The interval to which a may belong so that the function `f(x)=(1-(sqrt(21-4a-a^2))/(a+1))x^3+5x+100` is increasing for `x in R`

Answer

To determine the interval to which \( a \) may belong so that the function

\[ f(x) = \left(1 - \frac{\sqrt{21 - 4a - a^2}}{a + 1}\right)x^3 + 5x + 100 \]

is increasing for all \( x \in \mathbb{R} \), we need to analyze the derivative of the function and set conditions for it to be positive.

### Step 1: Rewrite the function
We can express the function \( f(x) \) in a simpler form by letting \( A = 1 - \frac{\sqrt{21 - 4a - a^2}}{a + 1} \). Thus, we can rewrite:

\[ f(x) = Ax^3 + 5x + 100 \]

### Step 2: Differentiate the function
To find when the function is increasing, we differentiate \( f(x) \):

\[ f'(x) = 3Ax^2 + 5 \]

### Step 3: Set the derivative greater than zero
For \( f(x) \) to be increasing for all \( x \in \mathbb{R} \), we need:

\[ f'(x) > 0 \quad \text{for all } x \]

This implies:

\[ 3Ax^2 + 5 > 0 \quad \text{for all } x \]

### Step 4: Analyze the quadratic term
The term \( 3Ax^2 + 5 \) is a quadratic function in \( x \). Since the coefficient of \( x^2 \) is \( 3A \), we need \( 3A \) to be non-negative for the quadratic to be positive for all \( x \):

\[ 3A > 0 \implies A > 0 \]

### Step 5: Substitute back for \( A \)
Substituting back for \( A \):

\[ 1 - \frac{\sqrt{21 - 4a - a^2}}{a + 1} > 0 \]

### Step 6: Rearranging the inequality
Rearranging gives:

\[ \frac{\sqrt{21 - 4a - a^2}}{a + 1} < 1 \]

### Step 7: Cross-multiply
Cross-multiplying (valid since \( a + 1 \) must be positive):

\[ \sqrt{21 - 4a - a^2} < a + 1 \]

### Step 8: Square both sides
Squaring both sides gives:

\[ 21 - 4a - a^2 < (a + 1)^2 \]

Expanding the right side:

\[ 21 - 4a - a^2 < a^2 + 2a + 1 \]

### Step 9: Rearranging the inequality
Rearranging yields:

\[ 21 - 4a - a^2 - a^2 - 2a - 1 < 0 \]

This simplifies to:

\[ -2a^2 - 6a + 20 < 0 \]

### Step 10: Factor the quadratic
Dividing through by -2 (and reversing the inequality):

\[ a^2 + 3a - 10 > 0 \]

### Step 11: Finding the roots
Finding the roots of \( a^2 + 3a - 10 = 0 \) using the quadratic formula:

\[ a = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{49}}{2} = \frac{-3 \pm 7}{2} \]

This gives:

\[ a = 2 \quad \text{and} \quad a = -5 \]

### Step 12: Test intervals
We test the intervals \( (-\infty, -5) \), \( (-5, 2) \), and \( (2, \infty) \):

- For \( a < -5 \): \( a^2 + 3a - 10 > 0 \) (True)
- For \( -5 < a < 2 \): \( a^2 + 3a - 10 < 0 \) (False)
- For \( a > 2 \): \( a^2 + 3a - 10 > 0 \) (True)

### Step 13: Combine intervals
Thus, the solution is:

\[ a \in (-\infty, -5) \cup (2, \infty) \]

### Final Answer
The intervals to which \( a \) may belong so that the function \( f(x) \) is increasing for \( x \in \mathbb{R} \) are:

\[ (-\infty, -5) \cup (2, \infty) \]