Question

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

• A. $\pm 2$
• B. $\pm 3$
• C. $\pm 4$
• D. $\pm 1$

Answer 1

Answer: (D)

$\frac{dy}{dx}=\left|x\right|=2\therefore x=\pm 2$
We cari solve for y to get
$y_1=\underset{0}{\overset{2}{\int }}\left|t\right|dt=\underset{0}{\overset{2}{\int }}tdt=\frac{t^2}{2}{|}_0^2=2$
and $y_3\underset{0}{\overset{-2}{\int }}\left|t\right|dt=-\underset{0}{\overset{-2}{\int }}tdt=-2$
Tangents are $y-2=2(x-2)$ and $y+2=2(x+2)$ .
Then the $x$ intercepts are obtained by putting $y=0.$
We then get $x=\pm 1$