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Explanation:

f(–r) = 1 + (–r) = 1 – r < 0 (∵ r ≥ 2)

f2(–r) = f(f(–r) = f(1 – r) = 1 + (1 – r) = 2 – r

f2(–r) = 2 – r = 0,     if r = 2

                   < 0,     if r > 2

Similarly,

f3(–r) = 3 – r = 0,     if r = 3

                   < 0,     if r > 3

and so on.

i.e. fr–1(–r) < 0 and fr(–r) = 0

∴ fr+1(–r) = 1/(1+0) = 1

∴ fr – 1 (–r) +  fr (–r) + f r + 1 (–r)  = –1 + 0 + 1 = 0.

Hence, option (b).

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