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Previous Year Question (PYQs)
1
If $f(x)+f(1-x)=2$, then the value of
$f\left(\dfrac{1}{2001}\right)+f\left(\dfrac{2}{2001}\right)+\cdots+f\left(\dfrac{2000}{2001}\right)$ is
Solution
Terms pair as
$f\left(\frac{k}{2001}\right)+f\left(1-\frac{k}{2001}\right)=2$
There are $1000$ such pairs.
Sum $=1000\times2=2000$
Answer: $\boxed{2000}$
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