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.