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AP EAMCET Engineering 2013 Solved Paper
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© examsnet.com
Question : 90
Total: 160
1
2
.
3
+
1
4
.
5
+
1
6
.
7
+
1
8
.
9
+
.
.
.
is equal to
log
(
2
e
)
log
(
e
2
)
log
(
2
e
)
e
−
1
Validate
Solution:
1
2
.
3
+
1
4
.
5
+
1
6
.
7
+
1
8
.
9
+
.
.
.
∞
=
(
1
2
−
1
3
)
+
(
1
4
−
1
5
)
+
(
1
6
−
1
7
)
+
(
1
8
−
1
9
)
+
.
.
.
.
.
∞
=
1
−
log
e
2
=
log
e
e
−
log
e
2
=
log
(
e
2
)
© examsnet.com
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