Question

If the inequality $\frac{mx^{2} + 3x + 4}{x^{2} + 2x + 2} < 5$ is satisfied for all x ∈ R, then

• A. 1 < m < 5
• B. –1 < m < 5
• C. 1< m < 6
• D. m <$\frac{71}{24}$

Answer 1

Answer: (D)

m <$\frac{71}{24}$

Answer 2

We have x² + 2x + 2 = (x + 1)² + 1 > 0, ∀ x ∈ R.

Therefore, $\frac{mx^{2} + 3x + 4}{x^{2} + 2x + 2} < 5$ ⇒ mx² + 3x + 4 < 5(x² + 2x + 2)

⇒ (m – 5)x² – 7x – 6 < 0, ∀ x ∈ R.

This is possible if D = b² – 4ac = 49 + 24(m – 5) < 0 and

m – 5 < 0

⇒ m < $\frac{71}{24}$