Official solution
y = f (x) = xloge x
\[ \Rightarrow \] \[{{dy} \over {dx}} = \] 1 + loge x
\[ \Rightarrow \] \[{\left. {{{dy} \over {dx}}} \right|_{\left( {c,f\left( c \right)} \right)}}\] = 1 + loge e = m1
This tangent parallel to the
line-segment
joining the points (1, 0) and
(e, e).
\[ \therefore \] Slope of line-segment joining the points (1, 0) and
(e, e) = m1
\[ \Rightarrow \] \[{{e - 0} \over {e - 1}}\] = 1 + loge e
\[ \Rightarrow \] loge e = \[{{e - 0} \over {e - 1}}\] - 1 = \[{1 \over {e - 1}}\]
\[ \Rightarrow \] c = \[{e^{\left( {{1 \over {e - 1}}} \right)}}\]