JEE Math Practice Question
If the solution curve of the differential equation $$(({\tan ^{ - 1}}y) - x)dy = (1 + {y^2})dx$$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is
- A.2e
- B.$${2 \over e}$$
- C.2
- D.$${1 \over e}$$
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