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The intercepts on x- axis made by tangen...

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

A

`+-1`

B

`+-2`

C

`+-3`

D

`+-4`

Text Solution

Verified by Experts

The correct Answer is:
A
Given , y `= int_(0)^(x) |t| dt`
`:. (dy)/(dx) = |x|*1-0=|x|` [ by Leibnitz ' s rule]
`:'` Tangent to the curve y `=int_(0)^(x)|t| dt , x in R` are paralled to the line y = 2x
`:.` Slope of both are equal `rArr x = +-2`
Points, ` y = int _(0)^(+-2) |t| dt = +-2`
Equation of tangent is
`y-2=2(x-2)and y +2 =2 (x+2)`
For x intercept put y =0 we get
`0-2=2(x-2) and 0+ 2 =2 (x+2)`
`rArr x=+-1`
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