Solve
Guides
Join / Login
Use app
Login
0
You visited us
0
times! Enjoying our articles?
Unlock Full Access!
Standard XI
Mathematics
Question
If
x
+
1
<
0
then
|
x
|
is
0
x
+
1
−
x
−
x
−
1
x
A
0
B
−
x
−
1
C
−
x
D
x
E
x
+
1
Open in App
Solution
Verified by Toppr
x
+
1
<
0
, which gives
x
<
−
1
, therefore
x
<
0
Since
x
is negative, we get
|
x
|
=
−
x
>
0
.
Was this answer helpful?
0
Similar Questions
Q1
If:
f
(
x
)
=
x
(
e
1
/
x
−
e
−
1
/
x
)
e
1
/
x
+
e
−
1
/
x
,
.
.
.
.
x
≠
0
=
0
,
.
.
.
.
.
x
=
0
then
f
(
x
)
is
View Solution
Q2
If
f
(
x
)
=
e
1
/
x
−
1
e
1
/
x
+
1
,
x
≠
0
and
f
(
0
)
=
0
, then
f
(
x
)
is
View Solution
Q3
If
f
(
x
)
=
{
x
−
1
,
x
≥
1
2
x
2
−
2
,
x
<
1
,
g
(
x
)
=
{
x
+
1
,
x
>
0
−
x
2
+
1
,
x
≤
0
, and
h
(
x
)
=
|
x
|
, then
lim
x
→
0
f
(
g
(
h
(
x
)
)
)
is
View Solution
Q4
Let
f
(
x
)
=
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎩
(
1
+
a
x
)
1
x
,
if
x
<
0
b
,
if
x
=
0
(
x
+
c
)
1
3
−
1
(
x
+
1
)
1
2
−
1
,
if
x
>
0
is continuous at
x
=
0
. Then
View Solution
Q5
If
lim
x
→
0
[
1
+
x
+
f
(
x
)
x
]
1
x
=
e
3
, then the value of
l
n
⎛
⎜ ⎜ ⎜
⎝
lim
x
→
0
[
1
+
f
(
x
)
x
]
1
x
⎞
⎟ ⎟ ⎟
⎠
is .............
View Solution