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Byju's Answer
Standard XII
Mathematics
Global Maxima
Consider the ...
Question
Consider the functions
f
(
x
)
=
{
x
+
1
,
x
≤
1
2
x
+
1
,
1
<
x
≤
2
g
(
x
)
=
{
x
2
,
−
1
≤
x
<
2
x
+
2
,
2
≤
x
≤
3
Domain of
f
(
g
(
x
)
)
is
A
[
−
1
,
2
]
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B
[
0
,
√
2
]
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C
[
√
2
,
√
2
]
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D
[
−
1
,
√
2
]
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Solution
The correct option is
D
[
−
1
,
√
2
]
f
(
g
(
x
)
)
=
{
g
(
x
)
+
1
g
(
x
)
≤
1
2
g
(
x
)
+
1
1
<
g
(
x
)
≤
2
Now from graph of
g
(
x
)
,
g
(
x
)
≤
1
⇒
x
∈
[
−
1
,
1
]
and
g
(
x
)
∈
(
1
,
2
]
⇒
x
∈
(
1
,
√
2
]
∴
f
(
g
(
x
)
)
=
{
g
(
x
)
+
1
,
x
∈
[
−
1
,
1
]
2
g
(
x
)
+
1
,
x
∈
(
1
,
√
2
]
=
{
x
2
+
1
,
x
∈
[
−
1
,
1
]
2
x
2
+
1
,
x
∈
(
1
,
√
2
]
So, domain of
f
(
g
(
x
)
)
=
[
−
1
,
√
2
]
Suggest Corrections
0
Similar questions
Q.
Consider the functions
f
(
x
)
=
{
x
+
1
,
x
≤
1
2
x
+
1
,
1
<
x
≤
2
g
(
x
)
=
{
x
2
,
−
1
≤
x
<
2
x
+
2
,
2
≤
x
≤
3
Domain of
f
(
g
(
x
)
)
is
Q.
Consider the functions
f
(
x
)
=
{
x
+
1
,
x
≤
1
2
x
+
1
,
1
<
x
≤
2
g
(
x
)
=
{
x
2
,
−
1
≤
x
<
2
x
+
2
,
2
≤
x
≤
3
Range of the function
f
(
g
(
x
)
)
is
Q.
Consider the functions
f
(
x
)
=
{
x
+
1
,
x
≤
1
2
x
+
1
,
1
<
x
≤
2
g
(
x
)
=
{
x
2
,
−
1
≤
x
<
2
x
+
2
,
2
≤
x
≤
3
The number of roots of the equation
f
(
g
(
x
)
)
=
2
is
Q.
f
(
x
)
=
{
e
−
1
/
x
2
,
x
>
0
0
,
x
≤
0
, then
f
(
x
)
is
Q.
Let
f
(
x
)
=
{
x
3
−
x
2
+
10
x
−
5
,
x
≤
1
−
2
x
+
log
2
(
b
2
−
2
)
,
x
>
1
the set of values of
b
for which
f
(
x
)
has greatest
value at
x
=
1
is given by:
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