Find the complete set of values of m in terms of c such that the graphs of y = mx + c and y = √x have two points of intersection.
- A0 < m < 14c
- B0 < m < 4c2
- Cm > 14c
- Dm < 14c
- Em > 4c2
- Fm < 4c2
Show the answer and worked solution
answer · A
- A0 < m < 14c
- B0 < m < 4c2
- Cm > 14c
- Dm < 14c
- Em > 4c2
- Fm < 4c2
Substituting t = √ x, where t ≥ 0, turns √ x = mx + c into mt2 − t + c = 0, and each root t ≥ 0 gives exactly one intersection point. Two intersections therefore need two distinct non-negative roots. The discriminant condition is 1 − 4mc > 0, the sum of roots 1m must be positive, forcing m > 0, and the product cm must be non-negative, forcing c ≥ 0. With m > 0 and c > 0, 1 − 4mc > 0 rearranges to m < 14c, so the answer is 0 < m < 14c. Dropping the requirement m > 0 gives the tempting option D, but a line with negative gradient cuts y = √ x at most once.