抛物线y=x²-2bx+1和直线y=-1/2x+m/2,不论b为何实数时总有交点,求实数m的取值范围
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![抛物线y=x²-2bx+1和直线y=-1/2x+m/2,不论b为何实数时总有交点,求实数m的取值范围](/uploads/image/z/13899835-19-5.jpg?t=%E6%8A%9B%E7%89%A9%E7%BA%BFy%3Dx%26%23178%3B-2bx%2B1%E5%92%8C%E7%9B%B4%E7%BA%BFy%3D-1%2F2x%2Bm%2F2%2C%E4%B8%8D%E8%AE%BAb%E4%B8%BA%E4%BD%95%E5%AE%9E%E6%95%B0%E6%97%B6%E6%80%BB%E6%9C%89%E4%BA%A4%E7%82%B9%2C%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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抛物线y=x²-2bx+1和直线y=-1/2x+m/2,不论b为何实数时总有交点,求实数m的取值范围
抛物线y=x²-2bx+1和直线y=-1/2x+m/2,不论b为何实数时总有交点,求实数m的取值范围
抛物线y=x²-2bx+1和直线y=-1/2x+m/2,不论b为何实数时总有交点,求实数m的取值范围
y=x²-2bx+1和y=-1/2x+m/2联立成方程组,有x²-2bx+1=-1/2x+m/2,即x²-(2b-1/2)x+1-m/2=0
于是可得:无论b 为何实数,此方程一定有解.
有△=(2b-1/2)^2-4(1-m/2)≥0,由于b可取一切实数不等式都成立,因此1-m/2≤0即可,得出m≥2