If `1+2x +3x^2+4x^3 +…..oo ge 4 ` then

Options

A. least value of x is 1/2
B. greatest value of x is 4/3
C. least value of x is 2/3
D. greatest value of x does not exist

Explanation

To solve the inequality \( 1 + 2x + 3x^2 + 4x^3 + \ldots \geq 4 \), we can follow these steps:

### Step 1: Identify the series
The series can be expressed as:
\[ S = 1 + 2x + 3x^2 + 4x^3 + \ldots \]
This series can be recognized as a power series.

### Step 2: Use the formula for the sum of the series
The sum of the series \( S \) can be derived using the formula for the sum of an infinite series. The series \( S \) can be represented as:
\[ S = \sum_{n=1}^{\infty} n x^{n-1} \]
Using the formula for the sum of the series, we have:
\[ S = \frac{1}{(1-x)^2} \quad \text{for } |x| < 1 \]

### Step 3: Set up the inequality
We need to solve the inequality:
\[ \frac{1}{(1-x)^2} \geq 4 \]

### Step 4: Manipulate the inequality
To solve the inequality, we can first rearrange it:
\[ 1 \geq 4(1-x)^2 \]
Expanding the right side:
\[ 1 \geq 4(1 - 2x + x^2) \]
This simplifies to:
\[ 1 \geq 4 - 8x + 4x^2 \]
Rearranging gives:
\[ 0 \geq 4x^2 - 8x + 3 \]

### Step 5: Solve the quadratic inequality
We can rewrite the inequality:
\[ 4x^2 - 8x + 3 \leq 0 \]
To find the roots of the quadratic equation \( 4x^2 - 8x + 3 = 0 \), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 4 \cdot 3}}{2 \cdot 4} \]
Calculating the discriminant:
\[ = \frac{8 \pm \sqrt{64 - 48}}{8} = \frac{8 \pm \sqrt{16}}{8} = \frac{8 \pm 4}{8} \]
This gives us the roots:
\[ x = \frac{12}{8} = \frac{3}{2} \quad \text{and} \quad x = \frac{4}{8} = \frac{1}{2} \]

### Step 6: Analyze the intervals
The roots divide the number line into intervals. We need to test the sign of the quadratic in the intervals:
1. \( (-\infty, \frac{1}{2}) \)
2. \( (\frac{1}{2}, \frac{3}{2}) \)
3. \( (\frac{3}{2}, \infty) \)

Testing a point in each interval:
- For \( x = 0 \) in \( (-\infty, \frac{1}{2}) \): \( 4(0)^2 - 8(0) + 3 = 3 > 0 \)
- For \( x = 1 \) in \( (\frac{1}{2}, \frac{3}{2}) \): \( 4(1)^2 - 8(1) + 3 = -1 < 0 \)
- For \( x = 2 \) in \( (\frac{3}{2}, \infty) \): \( 4(2)^2 - 8(2) + 3 = 3 > 0 \)

Thus, the quadratic \( 4x^2 - 8x + 3 \leq 0 \) holds true in the interval:
\[ \frac{1}{2} \leq x \leq \frac{3}{2} \]

### Step 7: Combine with the convergence condition
Since the series converges for \( |x| < 1 \), we also need to consider this condition. Therefore, we combine the intervals:
\[ \frac{1}{2} \leq x < 1 \]

### Conclusion
The solution to the inequality \( 1 + 2x + 3x^2 + 4x^3 + \ldots \geq 4 \) is:
\[ \frac{1}{2} \leq x < 1 \]