If f(x) = 1 - f(1-x) and then f(1100)+f(2100)+.....+f(98100)+f(99100).

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If f(x) = 1 - f(1-x) and then

f (\frac{1}{100}) + f (\frac{2}{100}) + . . . . . + f (\frac{98}{100}) + f (\frac{99}{100}) .

Answer: Option B
:
B

f (x) = 1 - f (1 - x) \Rightarrow f (x) + f (1 - x) = 1

f (\frac{1}{2}) = 1 - f (\frac{1}{2}) \Rightarrow f (\frac{1}{2}) = \frac{1}{2}

f (\frac{99}{100}) = f (1 - \frac{1}{100}) = 1 - f (\frac{1}{100})

f (\frac{98}{100}) = f (1 - \frac{2}{100}) = 1 - f (\frac{2}{100})

So,

f (\frac{1}{100}) + f (\frac{2}{100}) + . . . . . + f (\frac{98}{100}) + f (\frac{99}{100})

= f (\frac{1}{100}) + f (\frac{2}{100}) + . . . . . + 1 - f (\frac{2}{100}) + 1 - f (\frac{2}{100})

[f (\frac{50}{100})] = f (\frac{1}{2}) = \frac{1}{2}

= 49 + f (\frac{50}{100})

= 49 + \frac{1}{2} = 49.5

Hence option (b).

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