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Mathematics
Problem
Let
f
(
x
)
=
e
(
p
+
1
)
x
−
e
x
f(x) = e^{(p+1)x} - e^x
f
(
x
)
=
e
(
p
+
1
)
x
−
e
x
for real number
p
>
0
p > 0
p
>
0
, then The value of
x
=
s
p
x = s_p
x
=
s
p
for which
f
(
x
)
f(x)
f
(
x
)
is minimum, is a.
−
log
e
(
p
+
1
)
p
\frac{-\log_e(p+1)}{p}
p
−
l
o
g
e
(
p
+
1
)
b.
−
log
e
(
p
+
1
)
-\log_e(p+1)
−
lo
g
e
(
p
+
1
)
c.
−
log
e
p
-\log_e p
−
lo
g
e
p
d.
log
e
(
p
+
1
p
)
\log_e\left(\frac{p+1}{p}\right)
lo
g
e
(
p
p
+
1
)
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