Mth001 Assignment

MTH001 - Elementary Mathematics first assignment fall 2010

MTH001 - Elementary Mathematics first assignment Solution November fall 2010
Using a truth table determine whether the following argument is VALID or INVALID:

If you practice a lot, then you get very good marks in Mathematics.
You did not practice a lot.
Therefore, you did not get very good marks.

Also formulate the above arguments symbolically and mention critical row (or rows) in the truth table.

Question2)
a)
For the conditional:
“If a>3 then a>0”. Determine the truthness of its converse.
b)
Justify the truthness of the biconditional: a> 0<=>a^2 > 0.

Question3)
Let A = {1, 2, 3, 4} and R = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (3, 3), (4, 1), (4, 4)}
be a binary relation on A. Then
(a) Decide whether the given relation R is reflexive, irreflexive, symmetric, and/or transitive. Justify with reasons.
(b) Draw the directed graph of R.
(c) Write matrix representation of the given relation R on set A

Mth202 Online Quiz No. 1

Online Quiz Announcement

Quiz will cover lecture no. 1 to 13

Schedule

Opening Date and Time 

November 03, 2010 At 12:01 AM (Mid-Night)  

 

Closing Date and Time 

November 04, 2010 At 11:59 PM (Mid-Night)

24 hours extra time is not available

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MTH301 Assignment No. 1 solution

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Mth301 Assignment No. 1 solution

Mth202 Assignment 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

Assignment #1

 MTH101 (Fall 2010)

                                                                                              Maximum Marks: 30                                                                                       

                                                                                                     Due Date: Nov 03, 2010

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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