if the relation R on the set of natural numbers N is defined as R = {(a,b}:a+5b=27,a€N ,b€N}then find R1 R-1 ,domain of R and range of R
Answers
Answer:
The domain is
x
∈
(
−
2
,
2
)
. The range is
[
1
2
,
+
∞
)
.
Explanation:
The function is
f
(
x
)
=
1
√
4
−
x
2
What'under the
√
sign must be
≥
0
and we cannot divide by
0
Therefore,
4
−
x
2
>
0
⇒
,
(
2
−
x
)
(
2
+
x
)
>
0
⇒
,
{
2
−
x
>
0
2
+
x
>
0
⇒
,
{
x
<
2
x
>
−
2
Therefore,
The domain is
x
∈
(
−
2
,
2
)
Also,
lim
x
→
2
−
f
(
x
)
=
lim
x
→
2
−
1
√
4
−
x
2
=
1
O
+
=
+
∞
lim
x
→
−
2
+
f
(
x
)
=
lim
x
→
−
2
+
1
√
4
−
x
2
=
1
O
+
=
+
∞
When
x
=
0
f
(
0
)
=
1
√
4
−
0
=
1
2
The range is
[
1
2
,
+
∞
)
graph{1/sqrt(4-x^2) [-9.625, 10.375, -1.96, 8.04]}
Domain is {2,7,12,17,22} and range is {5,4,3,2,1}
Step-by-step explanation:
Given relation,
R = {(a,b}:a+5b=27,a∈N ,b∈N}
Solve the equation a+5b=27 for b.
5b=27-a
Since b∈N, so,
Also, a∈N, thus a ≥ 1.
So, check natural values of a from 1 to 22,
We observed that only at a=2,7,12,17 and 22, b is a natural number.
And has values 5,4,3,2 and 1 respectively.
Thus, R = {2,5),(7,4),(12,3),(17,2),(22,1)}
Now, domain is the set of all possible input values,
Hence, domain = {2,7,12,17,22}
Also, range is the set of all possible output values,
Hence, range = {5,4,3,2,1}
#Learn more:
Show that the relation xy=-2 is a function for a suitable domain find its domain and range
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