If `f(x)=x^3+4x^2+lambdax+1` is a monotonically decreasing function of `x` in the largest possible interval `(-2,-2/3)dot` Then `lambda=4` (b) `lambda=2` `lambda=-1` (d) `lambda` has no real value

Options

A. `lambda=4`
B. `lambda=2`
C. `lambda=-1`
D. `lambda` has no real value

Explanation

To determine the value of \( \lambda \) for which the function \( f(x) = x^3 + 4x^2 + \lambda x + 1 \) is monotonically decreasing in the interval \( (-2, -\frac{2}{3}) \), we can follow these steps:

### Step 1: Find the First Derivative
To check if the function is monotonically decreasing, we need to find the first derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 + 4x^2 + \lambda x + 1) = 3x^2 + 8x + \lambda \]

### Step 2: Set the Condition for Monotonicity
For \( f(x) \) to be monotonically decreasing, the first derivative \( f'(x) \) must be less than or equal to zero in the interval \( (-2, -\frac{2}{3}) \):
\[ f'(x) \leq 0 \]

### Step 3: Analyze the Quadratic Equation
The expression \( f'(x) = 3x^2 + 8x + \lambda \) is a quadratic function. For this quadratic to be non-positive in the interval \( (-2, -\frac{2}{3}) \), it must have its roots within this interval.

### Step 4: Find the Roots of the Quadratic
The roots of the quadratic equation \( 3x^2 + 8x + \lambda = 0 \) can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot \lambda}}{2 \cdot 3} \]
\[ = \frac{-8 \pm \sqrt{64 - 12\lambda}}{6} \]

### Step 5: Ensure Roots are in the Interval
For the roots to lie within the interval \( (-2, -\frac{2}{3}) \), the following conditions must hold:
1. The discriminant must be non-negative:
\[ 64 - 12\lambda \geq 0 \implies \lambda \leq \frac{64}{12} \implies \lambda \leq \frac{16}{3} \]

2. The roots must be less than -2 and greater than -\(\frac{2}{3}\). We can use the product of the roots condition:
\[ \text{Product of roots} = \frac{c}{a} = \frac{\lambda}{3} \]
The product of the roots must be greater than \( -2 \times -\frac{2}{3} = \frac{4}{3} \):
\[ \frac{\lambda}{3} > \frac{4}{3} \implies \lambda > 4 \]

### Step 6: Combine the Conditions
From the conditions derived:
- \( \lambda \leq \frac{16}{3} \)
- \( \lambda > 4 \)

This means \( \lambda \) must be in the range \( (4, \frac{16}{3}] \).

### Conclusion
Since the only integer value of \( \lambda \) that satisfies these conditions is \( \lambda = 4 \), we conclude that:
\[ \lambda = 4 \]

### Final Answer
The correct option is (a) \( \lambda = 4 \).

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