Question 10. For \(0

Solution:

On each dyadic interval $x\in[2^{k},\,2^{k+1})$ with $k\le 0$, the value $\lfloor \log_2 x \rfloor = k$ is constant, so $$f(x)=\left(\tfrac{3}{4}\right)^{k}\quad \text{for }x\in[2^{k},2^{k+1}).$$ Therefore, $$\int_{2^{k}}^{2^{k+1}} f(x)\,dx = \left(\tfrac{3}{4}\right)^{k}\!\left(2^{k+1}-2^{k}\right) = \left(\tfrac{3}{4}\right)^{k} 2^{k} = \left(\tfrac{3}{2}\right)^{k}.$$

Summing these pieces for all $k\le 0$ (since \(0

Answer: (C) $3$