在同一平面直角坐标系内，若直线y=3x-1与直线y=x-k的交点在第四象限，则k的取值范围是（）A.k＜13B.13＜k＜1C.k＞1D.k＞1或k＜13

Question

(1)/(3)
B. (1)/(3)＜k＜1
C. k＞1
D. k＞1或k＜(1)/(3)(1)/(3)1133
(1)/(3)＜k＜1
C. k＞1
D. k＞1或k＜(1)/(3)(1)/(3)1133

(1)/(3)(1)/(3)1133

Answer 1

解关于x，y的方程组

y＝3x−1

y＝x−k

解得：

x＝(1−k)/(2)

y＝(1−3k)/(2)

∵交点在第四象限
∴得到不等式组

(1−k)/(2)＞0

(1−3k)/(2)＜0

解得(1)/(3)＜k＜1
故选B．

y＝3x−1

y＝x−k

y＝3x−1

y＝x−k

y＝3x−1

y＝x−k

y＝3x−1

y＝x−k

y＝3x−1y＝3x−1y＝3x−1y＝x−ky＝x−ky＝x−k
解得：

x＝(1−k)/(2)

y＝(1−3k)/(2)

∵交点在第四象限
∴得到不等式组

(1−k)/(2)＞0

(1−3k)/(2)＜0

解得(1)/(3)＜k＜1
故选B．

x＝(1−k)/(2)

y＝(1−3k)/(2)

x＝(1−k)/(2)

y＝(1−3k)/(2)

x＝(1−k)/(2)

y＝(1−3k)/(2)

x＝(1−k)/(2)

y＝(1−3k)/(2)

x＝(1−k)/(2)x＝(1−k)/(2)x＝(1−k)/(2)1−k1−k1−k222y＝(1−3k)/(2)y＝(1−3k)/(2)y＝(1−3k)/(2)1−3k1−3k1−3k222
∵交点在第四象限
∴得到不等式组

(1−k)/(2)＞0

(1−3k)/(2)＜0

解得(1)/(3)＜k＜1
故选B．

(1−k)/(2)＞0

(1−3k)/(2)＜0

(1−k)/(2)＞0

(1−3k)/(2)＜0

(1−k)/(2)＞0

(1−3k)/(2)＜0

(1−k)/(2)＞0

(1−3k)/(2)＜0

(1−k)/(2)＞0(1−k)/(2)＞0(1−k)/(2)1−k1−k1−k222＞0(1−3k)/(2)＜0(1−3k)/(2)＜0(1−3k)/(2)1−3k1−3k1−3k222＜0
解得(1)/(3)＜k＜1
故选B． (1)/(3)111333＜k＜1
故选B．