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The intercepts on x-axis made by tangents to thecurve, `y=int_0^x|t|dt , x in R ,`which are parallel to the line `y""=""2x`, are equalto(1) `+-2`(2) `+-3`(3) `+-4`(4) `+-1`A. `pm 1`B. `pm 2`C. `pm 3`D. `pm 4` |
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Answer» Correct Answer - A 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. ` |
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