If f(x) is continuous for all real values of x and f(x) takes on only rational values, then if f(1)=1, the value of f(0) is
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
16% of 250 + 32% of 500 = ?
√(2670+ √(1141+ √(260- √(1251- √(637+ √1521) ) ) ) ) =?
(1/2) – (3/5) + 3(1/3) = ? + (5/6)
(7 ÷ 0.125) x (10 ÷ 10/3 ) = ?
360 × 9 ÷ 3 + 120 + 900 ÷ 5 = ?
50 ÷ 2.5 × 64 + ? = 1520
756 + 432 – 361 + ? = 990
(0.125)³÷ (0.25)² x (0.5)² = (0.5)?-3
(1296) -3/4 = ?
2(1/3) + 2(5/6) – 1(1/2) = ? – 6(1/6)