Question

15．已知函數(shù)f（x）=sinx-a（0$≤x≤\frac{5π}{2}$）的三個零點成等比數(shù)列，則log${\;}_{\sqrt{2}}$a=-1．

Answer 1

分析 不妨設函數(shù)f（x）=sinx-a，（0$≤x≤\frac{5π}{2}$）的三個零點從小到大依次為x₁，x₂，x₃，從而由三角函數(shù)的性質(zhì)及等比數(shù)列可得$\left\{\begin{array}{l}{{x}_{1}+{x}_{2}=π}\\{{x}_{2}+{x}_{3}=3π}\\{{{x}_{2}}^{2}={x}_{1}{x}_{3}}\end{array}\right.$，從而解得x₂=$\frac{3π}{4}$，
從而求出a的值，再求對數(shù)即可．

解答 解：設函數(shù)f（x）=sinx-a，（0$≤x≤\frac{5π}{2}$）的三個零點從小到大依次為x₁，x₂，x₃，
則$\left\{\begin{array}{l}{{x}_{1}+{x}_{2}=π}\\{{x}_{2}+{x}_{3}=3π}\\{{{x}_{2}}^{2}={x}_{1}{x}_{3}}\end{array}\right.$，
解得，x₂=$\frac{3π}{4}$，
∴a=sin$\frac{3π}{4}$=$\frac{\sqrt{2}}{2}$，
∴l(xiāng)og${\;}_{\sqrt{2}}$a=log${\;}_{\sqrt{2}}$$\frac{\sqrt{2}}{2}$=-1；
故答案為：-1．

點評 本師考查了三角函數(shù)的性質(zhì)及等比數(shù)列的性質(zhì)應用，屬于基礎題．