The intercepts on x-axis made by tangents to the curve, `y=int_0^x|t|dt , x in R ,` which are parallel to the line `y""=""2x` , are equal to (1) `+-2` (2) `+-3` (3) `+-4` (4) `+-1`

Options

A. `+-1`
B. `+-2`
C. `+-3`
D. `+-4`

Explanation

To solve the problem, we need to find the x-intercepts of the tangents to the curve defined by \( y = \int_0^x |t| dt \) that are parallel to the line \( y = 2x \).

### Step 1: Determine the curve \( y = \int_0^x |t| dt \)

The integral \( \int_0^x |t| dt \) can be evaluated based on the sign of \( x \).

1. **Case 1:** When \( x \geq 0 \):
\[ y = \int_0^x t \, dt = \left[ \frac{t^2}{2} \right]_0^x = \frac{x^2}{2} \]

2. **Case 2:** When \( x < 0 \):
\[ y = \int_0^x -t \, dt = -\left[ \frac{t^2}{2} \right]_0^x = -\frac{x^2}{2} \]

Thus, the piecewise function can be expressed as:
\[ y = \begin{cases} \frac{x^2}{2} & \text{if } x \geq 0 \\ -\frac{x^2}{2} & \text{if } x < 0 \end{cases} \]

### Step 2: Find the derivative \( \frac{dy}{dx} \)

Next, we compute the derivative of both cases:

1. **For \( x \geq 0 \):**
\[ \frac{dy}{dx} = x \]

2. **For \( x < 0 \):**
\[ \frac{dy}{dx} = -x \]

### Step 3: Set the slope equal to the slope of the line \( y = 2x \)

The slope of the line \( y = 2x \) is 2. We set the derivatives equal to 2 to find the points where the tangents are parallel to this line.

1. **For \( x \geq 0 \):**
\[ x = 2 \implies x = 2 \]

2. **For \( x < 0 \):**
\[ -x = 2 \implies x = -2 \]

### Step 4: Find the y-values at these x-values

Now we calculate the corresponding y-values at \( x = 2 \) and \( x = -2 \):

1. **For \( x = 2 \):**
\[ y = \frac{2^2}{2} = 2 \]

2. **For \( x = -2 \):**
\[ y = -\frac{(-2)^2}{2} = -2 \]

### Step 5: Find the equations of the tangents at these points

Using the point-slope form of the line equation \( y - y_1 = m(x - x_1) \):

1. **At \( (2, 2) \):**
\[ y - 2 = 2(x - 2) \implies y = 2x - 2 \]

2. **At \( (-2, -2) \):**
\[ y + 2 = 2(x + 2) \implies y = 2x + 2 \]

### Step 6: Find the x-intercepts of these tangent lines

To find the x-intercepts, set \( y = 0 \):

1. **For \( y = 2x - 2 \):**
\[ 0 = 2x - 2 \implies 2x = 2 \implies x = 1 \]

2. **For \( y = 2x + 2 \):**
\[ 0 = 2x + 2 \implies 2x = -2 \implies x = -1 \]

### Conclusion

The x-intercepts of the tangents to the curve that are parallel to the line \( y = 2x \) are \( x = 1 \) and \( x = -1 \).

Thus, the intercepts on the x-axis made by the tangents are equal to \( \pm 1 \).

**Final Answer: (4) \( \pm 1 \)**

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