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TS EAMCET 19-July-2022 Shift 2 Solved Paper
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© examsnet.com
Question : 2
Total: 160
If
D
is the domain and
G
is the range of the real valued function
f
(
x
)
=
√
1
−
x
2
1
+
x
2
, then
D
∩
G
=
[
0
,
∞
)
[
0
,
1
]
[
0
,
1
2
]
[
−
1
,
1
]
Validate
Solution:
Given,
f
(
x
)
=
√
1
−
x
2
1
+
x
2
1
−
x
2
1
+
x
2
≥
0
⇒
1
−
x
2
≥
0
[
∵
1
+
x
2
>
0
]
⇒
x
2
≤
1
⇒
x
∈
[
−
1
,
1
]
∴
D
=
[
−
1
,
1
]
Let
y
=
1
−
x
2
1
+
x
2
y
+
x
2
y
=
1
−
x
2
⇒
x
2
(
y
+
1
)
=
1
−
y
⇒
x
2
=
1
−
y
1
+
y
⇒
x
=
±
√
1
−
y
1
+
y
⇒
1
−
y
1
+
y
≥
0
As we know that
y
≥
0
,
1
+
y
>
0
.
⇒
1
−
y
≥
0
y
≤
1
y
∈
[
0
,
1
]
G
=
[
0
,
1
]
D
∩
G
=
[
0
,
1
]
© examsnet.com
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