The number of integral value of `a,a, in [-5, 5]` for which the equation: `x ^(2) +2 (a-1) x+a+5=0` has one root smaller than 1 and the other root is greater than 3 :

Answer

To solve the problem, we need to find the integral values of \( a \) in the interval \([-5, 5]\) for which the quadratic equation

\[ x^2 + 2(a-1)x + (a+5) = 0 \]

has one root less than 1 and the other root greater than 3.

### Step 1: Ensure the quadratic has real roots

For the quadratic equation to have real roots, the discriminant must be greater than zero. The discriminant \( D \) for the quadratic equation \( ax^2 + bx + c = 0 \) is given by:

\[ D = b^2 - 4ac \]

In our case, \( a = 1 \), \( b = 2(a-1) \), and \( c = a + 5 \). Thus, we have:

\[ D = [2(a-1)]^2 - 4 \cdot 1 \cdot (a + 5) \]

Calculating this gives:

\[ D = 4(a-1)^2 - 4(a + 5) \]
\[ D = 4[(a-1)^2 - (a + 5)] \]
\[ D = 4[a^2 - 2a + 1 - a - 5] \]
\[ D = 4[a^2 - 3a - 4] \]

For the roots to be real, we need:

\[ 4(a^2 - 3a - 4) > 0 \]

This simplifies to:

\[ a^2 - 3a - 4 > 0 \]

### Step 2: Solve the quadratic inequality

To solve \( a^2 - 3a - 4 > 0 \), we first find the roots of the equation \( a^2 - 3a - 4 = 0 \) using the quadratic formula:

\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
\[ = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2} \]

Calculating the roots gives:

\[ a = \frac{8}{2} = 4 \quad \text{and} \quad a = \frac{-2}{2} = -1 \]

The quadratic \( a^2 - 3a - 4 \) opens upwards (since the coefficient of \( a^2 \) is positive). Thus, the solution to the inequality \( a^2 - 3a - 4 > 0 \) is:

\[ a < -1 \quad \text{or} \quad a > 4 \]

### Step 3: Consider the conditions for the roots

We need one root \( \alpha < 1 \) and the other root \( \beta > 3 \). By Vieta's formulas, we know:

1. The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{2(a-1)}{1} = -2(a-1) \)
2. The product of the roots \( \alpha \beta = \frac{c}{a} = a + 5 \)

From the conditions \( \alpha < 1 \) and \( \beta > 3 \), we can derive:

\[ \alpha + \beta < 1 + 3 = 4 \quad \Rightarrow \quad -2(a-1) < 4 \]
\[ -2a + 2 < 4 \quad \Rightarrow \quad -2a < 2 \quad \Rightarrow \quad a > -1 \]

And for the product:

\[ \alpha \beta > 1 \cdot 3 = 3 \quad \Rightarrow \quad a + 5 > 3 \quad \Rightarrow \quad a > -2 \]

### Step 4: Combine the conditions

From the discriminant condition, we have:

1. \( a < -1 \) or \( a > 4 \)

From the root conditions, we have:

2. \( a > -1 \)

Combining these gives us:

- The only valid interval for \( a \) is \( a > 4 \).

### Step 5: Find integral values in the range \([-5, 5]\)

The only integral value of \( a \) in the range \([-5, 5]\) that satisfies \( a > 4 \) is:

\[ a = 5 \]

### Conclusion

Thus, the number of integral values of \( a \) in the interval \([-5, 5]\) for which the equation has one root less than 1 and the other root greater than 3 is:

\[ \boxed{1} \]