If f(x)+f(1-x)=2, then the value of f(\frac{1}{2001})+f(\frac{2}{2001})+. . .+f(\frac{2000}{2001}) is

NIMCET 2008 Question Paper with solutions

Question : 8 of 120

Marks: +1, -0

f(x)+f(1-x)=2,

then the value of

f(\;{1}/{2001})+f(\;{2}/{2001})+. . .+f(\;{2000}/{2001})

Solution:

f(x)+f(1-x)=2

⇒ f( {1}/{2001})+f( {2000}/{2001})=2

⇒ f( {2}/{2001})+f( {1999}/{2001})=2

⇒ f( {1000}/{2001})+t( {1001}/{2001})=2

which are 1000 pairs in all.
So,

f( {1}/{2001})+f( {2}/{2001})+. . .+t( {2000}/{2001})=2000

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