If 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 :

y = f (x) = xlog_e x <br><br>$\Rightarrow$ ${{dy} \over {dx}} =$ 1 + log_e x <br><br>$\Rightarrow$ ${\left. {{{dy} \over {dx}}} \right|_{\left( {c,f\left( c \right)} \right)}}$ = 1 + log_e e = m₁ <br><br>This tangent parallel to the line-segment<br> joining the points (1, 0) and (e, e). <br><br>$\therefore$ Slope of line-segment joining the points (1, 0) and (e, e) = m₁ <br><br>$\Rightarrow$ ${{e - 0} \over {e - 1}}$ = 1 + log_e e <br><br>$\Rightarrow$ log_e e = ${{e - 0} \over {e - 1}}$ - 1 = ${1 \over {e - 1}}$ <br><br>$\Rightarrow$ c = ${e^{\left( {{1 \over {e - 1}}} \right)}}$