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KVPY SX SB Exam 2015 Question Paper
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© examsnet.com
Question : 9
Total: 120
Let p(x) be a polynomial such that
p
(
x
)
–
p
′
(
x
)
=
x
n
, where n is a positive integer. Then p(0) equals -
n
!
(
n
−
1
)
!
1
n
!
1
(
n
−
1
)
!
Validate
Solution:
Let
P
(
x
)
=
a
0
x
n
+
a
1
x
n
−
1
+
a
2
x
n
−
2
+
…
+
a
n
P
′
(
x
)
=
n
a
0
x
n
−
1
+
(
n
–
1
)
a
1
x
n
−
2
+
…
+
a
n
−
1
P
(
x
)
P
′
(
x
)
=
a
0
x
n
+
(
a
1
−
n
a
0
)
x
n
−
1
+
(
a
2
−
(
n
−
1
)
a
1
)
x
n
−
2
+
…
+
(
a
n
−
a
n
−
1
)
given
P
(
x
)
−
P
′
(
x
)
=
a
0
x
n
.
so that
a
1
−
n
a
0
=
0
a
1
a
0
=
n
a
2
−
(
n
−
1
)
a
1
=
0
a
2
a
1
=
(
n
−
1
)
a
n
−
a
n
−
1
=
0
a
n
a
n
−
1
=
1
P
(
0
)
=
a
n
(
a
n
a
n
−
1
)
(
a
n
−
1
a
n
−
2
)
.
.
.
(
a
1
a
0
)
= 1 × 2 × 3 × …× n
= n!
© examsnet.com
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