The number of integral solutions of `2(x+2)gt x^(2)+1`, is

Options

A. 2
B. 3
C. 4
D. 5

Answer

To find the number of integral solutions of the inequality \( 2(x + 2) > x^2 + 1 \), we can follow these steps:

### Step 1: Rewrite the Inequality
Start by rewriting the given inequality:
\[ 2(x + 2) > x^2 + 1 \]
Distributing the 2 on the left side:
\[ 2x + 4 > x^2 + 1 \]

### Step 2: Rearrange the Inequality
Next, we rearrange the inequality to bring all terms to one side:
\[ 2x + 4 - x^2 - 1 > 0 \]
This simplifies to:
\[ -x^2 + 2x + 3 > 0 \]
Multiplying through by -1 (and reversing the inequality):
\[ x^2 - 2x - 3 < 0 \]

### Step 3: Factor the Quadratic Expression
Now, we need to factor the quadratic expression \( x^2 - 2x - 3 \):
\[ x^2 - 2x - 3 = (x - 3)(x + 1) \]
Thus, the inequality becomes:
\[ (x - 3)(x + 1) < 0 \]

### Step 4: Determine the Critical Points
The critical points from the factors are:
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]
These points divide the number line into intervals. We will test the sign of the product in each interval.

### Step 5: Test the Intervals
The intervals to test are:
1. \( (-\infty, -1) \)
2. \( (-1, 3) \)
3. \( (3, \infty) \)

- For \( x < -1 \) (e.g., \( x = -2 \)):
\((x - 3)(x + 1) = (-5)(-1) = 5 > 0\)

- For \( -1 < x < 3 \) (e.g., \( x = 0 \)):
\((x - 3)(x + 1) = (-3)(1) = -3 < 0\)

- For \( x > 3 \) (e.g., \( x = 4 \)):
\((x - 3)(x + 1) = (1)(5) = 5 > 0\)

Thus, the inequality \( (x - 3)(x + 1) < 0 \) holds true in the interval:
\[ (-1, 3) \]

### Step 6: Identify Integral Solutions
Now we need to find the integral solutions within the interval \( (-1, 3) \). The integers in this interval are:
- 0
- 1
- 2

### Final Count of Integral Solutions
Thus, the total number of integral solutions is:
\[ \text{Total integral solutions} = 3 \]

### Conclusion
The number of integral solutions of the inequality \( 2(x + 2) > x^2 + 1 \) is **3**.

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