A) 1
B) -2
C) 2
D) None of these
Correct Answer: A
Solution :
Given,\[y=\int_{0}^{x}{|t|}\,\,dt\] On differentiating w.r.t. x, we get \[\frac{dy}{dx}=|x|\] Since, the tangent is parallel to the line\[y=2x\] \[\therefore \] \[|x|\,\,=2\Rightarrow x=\pm 2\] When\[x=2,\,\,y=\int_{0}^{2}{|t|}\,\,dt=\int_{0}^{2}{t\,\,dt}\] \[=\left[ \frac{{{t}^{2}}}{2} \right]_{0}^{2}=2\] When\[x=-2\], \[y=\int_{0}^{-2}{|t|}\,\,dt=\int_{-2}^{0}{t\,\,dt}\] \[=\left[ \frac{{{t}^{2}}}{2} \right]_{-2}^{0}=-2\] Then, tangent at (2, 2) is \[y-2=2(x-2)\] \[\Rightarrow \] \[2x-y-2=0\] For,\[x-\]intercept, put\[y=0\], then\[x=1\] Also, tangent at (-2,-2) is \[y+2=2(x+2)\] \[\Rightarrow \,\,\,\,2x-y+2=0\] For\[x-\]intercept, put\[y=0\], then\[x=-1\].
You need to login to perform this action.
You will be redirected in
3 sec
