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
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