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