r is an integer > 2. Then, what is the value of f r – 1(–r) + f r(–r)+ f r + 1 (–r)?
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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