A tangent to the curve $y=\int\limits_0^x|x| d t$, which is parallel to the line y = x, cuts off an intercept from the y-axis is equal to

$\frac{-1}{2}, \frac{1}{2}$

We have,

$y=\int\limits_0^x|x| d t$            ......(i)

Differentiating w.r.t. x, we get

Let $P\left(x_1, y_1\right)$ be a point on the curve (i) such that the tangent at P is parallel to the line y = x. Then,

Slope of the tangent at $\left(x_1, y_1\right)=1$

$\Rightarrow \left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=1 \Rightarrow\left|x_1\right|=1 \Rightarrow x_1= \pm 1$

Now,

$y=\int\limits_0^x|t| d t \Rightarrow y= \begin{cases}\int\limits_0^x t d t=\frac{x^2}{2}, & \text { if } x \geq 0 \\ \int\limits_0^x-t d t=-\frac{x^2}{2}, & \text { if } x<0\end{cases}$

∴  $x_1=1 \Rightarrow y_1=\frac{1}{2}$          [Putting $x_1=1$ in $y_1=\frac{x_1^2}{2}$]

and,

$x_1=-1 \Rightarrow y_1=-\frac{1}{2}$           [Putting $x_1=-1$ in $y_1=-\frac{x_1{ }^2}{2}$]

Thus, the two points on the curve are $(1,1 / 2)$ and $(-1,-1 / 2)$

The equations of the tangents at these two points are $y-\frac{1}{2}=1(x-1)$ and $y+\frac{1}{2}=1(x+1)$ respectively.

$\Rightarrow 2 x-2 y-1=0$ and $2 x-2 y+1=0$

Clearly, these tangents cut off intercepts $-\frac{1}{2}$ and $\frac{1}{2}$ respectively on y-axis.