Consider a quadratic expression `f (x) =tx^(2) -(2t -1) x+ (5x -1)`
If `f (x)` is non-negetive `Aax ge 0` then t lies in the interval

Options

A. `[(1)/(5),(1)/(4)]`
B. ` [(1)/(4), oo)`
C. `[(1-)/(4), (1)/(4)]`
D. `[(1)/(5), oo)`

Answer

To solve the quadratic expression \( f(x) = tx^2 - (2t - 1)x + (5t - 1) \) and determine the interval for \( t \) such that \( f(x) \) is non-negative, we will follow these steps:

### Step 1: Identify the coefficients
The quadratic expression can be rewritten in the standard form \( ax^2 + bx + c \) where:
- \( a = t \)
- \( b = -(2t - 1) = -2t + 1 \)
- \( c = 5t - 1 \)

### Step 2: Determine the condition for non-negativity
For the quadratic \( f(x) \) to be non-negative for all \( x \), the discriminant \( D \) must be less than or equal to zero:
\[ D = b^2 - 4ac \leq 0 \]

### Step 3: Calculate the discriminant
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (-2t + 1)^2 - 4(t)(5t - 1) \]
Expanding this:
\[ D = (4t^2 - 4t + 1) - (20t^2 - 4t) \]
\[ D = 4t^2 - 4t + 1 - 20t^2 + 4t \]
\[ D = -16t^2 + 1 \]

### Step 4: Set the discriminant less than or equal to zero
Now, we set the discriminant less than or equal to zero:
\[ -16t^2 + 1 \leq 0 \]
Rearranging gives:
\[ 16t^2 \geq 1 \]
\[ t^2 \geq \frac{1}{16} \]
Taking the square root:
\[ |t| \geq \frac{1}{4} \]
This gives us two intervals:
\[ t \leq -\frac{1}{4} \quad \text{or} \quad t \geq \frac{1}{4} \]

### Step 5: Consider the condition \( t > 0 \)
Since we are given that \( t > 0 \), we only consider the interval:
\[ t \geq \frac{1}{4} \]

### Step 6: Verify if \( f(0) \geq 0 \)
We also need to ensure that \( f(0) \geq 0 \):
\[ f(0) = 5t - 1 \geq 0 \implies 5t \geq 1 \implies t \geq \frac{1}{5} \]

### Step 7: Combine the intervals
Now we combine the conditions:
1. From the discriminant, we have \( t \geq \frac{1}{4} \).
2. From \( f(0) \geq 0 \), we have \( t \geq \frac{1}{5} \).

The more restrictive condition is \( t \geq \frac{1}{4} \).

### Conclusion
Thus, the interval for \( t \) such that \( f(x) \) is non-negative is:
\[ \boxed{\left[\frac{1}{4}, \infty\right)} \]