Question

The set of all real values \( a \) for which \[ -1<\frac{2x^2 + ax + 2}{x^2 + x + 1}<3 \] holds for all real values of \( x \) is:

• A. \( (-7,5) \)
• B. \( (5, \infty) \)
• C. \( (1,5) \)
• D. \( (-\infty, 1) \)

Hint:

For rational inequalities to hold for all real values, ensure that the discriminant conditions lead to a valid range for parameters.

Answer 1

The Correct Option is C

Step 1: Understanding the given inequality 
We need to find the values of \( a \) such that the given inequality: \[ -1<\frac{2x^2 + ax + 2}{x^2 + x + 1}<3 \] holds for all \( x \in \mathbb{R} \).

Step 2: Setting up the inequality constraints 
Rewriting the given fraction: \[ -1<\frac{2x^2 + ax + 2}{x^2 + x + 1}<3. \] Multiply both sides by \( x^2 + x + 1 \) (which is always positive for all real \( x \) since the discriminant is negative): \[ - (x^2 + x + 1)<2x^2 + ax + 2<3(x^2 + x + 1). \] Expanding both inequalities: \[ - x^2 - x - 1<2x^2 + ax + 2<3x^2 + 3x + 3. \] 

Step 3: Solving for \( a \) 
Rearrange both inequalities: 1. \( 2x^2 + ax + 2 + x^2 + x + 1>0 \Rightarrow 3x^2 + (a+1)x + 3>0 \). 2. \( 2x^2 + ax + 2<3x^2 + 3x + 3 \Rightarrow -x^2 + (a - 3)x - 1<0 \). For these quadratic inequalities to hold for all real \( x \), the discriminants must be negative: 1. \( (a+1)^2 - 4(3)(3)<0 \Rightarrow a^2 + 2a + 1 - 36<0 \Rightarrow a^2 + 2a - 35<0 \). 2. \( (a - 3)^2 - 4(-1)(-1)<0 \Rightarrow a^2 - 6a + 9 - 4<0 \Rightarrow a^2 - 6a + 5<0 \). Solving these quadratic inequalities gives: 1. \( (a - 5)(a + 7)<0 \Rightarrow -7<a<5 \). 2. \( (a - 1)(a - 5)<0 \Rightarrow 1<a<5 \). The intersection of these intervals is \( (1,5) \). 

Step 4: Conclusion 
Thus, the correct set of values for \( a \) is: \[ \boxed{(1,5)} \]