3 1 6 22 2 2 1 33 3 3 6 22 2 2 1 4 1 4 b a ba abxx a b b a aaa b a bb b b a a a b ab b b a a a b ab b ba −+ ≈++   =− ++ + ++ = − +++  = − +++ = − ∫ (iii) The Simpson’s rule is a quadratic estimation of the function, and it will give the correct value for any quadratic polynomial. Since any cubic polynomial can be written as 3ax + ( )gx , where ( )gx is a quadratic polynomial, together with (ii), the Simpson’s rule will give the exact value for any cubic polynomial.

3 (i) 00 1 , 1, 1.53hxy= = = ( )10 1 sin 1 1.5 1.8325 (4dp)3yy= + ⋅= ( )21 14sin 1.8325 2.0468 (4dp)33yy = +⋅ =  ( )32 15sin 2.0468 1.9580 (4dp)33yy = +⋅ =  when 2, 1.9580xy∴= ≈ (ii)(a) They suggest that there is a maximum point between 1x= and 2x= . (ii)(b) Let d 0dyx = sin 0 where xy xy k kπ = = ∈  Using 2y from (i), where 2.0468y≈ , and 12 x<< , 2.0468 4.0936xy<< 1,k xy π⇒= = 1.5348 1.53 (3sf)2.0468x π∴≈ ≈ = 4 (i) 2 2300 65 (2 )(5 ) 18 0 80 81 7 λλ λ λλ λλ λ −= ⇒= − − −−= = − − − − = I or M 6 3 33, 6 3 66 63 0 33 1, 0 8 2 xx yy xy xy yx xy λλ −     = =     −      −+= += = =− = =−0 0

1 2   is the corresponding eigenvector 11− is the corresponding eigenvector 1 1 1 1 11 80,, 21 01 1180 11 21 0121 11 1 180 2 1 210 ( 1) 11 118 1 3 0 21 21 0 ( 1) 11 881 213 2( 1 ( 3 ) 1) 1 n nn n n n n nn nn − − − + = = = −−   = =   −− −   −−  =−   −− −     =  −− −   =  − −−  M QDQ Q D M QD Q 1 1 8 2( 1) 8 ( 1)1 3 2(8 ) 2( 1) 2(8 ) ( 1) n nnn n nn n + +   + − +−=  + − +−

5 (i) 1 11 11 11 2 ( )3 2 12( 1) ( 2 2( 1)) 32 1 2 1 3 2( 1)2 1 2 11 022 1 1 1 2 1 2 12 2 n nn nn n nn n nn n nn P PQ n Q n Q nQ n Q QQ n n Q QQ m m PP m or − +− +− +− + = + ++ = + + +− + + −+− + − −= =− = + = + = + 0 1 1)2 12) 2 5: 5 110 : )2 2, 7 1) 722 ( ( 10 2 ( 2( n n n n n n QA A B Pn B P AB P A P B AB n = − −− −= + = + = = + = + = −= += − + − − As 1,( ) 2 0,nn − →∞ → nP behaves asymptotically like 7 + 2n 6 (i) ( ) ( ) 2 2 f 1 cos 1 f' sin 1 x xx xxx x =−+ − −= − −

( ) 2 2 22 2 2 1 cos 1g sin 1 1 1 cos 1 1 sin xxxx x x x x xx xx x xx −+ −= − −− − −+− −−= + +− (ii)(a) When ( )0, f ' 0 0x= = . ( )g x will be undefined. (ii)(b) When 1x= ( )f'1 is undefined. (ii)(c) 1 0.5x = , 2 0.5 0.703657 1.2036x ≈+ ≈ However, f is only defined in 01 x≤≤ . Therefore the iteration process fails. (ii)(d) Set 1 0.7x = , 2 3 4 5 6 0.99487 0.96167 0.92616 0.91941 0.91928 u u u u u = = = = = 0.919 (3sf)α∴= 7 (i) The matrix representation is given by 13 2 3321 2 3133 2 3 1 2 10 200 1 1 1 1 66 1 1 1 1 66 1 1 1 1 66 1 2 3 1 100 1 2 3 1 100 0 1 2 0 34 0 1 2 0 34 1 1 1 1 66 3 1 2 10 200 0 2 1 7 2 0 0 3 7 70 RR R RR R RRR RR ↔ +→ → − − →           →   →  →         −−    

(ii) 1 (70 7 ) 2 34 34 2 (14 38) 166 66 ( 166 10 )3 3 7 70 3 1 3 w yz y zw xyzw x yzw w zw z − += =−= − +++= ⇒= −− = −= − + = w x y z Remarks 1 52 -8 21 Rejected since y has to be at least 0 2 48.7 -10/3 18.7 Reject since ,,xyz +∈  3 45.3 4/3 16.3 Reject since ,,xyz +∈  4 42 6 14 5 38.7 10.7 11.7 Reject since ,,xyz +∈  6 35.3 15.3 9.3 Reject since ,,xyz +∈  7 32 20 7 8 28.7 24.7 4.7 Reject since ,,xyz +∈  9 25.3 29.4 2.3 Reject since ,,xyz +∈  10 22 34 0

(iii) 1 (166 10 )) 3( (14 4, 2323 112( 33 358 17 33 7, 32, 20, 7, 159 10, 22, 34, 0, 17 38)) 2( (70 7 )) 33 142 6 w w ww w P xyzw wx y z wz w P xy P −+ = =+++ = = = = = = = = = = = ++ = = −− + Thus 176maxP = with 10, 22, 34, 0wx y z= = = = . Company may cease production of ‘z’ 8 (i) 2 1 e , 1, 2, , 22cos isin n k n z zk n kk nn π ππ = = =  = +     22cos ,sinr rrP nnππ∴ 

(ii) ( ) ( ) 11 2 i 1 22 i 2 2 i 2 2 2 22cos Re cos isin Re e e 1e Re 1e e 1e Re 1e Re 0 0 nn rr rn n r n inn in in in r rr n nn π ππ π π π π π ππ = = = = + =      − =  −  −=   − = = ∑∑ ∑

( ) ( ) 4 i 11 4 i 1 4i 4 4 4i 4 4 4cos Re e Re e e 1e Re 1e e 1e Re 1e Re 0 0 rnn n rr rn n r n inn in in in r n π π ππ π π π π π = = =  =    =      − =  −  −=   − = = ∑∑ ∑ (iii) 2 2 2 22 1 2 12 2cos sin2 12 2 2cos cos sin4 52 cos4 rr r dz rr nn r rrd n nn r n ππ πππ π = − = −+ = −+ + = −

( ) 2 4 11 2 1 2 1 2 11 1 52 cos4 25 5 2 2cos cos16 2 33 5 2 2 1cos cos16 2 2 33 5 2 1 4cos cos16 2 2 33 16 nn r rr n r n r nn n rr r rd n rr nn rr nn rr nn n π ππ ππ ππ = = = = = = = = −  = −+ = − +− = −+ = ∑∑ ∑ ∑ ∑∑ ∑ 9 (i) d , 0dTH kA k x= > (ii) 2 2 d dd d d dd d H AT Tk kAx xx x= ⋅ +⋅ d 0dHx = ⇒ 2 2 2 2 dd d 0dd d d 1d d 0d dd AT Tk kAxx x T AT x Ax x ⋅ +⋅ = += (iii) Cylinder would imply that A is a constant. d 0dAx⇒= 2 2 d 0d d d T x T Cx T Cx D = = = +