Question

Find all values of m for which mx² + (m − 3)x + 1 < 0 for at least one positive real x. 

Answer 1

Let f(x) = mx² + (m − 3) x + 1.

Case (i):  f(x) < 0 trivially if m < 0, as parabola will be concave downwards.

Case (ii): If m > 0, then the given condition is satisfied if f(x) has distinct roots and at least one of them is positive real root 

Fig.

D > 0 ⇒ (m − 3)² − 4m > 0

⇒ m < 1 or m > 9

At least one root is positive, that is, R − intervals when both are non-positive.

So m < 3 (as sum ≤ 0 and product ≥ 0)

Their intersection gives m < 1. Hence, from the above two cases m ∈ (−∞, 1).