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 \( |x + 2y| = 3 \), we start by rewriting the absolute value equation into two separate linear equations:

1. \( x + 2y = 3 \)
2. \( x + 2y = -3 \)

### Step 1: Express the equations of the lines

From the first equation, we can express \( y \):
\[ y = \frac{3 - x}{2} \]

From the second equation, we can also express \( y \):
\[ y = \frac{-3 - x}{2} \]

### Step 2: Determine the conditions for the point to lie between the lines

For the point \( (\lambda, \lambda^2) \) to lie between the two lines, it must satisfy the following inequalities:
\[ \frac{-3 - \lambda}{2} < \lambda^2 < \frac{3 - \lambda}{2} \]

### Step 3: Solve the inequalities

#### Inequality 1: \( \lambda^2 < \frac{3 - \lambda}{2} \)

Multiply both sides by 2 (note that 2 is positive, so the inequality direction remains the same):
\[ 2\lambda^2 < 3 - \lambda \]
Rearranging gives:
\[ 2\lambda^2 + \lambda - 3 < 0 \]

#### Inequality 2: \( \frac{-3 - \lambda}{2} < \lambda^2 \)

Again, multiply both sides by 2:
\[ -3 - \lambda < 2\lambda^2 \]
Rearranging gives:
\[ 2\lambda^2 + \lambda + 3 > 0 \]

### Step 4: Analyze the first inequality \( 2\lambda^2 + \lambda - 3 < 0 \)

To find the roots of the quadratic equation \( 2\lambda^2 + \lambda - 3 = 0 \), we can use the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = 1, c = -3 \).

Calculating the discriminant:
\[ D = 1^2 - 4 \cdot 2 \cdot (-3) = 1 + 24 = 25 \]
Thus, the roots are:
\[ \lambda = \frac{-1 \pm 5}{4} \]
Calculating the roots:
1. \( \lambda_1 = \frac{4}{4} = 1 \)
2. \( \lambda_2 = \frac{-6}{4} = -\frac{3}{2} \)

The quadratic \( 2\lambda^2 + \lambda - 3 < 0 \) is satisfied between the roots:
\[ -\frac{3}{2} < \lambda < 1 \]

### Step 5: Analyze the second inequality \( 2\lambda^2 + \lambda + 3 > 0 \)

The quadratic \( 2\lambda^2 + \lambda + 3 \) has a discriminant:
\[ D = 1^2 - 4 \cdot 2 \cdot 3 = 1 - 24 = -23 \]
Since the discriminant is negative, the quadratic is always positive for all \( \lambda \).

### Step 6: Combine the results

The point \( (\lambda, \lambda^2) \) lies between the lines when:
\[ -\frac{3}{2} < \lambda < 1 \]

Thus, the final answer is:
\[ \lambda \in \left(-\frac{3}{2}, 1\right) \]