Consider the following functions defined for x > 1: f(x) = log2(log2 √x) g(x) = log2(√log2 x) Which one of the following is true for all values of x > 1?
- A0 ≤ f(x) ≤ g(x) or g(x) ≤ f(x) ≤ 0
- B0 ≤ g(x) ≤ f(x) or f(x) ≤ g(x) ≤ 0
- C12 ≤ f(x) ≤ g(x) or g(x) ≤ f(x) ≤ 12
- D12 ≤ g(x) ≤ f(x) or f(x) ≤ g(x) ≤ 12
- E1 ≤ f(x) ≤ g(x) or g(x) ≤ f(x) ≤ 1
- F1 ≤ g(x) ≤ f(x) or f(x) ≤ g(x) ≤ 1
Show the answer and worked solution
answer · F
- A0 ≤ f(x) ≤ g(x) or g(x) ≤ f(x) ≤ 0
- B0 ≤ g(x) ≤ f(x) or f(x) ≤ g(x) ≤ 0
- C12 ≤ f(x) ≤ g(x) or g(x) ≤ f(x) ≤ 12
- D12 ≤ g(x) ≤ f(x) or f(x) ≤ g(x) ≤ 12
- E1 ≤ f(x) ≤ g(x) or g(x) ≤ f(x) ≤ 1
- F1 ≤ g(x) ≤ f(x) or f(x) ≤ g(x) ≤ 1
Substitute to strip the nesting. Put t = log2 x, which runs over all positive reals as x runs over x > 1, and then s = log2 t, which runs over all reals. Since log2√x = t2, we get f = log2t2 = s − 1, and since log2√t = 12log2 t, we get g = s2. Now compare: f − g = s2 − 1, so if s ≥ 2 then g = s2 ≥ 1 and f ≥ g, giving 1 ≤ g ≤ f; and if s < 2 then f < g and g = s2 < 1, giving f ≤ g ≤ 1. That is exactly the last option, and the cutoff at 1 rather than 0 or 12 is what rules the others out.