If `kx ^(2)-4x+3k + 1 gt 0` for atleast one `x gt 0,` then if `k in S` contains :

Options

A. `(1,oo)`
B. ` (0,oo)`
C. ` (-1, oo)`
D. ` (-(1)/(4), oo)`

Answer

To solve the problem, we need to analyze the quadratic inequality given by \( kx^2 - 4x + (3k + 1) > 0 \) for at least one \( x > 0 \).

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The quadratic can be expressed in the standard form \( ax^2 + bx + c \) where:
- \( a = k \)
- \( b = -4 \)
- \( c = 3k + 1 \)

2. **Condition for \( a \)**:
For the quadratic to be positive for at least one \( x > 0 \), the coefficient \( a \) must be positive:
\[ k > 0 \]

**Hint**: The leading coefficient of a quadratic must be positive for the parabola to open upwards.

3. **Discriminant Condition**:
The discriminant \( D \) of the quadratic must be less than or equal to zero for the quadratic to not have real roots (or to have a double root). The discriminant is given by:
\[ D = b^2 - 4ac = (-4)^2 - 4(k)(3k + 1) = 16 - 4k(3k + 1) \]
Simplifying this:
\[ D = 16 - (12k^2 + 4k) = -12k^2 - 4k + 16 \]

4. **Set the Discriminant Less Than or Equal to Zero**:
We need to find when \( D \leq 0 \):
\[ -12k^2 - 4k + 16 \leq 0 \]
Multiplying through by -1 (which reverses the inequality):
\[ 12k^2 + 4k - 16 \geq 0 \]

5. **Factor the Quadratic**:
We can factor the quadratic:
\[ 12k^2 + 4k - 16 = 4(3k^2 + k - 4) \]
To find the roots of \( 3k^2 + k - 4 = 0 \), we can use the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 48}}{6} = \frac{-1 \pm 7}{6} \]
This gives us:
\[ k = 1 \quad \text{and} \quad k = -\frac{4}{3} \]

6. **Determine Intervals**:
The roots divide the number line into intervals. We need to test the sign of \( 12k^2 + 4k - 16 \) in the intervals:
- \( (-\infty, -\frac{4}{3}) \)
- \( (-\frac{4}{3}, 1) \)
- \( (1, \infty) \)

Testing these intervals:
- For \( k < -\frac{4}{3} \): Positive
- For \( -\frac{4}{3} < k < 1 \): Negative
- For \( k > 1 \): Positive

Thus, \( 12k^2 + 4k - 16 \geq 0 \) for:
\[ k \in (-\infty, -\frac{4}{3}] \cup [1, \infty) \]

7. **Combine Conditions**:
We have two conditions:
- \( k > 0 \)
- \( k \in (-\infty, -\frac{4}{3}] \cup [1, \infty) \)

The intersection of these two conditions gives:
\[ k \in [1, \infty) \]

### Final Answer:
Thus, the set \( S \) contains:
\[ k \in [1, \infty) \]