If f(x) is continuous for all real values of

Solution

Property: - Let f:[0,1]→ R be continuous such that f(x)∈Q for any x∈[0,1] then f (x) is constant. Proof Suppose f isn't constant. Then for some a,b∈[0,1], f(a)≠f(b); Without Loss of Generality, f(a) Since f is continuous, by the Intermediate Value Theorem, it must take every value in the interval [f(a), f(b)]. But this interval contains an irrational number (in fact, uncountably many of them). Contradiction. Hence, f is constant and equal to 1. Therefore, Since f(x) can take only rational values, option c