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AIEEE 2004 Math Solved Paper
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© examsnet.com
Question : 33
Total: 68
A function
y
=
f
(
x
)
has a second order derivative
f
′
′
(
x
)
=
6
(
x
−
1
)
. If its graph passes through the point
(
2
,
1
)
and at that point the tangent to the graph is
y
=
3
x
−
5
, then the function is
(
x
+
1
)
2
(
x
−
1
)
3
(
x
+
1
)
3
(
x
−
1
)
2
Validate
Solution:
f
′
′
(
x
)
=
6
(
x
−
1
)
. Integrating
we get
f
′
(
x
)
=
3
x
2
−
6
x
+
c
Slope at
(
2
,
1
)
=
f
′
(
2
)
=
c
=
3
[ As slope of tangent at
(
2
,
1
)
is 3 ]
∴
f
′
(
x
)
=
3
x
2
−
6
x
+
3
=
3
(
x
−
1
)
2
Integrating again, we get
f
(
x
)
=
(
x
−
1
)
3
+
D
The curve passes through
(
2
,
1
)
⇒
1
=
(
2
−
1
)
3
+
D
⇒
D
=
0
∴
f
(
x
)
=
(
x
−
1
)
3
© examsnet.com
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