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If the function f : N → N is defined by f (x) = \(\sqrt{x}\), then\(\frac{f(25)}{f(16)+f(1)}\) is equal to (a) \(\frac{5}{6}\)(b) \(\frac{5}{7}\)(c) \(\frac{5}{3}\)(d) 1 |
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Answer» Answer : (d) 1 f : R → R given by f (x) = \(\sqrt{x}\) ∴ \(\frac{f(25)}{f(16)+f(1)} = \frac{5}{4+1} = \frac{5}{5}\) = 1. |
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