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Mathematics
Problem
The function defined by
f
(
x
)
=
{
(
x
2
+
e
1
2
−
x
)
−
1
,
x
≠
2
k
,
x
=
2
f(x) = \begin{cases} \left( x^2 + e^{\frac{1}{2-x}} \right)^{-1}, & x \neq 2 \\ k, & x = 2 \end{cases}
f
(
x
)
=
⎩
⎨
⎧
(
x
2
+
e
2
−
x
1
)
−
1
,
k
,
x
=
2
x
=
2
from right at the point
x
=
2
x = 2
x
=
2
, then
k
k
k
is equal to (a) 0 (b)
1
/
4
1/4
1/4
(c)
−
1
/
4
-1/4
−
1/4
(d) None of these
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