我看了之後有個問題請問:\( a=-cos x+\sqrt{1-cos x} \)可推出\( 1\le a \le 1+\sqrt{2} \) 這是如何算出來的???
我用我的方式算~算出\( \displaystyle -\frac{5}{4} \le a \le 1-sqrt{2} \) ,可以請各位幫我看看!我哪裡出了問題嗎?
\( sin^2 x-(2a+1)cosx-a^2=0 \)
\( 1-cos^2 x-(2a+1)cosx-a^2=0 \)
\( cos^2 x+(2a+1)cos x+(a^2-1)=0 \)
令\( t=cos x \),\( t^2+(2a+1)t+(a^2-1)=0 \),\( t \in R \),\( -1\le t \le 1 \)
1.
\( D\ge 0 \),\( \displaystyle (2a+1)^2-4(a^2-1)\ge 0 \Rightarrow a\ge -\frac{5}{4} \)
2.
令\( f(t)=t^2+(2a+1)t+(a^2-1) \)
(1)\( f(1)\ge 0 \Rightarrow a^2+2a+1 \ge 0 \Rightarrow a \in R \)
(2)\( f(-1)\ge 0 \Rightarrow a^2-2a-1 \ge 0 \Rightarrow a\ge 1+\sqrt{2} \)或\( a \le 1-\sqrt{2} \)
(3)\( \displaystyle -1<\frac{-2a-1}{2}<1 \Rightarrow -\frac{3}{2}\le a \le \frac{1}{2} \)
故\( \displaystyle -\frac{5}{4}<a \le 1-\sqrt{2} \)