exercises font latex

8/9/2019 Exercises FonT Latex

1/56

Exercise:

Compute ,

1 1

dx

x

Solution: Let

11 2

2 1u x du dx udu dx

x

211 1

dx u du u x

21

u du u

1 121

u du u

12 11

du u

2 2 ln 1 2 1 2 ln 1 1u u c x x c Exercise:

Compute , lnn x x dx

Solution: Let lnu x and n dv x dx 1

du x

and1

1

n x v

n

1 1ln 1ln

1 1

n n n x x x x x dx dx

n n x

11 11ln

1 1

n n x x x x dx

n n

1 1ln

1 1

n n x x x dx

n n

1 11

ln1 1 1

n n x x x c

n n n

1 1

ln1 1

n x x c

n n

Exercise:

Show that,1

limn

i n i

i

a

2

1

a

a

where a

Solution: Let1

n

i i

i S

a 2 3

1 2 3.... ...

n

n

a a a a (1)

2 3 4 1

1 1 2 30 .... ...

n

n S

a a a a a (2) subtracting (1) from (2) to get

2 2 3 3

1 1 2 1 3 2(1) (2) : 0 ....S S

a a a a a a

2 3

1 1 1....

a a a

1

1i

i a

1

1 11

i i

S a a

1

1 1 i

i

a S

a a

11

11

a S a a

a

1

a a

1a

a

1

a

a

Thus

2

11 1

a a a S

a a

i.e

2

1

lim1

n

i n i

i a

a a

, a


2/56

Exercise:

Show that ,0

2

m x m

x m x

where m

Solution: Let u m x du dx if 0x u m and if 0x m u

0

0 0

m m

m

x m u m u

I dx du du x m x m u u m u u

0 0

2m m

m x x I I I dx dx m

m x x

thus 2

m I

Exercise:

Show that , b b

a a

f a b x dx f x dx

Solution: Let f be continuous function in ,a b and let u a b x du dx

if x a u b and if x b u a a a b

b b a

f a b x dx f u du f u du

Exercise:

Solve in 1 27y x and 2 5 1

3

y x

Solution: we have 2 5 1

3

y x 2

22 5 5

5

1 13

3 3

y y y

x x x x x

x

also we have 1 33y x 3

3 33y y

x x x x

5

32 33 y

y x

x

Let 15 7

2 7 15 7 14 2 275

33 3 3 3 3 3y t x t t t t

t

Thus 23 3y x x and 2y


3/56

Exercise:

Show that , 0

2

log sin log 22

x dx

Solution: Let 0

2

log sinJ x dx

and Let2

u x

du dx if 0

2

x u

and

if 02

x u

hence 0 02

0

2 2

log sin log sin log cos2

J x dx u du u du

Thus 0 0

2 2

2 log sin log cosJ x dx x dx

0

2

log sin log cosx x dx

0

2

log si2

n2

cosx x dx

0

2

1log sin 2

2x dx

0 0

2 2

1log log sin 2

2dx x dx

So we obtain that 1 22J J J where0

1

2

1log

2J dx

and

0

2

2

log sin2J x dx

0 0

1

2 2

1log log 1 log 2

2J dx dx

00

22

log 2 log 2dx x

log 2 0 log 2 log 22 2

Thus 1 log 22

J

0

2

2

log sin2J x dx

Let 2v x 2 2dv

dv dx dx so v if 2

x and 0v if

0x so

0

2

2

log sin2J x dx

01

log sin2

v dv

0

2

12 log sin

2x dx J

Thus 2 log 2 log 22 2

J J J

Q.E.D


4/56

Exercise:

Compute, sin ln cos lnx x dx

Solution: Let ln t t t x e x e dt dx so sin ln cos ln sin cos t x x dx t t e dt

sin cost t t e dt t e dt sin sin lnt t e c x x c

Let sin sin cost t t U t e dt t e t e dt sin cos sint t t t e dt t e dt t e

Exercise:

Solve in , 4 4

1 1 1 1 16x x

Solution: First let’s expand 2

1 1 1 1 2 1 2 1x x x x x and

2

1 1 1 1 2 1 2 1x x x x x

Hence 4 4 2 2

1 1 1 1 16 2 1 2 1 16x x x x x x

2 24 1 4 1 4 1 4 1 16x x x x x x 22 8 8 16x x 22 8 24 0x x 2 4 12 0x x 2 4 42 2 12 0x x

2

2 16 2 4x x 4 2 2x or 4 2 6x which is rejected

Exercise:

Solve in , 2 22 | | 5 3 0z z i where 2 1i

Solution: Let ,z z x iy 2

2 2 2 2 2| |z x y x y

and 2z x iy x iy

2 2 2 2z x ixy ixy i y 2 2 2x y ixy

thus 2 2 2 2 2 22 | | 5 3 0 2 2 5 3 0z z i x y x y i xy i

2 2 2 22 2 2 5 3 0x y x y i xy i 2 23 2 5 3x y i xy i 2 23 5x y

and 2 3xy 3

2y

x

so

2

2 33 52

x x

22

33 5

4x

x 4 212 3 20x x

4 212 20 3 0x x Let 2 2 1 3

12 20 3 06 2

t x t t t or t

thus1 3

26x or x hence

2 3 2

2 2y or so

1 3 2 3 2

2 2 26z i or z i


5/56

Exercise:

Compute ,1

1

x dx

x

Solution: Let1

22

u x du dx udu dx x

so1 1

211

x u dx udu

u x

2 1

1

u u du

u

22 1

1 1

u u du

u u

2

2

12

1

u u du

u

2

2 2

12 2

1 1

u u u u du du

u u

2

2 22

1

1 12

1

u u du du

u u

22 2

1 12 2

1 1

u u du du

u u

22

1

u du

u

2

2 2

12 2

1 1

u du du

u u

2

2 2

22 1 2

1 1

udu du du u du

u u

let’s compute each

term aside as follows: Let 21 2 2t u dt udu dt udu

1/21/2 1/2 2

2

22 2 1

112

u dt t du t dt t c u c

t u

thus 22

22 1

1

u du u c

u

also 2

arcsin1

du u c

u

,Let sin cosu du d so2 2

1 sin cos

so 2 2 21 cos cos cosu du d d 1

1 cos 22

d 1 1

sin 22 4

c

thus 1 12 1 arcsin sin 2 arcsin21x

dx x x x c x

‚ 1 1 1u u u since 1 0u ‚


6/56

Exercise:

Solve in the following system :

3

5

4

x y y z

y z z x

z x x y

Solution: Adding the above equations we get 2 12x y y z z x

6x y y z z x 3 6 3z x z x , 5 6x y

1x y and 4 6 2y z y z S.B.S for the box equation we get:

9, 1z x x y and 4y z Adding the pervious equations we get: 2 14z x y

7x y z but 9 7 9 2z x y thus 2y , 6z and 3x

Exercise:

Compute,3

8

7 3lim

8x

x

x

Solution: Let 3 3t x t x as 8x , 2t hence3

38 2

7 3 7 3lim lim

8 8x t

x t

x t

3 3 32 2

7 3 7 3 7 9lim lim

8 7 3 2 7 3t t

t t t

t t t t

3 32

2lim

2 7 3t

t

t t

but 3 3 22 2 2 4t t t t 23 32

2 42

t t t

t

Thus

3 3 22 22 1 1 1

lim lim12 6 722 7 3 2 4 7 3t t

t

t t t t t


7/56

Exercise:

Show that

647

2

1 16lim 2

2x

x x

x

Solution: Let 6 3 34 2 21 16 1 4 1 4g x x x x x x x

3 32 264

2 2

1 4 1 41 16lim lim

2 2x x

x x x x x x

x x

3232

2 2

1 4lim lim 1 4

2x x

x x x x

x

but

32

2lim 1 4 4 1 4 8x

x x

32

2

1 48lim

2x

x x

x

Now let 1 1t x t x as 2, 1x t

3 2 22 3 3

2 1 11 4 1 4 1 48 lim 8 lim 8 lim

2 1 2 1x t t x x t t t t

x t t

but 23 3 2 5 4 31 4 2 1 4 2 4t t t t t t t t 5 4 32 3 1t t t

23 5 4 3

1 1

1 4 1 2 38 lim 8 lim

1 1t t

t t t t t

t t

5 4 3

1 1

1 2 38 lim 8 lim

1 1t t

t t t

t t

but 5 4 3 21 1 1t t t t t t 5

4 3 2

1 1

1lim lim 1 5

1t t

t t t t t

t

hence5

1

18 lim 8 5 401t

t

t

we have 4 3 4 3 4 32 3 2 2 1 2 1 1t t t t t t

but 4 3 21 1 1t t t t t and 3 21 1 1t t t t

so 4 34 3

1 1

2 1 12 3lim lim

1 1t t

t t t t

t t

4 3

1 1

1 12 lim lim

1 1t t

t t

t t

3 2 21 1

2 lim 1 lim 1t t

t t t t t

2 4 3 8 3 11

thus 6

4 7

2

1 16lim 40 8 11 128 2

2x x x

x


8/56

Exercise:

Show that ,

5

4

4 2 cos sinlim 5 2

1 sin 2x

x x

x

Solution: We know that cos sin2

x x

and cos cos 2 cos cos2 2

a b a ba b

2 2cos sin cos cos 2 cos cos2 2 2

x x x x x x x x

222 cos cos

4 2

x

2

2 cos 2 cos2 4 4

x x

hence

5

4

4 2 2 cos4

lim1 sin 2

x

x

x

5

4

4 2 1 cos4

lim1 sin 2

x

x

x

Now we have 04 4

x x

so let | | 04

h x h

5

5

0

4

4 2 1 cos4 1 cos

lim 4 2 lim1 sin 21 sin 2

2

h x

x h

x h

5

0

1 cos4 2 lim 1 cos 2h

h

h

but 22 sin 1 cos2h h 5 5

2 20 0

1 cos 1 1 coslim lim

22 sin 1 cosh h

h h

h h

Let cost h as 0h , 1t

55 5 5

2 2 220 1 1 1

11 cos 1 1lim lim lim lim

1 cos 1 11h t t t

t h t t

h t t t

but

4 3 25

2

1 11

1 11

t t t t t t

t t t

5 4 3 2

21 1

1 1 5lim lim1 21t t

t t t t t

t t

Thus

5

4

4 2 cos sin 5lim 4 21 sin 2 4

x

x x

x

5 2


9/56

Exercise:

Compute, 2

lim 2 tanx

x x

Solution: Let t tx x

as 2,

2x t

2

2

lim 2 tan lim 2 tanx

t

x t x t

2

2lim tant

t t

t

but cot tan

2t t

and

1cot

2tan

2

t

t

2 2 2 2

2 22 2lim 2 tan lim lim lim

tan tan2 22 2

t t t t

t t t t t

t t t t t

but

2

2lim 1

tan2

t

t

t

so

2 2

22 2 4

lim lim2

2 2t t

t

t

t t

Exercise:

Compute,

5

1

2 3 3 2lim

1x

x x

x

Solution: we have 1 1 0x x so by taking 1h x we get 0h

5

0

2 1 3 3 1 2limh

h h

h

5 5

0 0

2 1 3 1 2 1 1 3lim limh h

h h h h

h h

5

0 0

2 1 1 3lim limh h

h h

h h

5

0

2 1 1lim 3h

h

h

Now Let 2 1t h as 0, 1h t

So

5

54 3 2

1 1 1

2 11lim lim 2 lim 1 2 5 10

1 1

2

t t t

t t t t t t

t t

thus

5

1

2 3 3 2lim 13

1x

x x

x


10/56

Exercise:

Solve in ,224 40log 64 2 0x x

Solution: we know that log 1 0 hence224 40log 64 2 log1x x

224 4064 2 1x x

2

1

244064 2 1x x

224

40 1264

x x

2 40

24 1446

1 12

22

x x 2 40 1442 2x x

2 40 144 0x x 2 2 20 400 400 144 0x x 2

20 256x 20 16x

16 20 36x or 20 16 4x so the roots are 36 4x or x

Exercise:

Solve in ,25 124 5 125x x

Solution: we know that 2

225 5 5x

x x so by letting 5 x t we will obtain the

following 2 124 125t t 2 2 22 62 62 62 125 0t t 2 2

62 3969 63t

62 63t thus 125 1t or but 1t is rejected since 0t thus 35 125 5x

therefore 9x

Exercise:

Solve the following ODE system4

2 4 t

t

dy x y e dx dy

x e dx

Solution: We have 2 4 22

t t dy dy e x y e x y

dx dx but 4t

dy x e

dx 4 2

2

t t e e y

442

2 4 2

t t t t e e e y e y now using the second equation we get

44

4 2

t t t d e e x e

dx

4 4 4 424 2 4

t t t t t t d e e dt e dt x e x e e

dt dx dx

424

t t t e x e dt e dx

4

4

24

t t

t

e e

dx dt

x e

3

12

4t dx e dt

x

31

ln 212

t x e t c 31 2

12t

e t c x e

31 212

t e t

x ce


11/56

Chapter 4: Logarithm

Logarithm to base a:

The logarithm to the base a where 0, 1a a of a number 0x is the number y

such that y a x thus loga

y x

Remark: 10if a , we called 10log x the common logarithm where

if a e , we called loge x the natural logarithm

Remark: ln loge

x x

Properties:

1) log 1 0 , 0, 1a

a a

2) log 1, 0, 1a

a a a

3)

log

x

a e x x and

log ln

0e x x

e e x x

4) log log loga a a xy x y , 0 , 0, 1x y a a

5) log log log , 0 , 0, 1a a a

x x y x y a a

y

6) 1

log log , 0, 0, 1a a

x x a a x

7) if a and 0x we define log ln

0 0

log lne a x n n a a x

e n n

x e e a x a x

8) log logy x y x 0,x y

9) log

, 0, 1a y x y x a a a

10) 0, 1if a a thenlog ln

loglog ln

e a

e

x x x

a a ( Change of the base )

11) , , 0if a b c and , 1a c thenlog

loglog

c a

c

bb

a ( Change of the base )

12) 0if x y then log logx y so log is an increasing function

0 1if x then log 1x , in particular 1if x then log 0x

13) 1if a and 0 x y then log loga a

x y and x y a a

0 1if a and 0 x y then log loga a

x y and x y a a

14) log

, 0, 1a x

a x a a


12/56

Properties for logarithm inequalities:

1) 1 , log log 0x x for x f x g x f x g x

2) ‘ 0 1 , log log 0x x for x f x g x f x g x

3)

1 , log x

x for x f x a f x a

4)

0 1 , log 0 x

x for x f x a f x a

Remark: , , 0 , , 1if a b c a b then log

log log loglog

ba b a

b

c c c b

a

Exercises:

Solve in , 22log log

4 log log 1x

x x


2 2log log 2 log log log log

4 2 2 log

x x x

x

thus 2 2

log log log 1x x x 2

2 log log 1 0x x 2 1 1

log log 02 2

x x

Let logt x 2 1 1

02 2

t t 1

1 02

t t

hence log 1x or1

log2

x rejected

since log 0x thus x e

Exercise:

Determine the solution for this inequality, 2sin2

3log 8 23log sinx

x x x

Solution: we need first to the domain of definition for the above inequality2

log sin 0x

sin 1 sin 22 2

x x k

thus 2 1

2

n x

,n where n

also 222

sin

2 2

log 8 23 3log 8 23

log sin log sinx

x x x x

x x

as

2log sin 0x we get

22log 8 23 3x x 2 38 23 2 8x x 2 8 15 0x x after studying the sign

we get 3 5x So the solution is 3 3

3, , ,52 2

x


13/56

Exercise:

Determine the solution 3 1/3 31

log 2 4 log 2 log 72

x x x (*)

Solution: (*) is defined when 2 4 0, 2 0x x x 2x

so we can write

3 33

33

log 2 log 71

log 2 4 21 log 3log3

x

x x

but 3 31

log log 3 13

33 3 33

log 7log 2 4 log 2 log 7

log 3x x x 3 3log 4 log 7x 4 7x

3x

Exercise:

Determine the solution for this inequality

20.3 0.3log 20 log 4 0x x x (**)

Solution: (**) is defined when 2 20 0x x , 4 0x 4 5 0x x

So 2 20 0x x whenever 4 5x or x and we have 4 0 4x x

thus (**) is defined when 5x

But the base is 0.3 which is between 0 and 1 thus 0.3log 5 0x 5 1 6x x

So the solution is 5 6x

Exercise:

Determine the solution 5log

5x

x

Solution: clearly that 0x 5log

5 5log log 5 1

x x 5 5log log 1x x

2

5 5 5log 1 log 1 log 1 0x x x

5 5log 1 log 1x or x

5log 1 5x x or 1

5 5 5log 1 log 5 log 5x 15x

Thus the solution is 10, 5,5x

5

2

5

log 1 1

log 1 0 0

x

x


14/56

Exercise:

Determine the solution for2

0.7 6log log 0

4

x x

x

(***)

Solution:2

0.7 6log log 0

4

x x

x

for 0x , log

a x is increasing function 1a

so we can write (***) as2

6log 1

4

x x

x

2

64

x x

x

2 6 24 0x x x

2 5 24 0x x now solving we get 2 5 24 8 3 0x x x x 8 3x or x

thus 3 8x or x from the domain we have2

04

x x

x

4x and

2 1 0x x x x 0, 1x x

So the solution for this is

: 0 4 1S x x or x thus 4 3x or 8x

2

4 1 0

0 0

4 0 0

0 0

x

x x

x

f

thus 4, 3 8,S

Exercise:

Solve in 3 3sin sin 2 sin 32

x x x

Solution: we know that3 3

sin sin 3 2 sin cos2 2

x x x x x x

2 sin2 cosx x

hence3 3

sin sin 2 sin 3 2 sin 2 cos sin 22

x x x x x x

3 3

sin 2 2 cos 12

x x

but3 3 2 1 3 1 3

1 sin sin sin2 2 2 2 2 6 3


15/56

Exercise:

Determine the differentiable function f x defined for 0x such that for f x and the

inverse function g x which satisfied 3

2

1

18

3

f x

g t dt x

Solution: Let y f x take inverses both sides we get 1 1

f y f f x x

g y x

3

2

1

18

3

y d d

g t dt x dx dx

by Chain rule

3

2

1'

18

3

y

y x g y

d dy d g t dt x

dy dx dx

'2

x g y f x but '

2

x g y x xf x '

2

x f x

x

hence 1

1 1/221 12 2

f x x dx x dx x c

f x x c let’s find c

2

f x c x y c x y c x g y thus 2

g t t c

32 3 3 3

1 1 1

1 1 11 1

3 3 3

f x x c x c

g t dt t c dt t c x c c c x x c

So our result is more same as above equation thus 3 31 8 1 8 2 1c c c

Therefore 1 f x x

Exercise:

Compute,cos

1 cos

x dx

x

Solution: we know that 2 21 cos 2

cos 2 cos 1 cos 22

x x x x

22 cos 1 cos

2

x x

so

2

2

2 cos 12cos

1 cos2cos

2

x

x dx dx

x x

21 sec

2 2

x dx dx

tan 2

x x d

tan2

x x c

since 2

1tan sec

2 2 2

x x d


16/56

Exercise:

Determine the values of ,a c such that for all 0x , 3

ln 12 3

x x a x

x bx c

holds true.

Then determine the range of b .

Solution: Let 3

,

2 3

x x a f x g x

x bx c

and ln 1h x x

thus 0 f x g x h x x so in particular 0 0 0 0 0 0 f g h g

0 0g 0 0 0a

g a c

Notice that the two curves & f h have same

tangent point at 0x and below the liney x

thus f x h x y x we are interesting in g x

so g must have y x as a tangent line

at 0x

2 2 2

1'

bx c b x a bx c bx ab c abg x

bx c bx c bx c

so

2' 0

c abg

c

Thus 2 2

0 1' 0

c c g

c c c

1' 0g

c but ' ' ' f x g x h x thus 1 ' 0 1g

11 1c

c so the obtain expression is

3ln 1

2 3 1

x x x

x bx

Now we have h x g x 0h x g x Let 0 0w x h x g x x

hence ln 11

x w x x

bx

but

2 2

11 1 1'

1 11 1

bx b x w x

x x bx bx

and 0 0w g is increasing for every 0x

2

1 10

1 1x bx

2

2

1 1

01 1

bx x

x bx

22 2

2 2

2 11 2 1

01 1 1 1

x xb bb x bx x

x bx x bx

1

2 1 0 2b b

Now suppose that f x g x with 0, 0b x 0 f x g x 3

02 3 1

x x

x bx

2 2

3 2 3 2 3 10

2 3 1

x x bx b x

x bx

2 2

6 9 6 10

2 3 1

x x bx bx

x bx


17/56

3 10

2 3 1x bx

, 0x

3 1 2 30

2 3 1

bx x

x bx

3 3 3 2

2 3 1

bx x

x bx

3 20

2 3 1

x b

x bx

3 20

1

b

bx

thus

1 23 2 0

2 3b b

Remark:

‚

2 2

2 1 2 1

11 1 1

x bx b x bx b

x x bx bx

but 0

1

x

x

0x thus 2 1 0bx b 0x

hence 2 1 0b so1

2b and similarly for

3 2 3 20 0

2 3 12 3 1

x b x bx

x bx x bx

but 0 02 3

x x

x

hence

23 2 0

3b b

Exercise:

Compute,1

1

x dx

x x

Solution: Let1

1 22 1

u x du dx udu dx x

So1 1

1 1 1

x x dx dx dx

x x x x x x

1 1

1 1dx dx

x x x

hence 1 2 2 2 2 11

u dx du du u c x c u x

so1 2 2

1

u dx du du

xu x x x

but

2 21 1 1u x u x u x

thus 12

12 2 tan

11

du dx u c

u x x

1

2 1 2 arctan 11

x dx x x c

x x


18/56

Exercise:

Compute,2

4x dx

x

Solution: Let 24tanx t 28 tan secdx t t dt

so 2

22 2

2 4 4

4 tan 14 4 tan 4 8 tan sec 8 tan sec

16 tan 16 tan

t x t dx t t dt t t dt x t t

but 2 2sec tan 1t t 2

2

2 4

4 4 sec8 tan sec

16tan

x t dx t t dt

x t

242sec

8 tan sec16tan

t t t dt

t

33

3

sec sec

tantan

t t dt dt

t t

but

1

sec cos

tan

t t

t

sin

cos

t

t

1csc

sint

t

3 2

2

4csc csc csc

x dx t dt t t dt

x

let

2csc & cscu t dv t dt

csc cot & cotdu t t dt v t

so 3csc csc cot cot csc cott dt t t t t dt 2csc cot cot csct t t t dt but

2 2cot csc 1t t

hence 3 2 3csc cot csccs 1 csc csc coc csct csct t t t dt t t t d t dt d t t t

32 csc csc cot csct dt t t t dt but csc ln csc cott dt t t c

Thus 3 1 1

csc csc cot ln csc cot2 2

t dt t t t t c but2 24 tan tan

4

x x t t

2tan cot

2

x t t

x and we know that 2 2csc 1 cott t 2csc 1 cott t

4csc 1t

x

4x

x

thus

2

4 1 4 2 1 4 2ln

2 2

x x x dx c

x x x x x


19/56

Exercise:

Show that

2 2

2 2

2 2

det det

bc c b a c b

c ca a abc c b a

b a c b a ab

Solution:

2 2

2 2 4 2 2 2 2 2 2 2 2

2 2

det

bc c b

c ca a bc caab a c c ab b a b c a cab

b a ab

2 23 4 4bc a cba a c ab cab bca cab

3 3 3abc cab a abc c cab abc abc b

3 3 3abc abc a c abc abc b

2 2 2abc a bc a c c ab b ac b b a c b a b

abc a c ba c a c b c

deta c b

abc c b a

b a c


20/56

Exercise:

Suppose that1

2x x

then what is value of1

9x x

Solution: Let 20w x w x hence 21 1

2x w w x

multiply by w both sides

3 31 2 2 1 0w w w w 21 1 0w w w 21 1 0w or w w

but 1w is rejected since 0w thus 2 1 1

14 4

w w

21 5 1 5

2 4 2w w

so1 5

2w x

2

1 5

4x

2 3

1 5 1 5 8 36 1 529

4 1 5 4 1 5

1 5 1 5 2 5 36 8 2 5 30 2 1 5

2 5 30 24 41 54 1 5

4 5 82 5 30 2 2 55 8

4 4

Exercise:

Show that0

1lim ln

x

x

a a

x

where 0a

Solution: We know that ln lnx x a x a a e e thus

ln

0 0

1 1lim limx x a

x x

a e

x x

Now let ln ln1 1

1 1x a x a e e t t

take ln both sides1

ln 1 lnx a t

thus

1ln 1

ln

t x

a

as1

0 , 0x t

hence t thus0

11

lim lim1

ln 1

ln

x

x t

a t x

t

a

ln lnlim lim

1 1ln 1 ln 1

t t t

a a

t t t

but the natural logarithm is a continuous function

so1 1

lim ln 1 ln lim 1 ln 1

t t

t t e

t t

thus0

1lim ln

x

x

a a

x


21/56

Exercise:

Let 1 2,z z such that 1z x a iy b b and 2z x a a iy b where2, 1i i if the product of

1z and

2z is equal to 1 4 3i and 1 2Re z z is equal to the

curve that passes through two points are 2,1 and 3, 2 1)

Write the equation of that curve

2) Deduce1

z and2

z

Solution:

1) 1 2z z x a iy b b x a a iy b 2 2

xa iyx ab iyx ab ab yb

2 2

xa yb i yx ab ab ab

2 2

1 1 4 3xa yb ixy ab ab i

Now by comparison we get 2 2 1xa yb and 1 4 3xy ab ab since 1 2Re z z is

a curve passes through two points 2,1 & 3, 2 thus 2 2

2 1a b …(1) and

2 2

3 2 1a b …(2) 2 2 2 22 1 & 3 2 1a b a b from equation (1) we obtain

2 22 1b a from equation (2) we get 2 23 2 2 1 1a a 2 2 23 4 2 1 1a a a

1a thus 1b but , 0a b thus the accepting values are 1, 1a b hence

1 2Re z z is a curve whose equation is2 2 1x y …(1*) which is an hyperbola curve

2) we already have 1ab thus the imaginary part

for1 2

z z will be in this form 2 4 3xy 2 3xy ..(2*)

From (2*) we get 22

2 3 12y y

x x now back to the

equation (1*) and substitute to get :

2 4 2 4 2

2

12

1 12 12 0x x x x x x

let2 2 12 0t x t t

21 1

122 4

t

1 7 1 7

2 2 2t t

4, 3t

but 2t x x t So 2x where 3t will give two complex numbers are

3x i thus 1 22 3 & 2 3z i z i


22/56

Exercise:

Solve the following differential equation1 1 1

' x y y e and 0 2y

Solution: 1 1 ' 1

' '' '

x

x

y y e y y yy

y y y y e

' 'x x e y e y yy x x

dy dy e y e y

dx dx

2

2x d d y e y

dx dx

2

2x y e y thus

2

2x y e y since 0 2y 0

so2

22 22

x x x y e y e y y y e

Or we can solve it using exact and none exact differential equation:

' 'x x e y e y yy ' 'x x e y yy e y x x x x dy e y y e e ydx y e dy dx

0x x e ydx y e dx

,,

0x x

M x y N x y

e y dx e y dy , , 0M x y dx N x y dy

x

x

M N M e

x y

hence exact differential equation , suppose that the solution y is in the

form , f x y where thus f

M x

and

f N

y

x

f e y

x

and x

f e y

y

, 0x f x y e y g y and

2

, 02

x y f x y e y h x 2

2

x y e y for

0 2y

we get 0 hence 2 x y e

Exercise:

Compute,1

1

x dx

x

Solution: Let1

2t x dt dx

x

12

2dt dx tdt dx

t

So 21 1 2 2

21 11

x t t t dx t dt dt

t t x

2

21

t t dt

t

2

2 21

t dt t

4 2 41

dt dt t dt

t

24 4 ln 1t t t c 4 4 ln 1x x x c


23/56

Exercise:

Show that,4

2

0

4 16x dx

Solution: 2 24 0 4 2x x x

4 2 4

2 2 2

0 0 2

4 4 4x dx x dx x dx

We have 2 20 2 0 4 4 4 0x x x thus 20 4 4x

thus 22 2 2 3

2 2 2

0 0 0 0

84 4 4 4 8

3 3

x x dx x dx x dx x

for 2 22 4 4 16 0 4 12x x x thus 4 4

2 2

2 2

4 4x dx x dx

43

2

64 84 16 8

3 3 3

x x

hence4

2

0

8 64 84 8 16 8 16

3 3 3x dx

Exercise:

Show that,2 2

0 4

a dx

a x a

Solution: we have2

2 2 2

2 1

x x a a

a

hence2 2 2 2

0 0

1

1

a a dx dx

x a a x

a

Now letx

t ta x adt dx a

where 0t if 0x and 1t if x a

so 1

11

2 2 2 2 20 00 0

1 1 1 1arctan arctan 1 arctan 0

41 1

a dx adt dt

t a a a ax a a t t


24/56

Elementary Row operation for Determinant

We use elementary row operations to reduce the given matrix to an upper or lower triangular

matrix.

Theorem:

If A is n n upper or lower triangular matrix then

11 12 1 11

22 2 21 2211 22

1

1 1

... 0 ... 0

0det ....

0

0 0

n

n n

nn ii i

nn n nn nn

a a a a

a a a a A A a a a a

a a a a

Properties for determinant operations:

Let ,A B are n n square matrices, then

1) det det T A A 2) det det detAB A B

3) If A has a row or column of zeros then det 0A

If two rows or two columns of A are identical then det 0A

4) If B is obtained by interchanging two rows of A then

det detB A

5) If B is obtain by adding a multiple of one row ( or column ) of A to another ,then

det detB A

6) If the matrix B is obtained by multiplying single row or single column by a scalar

then det detB A

If all n -rows (or all columns) of A are multiplied by any scalar to obtain B ,then

det detn

B A

Summary:

Interchanging two rows will change the sign of the determinanti j

R R

Multiply a row or a column by scalar will also multiply the determinant by same scalar


25/56

Theorem:

Let 1 2, , ..., n a a a denote the row of A If thethi row vector of A is the sum of the two row

vectors i.e.i i i

a b c then det det detA B C

where

1

1

1

i

i

i

n

a

a

bB

a

a

and

1

1

1

i

i

i

n

a

a

c C

a

a

Exercises:

1)

Evaluate

2 1 3 7

1 2 4 3

2 4 2 1

2 2 8 4

2) Use row reduction to compute the determinant of

2 3 3 1

0 4 3 32 1 1 3

0 4 3 2

A

3) Show that ,

2 4 4 2

5 3 3 2 0

1 2 2 1

x

x

x

4)

Find all x satisfying

2 1

1 1 1 04 2 1

x x


26/56

Solution:

1) We have 2 13 1

4 1

1 3 71

2 2 21 2 4 3 1 1 2 4 3det2 4 2 1 22 2 4 2 12 2 8 4 22 2 8 4

2 1 3 7

R RA

R R

R R

3 5 1

2 2 2

1 3 71

2 2 21 02

0 5 1 8

0 1 5 11

3 3 2

4 4 2

1 3 71

2 2 25 11 2 0 1

3 3 52 30 5 1 8

0 1 5 11

R R R

R R R

1 3 71

2 2 25 1

0 11 3 322 293

0 03 3

10 320 0

3 3

4 4 3

1 3 712 2 2

5 10 11 3 3 3

293 220 0 1

2210 32 100 03 3 3

R R R

1 3 712 2 2

5 10 11 3 3

29220 0 1

2269

0 0 011

1 3 71

2 2 25 11 11 10 1

det det3 322 69 13829

0 0 122

0 0 0 1

A B

since det 1B

thus det 138A

2)


27/56

Exercise:

Compute, cos lnx dx

Solution: let lnt x 1

dt dx xdt dx x

but t e x thus t e dt dx

so cos ln cos t x dx t e dt , Let & cost u e dv t dt & sint du e dt v t

so cos sin sint t t t e dt e t t e dt and a same process take & sint U e dV t

& cost dU e dt V t so sin cos cost t t t e dt e t t e dt

thus sinc cos coso st t t t t e dt t e d e t e t t 2 cos sin cost t t t e dt e t e t c

1

cos sin cos2

t t t e dt e t t c 1

cos ln sin ln cos ln2

x dx x x x c

Exercise:

Solve in ,

252

5

log 8log 8 log 2

log 2

x x x

x

Solution: we know thatlog

logloga

bb

a hence

22

5

log 8log 8

log5

x x

and

5

log 2log 2

log5

x x

thus

2 25

5

log 8 log 8

log 2 log 2

x x

x x

so

22

log 8log 8 log 2

log 2

x x x

x


28/56

Exercise:

Solve in , 2log 3

33 9 3x

x x x x

Solution: Let 3 & 3u x v x

2log 3

33 9 3x

x x x x

2log 9v u u x v but 2 9 3 3x x x vu

2logv u u uv 2loglog logv v v u u uv 2log log log logv v v v u u u v

log log log 2 logv v v v u u u v but log 1v v now take logv t u to get:

2 22 2 0t t t t

21 9 1 3 1 3

2 4 2 2 2t t t

2 1t or

If 22 22 log 2 3 3 7 6 0

v t u u v x x x x 6x or 1x

Since 0 3 0 3v x x so the accepted solution is 6x hence 1x is rejected

If 1 2 21

1 log 1 3 9 1 10 103v

t u u v x x x x x

But we are interesting in 0v which means 3x thus the accepted solution is 10x

where as 10x is rejected. So the solution set is 6, 10S Exercise:

Find the general solution for this differential equation 2 ln 0dy

y x y x dx

Solution: we have 2 2ln 0 ln 0dy x dy y x y x x y dx y dx

2 ln 0x x y dx dy y

, , 0M x y dx N x y dy so1M N

y y x

hence the obtain

differential equation is exact, so the solution will be in the form , f x y

Thus & f f

M N x y

32

, ln ln ( ) 03

x

f x y M x x y dx x y g y and

, ln ( ) 0x

f x y N y dy x y h x y

hence 3

3

x h x and 0g y

Therefore3

ln3

x x y is the General solution for this D.E


29/56

Exercise:

Compute,

2

1

sin cosdx

x x

Solution: we have 2

2 22sinsin cos cos 1 cos tan 1cos

x x x x x x

x

Let 22

tan 1 seccos

dx u x du x dx

x

so

2

2 2 2 22

sec 1

sin cos cos tan 1 tan 1

dx dx x du dx c

u u x x x x x

thus

2

1

tan 1sin cos

dx c

x x x

Exercise:

Compute,2 2

sin

cos

x dx

a x where a

Solution: Let cos sin sinu x du x dx du x dx so2 2 2 2

sin

cos

x du dx

a x a u

2

2 1

du

u a

a

2 21

1

du

a u

a

Let1u

w dw du adw du a a

So 2 2 2 2 2

1 1 1 1arctan

1 11

du adw dw w c

a a a a w w u

a

1

arctan u

c a a

Thus2 2

sin 1 cosarctan

cos

x x dx c

a a a x


30/56

Exercise:

Solve the following linear congruency system

11 5 7 mod20

6 3 8 mod20

x y

x y

Solution: We need to check if this system had a solution or not first

so gcd 11,5,20 1 | 7 and gcd 6,3,20 1| 8 thus there are exactly md solution module m

11 5 7 mod20 3

6 3 8 mod20 5

x y

x y

33 15 21 mod 20 1 mod 20

30 15 40 mod20 0 mod20

x y

x y

3 1 mod 20x but gcd 3, 20 1 thus 13 exits which is 13 7 mod 20

hence 7 mod20x Now just substitute x in (2) to get :

6 7 3 8 mod20 3 8 42 mod20 34 mod20 6 mod20y y thus 3 6 mod20y hence 6 7 mod 20 42 mod 20y 2 mod20y

Exercise:

Solve in , 3log 8 log 3 6x x

Solution: we know that3

3

ln ln 3 1 1log & log 3

ln 3 ln ln log

ln 3

x

x x

x x x

so3 3

3

8log 8 log 3 6 log 6

logx x x

x let 3logw x

86w

w

22 28 6 2 3 9 9 8 0 3 1w w w w w 3 1w 2 4w or

So 2 43

log 2 4 3 3x or x or x


31/56

Exercise:

Solve in , 2 6x x

Solution: we have 2 6 2 6 2 6x x x x or x x

6 2 6 2x x or x x S.B.S

2 2 2 236 12 4 4 36 12 4 4x x x x or x x x x

36 4 4 12 36 4 4 12x x or x x 32 16 32 8x or x 2 4x or x

If 2 2 2 2 2 4 2 2 6x thus rejected hence the acceptable solution is

4x or we can solve this exercise as follows :

let 2y x 6x y 22 2y x and

236x y

2 2 2

2 36x y x y xy 22 2 2

2 2 36 4 4 2 36x x xy x x x xy

2 22 4 2 32 2 16x x xy x x xy 216 2xy x x 2 2 16x x

y x

2 16 162

x x y y x

x x

but 16 1, 2, 4, 8, 16D

If 1 13x y rejected If 1 17x y rejected

If 2 4x y rejected If 2 8x y rejected

If 4 2x y rejected If 4 2x y accepted since 2 4 2 2 2

If 8 8x y rejected If 8 4x y rejected If 16 17x y rejected

If 16 13x y rejected thus the accepted solution is 4x

Exercise:

Solve in , x x x

Solution: we have x x x x x x thus x x x x take ln both sides we get :

1 1ln ln ln ln ln 0x x x x x x x x x x x

ln 0 ln1x 21 0 1 0x x x

1x or 2 1 1 1 0x x x 1x or 1x thus the solution set is 1,1S


32/56

Exercise:

Solve in ,

2 2 2 11log 4 1 log 2 1

2 4 3 4x x

x x x x

Solution: Let 2 & 4a x b x thus 2 6 2 3 32

a ba b x x x

2 2 2 11log 1 log 1 3b a a b x b 2

11

4a b b thus 2 2

2

11log 1 log 1

4b a a ba b b

We have 2 2 2 22 2 2 2 2 2log log log log

log log log log log loga b a b

b a a b b a b a w

thus

2

11

4

a bw wa b

b

2 12

4

a b bw ab w

2 12

14

a b bw ab

but 2

221 4 2 1 6 8 1 32

a bab x x x x x

thus

2

12

a bab

2 212

2 2

a b a bw b

2

12 02

a bw b

2

02

a b

or 12w b

0a b a b thus 2 4 2 6 3x x x x

212 12 12log bw b w b a b 2 122 2loglog logba b

2 2 2 2 2log log 12 log log log 12 0b a b b a thus

02

log 0 2 1 1b b b

2log 12a 122a thus 122a so 4 1 3b x x or

12 12 11 112 2 2 2 2 2 2 2 2 1 2 2047 4094a x x but 3x is rejected

since 2x thus the accepted solution is 4094x

Exercise:

Compute ,2

1

x

x

e dx

e

Solution: Let x x u e du e dx so2

1 11 11

x

x e udu u dx du

u u e

ln 1 ln 11

x x du du u u c e e c u


33/56

Exercise:

Compute,

2

ln

1 ln

x dx

x

Solution: Let1

lnu x du dx x

but u u x e e du dx

So

2 2 2

ln

1 ln 1 1

u u x ue u dx du e du

x u u

Let

2

1&1

u U ue dV du

u

1

1 &1

u dU e u du V u

so

2

11

1 11

u u u ue ue du u e du

u u u

ln ln

2

ln ln

1 ln1 ln

x x x x dx e e c x x

ln

1 ln

x x x c

x

Exercise:

Compute,cot tan

1 3 sin 2

x x dx

x

Solution: Let 2 2tan arctant x x t 42

1

t dx dt

t

but 22

1 1cot cot

tanx x t

x t

and we know that2

2 4

2 tan 2sin2

1 tan 1

x t x

x t

So

2 2

2 4

4

cot tan 21 3 sin 2 2 1

1 31

x x t t t dx dt x t t

t

1

2 4

4

26 1

11

t t t dt t t

t

2

4 2 4

4

2 2 1

1 6 1

1

t dt

t t t

t

2

4 2

12

1 6

t dt

t t

2

4 2 2

12

2 4 1

t dt

t t t

2

22 2

12

1 4

t dt

t t

2

22 2

12

1 4

t dt

t t

2

2

22 2 2

11

2

1 4

t t

dt

t t t

2

22 2

12

1 4

t dt

t t

2

22 2 2

12

1 4

t dt

t t t

2

22 2

12

1 4

t dt

t t

2

21

12

4

t dt

t t

1

arctan2

t t c

tan cotarctan

2

x x c


34/56

Exercise:

Solve the following 4 28 cos 1 8 cos sin 2 where 0,2

Solution:

222

4 2 1 cos 21 cos 2

cos cos2 4

hence 2

24 1 cos 28 cos 8 2 1 cos 24

so 2 22 1 cos 2 1 8 cos sin 2

2 22 1 cos 2 2 cos 2 1 8 cos sin 2

2 1 cos 22 2 cos 2 4 cos2 1 8 sin 22

23 2 cos 2 4 cos 2 4 4 cos 2 sin2 21 2 cos 2 sin 2 0

22 cos 2 1 sin 2 22 cos 2 2 1 sin 2 22 1 cos 2 1 sin 2 22 sin 2 1 sin 2 22 sin 2 sin 2 1 0 Let 2sin 2 2 1 0t t t

2 1 1 02 2

t t 1

12

t or

so1

sin 2 12

or

sin 2 sin sin2 6

or

2 2 2 22 6

k or k

4 12

k or k

on the other hand we

have 2 2 2 22 6

k or k

7

4 12k or k

where k

thus the solution set is7

, ,4 12 12

Exercise:

Solve in , 3 3 7x y and 1x y

Solution: we have 3 3 2 2x y x y x xy y but 3 3 2 21x y x y x y xy

but 22 2 2 2

2 2 2 3x y xy x y xy xy xy x y xy

3 3 67 1 3 7 23

x y xy xy xy so 22

1 2 0x x x x

2

2 1 1 1 9 1 3 1 32 0 2 14 4 2 4 2 2 2

x x x x x x or

thus , 2, 1 , 1,2x y


35/56

Exercise:

Solve in 2 3 2 3 4x x

Solution: Let 1 1 2 3 2 3

2 3 2 34 32 3 2 3

t t

thus 2 3t and

12 3

t so 1 4

x x

t t

14

x

x t

t

let x

y t 22 1 4 0 2 3 2 3y y y y

if 22 3 2 3 2 3x x

y t t 2x

t t 12

x 2x

If 12 1

2 3 2 3

x x

y t t t t

2x

Exercise:

Find the General solution for 2' x y xy x e

Solution: we have 2' x y xy x e multiply by

2

2

x

e we get

2 2 2

22 2 2'x x x

x e y xe y x e e

2 2 2 2 2

2 22 2 2 2 2

'

'

1 1 1 1x x x x x

x x x x

v v

v u

d e y x e x e x x e e

dx

2 2

2 21x x

x d d e y x e

dx dx

so 2 2

2 21x x

x

e y x e c

2 2 2

2 2 21x x x

x

y x e e ce

2

21x

x y x e ce


36/56

Exercise:

Solve the following system

2 4 4

9 3 9

16 16 4

log log log 2

log log log 2

log log log 2

x y z

x y z

x y z

Solution: (1) 2 4log log 2x yz , (2) 9 3log log 2xz y and (3) 16 4log log 2xy z

(2)9 3 3 3

1log log 2 log log 2

2xz y xz y 3 3log 2 log 4xz y

23 3

log 4 log 3xzy

2 81xzy (*) so (1) 2 42 4 2 2 2 2

log log 2 2 log log 4 log log 2x yz x yz x yz

2 16x yz (**) so (3)16 4 4 4

log log 2 log 2 log 4xy z xy z

2 44 4 4

log 4 log 4 log 4xyz 2 256xyz (***)2

2

(* * *) 256: 16

(**) 16

xyz z

x x yz

So 22 2 2 2 8181 16 81 16 81

16xzy x x y x y xy

9

4xy

Note that loga u x is defined in only when 0u x thus , & 0x y z hence 0xy so

9

4xy from (***)

2

2 2 2 29 256 4 32 32256 2564 9 3 3

xyz z z z z

Now back

32

32 231616 16 48 3z z x x x so 9 2 9

4 3 4xy y 9 3 27

4 2 8y


37/56

Exercise:

Determine the value of x that satisfied 1 11

sin sin cos 15

x

Solution: Let 1 1 1

sin sin5 5

and let 1cos cosx x

So sin 1 sin2

2

Now expanding sin sin cos sin cos 1 1

sin cos 15

x

But by Pythagoras theorem we get1 24

cos 125 5

and 2sin 1 x

224 115 5

x x

2

24 1 5x x 2 2

5 24 1x x

2 2 210 25 24 24 25 10 1 0x x x x x 2225 2 5 1 0 5 1 0x x x

15 1 0

5x x

Exercise:

Solve in , 3 3 56x y

Solution: First we factorize 3 3 2 2

x y x y x xy y so3 3

56x y

2 2 56x y x xy y but56 2 28 2 2x y x y

So 2 22 2 28y y y y 2 2 24 4 2 28y y y y y

2 23 6 24 0 2 8 0y y y y 2 2 1 1 8 0y y

2

1 9 0 1 3y y thus 3 1 2 4y or y

therefore , 4,2 , 2, 4x y


38/56

Exercise:

Show that,0

1 sinarctan 0

cos

x dx

x

Solution: 20 0

2

1 sin 1 sin 1 sinarctan arctan arctan

cos cos cos

x x x dx dx dx

x x x

Let1 sin 1

arctan & &cos 2

x u dv dx du dx v x

x

so2 2

20

0

1 sin 1 sinarctan arctan

cos cos 4

x x x dx x

x x

21 sin / 2 / 4arctan

2 4cos / 2

21 1arctan

2 0 16

but

2 22

1 sin 1 sinlim arctan arctan lim arctan

cos cos 2x x

x

x x

x x

thus 2

21 0

3

16I f x

on the other hand 2

22

2

1 sinarctan

cos 4

x x I f x dx x

x

But 1 sin

lim arctan arctan 1cos 4x

x

x

2

1 sinlim arctan arctan

cos 2x

x

x

So

2 2 2 2 2

2

3

4 4 4 16 16I

so 0 0 f x dx

where 1 sin

arctan cos

x f x x

Even and odd functions:

A function is said to be odd if f x f x then

the element of symmetry is O and 0a

a f x dx

A function is said to be even if f x f x then

The element of symmetry is 'y oy and 02a a

a f x dx f x dx Properties:

1) The product of even and odd function is odd

2) The product of two odd functions is always even

3) The product of two even functions is always even

4) Any scalar multiplied by odd respectively even is odd respectively even


39/56

Properties:

1) The sum of two even functions is even

2) The sum of two odd functions is odd

Exercise:

Compute,

2 2

4

1 1

1

x x

dx x

Solution: we have 4 2 21 1 1x x x

2 2 2 2 2 2

4 2 2 2 2 2 2

1 1 1 1 1 1

1 1 1 1 1 1 1

x x x x x x dx dx dx dx

x x x x x x x

2 2

1 1

1 1

dx dx

x x

2

1arcsin

1

d x dx

x

arcsin arcsinhd x d x arcsin arcsinhx x c

Remark that 2 2

1 1arcsin & arcsinh

1 1

d d x x

dx dx x x

Exercise:

Compute,20

sin

1 cos

x x dx

x

Solution: 2 20 0

sinsin1 cos 1 cos

x x x x dx dx x x

20 sin1 cos

x x dx x

2 20 0

sin sin

1 cos 1 cos

x x x dx dx

x x

2 20 0

sin sin2

1 cos 1 cos

x x x dx dx

x x

20

sin

1 cos

x dx

x

Let cos sinu x du x dx so

2 20 0

sin

1 cos 1

u

u

x du dx

x u

1

21 1

du

u

1

1arctan arctan 1 arctan 1

4 4 2u

So2

20

sin

2 2 41 cos

x x dx

x


40/56

Exercise:

Compute,cos sin

2 sin 2

x x dx

x

Solution: we have 2 2

1

2 sin 2 2 2 cos sin 1 2 cos sin cos sinx x x x x x x

2

1 cos sinx x 2

cos sin cos sin

2 sin 2 1 cos sin

x x x x

dx dx x x x

Take cos sin sin cosu x x du x x dx

Thus 2

cos sinarctan arctan cos sin

2 sin 2 1

x x du dx u c x x c

x u

Notation:

n

n powers

a a a a a

& ....n

n times

a a a a

where n

Example: 5

553 312555 5 55

where as 35 5 5 5 125

Example: Perform3

3

3

3

Solution:33 2733 33 so

3 2727 3 24

3 3

3 33 3

3 3

Exercise:

Solve in , 723 1x x

Solution: we have 72 03 1 3x x 72 0x x , let 2u x u x

272 0u u Now finding the roots 2 2 1 1

72 0 72 04 4

u u u u

21 1 288

2 4u

1 17 1 179 8

2 2 2u u u or

9 8

0 0

u

E

So 9 8u but 0u 20 8 0 64 0 64u u x

also its acceptable if72 00 3 3x as 72 0 So 0 64x


41/56

Exercise:

Show that,0

tan 4 1lim

1 cos 4 2x

x x

x

Solution: we have

2 2

tan 4 1 cos 4 tan 4 1 cos 4tan 4 1 cos 4

1 cos 4 1 cos 4 1 cos 4 sin 4

x x x x x x x x x

x x x x

So 20 0 0

tan 4 tan 4lim lim lim 1 cos 4

1 cos 4 sin 4x x x

x x x x x

x x

but

20

tan4lim

sin 4x

x x

x

2

20

2

tan4lim

sin 4x

x x x

x

x

2 2 2 20 0 0

2

tan 4 tan 4 sin 44 1cos4lim lim lim

4sin 4 4 4sin4x x x

x x x

x x x x

x x

x x

we know that

0 0sin sinlim lim

x u ax u a

x u

a

Let sin4cos4x f x x x 0limx f x 0 01lim lim sin 4cos4x x x x x

now let 44

y y x x so as 0, 0x y

0 0 0 0 0

1 1 sinlim lim lim sin lim lim 1 4

coscos

4 4

x y y y y

y f x y

y y y y

Thus0

tan 4 1 2 1lim 21 cos 4 4 4 2x x x

x

Exercise:

Show that,sin2 2

limsin 3 3x

x

x

Solution: we have 0x x take 0y x y

So

0

sin2sin2lim lim

sin3 sin3x y

y x

x y

2 30 0

sin 2 2 2cos sinlim lim

sin 3 3 3 cos sin siny y

y y y

y y y y

0

sin2lim

sin3y

y

y

0

sin2

2lim

sin 3 3y

y

y

y

y


42/56

Exercise:

Let a such that 0a which satisfies2

2

3

4ln 4 ln 1

3

a

a

a x dx

x

, Show that 1a


2 2ln ln ln lna a x

x a x x x x

ln lna x a x x

ln ln lna x a x x

So 2

2 2 2 2

3 3 3 3ln ln ln ln

a a a a

a a a a

a x dx a x dx a x dx x dx

x

Let ln & &dx

u a x dv dx du v x a x

Hence 2

3ln

a

a a x dx

2 2

33ln

a a

a a

x x a x dx

a x

2 2

33ln

a a

a a

x a x a x dx

x

a

a

But 2 2 2 2233 3 3 3

a a a a a

a a a a a

a x a a dx dx dx dx x a

a x a x a x

So 22

3 3ln ln ln

a a

a a a x dx x a x x a a x

2 ln 3 2 ln 3 3 ln 4 3 ln 4a a a a a a a a a a

2 ln 3 2 ln 3 3 ln 4 3 ln 4a a a a a a a a a a

3 ln 3 4 ln 4a a a a a

Similarly we work for 22

3 3ln ln ln

a a

a a a x dx x a x x a a x

2 ln 2 ln 3 ln 2 3 ln 2a a a a a a a a a a

ln 2 ln 2a a a a a

2 2

33ln ln 2 ln 2 2 3 ln 3 3

a a

a a x dx x x x a a a a a a

So2

2

3ln

a

a

a x dx

x


43/56

Exercise:

Let 3 2g x ax bx cx d be numerical function

defined over By using the adjacent figure

Determine the values for , , &a b c d

Solution: Graphically the obtain data are :

1 0, 1 4, 0 2, 2 0g g g g

' 1 0 & ' 1 0g g

1 0 0g a b c d (1) 1 4 4g a b c d (2) 0 2 2g d

2 0 8 4 2 0g a b c d (3) but 2' 3 2g x ax bx c

' 1 0 3 2 0g a b c (4) and ' 1 0 3 2 0g a b c (5)

By using 2d we get : 8 4 2 2a b c , (1) 2a b c & 2a b c

8 4 2 2

2

2

a b c

a b c

a b c

2 0 0b b 8 2 2

2 (2)

a c

a c

8 2 2

2 2 4

a c

a c

6 6 1a a & 3c thus 3 3 2g x x x

Exercise:

Let : f be a numerical function defined as

2 f x ax bx c

By using the adjacent figure determine the values for , &a b c

Solution: Graphically we obtain the following data

0 1 f , ' 1 0 f and 1, 1 2 f

So we have 3 equation of 3 unknowns are:

0 1 1 f c ….(1)

' 2 f x ax b ' 1 0 0 2 2b f a b a (2)

We also have point ( Local Max ) 2 a b c 2

2 1 1 22 2

b b bb b

(3)

Thus our function will be 2 2 1 f x x x


44/56

Exercise:

Compute,2

20

4

4

x x dx

x

where

Solution: we have

22 2

0 0 0 02 2 2 2

24

4 4 1 22

24 4 4 4

x x x x x x

dx dx dx dx

x x x x

Let 24 2u x du xdx 2du xdx 20 4 & 4u u

Put

2

2

41 1

( ) 42 2

1 (0) 4

4

2u

u

du I u du u

u

241

21

4

1

2I u

24 2

2 222

2 20 0 02

4 444

44

x x x I dx dx x dx

x x

Now let 2 sin 2 cos & arcsin2

x x dx d

24 4 sin 2 cos d

24 1 sin 2 cos d 22 cos 2 cos d 22 2 cos d 24 cos d

1

4 1 cos 2 2 2 cos22

d d d 2 sin2 c 2 arcsin sin 22x

c

2 arcsin 2 sin cos

2

x c

2 arcsin 2 cos

2

x x c

But 2 2 2 2 2sin cos 1 cos 1 sin cos 1 sin , but

22sin sin

2 4

x x so

22cos 1 4

4

x x

So 2 24 2 arcsin 2 42

x x dx x x c

thus

2 2

04 2 arcsin 2 4

2x dx

Therefore

22

0 2

4

4 1 2 2 arcsin 224

x x

dx x


45/56

Exercise:

Let 3,U V X by Using the Standard Euclidean Basis for 3

Drive the formula for ,U V in terms of , ,x y z

Solution: We have 3

U U a b c x y z

X

where 3

, ,a b c C

also 3V V x y z

X where 3, , C

So , ,U V a b c x y z x y z

since .,. is bilinear then

,, ,, ,, b ba a a x

by x y y y z x x y x z

, , ,c c c z x z y z z

a a a a a a x x x x x y y x x z z x

b b b b b by x x y y y y y y z z y

c c c c c c

z x x z z y y z z z z z

2 2 2 2

2 2

a a a a a a

x x x x x y xy y x yx x x

2 2a a a

x z xz z x zx

2 2bb b

y x yx x y xy

2 2 2 2

2 2

b bb b b b

y y y y y z yz z y zy y y

2 2 2 2c c

c c c c z x zx x z xz z y zy y z yz

2 2

2 2

c c c

z z z z z z


46/56

a a a ba a a b

x x x y x z y x x x x x x y z y

y

b b

y z b

y z y b

y y

x

c c

z y c

z y z c

x z

c c

z z z z

a a a b b ba b c a b c

x x y z y z x x x y y z z

c c c

a b c x y x y z z z

a a a a b c

x y y x y z x

b b b

a b c x y y x y z y

c c c a b c

x y y x y z z

So ,U V V U a V U b V U c x y z

Exercise:

Solve this inequality in , 0.1 0.01log 2 log 2x x

Solution loga u x is defined when 0, 1 & 0a a u x thus 2x

We know thatln

loglna

bb

a so

0.01 2 1

ln 2 ln 2 ln 2log 2

ln0.01 ln10 2 ln10

x x x x

0.11

log 22

x thus 0.1 0.1 0.1 0.11

log 2 log 2 2 log 2 log 22

x x x x

2

0.1 0.1log 2 log 2x x

0 1 , log log 0x x x f x g x f x g x


47/56

2

2 2 2 2 1 0x x x x 2 3 0x x so the roots are

2 3x or x using the sign table to obtain2 3x

2 3

2 0

3 0

0 0

x

x

x

P

Exercise:

Show that,2

23

4 1lim

6

n

n i

i

i

Solution: 2

2 23 3 3 3 3

2 24 2 2 2 2n n n n n

i i i i i

i i i i i i i

i i i i i i

1 2 3 4 5 6 7 8... .....

3 4 5 6 3 4 5 6

5 3

1 2 2 2

3 4

n n

i i

i i

i i

but

3 5

2

2

n n

i i

i i

i i

5 5

2 2 1 11

12 2 6 6

n n

i i

i i

i i

so

2

23

4 1lim

6

n

n i

i

i

Exercise:

If 8 27m then find the value of 4m

Solution: We have 3

33 3 338 27 2 3 2 3 2 3m

m m m S.B.S 4 9m


48/56

Exercise:

Solve the following linear system using Gauss Jordan Elimination

1 2 3

1 2 3

1 2 3

2 3 3 15

3 2 5 19

5 4 2 2

x x x

x x x

x x x

Solution: Let 31 2 3 2 2, , , & 15,19, 2X x x x A M R b then AX b

1

2 2 1

3 3 1

2 3 3 : 152

3 2 5 : 19

exercises font latex

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