04-14-2012, 05:23 PM
Idea Solution
Solve the two sided inequality and show the solution on real line7 < 1-2x ≤ 10Step 1: subtract 1 from both sides and we get6<-2x<9Step2: 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
--
|---|-----
|
-----|----|---0--------------------------------
-5 -4 -3 -2 -1
Question 2; Marks: 10
Given two functions as:f(x) =
2
1
x x
− −
and g(x) =3
x
Find fog(x) also find the domain of
f
,
g
and
fogSolution:
Fog(x)=f(g(x))fog(x)=(g(x))
2
-(g(x))-1fog(x)=(3/x)
2
-(3/x)-1fog(x)=(9/x
2
)-3/x-1fog(x)= -(x
2
-3x+9)/x
2
(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
3 232
56lim9
x
x x x x
→
− +−
Solution:
Lim(x->3) x(x
2
-5x+6) / x
2
-3
2
Lim(x->3) x(x
2
-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
Solve the two sided inequality and show the solution on real line7 < 1-2x ≤ 10Step 1: subtract 1 from both sides and we get6<-2x<9Step2: 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
--
|---|-----
|
-----|----|---0--------------------------------
-5 -4 -3 -2 -1
Question 2; Marks: 10
Given two functions as:f(x) =
2
1
x x
− −
and g(x) =3
x
Find fog(x) also find the domain of
f
,
g
and
fogSolution:
Fog(x)=f(g(x))fog(x)=(g(x))
2
-(g(x))-1fog(x)=(3/x)
2
-(3/x)-1fog(x)=(9/x
2
)-3/x-1fog(x)= -(x
2
-3x+9)/x
2
(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
3 232
56lim9
x
x x x x
→
− +−
Solution:
Lim(x->3) x(x
2
-5x+6) / x
2
-3
2
Lim(x->3) x(x
2
-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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