ASRJC 9649 2024 Prelim P1 Solutions

1 (a) Volume 3 6 2d xy x  =  3 5 6 3 32 6 33 2 3 2 66 3 3 3 2 3 2 6 6 6 42 3 6 2 tan d 2 tan (sec 1)d 2 (sec ) tan d (sec 1) tan d 2 (sec ) tan d sec tan d tan d tan tan2 ln sec42 9 3 1 12 ln 24 2 36 xx x x x x x x x x x x x x x x x x x xx x                     = =−  = − −   = − +  = − + = − + − −       2ln6 3 16 ln 39   +  =+  (b) Let 5tanf ( ) sin xxx x=− . Estimate = 0.5f (0.8) 0.8 f (0.5) 0.6031511 0.603 (3 s.f.)f (0.8) f (0.5) − =− (c) From GC, f (0.5) 0.0843 = , which is close to zero. The tangent at 0.5x= has a gentle gradient and therefore cuts the x-axis far from the actual root required, which is undesirable. 2 (a) Auxiliary equation: 2 4 5 0mm− + = 2mi =  22 12 1 5, tan 2r = + = = ( ) ( )22 nn nu C i D i= + + − ( ) ( )5 cos sin n nu A n B n =+ 0 1 00 11 1 5 sin 5 1 5 uA u B B B  =  =   =  = =  =   

( ) 2 1 1 25 sin tan n nun −= (b) If 0.1,a= then ( ) 2 20.1 5 sin . n n nvn = As ( ) 2 , 0.5 0. n nnv→ → oscillates and converges to 0. If 1 50.2 ,a== then ( ) 2 21 5 5 sin sin . n n nv n n =  = As , so does not convergennv→ , is bounded between 1 and 1 and oscillatory.− 3 (a) The displacement decreases from 0x and approaches zero in the long run. (b) 1 1 1 ( ) (1 )n n n nx x h ax ah x− − −= + − = − 2 2 0 (1 ) (1 ) n n ah x ah x −=− =− (c) Estimates will not decay to zero if 11 ah− 11 ah −  or 11 ah− − 2 (rej, since 0)hh a−  2h a (d) ( )1 1 1 12 n n n n hx x ax ax ah− − −    = + − − − 22 1 1 1 22 1 22 0 2 1 2 1 2 n n n n n ahx ahx x ahah x ahah x − − − −   = − +  = − + = − + 4 (a) ( ) 2 22 2 (since 0)x x x x x x x x x x+ = + +  +  Taking square root, 2x x x x+  + (shown) (b) Let nP be the proposition: !nun for ,0nn . For n = 1, LHS = 0 1u = , RHS = 0! 1= . For n = 2, LHS = 1 1u = , RHS = 1! 1= . Since LHS  RHS for both base cases, 01,PP are true.

Assume 1,kkPP + are true for some non-negative integer k, i.e. !kuk and 1 ( 1)!kuk+ + , For 2nk=+ , ( ) 21 ( 1) ( 1)! ( 1) ! ( 1) ! ( 1) ! ! 1 ( 1) ! ( 1)(1 1) [using (a)] ( 2)! k k ku u k u k k k k k k k k k k k k k k ++ = + +  + + + = + + + = + + +  + + + =+ So 12, true is true.k k kP P P++  Since 01,PP are true, and 12, are true is truek k kP P P++  , by induction nP is true for all ,0nn . 5 (a) Let 1 1 0 0 0 1 0 000  = M , 2 000 000 0 0 1  = M . Since 12det( ) det( ) 0==MM , 12, UMM . However, 12 1 0 0 0 1 0 0 0 1  +=  MM , where 12det( ) 1 0+ = MM . Not closed under addition. Therefore, A is not a subspace of U. (b) The zero polynomial 0 is in S, since 0(1) = 2 (3) 0 = 0. Let 12f , f S . Then 11f (1) 2f (3)= and 22f (1) 2f (3)= . ( )1 2 1 2 1 2 1 2 1 2(f f )(1) f (1) f (1) 2f (3) 2f (3) 2 f (3) f (3) 2(f f )( 3)+ = + = + = + = + For any  , 1 1 1 1( f )(1) f (1) 2f (3) 2( f )(3)   = =  = Closed under addition and scalar multiplication. S is a subspace of V. ( )0 1 2 3 0 1 2 3 0 1 2 3 2 3 9 27 0 5 17 53 a a a a a a a a a a a a + + + = + + + = + + + 2 3 2 3 0 1 2 3 1 2 3 1 2 3 23 1 2 3 5 17 53 ( 5) ( 17) ( 53) a a x a x a x a a a a x a x a x a x a x a x + + + =− − − + + + = − + − + − A basis is   235, 17, 53x x x− − − .

(c) Matrix = 0 1 0 0 0 0 2 0 0 0 0 3 0 0 0 0       Null space = 1 0 ,0 0 kk    (d) Integrating 77 x+ twice yields 2377 26a bx x x+ + + . Required set , where ,72 76 a b ab    . 6 (a)(i) 0 100 000T = is the initial population as given ( )1 1kkT r T+ =− because a death rate of r means that ( )1 r− of the population is left after each week 0 11k because the model given is only valid for 12 weeks. (ii) ( ) 12 12 0 1T r T=− ( ) ( ) 12 123551 1 0.00355 0.62496 0.375 3 sf100000r r r− =  − = =  = (b)(i) After 16 weeks, the number of frogs is 0.62496 100000 54.154= So 54.154 30 p ( )30 0.5539 0.554 3 sf54.154p  = = (ii) There are no external threats to the population, like disease or predators which may reduce the population OR The same weekly death rate continues unchanged. (c) 30 surviving females would produce 75000 eggs, so the population is smaller than it was to start with, so each ‘round’ will result in smaller and smaller population.

7 (a) i(5 ) i 5 5 5 4 3 2 2 3 4 5 e cos5 isin 5 and (e ) (cos isin ) cos 5i cos sin 10cos sin 10i cos sin 5cos sin isin               =+ =+ = + − − + + Comparing real and imaginary parts, 5 3 2 4 4 2 3 5 cos5 cos 10cos sin 5cos sin sin 5 5cos sin 10cos sin sin             = − + = − + 4 2 3 5 5 3 2 4 35 24 5cos sin 10cos sin sintan 5 cos 10cos sin 5cos sin 5 tan 10 tan tan (shown)1 10 tan 5 tan               −+= −+ −+= −+ (b) Let 7  = and tan 7t = . Then 35 24 5 5 10tan 7 1 10 5 t t t tt  −+= −+ . Since 2 5 2 2tan tan (double angle formula)7 7 1 t t  =− =− − , 35 2 2 4 2 4 2 3 5 7 5 3 2 5 10 1 1 10 5 2 (1 10 5 ) ( 1)(5 10 ) 0 21 35 7 t t t t t t t t t t t t t t t t t t −+−=− − + − + = − − + = − + − Since tan 07t = , we can divide throughout by t and obtain 6 4 221 35 7t t t− + = . Therefore, tan 7t = is a root of this equation. (c) Any valid fixed-point iteration method with 0 1t = that yield an approximation 0.482, e.g. 4 1 4 21 7 35 n n n tt t + += + t1 t2 t3 t4 t5 t6 t7 t8 t9 0.882 0.743 0.617 0.534 0.498 0.486 0.483 0.482 0.482 e.g. 1 42 7 21 35 nt tt + = −+ t1 t2 t3 t4 t5 t6 0.683 0.524 0.488 0.483 0.482 0.482 (d) 5 9 13 17tan 5 1 5 , , , , 4 4 4 4 4     =  =

9 13 17, , , ,20 4 20 20 20      = Set tan 5 1 = in (a), 2 4 3 5 5 4 3 2 1 10 tan 5 tan 5 tan 10 tan tan 0 tan 5 tan 10 tan 10 tan 5 tan 1           − + = − + = − − + + − For the equation 5 4 3 25 10 10 5 1 0x x x x x− − + + − = , the roots are tan , 9 13 17, , , ,20 4 20 20 20      = . Sum of roots = ( 5) 51 −−= 1 9 13 17tan tan tan tan tan 520 4 20 20 20 973tan tan tan tan 4 (shown)20 20 20 20          = + + + + = + − − = 8 (a) 2 (2 1)i 2 1 2 1 11 2i 4i 2 24ii 2 44ii22 1 2 (2 1) 1sin Im e 1 e Im 11e ee Im ee nt T nn nn t T t T tt TT tt TT nt a T a a a aa aa         −   −− ==         −           −       − =   = −   −  =    −−        22ii3 42 eeIm 42 cos 1 tt TTaa taa T   −                    −=  −+  

3 42 2 4 2 2 2 2 22 22sin sin 42 cos 1 2( 1)sin 22 1 2sin 1 2( 1)sin 2( 1) 2 sin ttaa TT taa T taa T taa T taa T taa T       −   −      = −+  + = − − + + = −+  (b) When 01 a , the coefficients of the sine terms, 21 1 na − , are significant for large n values. This means that the remaining terms that are truncated off are significant, resulting in the significant difference in their graphs. (c)   2 2 2 2 2 2 2 00 22 2sin1 ( 1) f ( ) d d 2( 1) 2 sin TT t A a a TP A t t tTT taa T    + ==  −+   Let 2 2 22 2sin () 2( 1) 2 sin t Tyt taa T     = −+  . n 0 1 2 3 4 nt 0 1 4 T 1 2 T 3 4 T T ny 0 2 2 2 2 1 [( 1) 4 ]aa−+ 0 2 2 2 2 1 [( 1) 4 ]aa−+ 0 Using Simpson’s Rule,

 0 4 2 1 3 0 2 2 2 2 4 2 2 2 24 1d 2 4( )34 20 0 2(0) 412 [( 1) 4 ] 8 12 [ 2 1 4 ] 2 3( 1) T Ty t y y y y y T aa T a a a T a  + + + +  = + + +  −+ =  − + + = + 