MathematicsMedium175×since 2002Q2773If the tangent to the curve, y = f (x) = xlog_e x, (x > 0) at a point (c, f(c)) is parallel to the line-segment joining the points (1, 0) and (e, e), then c is equal to :Ae−1e{{e - 1} \over e}ee−1Be(11−e){e^{\left( {{1 \over {1 - e}}} \right)}}e(1−e1)Ce(1e−1){e^{\left( {{1 \over {e - 1}}} \right)}}e(e−11)D1e−1{1 \over {e - 1}}e−11Check answerSkip