Consider a quadratic expression `f (x) =tx^(2) -(2t -1) x+ (5x -1)`
If `f (x)` can take both positive and negative values then t must lie in the interval

Options

A. `((-1)/(4 ),(1)/(4))`
B. ` (-oo, (-1)/(4))uu ((1)/(4))`
C. `((1)/(4))={0}`
D. `(-4,4)`

Answer

To solve the problem, we need to analyze the quadratic expression given by:

\[ f(x) = tx^2 - (2t - 1)x + (5t - 1) \]

### Step 1: Identify coefficients
We can identify the coefficients \( a \), \( b \), and \( c \) from the standard form of a quadratic equation \( ax^2 + bx + c \):

- \( a = t \)
- \( b = -(2t - 1) = -2t + 1 \)
- \( c = 5t - 1 \)

### Step 2: Condition for both positive and negative values
For the quadratic expression \( f(x) \) to take both positive and negative values, the discriminant \( D \) must be greater than zero. The discriminant \( D \) is given by:

\[ D = b^2 - 4ac \]

### Step 3: Calculate the discriminant
Substituting the values of \( a \), \( b \), and \( c \):

\[ D = (-2t + 1)^2 - 4(t)(5t - 1) \]

Calculating \( D \):

\[ D = (4t^2 - 4t + 1) - (20t^2 - 4t) \]

\[ D = 4t^2 - 4t + 1 - 20t^2 + 4t \]

\[ D = -16t^2 + 1 \]

### Step 4: Set the discriminant greater than zero
To ensure that the quadratic can take both positive and negative values, we need:

\[ -16t^2 + 1 > 0 \]

### Step 5: Solve the inequality
Rearranging the inequality:

\[ 16t^2 < 1 \]

Dividing both sides by 16:

\[ t^2 < \frac{1}{16} \]

Taking the square root of both sides:

\[ -\frac{1}{4} < t < \frac{1}{4} \]

### Step 6: Conclusion
Thus, the values of \( t \) must lie in the interval:

\[ t \in \left(-\frac{1}{4}, \frac{1}{4}\right) \]