RVHS ACJC EJC NJC 2022 FM Prelim P1 (Solutions)

1 9649/01/2022 Solution 1. (i) ( ) 11 00 e 1 e1f ( ) d d e1 1 d 1 x xx x x u uuu −= + −= +   e e when 0, 1 when 1, e x x u du udx xu xu = == == == e 1 e 1 12 d1 2ln 1 ln 2ln(e 1) ln e 2ln 2 ln1 e12ln 1 (where 2 and 1)2 uuu uu ab −=+ + = + −  = + − − + += − = =−  (ii) By Simpson’s rule, ( ) ( ) ( ) 1 0 1 1 1 2 3f ( ) d f 0 4f 2f +4f f 13 4 4 4 4 0.24023 5 d.p. xx          + + +                 =  e12ln 1 0.2402332 e1ln 0.6202 + − +  Solution 2. ( ) 1 11 1 5 2 33 n n n nP P P − +−= + + ---(1) Applying the substitution ( ) 11 310 n nnPQ +=+ to (1): ( ) ( ) ( ) 2 1 1 11 1 1 1 13 5 3 2 3 310 3 10 10 n n n n n n nQ Q Q + + − +−     + = + + + +         ( ) ( ) ( ) ( )11 9 5 1 2 1 13 3 3 310 3 2 3 15 3 n n n n n n nQ Q Q+−+ = + + + + 11 11 52 33 3 5 2 0 n n n n n n Q Q Q Q Q Q +− +− =+  − − = Auxiliary Equation: 23 5 2 0xx− − = (3 1)( 2) 0xx+ − = 1 3x=− or 2x=

2 9649/01/2022 So ( )1 23 n n nQ A B = − + , where A and B are arbitrary constants ( ) ( ) 111 233 10 n n n nP A B + = − + + 0 30 10P A B=  + =− ---(2) 1 7 1 9 7 25 3 10 5P A B= − + + = 11 232AB− + = ---(3) Using GC to solve (2) and (3), 33 6,70 35AB=− = ( ) ( ) 133 1 6 1 2370 3 35 10 n n n nP + =− − + + Solution 3. (i) 2 cos5 5sin 5 r dr d   =+ =− 0dr d = sin 5 0 =  5 0, , 2 , 3 , 4    =  2340, , , , 5 5 5 5     = Hence, the polar coordinates of the required points are ( )3,0 , 1, 5   , 23, 5   , 31, 5   , 43, 5   . (ii) 2 5  =

3 9649/01/2022 (iii) Area = 4 2 0 1 2 rd   = ( ) 4 2 0 1 2 cos52 d  + = 1.51 units2 Solution 4. (a) ( ) ( ) ( ) 2 4 2 2 4 3 4 4 3 3 2 2 3 1 d 1 tan sec dtan sec dtan cos sin d sin 3 1 (shown)3 x xx x cx         − − + += = = = − + =− +       Alternatively, ( ) 2 4 2 32 32 3 2 2 3 2 2 3 1 d 11 d 1 d 1 11 32 2 1 (shown)3 x xx x xxx x x x x c x cx −− + += =+  + = − + + =− +      (b)(i) Integrating factor: ( ) 22 2 2 12 d d 211 1 ln 12 ln 1 2 ee e e 1. xx xx xx x x x ++ + +  = = = =+ 2 2 Let tan sec and sin 1 x dx d x x    = = = +

4 9649/01/2022 Thus, ( ) 2 2 4 3 2 2 3 11 d 1 (using part (i)'s result)3 xy x x x x cx ++= + =− +   Sub. 1x= and 0y= : ( ) 3 22 220. 33 cc=− +  = Therefore, we have ( ) 3 2 2 2 3 2 32 1 221 33 1 2 2 .3 3 1 x yx x xy xx + + =− + +=− + + (b)(ii) Euler method: ( )1 4 10 0.5 0 1 0.5. y = + −  = (b)(iii) The actual value of y when 1.5x= is ( ) ( ) ( ) 2 32 1 1.5 22 33 1.5 1 1 . 0.201988706048 .5 y +=− + + = Percentage error 0.201988706048 0.201 % 988706048 5 0.5 100% 147. 4 −= = which is large. Hence, there is a large discrepancy between the approximation and the actual value. (OR stating that the approximation more than doubled the actual value.) One reason for the large percentage error is that the value of d d y x at 1.5x= is 0.10430530756 which is much smaller than the value of d d y x at 1x= which is 1. OR

5 9649/01/2022 One reason for the large percentage error is that there is a rapid decrease in gradient value in the interval  1,1.5 as can be seen in the sketch below: Solution 5. (i) 2 1 2 3 1 3 4 1 4 R R R R 2R R R R R 2 3 6 2 2 3 6 2 2 3 3 3 0 0 3 1 4 6 9 5 0 0 3 1 2 3 3 0 0 3 2 aa +→ −→ −→ −−       − − − −    ⎯⎯⎯⎯⎯ →   −−    − − −    3 2 3 4 2 4 R R R R +R R 2 3 6 2 0 0 3 1 0 0 0 0 0 0 0 3 a +→ → −  −⎯⎯⎯⎯⎯ →  − 34RR 2 3 6 2 0 0 3 1 0 0 0 3 0 0 0 0 a  −  −⎯⎯⎯⎯ → −   1 1121 2231 333 3 RR 2 RR 1 RR 3 1 3 1 0 0 1 0 0 0 1 0 0 0 0 a− → → → −  −⎯⎯⎯⎯→   Since there are 3 non-zero rows in the row echelon form of M, dim( ) rank( ) 3R == M . Also, the 1 st, 3 rd and 4 th columns are pivot columns, thus the corresponding columns of M namely 2 2 4 2  −    , 6 3 9 3  −    and 2 3 5 a  −    are linearly independent.

6 9649/01/2022 Hence 262 2 3 3,,4 9 5 23 a            − − −                          is a basis of R. (ii) Since w x y z       belongs to R, w x y z       can be expressed as a linear combination of 2 2 4 2  −    , 6 3 9 3  −    and 2 3 5 a  −    . i.e. 1 2 3 2 6 2 2 3 3 4 9 5 23 w x c c cy za                − − −       = + +                        , where 1c , 2c and 3c   . Consider the augmented matrix 2 1 2 3 1 3 4 1 4 R R R R 2R R R R R 2 6 2 2 6 2 2 3 3 0 3 1 4 9 5 0 3 1 2 2 3 0 3 2 ww x x w y y w a z a z w +→ −→ −→        − − − − +    ⎯⎯⎯⎯⎯ →    −−       − − −    3 2 3 42 R R R R R R 4 2 6 2 0 3 1 0 0 0 0 0 3 w xw x y w a x z +→ +→   −+⎯⎯⎯⎯⎯ → +− −+ For the system to have consistent solution, we must have 0x y w w x y+ − =  = + (shown) (iii) Consider =Mx 0 where 1 2 3 4 x x x x   =   x . Using GC, 3 2 1 RREF 3 2 3 6 2 0 0 1 0 2 2 3 3 3 0 0 0 0 1 4 6 9 5 0 0 0 0 0 0 2 3 3 3 0 0 0 0 0 0  −    −    − − − −    ⎯⎯⎯ →   −      −   

7 9649/01/2022 So we have 1 2 4 3 22x x x=− , 34 1 3xx= . 3 242 2 241 43 4 362 2011 0123 03 xx x xxx x −−               = = +              x A basis for the null space of T is 36 20,01 03  −                     . Observe that 1 22 3 6 2 1 2 3 3 3 1 0 4 6 9 5 2 0 2 3 3 3 1 0 −        − − − −     =   −    −     . So 1 2 0 0 0       is a particular solution of 1 1 2 1  −=   Mx . 1 2 36 200 , where ,010 030     −               = + +               x  . Solution 6. Let 3f ( ) e 2 1 x xx= − + . f (1) 0.396 0= f (6) 3.61 0=−  f (9) 3.09 0= Hence, one root is between 1 and 6, while the other is between 6 and 9. (shown) 31f '( ) e 23 x x =− 1 f ( ) f '( ) n nn n xxx x + =− 0 6x = 1 13.7987x = 2 11.4600x = 0f '( ) f '(6) 0.463x ==

8 9649/01/2022 The gradient of the curve f ( )yx= at 0x is relatively gentle, resulting in the next approximation 1x (where tangent at 0x cuts the x-axis) being further away from the root α than 0x . Using the iterative formula 3 1 e1 2 nx nx + += , approximations diverge from α as seen in the diagram, regardless of whether the initial approximation is greater than or less than α. Using instead the iterative formula 1 3ln(2 1)nnxx+ =− , 1 2 3 4 5 6 7 8 9 10 7.5 7.9172 8.0908 8.1603 8.1876 8.1983 .20248 8.2040 8.2047 8.205 8.2049 8.205 x x x x x x x x x x = = = = = = = = = = 8.205 (to 3 dp)=

9 9649/01/2022 Solution 7. (i) 0p= , 3 5q= . (ii) 341 5 10 10 7 1 10 10 311 5 5 5 0    − − − − = − −− − − IM ( ) ( ) ( )( ) ( ) ( )( ) 7 3 3 74 1 1 1 1 5 10 5 50 10 50 10 5 10 2 3 28 3 3 1 4 1 21 2 50 5 50 250 500 50 500 32 69 721 10 50 25 32 det 0 0 0 0 50 105 69 14 0                 −=    − − − − + − − − =     − + − − + − − + = − + − = − + − = IM Using GC, the three eigenvalues of M are 7 2 1 2 3 10 51, ,   = = = . When 1 = : 3511 5 10 10 3 RREF3 11 10 10 3 1 1 2 5 5 5 0 1 0 0 0 0 0 1 0 0 0 0 0 0 − − −        − ⎯⎯⎯ → −      −−   