If `( lambda^(2) + lambda - 2)x^(2)+(lambda+2)x lt 1` for all ` x in R ` , then ` lambda ` belongs to the interval :

Options

A. (-2,1)
B. `(-2,(2)/5)`
C. `(2/5,1)`
D. None of these

Answer

To solve the inequality \((\lambda^2 + \lambda - 2)x^2 + (\lambda + 2)x < 1\) for all \(x \in \mathbb{R}\), we will follow these steps:

### Step 1: Rewrite the Inequality
We start by rewriting the inequality:
\[ (\lambda^2 + \lambda - 2)x^2 + (\lambda + 2)x - 1 < 0 \]
This is a quadratic inequality in \(x\).

### Step 2: Identify Coefficients
In the quadratic expression \(Ax^2 + Bx + C\), we identify:
- \(A = \lambda^2 + \lambda - 2\)
- \(B = \lambda + 2\)
- \(C = -1\)

### Step 3: Conditions for the Quadratic to be Negative
For the quadratic to be less than zero for all \(x\), the following conditions must be satisfied:
1. \(A < 0\) (the parabola opens downwards)
2. The discriminant \(D < 0\) (the quadratic has no real roots)

### Step 4: Solve \(A < 0\)
We need to solve:
\[ \lambda^2 + \lambda - 2 < 0 \]
Factoring gives:
\[ (\lambda + 2)(\lambda - 1) < 0 \]
Now we find the critical points: \(\lambda = -2\) and \(\lambda = 1\).

Using a sign chart:
- For \(\lambda < -2\), the product is positive.
- For \(-2 < \lambda < 1\), the product is negative.
- For \(\lambda > 1\), the product is positive.

Thus, the solution for \(A < 0\) is:
\[ -2 < \lambda < 1 \]

### Step 5: Solve the Discriminant Condition \(D < 0\)
The discriminant \(D\) is given by:
\[ D = B^2 - 4AC = (\lambda + 2)^2 - 4(\lambda^2 + \lambda - 2)(-1) \]
Calculating \(D\):
\[ D = (\lambda + 2)^2 + 4(\lambda^2 + \lambda - 2) \]
Expanding this:
\[ D = \lambda^2 + 4\lambda + 4 + 4\lambda^2 + 4\lambda - 8 \]
\[ D = 5\lambda^2 + 8\lambda - 4 \]
We need \(D < 0\):
\[ 5\lambda^2 + 8\lambda - 4 < 0 \]
Finding the roots using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{64 + 80}}{10} = \frac{-8 \pm \sqrt{144}}{10} = \frac{-8 \pm 12}{10} \]
Calculating the roots:
\[ \lambda_1 = \frac{4}{10} = \frac{2}{5}, \quad \lambda_2 = \frac{-20}{10} = -2 \]

Using a sign chart for \(5\lambda^2 + 8\lambda - 4\):
- For \(\lambda < -2\), the expression is positive.
- For \(-2 < \lambda < \frac{2}{5}\), the expression is negative.
- For \(\lambda > \frac{2}{5}\), the expression is positive.

Thus, the solution for \(D < 0\) is:
\[ -2 < \lambda < \frac{2}{5} \]

### Step 6: Combine the Results
Now we have two intervals:
1. From \(A < 0\): \(-2 < \lambda < 1\)
2. From \(D < 0\): \(-2 < \lambda < \frac{2}{5}\)

The common interval is:
\[ -2 < \lambda < \frac{2}{5} \]

### Final Answer
Thus, the value of \(\lambda\) belongs to the interval:
\[ \boxed{(-2, \frac{2}{5})} \]