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NIMCET 2008 Question Paper with solutions
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© examsnet.com
Question : 8 of 120
Marks:
+1
,
-0
If
f
(
x
)
+
f
(
1
−
x
)
=
2
,
f(x)+f(1-x)=2,
f
(
x
)
+
f
(
1
−
x
)
=
2
,
then the value of
f
(
1
2001
)
+
f
(
2
2001
)
+
⋯
+
f
(
2000
2001
)
f\;\left(\frac{1}{2001}\right)+f\;\left(\frac{2}{2001}\right)+\cdots+f\;\left(\frac{2000}{2001}\right)
f
(
2001
1
)
+
f
(
2001
2
)
+
⋯
+
f
(
2001
2000
)
is
2000
2001
1999
1998
Validate
Solution:
f
(
x
)
+
f
(
1
−
x
)
=
2
f(x)+f(1-x)=2
f
(
x
)
+
f
(
1
−
x
)
=
2
⇒
f
(
1
2001
)
+
f
(
2000
2001
)
=
2
\Rightarrow f\;\left(\frac{1}{2001}\right)+f\;\left(\frac{2000}{2001}\right)=2
⇒
f
(
2001
1
)
+
f
(
2001
2000
)
=
2
⇒
f
(
2
2001
)
+
f
(
1999
2001
)
=
2
\Rightarrow f\;\left(\frac{2}{2001}\right)+f\;\left(\frac{1999}{2001}\right)=2
⇒
f
(
2001
2
)
+
f
(
2001
1999
)
=
2
⇒
f
(
1000
2001
)
+
t
(
1001
2001
)
=
2
\Rightarrow f\;\left(\frac{1000}{2001}\right)+t\;\left(\frac{1001}{2001}\right)=2
⇒
f
(
2001
1000
)
+
t
(
2001
1001
)
=
2
which are 1000 pairs in all.
So,
f
(
1
2001
)
+
f
(
2
2001
)
+
⋯
+
t
(
2000
2001
)
=
2000
f\;\left(\frac{1}{2001}\right)+f\;\left(\frac{2}{2001}\right)+\cdots+t\;\left(\frac{2000}{2001}\right)=2000
f
(
2001
1
)
+
f
(
2001
2
)
+
⋯
+
t
(
2001
2000
)
=
2000
© examsnet.com
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