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Problems
f
(
x
)
=
{
(
e
x
2
−
1
)
sin
(
ln
(
1
+
2
x
)
)
tan
6
(
2
x
)
,
x
≠
0
k
,
x
=
0
f(x) = \begin{cases} \frac{(e^{x^2} - 1) \sin(\ln(1 + 2x))}{\tan^6(2\sqrt{x})}, & x \neq 0 \\ k, & x = 0 \end{cases}
f
(
x
)
=
{
t
a
n
6
(
2
x
)
(
e
x
2
−
1
)
s
i
n
(
l
n
(
1
+
2
x
))
,
k
,
x
=
0
x
=
0
…
Mathematics
Problem
f
(
x
)
=
{
(
e
x
2
−
1
)
sin
(
ln
(
1
+
2
x
)
)
tan
6
(
2
x
)
,
x
≠
0
k
,
x
=
0
f(x) = \begin{cases} \frac{(e^{x^2} - 1) \sin(\ln(1 + 2x))}{\tan^6(2\sqrt{x})}, & x \neq 0 \\ k, & x = 0 \end{cases}
f
(
x
)
=
{
t
a
n
6
(
2
x
)
(
e
x
2
−
1
)
s
i
n
(
l
n
(
1
+
2
x
))
,
k
,
x
=
0
x
=
0
The value of
k
k
k
for which
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
0
x = 0
x
=
0
, is (a)
1
2
4
\frac{1}{2^4}
2
4
1
(b)
1
2
5
\frac{1}{2^5}
2
5
1
(c)
1
2
3
\frac{1}{2^3}
2
3
1
(d)
1
2
6
\frac{1}{2^6}
2
6
1
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