Let f:(0, ∞) → R be a twice differentiable function. If for some a ≠ 0, ∫_0^1 f(λ x) d λ= a f(x), f(1)=1 and f(16)=1/8, then 16-f^′(1/16) is equal to…

Let $\displaystyle f:(0, \infty) \rightarrow \mathbf{R}$ be a twice differentiable function. If for some $\displaystyle \mathrm{a} \neq 0, \int_0^1 f(\lambda x) \mathrm{d} \lambda=\mathrm{a} f(x), f(1)=1$ and $\displaystyle f(16)=\frac{1}{8}$, then $\displaystyle 16-f^{\prime}\left(\frac{1}{16}\right)$ is equal to $\displaystyle \_\_\_\_$.