The possible greates integral value of a for which the expression ` (ax^(2)+3x+4)/(x^(2)+2x+2)` is less than 5 for all real x is ______

Answer

To solve the problem, we need to determine the greatest integral value of \( a \) such that the expression

\[ \frac{ax^2 + 3x + 4}{x^2 + 2x + 2} < 5 \]

for all real \( x \).

### Step 1: Rearranging the Inequality

Start by rearranging the inequality:

\[ \frac{ax^2 + 3x + 4}{x^2 + 2x + 2} < 5 \]

Multiply both sides by \( x^2 + 2x + 2 \) (which is always positive for all real \( x \)):

\[ ax^2 + 3x + 4 < 5(x^2 + 2x + 2) \]

### Step 2: Expanding the Right Side

Now, expand the right side:

\[ ax^2 + 3x + 4 < 5x^2 + 10x + 10 \]

### Step 3: Moving All Terms to One Side

Rearranging gives:

\[ ax^2 + 3x + 4 - 5x^2 - 10x - 10 < 0 \]

This simplifies to:

\[ (a - 5)x^2 - 7x - 6 < 0 \]

### Step 4: Analyzing the Quadratic Inequality

For the quadratic \( (a - 5)x^2 - 7x - 6 < 0 \) to hold for all \( x \), the following conditions must be satisfied:

1. The coefficient of \( x^2 \) must be negative:
\[ a - 5 < 0 \implies a < 5 \]

2. The discriminant of the quadratic must be less than zero:
\[ b^2 - 4ac < 0 \]
Here, \( b = -7 \), \( a = a - 5 \), and \( c = -6 \). Thus, we have:
\[ (-7)^2 - 4(a - 5)(-6) < 0 \]

### Step 5: Calculating the Discriminant

Calculating the discriminant:

\[ 49 + 24(a - 5) < 0 \]

This simplifies to:

\[ 49 + 24a - 120 < 0 \]

\[ 24a - 71 < 0 \implies 24a < 71 \implies a < \frac{71}{24} \approx 2.9583 \]

### Step 6: Finding the Greatest Integral Value of \( a \)

The greatest integral value of \( a \) that satisfies both conditions \( a < 5 \) and \( a < \frac{71}{24} \) is:

\[ a \leq 2 \]

Thus, the possible greatest integral value of \( a \) is:

\[ \boxed{2} \]