JEE Main 1984 Mathematics question on Complex Numbers and Quadratic Equations

Let y = (x-a)(x-b)/(x-c). Rearranging: (x-a)(x-b) = y(x-c), which gives x² - (a+b+y)x + (ab+yc) = 0. For real x, the discriminant must be ≥ 0 for ALL real y: (a+b+y)² - 4(ab+yc) ≥ 0, i.e., y² + 2(a+b)y + (a+b)² - 4ab - 4yc ≥ 0, i.e., y² + 2(a+b-2c)y + (a-b)² ≥ 0 for all real y. A quadratic in y is non-negative for all y iff its own discriminant ≤ 0: 4(a+b-2c)² - 4(a-b)² ≤ 0, i.e., (a+b-2c)² ≤ (a-b)², giving |a+b-2c| ≤ |a-b|. This means -(a-b) ≤ a+b-2c ≤ (a-b), which simplifies to b ≤ c ≤ a (assuming a ≥ b). So the condition is a ≥ c ≥ b, i.e., a ≤ c ≤ b is wrong only in labeling — it means c lies between a and b. Option 4 states a ≤ c ≤ b (so b ≥ c