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KEAM 2011 Math Paper
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© examsnet.com
Question : 45
Total: 120
If
y
=
(
x
+
1
)
(
x
+
2
)
(
x
+
3
)
(
x
+
4
)
(
x
+
5
)
,
then the value of
d
y
d
x
at
x
=
0
is equal to
374
742
472
247
274
Validate
Solution:
y
=
(
x
+
1
)
(
x
+
2
)
(
x
+
3
)
(
x
+
4
)
(
x
+
5
)
.....(i)
Taking log on both sides,
log
y
=
log
(
x
+
1
)
+
log
(
x
+
2
)
+
log
(
x
+
3
)
+
log
(
x
+
4
)
+
log
(
x
+
5
)
Differentiating w.r.t.
x
on both sides,
1
y
⋅
d
y
d
x
=
1
(
x
+
1
)
+
1
(
x
+
2
)
+
1
(
x
+
3
)
+
1
(
x
+
4
)
+
1
(
x
+
5
)
d
y
d
x
=
y
{
1
(
x
+
1
)
+
1
(
x
+
2
)
+
1
(
x
+
3
)
+
1
(
x
+
4
)
+
1
(
x
+
5
)
}
.......(ii)
[
∵
y
at
(
x
=
0
)
=
1
⋅
2
⋅
3
⋅
4
⋅
5
=
120
]
From Eq. (ii),
(
d
y
d
x
)
at
(
x
=
1
)
=
y
(
x
=
1
)
{
1
2
+
1
3
+
1
4
+
1
5
+
1
6
}
=
120
(
1
+
1
2
+
1
3
+
1
4
+
1
5
)
=
120
+
60
+
40
+
30
+
24
=
274
© examsnet.com
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