Value of `lamda` so that point `(lamda,lamda^(2))` lies between the lines `|x+2y|=3` is

Options

A. `(-1/2,2)`
B. `(-3/2,2)`
C. `(-3/2,1)`
D. `(-1,3/2)`

Explanation

To find the value of `lambda` such that the point `(lambda, lambda^2)` lies between the lines defined by the equation `|x + 2y| = 3`, we can follow these steps:

### Step 1: Understand the Lines
The equation `|x + 2y| = 3` represents two lines:
1. \( x + 2y = 3 \)
2. \( x + 2y = -3 \)

### Step 2: Substitute the Point
We need to check if the point \( (lambda, lambda^2) \) lies between these two lines. We substitute \( x = lambda \) and \( y = lambda^2 \) into the equations of the lines.

### Step 3: Set Up the Inequality
For the point to lie between the two lines, the following condition must hold:
\[ (x + 2y - 3)(x + 2y + 3) < 0 \]
Substituting \( x = lambda \) and \( y = lambda^2 \):
\[ (lambda + 2(lambda^2) - 3)(lambda + 2(lambda^2) + 3) < 0 \]

### Step 4: Simplify the Expressions
Calculating the expressions:
1. First expression: \( lambda + 2lambda^2 - 3 \)
2. Second expression: \( lambda + 2lambda^2 + 3 \)

Now we can rewrite the inequality:
\[ (2lambda^2 + lambda - 3)(2lambda^2 + lambda + 3) < 0 \]

### Step 5: Analyze the Quadratic
Let \( f(lambda) = 2lambda^2 + lambda - 3 \) and \( g(lambda) = 2lambda^2 + lambda + 3 \).

1. **For \( g(lambda) \)**: Since the leading coefficient (2) is positive and the constant term is positive, \( g(lambda) > 0 \) for all \( lambda \).
2. **For \( f(lambda) \)**: We need to find the roots of \( f(lambda) = 0 \) using the quadratic formula:
\[ lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4(2)(-3)}}{2(2)} \]
\[ = \frac{-1 \pm \sqrt{1 + 24}}{4} = \frac{-1 \pm 5}{4} \]
This gives us the roots:
\[ lambda_1 = 1 \quad \text{and} \quad lambda_2 = -\frac{3}{2} \]

### Step 6: Determine the Intervals
The quadratic \( f(lambda) \) opens upwards (since the coefficient of \( lambda^2 \) is positive). The roots divide the number line into intervals:
1. \( (-\infty, -\frac{3}{2}) \)
2. \( (-\frac{3}{2}, 1) \)
3. \( (1, \infty) \)

Since \( f(lambda) < 0 \) between the roots, the solution to the inequality is:
\[ -\frac{3}{2} < lambda < 1 \]

### Final Answer
The value of \( lambda \) such that the point \( (lambda, lambda^2) \) lies between the lines \( |x + 2y| = 3 \) is:
\[ \boxed{(-\frac{3}{2}, 1)} \]