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 determine the values of \( k \) such that the quadratic expression \( kx^2 - 4x + (3k + 1) > 0 \) for at least one \( x > 0 \).

### Step 1: Identify the quadratic function
The quadratic function can be expressed as:
\[ f(x) = kx^2 - 4x + (3k + 1) \]
This is a quadratic equation in the form \( ax^2 + bx + c \) where:
- \( a = k \)
- \( b = -4 \)
- \( c = 3k + 1 \)

### Step 2: Determine the condition for the quadratic to be positive
For the quadratic \( f(x) \) to be greater than 0 for at least one \( x > 0 \), we need to check the conditions on \( k \):
1. The leading coefficient \( a \) (which is \( k \)) must be positive, i.e., \( k > 0 \).
2. The discriminant \( D \) of the quadratic must be less than or equal to 0 for the quadratic to not have real roots (ensuring it does not cross the x-axis).

### Step 3: Calculate the discriminant
The discriminant \( D \) is given by:
\[ D = b^2 - 4ac = (-4)^2 - 4(k)(3k + 1) \]
Calculating this gives:
\[ D = 16 - 4k(3k + 1) = 16 - (12k^2 + 4k) \]
\[ D = 16 - 12k^2 - 4k \]

### Step 4: Set the discriminant less than or equal to zero
We want:
\[ 16 - 12k^2 - 4k < 0 \]
Rearranging this gives:
\[ 12k^2 + 4k - 16 > 0 \]
Dividing the entire inequality by 4:
\[ 3k^2 + k - 4 > 0 \]

### Step 5: Factor the quadratic inequality
To factor \( 3k^2 + k - 4 \), we can use the quadratic formula to find the roots:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ = \frac{-1 \pm \sqrt{1 + 48}}{6} = \frac{-1 \pm \sqrt{49}}{6} = \frac{-1 \pm 7}{6} \]
The roots are:
\[ k_1 = \frac{6}{6} = 1 \quad \text{and} \quad k_2 = \frac{-8}{6} = -\frac{4}{3} \]

### Step 6: Analyze the intervals
The quadratic \( 3k^2 + k - 4 \) opens upwards (since the coefficient of \( k^2 \) is positive). The roots divide the number line into intervals:
- \( (-\infty, -\frac{4}{3}) \)
- \( (-\frac{4}{3}, 1) \)
- \( (1, \infty) \)

Testing a point in each interval:
- For \( k < -\frac{4}{3} \) (e.g., \( k = -2 \)): \( 3(-2)^2 + (-2) - 4 = 12 - 2 - 4 = 6 > 0 \)
- For \( -\frac{4}{3} < k < 1 \) (e.g., \( k = 0 \)): \( 3(0)^2 + (0) - 4 = -4 < 0 \)
- For \( k > 1 \) (e.g., \( k = 2 \)): \( 3(2)^2 + (2) - 4 = 12 + 2 - 4 = 10 > 0 \)

### Step 7: Conclusion
The quadratic \( 3k^2 + k - 4 > 0 \) holds for:
\[ k < -\frac{4}{3} \quad \text{or} \quad k > 1 \]
Since \( k \) must also be positive for \( kx^2 \) to be a valid quadratic, we conclude:
\[ k \in (1, \infty) \]

### Final Answer
Thus, the set \( S \) containing \( k \) is:
\[ S = (1, \infty) \]