The intercepts on `x`- axis made by tangents to the curve, `y=int_(0)^(x)|t|dt, x epsilonR` which are parallel to the line `y=2x`, are equal to

Options

A. `pm 1`
B. `pm 2`
C. `pm 3`
D. `pm 4`

Answer

A

Explanation

We have,
`y=int_(0)^(x)|t|dt, x in R " "...(i)`
` therefore (dy)/(dx)=|x|, x in R `
Let `(x_(1),y_(1))` be a point on the curve `y =int_(0)^(x)|t|dt ` where tangent is parallel to the line `y=2x.` Then,
`((dy)/(dx))_((x_(1)","y_(1)))=2 rArr |x_(1)|=2 rArr x_(1)=pm 2 `
Since`(x_(1),y_(1))` lies on (i). Therefore,
`y_(1)= int_(0)^(x_(1))|t|d `
When `x=2, y_(1)=int_(0)^(2)|t|dt=int_(0)^(2)t dt =[(t^(2))/(2)]_(0)^(2)=2`
When `x=-2, y_(1)=int_(0)^(-2)-tdt=int_(0)^(-2)-t dt =[-(t^(2))/(2)]_(0)^(-2)=-2 `
So, points on the curve are (2,2) and (-2,-2). The equations of tangents at these points are
` y-2=2(x-2) " and " y+2=2(x+2) `
or, ` y=2x-2 ` and `y=2x+2 `
The x-intercepts of these tangents are ` pm 1. `