Prelims Answer Key

7/21/2019 Prelims Answer Key

1/8

Republic of the Philippines

BATANGAS STATE UNIVERSITYMain Campus II, Golden Country Subdivision, Alangilan, Batangas City

C!!"G" # "$GI$""RI$G, ARC%I&"C&'R" A$( #I$" AR&S

C%"MICA! A$( #( "$GI$""RI$G ("PAR&M"$&

Advanced Chemical "ngineering Mathematics )ith $umerical MethodCh" *+- #" *+.

PR"!IMI$AR/ "0AMI$A&I$A$S1"R 2"/

PAR& A3 M'!&IP!" C%IC"$

3

LETTE

R

S!'&I$-A$S1"R

43 C &he Set of Real $umbers

53 B S!'&I$

e

L(2t cost)= (s+2)

(s+2)2+1=

(s+2)

s2+4 s+4+1

= (s+2)

s2+4 s+5

A$S1"R

(s+2)

s2+4 s+5

.3 A S!'&I$

e

L(2tsin3 t)= 3

(s+2)2+9=

3

s2+4 s+4+9

= 3

s2+4 s+3

A$S1"R

3

s2+4 s+3

4+3 A S!'&I$

e

(3/2t)=0/5 e3 t/2

L1{ 12 s+3 }=12 L1{

1

s+3

2}=

1

2

A$S1"R

0/5e3 t/2

463 A S!'&I$

L1{ 2s+1 4s+3 }=2L1 { 1s+1}4L1 { 1s+3 }=2et4 e3 t

A$S1"R

2et4 e3 t

473 S!'&I$


2/8

Ay(s )=

200

s250 s+10625

= 200

s250 s+625+10000

y(s )= 200

s250s+10625

= 200

(s25)2+1002

y(s )= 200

s250s+10625

=2 100

(s25)2+1002

y(s )= 200

s250 s+10625

=2e25 tsin 100 t

A$S1"R

2e25 tsin100 t

553 B S!'&I$

3i26i+4

4 i3+2i2+3 i=

36 i+4

4 i+2+3 i=

16i

2i2+i

2i

3i26i+4

4 i3+2i2+3 i

=36 i+44 i+2+3 i

=2+i+12i+1642 i+2i+1

3i26i+4

4 i3+2i2+3 i

=36 i+44 i+2+3 i

=4+3 i

5

3i26i+4

4 i3+2i2+3 i

=36 i+44 i+2+3 i

=0.8+2.6 i

A$S1"R0.8+2.6 i

5.3 B/C S!'&I$

L1{3 s6s2+9}=L1{

3 s

s2+9 }L1{

6

s2+9 }

L1{3 s6s2+9}=3L

1 { ss2+32 }2L1 { 3s2+32 }

L

1

{3 s6

s2+9}=3cos3

t2sin3

t

L1{3 s6s2+9}=3cos3x2sin 3x

A$S1"R

3cos3x2sin3x

*+3 E S!'&I$

L1{ 3 s

2310(s1)( s+3)( s2) }=L1{ As1+ Bs+3 + Cs2 }3s

2310(s1)( s+3)( s2)

= A

s1+

B

s+3+

C

s2

3s2310=A ( s+3 ) (s2 )+B (s1) (s2)+C(s1)( s+3)


3/8

3 s2310=A(s2+s6)+B(s23 s+2)+C(s2+2 s3)

A+B+C=3

A3 B+2C=3

6A+2B3C=10

By elimination:

ABC=3

A3 B+2C=3

A4B+C=6

6A18 B+12C=18

6A+2B3C=10

A16 B+9C=28

B=13

10, C=

45

, A=5

2

L1{ 3 s

2310(s1)( s+3)( s2) }= 5/2s1+ 13/10s+3 4 /5s2

L1

{ 3 s

2310(s1)( s+3)( s2) }=

5

2 et

+13

10 e3t

4

5 e2 t

A$S1"R

5

2e

t+13

10e3 t

4

5e2 t

$one of the Choices*63 C S!'&I$

(3+4 i)1 /3

r=5 ; =53.13

r1 /3 (cosn+isinn )

51/3(cos( 13 ) (53.13 )+ isin( 13 )(53.13))

1.63+.52i

A$S1"R

1.63+.52i

*73 B S!'&I$

r=5 sin

r=5sin (30 )

r=2.5


4/8

sin 30=y

r=

y

2.5

y=1.25

cos30=

x

r=

x

2.5=2.17

A$S1"R

85349,435:653 A S!'&I$

(512 i)

r=5 ; =67.38

131/2

(cos (12 )(67.38 )+isin( 13 )(53.13 )

)1.63+0.52i3

A$S1"R

1.63+0.52 i

6.3 D S!'&I$

z1=1

4 4 4 5 6 ;7

4 5 6 7

4 5 6 7 +

z3+2z2+4z+8=0

z2=2, z

3=2 i , z

4=2 i

A$S1"R

4,;5, 5i,;5i

+3 C S!'&I$

ii . (x2 ) (x4 )=(x2+6x+8 )(x1i )=x2+6x8i x2+6 ix8i+x36x2

iii. (xi ) (x1 )=(x2xix+i )(x1+ i )=x2+x+ixi+i x2ix+x1+x3

x(1+i )(x (1i ) )=x22x+1=(x22x+2 )(x2 )iv . R"A!

A$S1"R

iiv

63 A S!'&I$

[(12j )+(1+j )](3+j )=(2j ) (3+j )

6+3j+2j+1

5+5j

A$S1"R

5+5j

73 S!'&I$


5/8

D z2+6z+10=0

(3+j)(3j)

A$S1"R

(3+j)(3j)

.53 B S!'&I$

(3j ) (z+42j )=10+10j

z+42j=10+20j3j

z=10+20j3j

z=1+7j4+2j

z=3+9j

A$S1"R

z=3+9j

..3 B S!'&I$

(a+bj)2=5+12j

a+bj=(5+12j )1/2

a+bj=3+2 i

r=13 ; =67.38

131/2 cos (67.38 )( 12 )+i sin(67.3)( 12 )3+2 i

A$S1"R

a=3,b=2

9+3 B S!'&I$

125

(cos25

+ isin25

)r=125 ; =25

25

cos(125)x=

x=113.2884734

25

sin(125)y=

y=52.82728272

A$S1"R

113.2884734+52.82728272 i


6/8

963 C 14(cos 225+ isin225)

PAR& B3 PRB!"M S!


7/8

1=A s3+Cs3+4A s2+B s2+8As+4 Bs+8B+Cs+D s2+D

A+C=0

4A+B+D=0

8A+4 B+C=0

8B+D=1

4ABD=0

8B+D=1

4A+7 B=1

A+C=0 ;72A+7C=4

A=465

, B= 4

65, C=

7

65, D=

9

65

L1 1(s2+1)( s2+4 s+8) =L1{

465

s+7

65

s2+1

+

4

65s+

9

65

s2+4 s+8

}465

L1 { ss2+1}+

7

65L

1{ 1s2+1 }+ 1

65L

1 { 4 s+9( s2+4 s+4 )+22 }

465

cost+ 7

65sint+

1

65L

1 { 4 (s+9

4)

((s+2)2 )+22}(s+2)+

1

4

(s+2)+1

4

18L1 { ((s+2)2 )+22 }

465

cost+ 7

65sint+

4

65L

1 { ((s+2)2)+22 }+ 465

465

cost+ 7

65sint+

4

65e2 t

cos2 t+ 1

130e2 t

sin 2t

*3 Solve

63 z1=6j4

z2=2+j

z3=3j5


8/8

2(3j5)

(6j4)z

1

z2

[2z3 ]2

=

(6j4)2+j

(36+120j100)

(6j4)(64+120j)

2+j

96+976j2+j

=4 (24+244j )

2+j =

784

5

2048

5j

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Prelims Answer Key