For `y=f(x)=overset(x)underset(0)int 2|t|`dt, the tangent lines parallel to the bi-sector of the first quadrant angle are

Options

A. `y=x+(3)/(4),y=x-(1)/(4)`
B. `y=-x+(1)/(4), y=-x+(3)/(4)`
C. `x-y=2, x-y=1`
D. None of these

Answer

To solve the problem, we need to find the tangent lines to the function \( y = f(x) = \int_0^x 2|t| \, dt \) that are parallel to the bisector of the first quadrant angle. The bisector of the first quadrant makes an angle of \( \frac{\pi}{4} \) with the x-axis, which means the slope of the tangent lines we are looking for is \( 1 \).

### Step 1: Define the function
The function is given as:
\[ y = f(x) = \int_0^x 2|t| \, dt \]
We need to evaluate this integral based on the value of \( x \).

### Step 2: Evaluate the integral for \( x \geq 0 \)
When \( x \geq 0 \), \( |t| = t \). Therefore, the integral becomes:
\[ f(x) = \int_0^x 2t \, dt \]
Calculating this integral:
\[ f(x) = 2 \left[ \frac{t^2}{2} \right]_0^x = [t^2]_0^x = x^2 \]
So, for \( x \geq 0 \):
\[ f(x) = x^2 \]

### Step 3: Evaluate the integral for \( x < 0 \)
When \( x < 0 \), \( |t| = -t \). Therefore, the integral becomes:
\[ f(x) = \int_0^x 2(-t) \, dt = -2 \int_0^x t \, dt \]
Calculating this integral:
\[ f(x) = -2 \left[ \frac{t^2}{2} \right]_0^x = -[t^2]_0^x = -x^2 \]
So, for \( x < 0 \):
\[ f(x) = -x^2 \]

### Step 4: Find the derivative
Now we find the derivative \( f'(x) \) for both cases.

1. For \( x \geq 0 \):
\[ f'(x) = \frac{d}{dx}(x^2) = 2x \]

2. For \( x < 0 \):
\[ f'(x) = \frac{d}{dx}(-x^2) = -2x \]

### Step 5: Set the derivative equal to the slope
We want the slope of the tangent line to be \( 1 \).

1. For \( x \geq 0 \):
\[ 2x = 1 \implies x = \frac{1}{2} \]

2. For \( x < 0 \):
\[ -2x = 1 \implies x = -\frac{1}{2} \]

### Step 6: Find the corresponding \( y \) values
Now we find the \( y \) values for both \( x \) values.

1. For \( x = \frac{1}{2} \):
\[ y = f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
So the point is \( \left(\frac{1}{2}, \frac{1}{4}\right) \).

2. For \( x = -\frac{1}{2} \):
\[ y = f\left(-\frac{1}{2}\right) = -\left(-\frac{1}{2}\right)^2 = -\frac{1}{4} \]
So the point is \( \left(-\frac{1}{2}, -\frac{1}{4}\right) \).

### Step 7: Write the equations of the tangent lines
Using the point-slope form of the equation of a line \( y - y_1 = m(x - x_1) \):

1. For the point \( \left(\frac{1}{2}, \frac{1}{4}\right) \):
\[ y - \frac{1}{4} = 1\left(x - \frac{1}{2}\right) \]
This simplifies to:
\[ y = x - \frac{1}{4} \]

2. For the point \( \left(-\frac{1}{2}, -\frac{1}{4}\right) \):
\[ y + \frac{1}{4} = 1\left(x + \frac{1}{2}\right) \]
This simplifies to:
\[ y = x + \frac{1}{4} \]

### Final Answer
The equations of the tangent lines parallel to the bisector of the first quadrant are:
\[ y = x - \frac{1}{4} \quad \text{and} \quad y = x + \frac{1}{4} \]