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Byju's Answer
Standard XII
Mathematics
Second Derivative Test for Local Maximum
f x = [ k - ...
Question
f
(
x
)
=
{
k
−
2
x
,
i
f
x
≤
−
1
2
x
+
3
,
i
f
x
>
−
1
}
,
if f has a local minimum at x=-1,then k =
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Solution
lim
x
→
1
+
f
(
x
)
=
1
f
(
−
1
)
=
k
+
2
Answer
f
has a local minimum at
x
=
−
1
f
(
−
1
+
)
≥
f
(
−
1
)
≥
f
(
−
1
−
)
⇒
1
≥
k
+
2
⇒
k
+
2
≤
1
∵
k
≤
−
1
k
=
−
1
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0
Similar questions
Q.
defined by
f
(
x
)
=
{
k
−
2
x
,
i
f
x
≤
1
2
x
+
3
,
i
f
x
>
−
1
}
,if has a local minimum at x= -1, then a pair
Q.
Let
f
:
R
→
R
be defined by
f
(
x
)
=
{
k
−
2
x
,
i
f
x
≤
−
1
2
x
+
3
,
i
f
x
>
−
1
If
f
has a local minimum at
x
=
−
1
, then the possible value of
k
is
Q.
Let f:
R
→
R
be defined by
f
(
x
)
=
{
k
−
2
x
,
i
f
x
≤
−
1
2
x
+
3
,
i
f
x
>
−
1
. If f has a local minimum at x = -1, then a possible value of k is
Q.
Let
f
:
R
→
R
be defined by
f
(
x
)
=
{
k
−
2
x
,
i
f
x
≤
−
1
2
x
+
3
,
i
f
x
>
−
1
be continous. then find possible value of
k
is
Q.
If
f
(
x
)
=
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎩
1
−
cos
4
x
x
2
,
i
f
x
<
0
a
,
i
f
x
=
0
√
x
√
16
+
√
4
−
4
,
i
f
x
>
0
, then what value of
a
,
f
is continuous at
x
=
0
?
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Standard XII Mathematics
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