The curve y−exy+x=0 has a curve tangent at the point:

The curve yexy+x=0 has a curve tangent at the point:

  1. A

    (1,1)

  2. B

    at no point

  3. C

    (0,1)

  4. D

    1, 0

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    Solution:

    The equation of the curve is

    yexy+x=0dydxexyy+xdydx+1=0dydx1xexy=yexy1dxdy=1xexyyexy1

    Clearly, dxdy=0 at (1, 0). So, required point is 1, 0.

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