(1)求点的坐标时,小明是这样想的:先设点坐标为,因为点在直线上,所以是方程的解;又因为点在直线上,所以也是方程的解,从而,满足,据此可求出点坐标为______.再求出点坐标为______;点坐标为______.(均直接写出结果)
(2)若线段上存在一点,使(为原点),求点坐标
(3)点是坐标平面内的动点,若满足,求的取值范围.
同类型试题
y = sin x, x∈R, y∈[–1,1],周期为2π,函数图像以 x = (π/2) + kπ 为对称轴
y = arcsin x, x∈[–1,1], y∈[–π/2,π/2]
sin x = 0 ←→ arcsin x = 0
sin x = 1/2 ←→ arcsin x = π/6
sin x = √2/2 ←→ arcsin x = π/4
sin x = 1 ←→ arcsin x = π/2
y = sin x, x∈R, y∈[–1,1],周期为2π,函数图像以 x = (π/2) + kπ 为对称轴
y = arcsin x, x∈[–1,1], y∈[–π/2,π/2]
sin x = 0 ←→ arcsin x = 0
sin x = 1/2 ←→ arcsin x = π/6
sin x = √2/2 ←→ arcsin x = π/4
sin x = 1 ←→ arcsin x = π/2