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If x ≠ 1 and \(f(x) = \frac{x+1}{x-1}\) is a real function, then \(fff\) (2) is(a) 1 (b) 2 (c) 3 (d) 4 |
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Answer» Answer : (c) 3 f(x) = \(\frac{x+1}{x-1}\) f f f (2) = f[f{f(2)}] = \(f\big[f\big(\frac{2+1}{2-1}\big)\big]\) = f [ f (3)] = f \(\big[\frac{3+1}{3-1}\big]\) = f(2) = \(\frac{2+1}{2-1}\) = 3 |
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