MTH301 Assignment No. 1 solution

Click on image for LARGE view

Mth301 Assignment No. 1 solution

MTH101 Assignment # 1 Solution

Question 1;                                                                                                          Marks:10        

   Solve the two sided inequality and show the solution on real line

7 < 1-2x ≤ 10

Step 1: subtract 1 from both sides and we get

6<-2x<9

Step2: divide by -2 on both sides and we get

-3>x>-5

(note: symbols are reversed with a negative operation)

So x can have a value between -3 and or equal to -5

ß--|---|-----|-----|----|---0--------------------------------à

     -5   -4     -3   -2   -1

 

Question 2;                                                                                                        Marks: 10

 Given two functions as:

f(x) =    and g(x) =  

Find fog(x) also find the domain of f, g  and  fog

Solution:

Fog(x)=f(g(x))

fog(x)=(g(x))²-(g(x))-1

fog(x)=(3/x)²-(3/x)-1

fog(x)=(9/x²)-3/x-1

fog(x)= -(x²-3x+9)/x²

(note: domain and ranges are to be found out yourselves, listen to lecture No. 6 for this)

Question 3;                                                                                                        Marks:10

Simplify, then apply the rules of limit to  evaluate

Solution:

Lim(x->3) x(x²-5x+6) / x²-3²

Lim(x->3) x(x²-3x-2x+6) / (x-3)(x+3)

Lim(x->3) x(x(x-3)-2(x-3)) / (x-3)(x+3)

Lim(x->3) x((x-3)(x-2)) / (x-3)(x+3)

Lim(x->3) x(x-2) /(x+3)

Now applying limits lim x->3

=3(3-2)/3+3

=3(1)/6

=3/6

=½

MTH101 Assignment # 1 Solution

Question 1;                                                                                                          Marks:10        

   Solve the two sided inequality and show the solution on real line

7 < 1-2x ≤ 10

Step 1: subtract 1 from both sides and we get

6<-2x<9

Step2: divide by -2 on both sides and we get

-3>x>-5

(note: symbols are reversed with a negative operation)

So x can have a value between -3 and or equal to -5

ß--|---|-----|-----|----|---0--------------------------------à

     -5   -4     -3   -2   -1

 

Question 2;                                                                                                        Marks: 10

 Given two functions as:

f(x) =    and g(x) =  

Find fog(x) also find the domain of f, g  and  fog

Solution:

Fog(x)=f(g(x))

fog(x)=(g(x))²-(g(x))-1

fog(x)=(3/x)²-(3/x)-1

fog(x)=(9/x²)-3/x-1

fog(x)= -(x²-3x+9)/x²

(note: domain and ranges are to be found out yourselves, listen to lecture No. 6 for this)

Question 3;                                                                                                        Marks:10

Simplify, then apply the rules of limit to  evaluate

Solution:

Lim(x->3) x(x²-5x+6) / x²-3²

Lim(x->3) x(x²-3x-2x+6) / (x-3)(x+3)

Lim(x->3) x(x(x-3)-2(x-3)) / (x-3)(x+3)

Lim(x->3) x((x-3)(x-2)) / (x-3)(x+3)

Lim(x->3) x(x-2) /(x+3)

Now applying limits lim x->3

=3(3-2)/3+3

=3(1)/6

=3/6

=½

MTH101 Assignment # 1 Solution

Assignment #1

 MTH101 (Fall 2010)

                                                                                              Maximum Marks: 30                                                                                       

                                                                                                     Due Date: Nov 03, 2010

DON’T MISS THESE: Important instructions before attempting the solution of this assignment:

 •     To solve this assignment, you should have good command over 1-10

lectures.

• Try to get the concepts, consolidate your concepts and ideas from these questions which you learn in the 01 to 10 lectures.

 •     Upload assignments properly through LMS, No Assignment will be accepted

        through email. 

 •     Write your ID on the top of your solution file.

• Don’t use colorful back grounds in your solution files.
• Use Math Type or Equation Editor etc for mathematical symbols.
• You should remember that if we found the solution files of some students are same then we will reward zero marks to all those students.
• Try to make solution by yourself and protect your work from other students, otherwise you and the student who send same solution file as you will be given zero marks.
• All steps are necessary to get full marks.
• Also remember that you are supposed to submit your assignment in Word format any other like scan images etc will not be accepted and we will give zero marks correspond to these assignments.

Question 1;                                                                                                           Marks:10        

  
Solve the two sided inequality and show the solution on real line

7 < 1-2x ≤ 10

Step 1: subtract 1 from both sides and we get

6<-2x<9

Step2: divide by -2 on both sides and we get

-3>x>-5

(note: symbols are reversed with a negative operation)

So x can have a value between -3 and or equal to -5

ß--|---|-----|-----|----|---0--------------------------------à

     -5   -4     -3   -2   -1

 

Question 2;                                                                                                         Marks: 10

 Given two functions as:

f(x) =    and g(x) =  

Find fog(x) also find the domain of f, g  and  fog

Solution:

Fog(x)=f(g(x))

fog(x)=(g(x))²-(g(x))-1

fog(x)=(3/x)²-(3/x)-1

fog(x)=(9/x²)-3/x-1

fog(x)= -(x²-3x+9)/x²

(note: domain and ranges are to be found out yourselves, listen to lecture No. 6 for this)

Question 3;                                                                                                         Marks:10

Simplify, then apply the rules of limit to  evaluate

Solution:

Lim(x->3) x(x²-5x+6) / x²-3²

Lim(x->3) x(x²-3x-2x+6) / (x-3)(x+3)

Lim(x->3) x(x(x-3)-2(x-3)) / (x-3)(x+3)

Lim(x->3) x((x-3)(x-2)) / (x-3)(x+3)

Lim(x->3) x(x-2) /(x+3)

Now applying limits lim x->3

=3(3-2)/3+3

=3(1)/6

=3/6

=½

                                                                                                                                                                                                                                 

MTH101 Assignment # 1 Solution

Assignment #1

 MTH101 (Fall 2010)

                                                                                              Maximum Marks: 30                                                                                       

                                                                                                     Due Date: Nov 03, 2010

DON’T MISS THESE: Important instructions before attempting the solution of this assignment:

 •     To solve this assignment, you should have good command over 1-10

lectures.

• Try to get the concepts, consolidate your concepts and ideas from these questions which you learn in the 01 to 10 lectures.

 •     Upload assignments properly through LMS, No Assignment will be accepted

        through email. 

 •     Write your ID on the top of your solution file.

• Don’t use colorful back grounds in your solution files.
• Use Math Type or Equation Editor etc for mathematical symbols.
• You should remember that if we found the solution files of some students are same then we will reward zero marks to all those students.
• Try to make solution by yourself and protect your work from other students, otherwise you and the student who send same solution file as you will be given zero marks.
• All steps are necessary to get full marks.
• Also remember that you are supposed to submit your assignment in Word format any other like scan images etc will not be accepted and we will give zero marks correspond to these assignments.

Question 1;                                                                                                           Marks:10        

  
Solve the two sided inequality and show the solution on real line

7 < 1-2x ≤ 10

Step 1: subtract 1 from both sides and we get

6<-2x<9

Step2: divide by -2 on both sides and we get

-3>x>-5

(note: symbols are reversed with a negative operation)

So x can have a value between -3 and or equal to -5

ß--|---|-----|-----|----|---0--------------------------------à

     -5   -4     -3   -2   -1

 

Question 2;                                                                                                         Marks: 10

 Given two functions as:

f(x) =    and g(x) =  

Find fog(x) also find the domain of f, g  and  fog

Solution:

Fog(x)=f(g(x))

fog(x)=(g(x))²-(g(x))-1

fog(x)=(3/x)²-(3/x)-1

fog(x)=(9/x²)-3/x-1

fog(x)= -(x²-3x+9)/x²

(note: domain and ranges are to be found out yourselves, listen to lecture No. 6 for this)

Question 3;                                                                                                         Marks:10

Simplify, then apply the rules of limit to  evaluate

Solution:

Lim(x->3) x(x²-5x+6) / x²-3²

Lim(x->3) x(x²-3x-2x+6) / (x-3)(x+3)

Lim(x->3) x(x(x-3)-2(x-3)) / (x-3)(x+3)

Lim(x->3) x((x-3)(x-2)) / (x-3)(x+3)

Lim(x->3) x(x-2) /(x+3)

Now applying limits lim x->3

=3(3-2)/3+3

=3(1)/6

=3/6

=½

                                                                                                                                                                                                                                 

Mth603 Assignment Solution

Q. No.  1.   (Marks 10)

Use the Method of False Position to find the solution accurate to within 10-4  for the following problem.

x - 0.8 - 0.2 sin x = 0;

é0, p ù

ê    2 ú

ë       û

x0	0	f ( x0 )	-0.8
x1	1.570796	f ( x1 )	0.570796
x  = x  -       x1  - x0                f ( x ) 2           1          f ( x ) -  f ( x  )      1 1                        0	0.916721	f ( x2 )	-0.042001
x  = x  -       x2  - x1               f ( x  ) 3            2          f ( x  ) -  f ( x )      2 2                       1	0.961551	f ( x3 )	-0.002465
x  = x  -       x3  - x1                f ( x  ) 4            3          f ( x  ) -  f ( x )      3 3                       1	0.964346	f ( x4 )	0.000011
x  = x  -       x4  - x1               f ( x  ) 5            4           f ( x  ) -  f ( x )      4 4                       1	0.964334	f ( x5 )	0
x  = x  -       x5  - x1                f ( x  ) 6            5          f ( x  ) -  f ( x )      5 5                       1	0.964334	f ( x6 )	0

Q. No.  2.    (Marks 10)

Solve  ex  - 3x2  = 0  for  0 £ x £ 1 and  3 £ x £ 5 by using the Secant Method up to four iterations. (Note: Accuracy up to four decimal places is required)

x0	0	f ( x0 )	1
x1	1	f ( x1 )	-0.2817
x  = x0  f ( x1 ) - x1 f ( x0 ) 2                f ( x ) -  f ( x  ) 1                       0	0.7802	f ( x2 )	0.3558
x  = x1 f ( x2 ) - x2  f ( x1 ) 3                f ( x  ) -  f ( x ) 2                       1	0.9029	f ( x3 )	0.0211
x  = x2  f ( x3 ) - x3  f ( x2 ) 4                 f ( x  ) -  f ( x  ) 3                       2	0.9106	f ( x4 )	-0.0018
x  =  x3  f ( x4 ) - x4  f ( x3 ) 5                 f ( x  ) -  f ( x  ) 4                       3	0.91	f ( x5 )	0

x0	3	f ( x0 )	-6.9145
x1	5	f ( x1 )	73.4132
x  = x0  f ( x1 ) - x1 f ( x0 ) 2                f ( x ) -  f ( x  ) 1                       0	3.1722	f ( x2 )	-6.3286
x  = x1 f ( x2 ) - x2  f ( x1 ) 3                f ( x  ) -  f ( x ) 2                       1	3.3173	f ( x3 )	-5.4277
x2  f ( x3 ) - x3  f ( x2 ) x4  = f ( x3 ) -  f ( x2 )	4.1915	f ( x4 )	13.4159
x  =  x3  f ( x4 ) - x4  f ( x3 ) 5                 f ( x  ) -  f ( x  ) 4                       3	3.5691	f ( x5 )	-2.7308