The intercepts on the $$x$$-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

We are given the curve

$$y \;=\; \int_{0}^{x} |t| \, dt, \qquad x \in \mathbb R.$$

We want the tangents to this curve that are parallel to the line

$$y \;=\; 2x.$$

Two tangents are parallel when they have the same slope. Thus every required tangent must have slope

$$m \;=\; 2.$$

To find the slope of the curve at any point, we differentiate. By the Fundamental Theorem of Calculus, if

$$y \;=\; \int_{0}^{x} f(t)\,dt,$$

then

$$\dfrac{dy}{dx} \;=\; f(x).$$

Here the integrand $$f(t)$$ is $$|t|$$. Therefore

$$\dfrac{dy}{dx} \;=\; |x|.$$

Now we demand that the derivative be equal to the desired slope:

$$|x| \;=\; 2.$$

Because the absolute‐value function is non-negative and symmetric, this equation gives two solutions:

$$x \;=\; 2 \quad \text{or} \quad x \;=\; -2.$$

Hence the points of tangency are at $$x=2$$ and $$x=-2$$. We must next find the corresponding $$y$$-coordinates on the curve.

First consider $$x \ge 0$$. On $$[0,x]$$ the variable $$t$$ is non-negative, so $$|t|=t$$. Then

$$y \;=\; \int_{0}^{x} t \, dt \;=\; \left[\tfrac{t^{2}}{2}\right]_{0}^{x} \;=\; \tfrac{x^{2}}{2}.$$

For $$x=2$$ we obtain

$$y(2) \;=\; \tfrac{2^{2}}{2} \;=\; \tfrac{4}{2} \;=\; 2.$$

Next take $$x\lt 0$$. On $$[x,0]$$ the variable $$t$$ is negative, so $$|t|=-t$$. We write

$$y \;=\; \int_{0}^{x} |t| \, dt \;=\; -\!\int_{x}^{0} |t| \, dt \;=\; -\!\int_{x}^{0} (-t) \, dt \;=\; -\!\left[ -\tfrac{t^{2}}{2} \right]_{x}^{0} \;=\; -\!\bigl(0 - (-\tfrac{x^{2}}{2})\bigr) \;=\; -\tfrac{x^{2}}{2}.$$

For $$x=-2$$ this yields

$$y(-2) \;=\; -\tfrac{(-2)^{2}}{2} \;=\; -\tfrac{4}{2} \;=\; -2.$$

So the two points of tangency are

$$(2,\;2)\quad\text{and}\quad(-2,\;-2).$$

Each tangent line has slope $$2$$. Using the point-slope form $$y - y_0 \;=\; m(x - x_0),$$ we now find the intercepts on the $$x$$-axis, i.e. the points where $$y=0$$.

For the point $$(2,2)$$:

$$y - 2 \;=\; 2\,(x - 2).$$

Set $$y=0$$:

$$0 - 2 \;=\; 2\,(x_{\text{int}} - 2).$$

So

$$-2 \;=\; 2x_{\text{int}} - 4 \;\;\Longrightarrow\;\; 2x_{\text{int}} \;=\; 2 \;\;\Longrightarrow\;\; x_{\text{int}} \;=\; 1.$$

For the point $$(-2,-2)$$:

$$y + 2 \;=\; 2\,(x + 2).$$

Set $$y=0$$:

$$0 + 2 \;=\; 2\,(x_{\text{int}} + 2).$$

Thus

$$2 \;=\; 2x_{\text{int}} + 4 \;\;\Longrightarrow\;\; 2x_{\text{int}} \;=\; -2 \;\;\Longrightarrow\;\; x_{\text{int}} \;=\; -1.$$

Therefore the two $$x$$-intercepts are $$1$$ and $$-1$$, i.e. they are equal in magnitude to $$\pm 1$$.

Hence, the correct answer is Option C.