Let `a in R` suppose f is defined by `f(x) =(x-1)/(a+ 1-x^(2))` If range of f does not contain the interval `[-1.-1/3]` then a lies in

Options

A. `(-1,infty)`
B. `(0, infty)`
C. `(-infty, -1/4)`
D. `(-infty, -1)`

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{x - 1}{a + 1 - x^2} \) and determine the conditions under which its range does not include the interval \([-1, -\frac{1}{3}]\).

### Step-by-Step Solution:

1. **Identify the function and the interval**: We have the function \( f(x) = \frac{x - 1}{a + 1 - x^2} \) and we want to find the values of \( a \) such that the range of \( f \) does not include the interval \([-1, -\frac{1}{3}]\).

**Hint**: Start by understanding the behavior of the function and how it can be manipulated.

2. **Set up the inequalities**: To find the values of \( a \), we need to analyze the function at the endpoints of the interval. We will set \( f(x) = -1 \) and \( f(x) = -\frac{1}{3} \) and find the conditions on \( a \).

**Hint**: Substitute the values into the function and simplify.

3. **Solve for \( f(x) = -1 \)**:
\[ \frac{x - 1}{a + 1 - x^2} = -1 \]
This leads to:
\[ x - 1 = - (a + 1 - x^2) \]
Rearranging gives:
\[ x^2 - x + (a + 2) = 0 \]
For this quadratic to have no real solutions, the discriminant must be less than zero:
\[ (-1)^2 - 4 \cdot 1 \cdot (a + 2) < 0 \]
Simplifying gives:
\[ 1 - 4(a + 2) < 0 \implies 1 < 4a + 8 \implies 4a > -7 \implies a > -\frac{7}{4} \]

**Hint**: Remember that for a quadratic to have no real solutions, the discriminant must be negative.

4. **Solve for \( f(x) = -\frac{1}{3} \)**:
\[ \frac{x - 1}{a + 1 - x^2} = -\frac{1}{3} \]
This leads to:
\[ 3(x - 1) = - (a + 1 - x^2) \]
Rearranging gives:
\[ x^2 - 3x + (a + 4) = 0 \]
For this quadratic to have no real solutions, the discriminant must be less than zero:
\[ (-3)^2 - 4 \cdot 1 \cdot (a + 4) < 0 \]
Simplifying gives:
\[ 9 - 4(a + 4) < 0 \implies 9 < 4a + 16 \implies 4a > -7 \implies a > -\frac{7}{4} \]

**Hint**: Again, check the discriminant for the quadratic equation.

5. **Combine the conditions**: Both conditions lead to the same inequality \( a > -\frac{7}{4} \). However, we also need to ensure that the range does not include the interval \([-1, -\frac{1}{3}]\).

**Hint**: Think about the intervals that are excluded based on the conditions you derived.

6. **Final conclusion**: Since we need the range of \( f \) to exclude \([-1, -\frac{1}{3}]\), we find that \( a \) must also satisfy \( a < -\frac{1}{4} \) for the quadratic to be positive. Therefore, the final answer is:
\[ a < -\frac{1}{4} \quad \text{or} \quad a > -\frac{7}{4} \]

**Hint**: The solution will be in the form of intervals based on the derived inequalities.

### Final Answer:
The values of \( a \) such that the range of \( f \) does not contain the interval \([-1, -\frac{1}{3}]\) are:
\[ a < -\frac{1}{4} \quad \text{or} \quad a > -\frac{7}{4} \]