InterviewSolution
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InterviewSolution
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A tangent to the curve `y= int_(0)^(x)|x|dt,` which is parallel to the line `y=x,` cuts off an intercept from the y-axis is equal toA. 1B. `(-1)/(2),(1)/(2)`C. `(1)/(2),1`D. -1 |
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Answer» Correct Answer - B We have, `y= int_(0)^(x)|x|dt " "...(i)` Differentiating w.r.t. x, we get ` (dy)/(dx)=|x| ` Let `P (x_(1),y_(1))` 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 `(x_(1),y_(1))=1` ` rArr ((dy)/(dx))_((x_(1)","y_(1)))=1 rArr |x_(1)|=1 rArr x_(1)= pm 1 ` Now, `y=int_(0)^(x)|t|dt rArr y={(int_(0)^(x)tdt=(x^(2))/(2) ",", "if " x ge 0),(int_(0)^(x) -tdt=-(x^(2))/(2) ",", "if " x lt 0):} ` ` therefore x_(1)=1 rArr y_(1)=(1)/(2) " "["Putting " x_(1)=1 " in " y_(1)=(x_(1)^(2))/(2)] ` and, ` x_(1)=-1 rArr y_(1)=-(1)/(2) " "[" Putting " x_(1)=-1 " in " y_(1)=-(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-(1)/(2)=1(x-1) " and " y+(1)/(2)=1(x+1) ` respectively. ` rArr 2x-2y-1=0 " and " 2x-2y+1=0 ` Clearly, these tangents cut off intercepts ` -(1)/(2) " and " (1)/(2) ` respectively on y-axis. |
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