$\text{f}\left(\text{x}\right)=\underset{\mathrm{\alpha }\to 0}{\mathrm{Lim}}\frac{{\mathrm{e}}^{\mathrm{x}}({\mathrm{e}}^{\mathrm{\alpha }}-1)·\mathrm{sin\alpha }·\mathrm{\ell n}(1+\mathrm{\alpha })(2017{\mathrm{\alpha }}^3+4)}{4{\mathrm{\alpha }}^3(2018{\mathrm{\alpha }}^3+1)}={\mathrm{e}}^{\mathrm{x}}$

$\mathrm{and}\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+\mathrm{x}\underset{-1}{\overset{1}{\int }}\mathrm{g}\left(\mathrm{y}\right)\mathrm{dy}+\underset{-1}{\overset{1}{\int }}\mathrm{yg}\left(\mathrm{y}\right)\mathrm{dy}$

$\mathrm{Let}\mathrm{A}=\underset{-1}{\overset{1}{\int }}\mathrm{g}\left(\mathrm{y}\right)\mathrm{dy}\mathrm{and}\mathrm{B}=\underset{-1}{\overset{1}{\int }}\mathrm{yg}\left(\mathrm{y}\right)\mathrm{dy}$

$\therefore$ g(x) = x² + Ax + B

$\therefore \mathrm{A}=\underset{-1}{\overset{1}{\int }}({\mathrm{y}}^2+\mathrm{Ay}+\mathrm{B})\mathrm{dy}=2{(\frac{{\mathrm{y}}^3}{3}+\mathrm{By})}_0^1=\frac{2}{3}+2\mathrm{B}$

$\Rightarrow \mathrm{A}-2\mathrm{B}=\frac{2}{3}$

$\text{and B }=\underset{-1}{\overset{1}{\int }}({\mathrm{y}}^3+{\mathrm{Ay}}^2+\mathrm{By})\mathrm{dy}$

$=2{\left(\frac{{\mathrm{Ay}}^3}{3}\right)}_0^1=\frac{2\mathrm{A}}{3}$

$\therefore \mathrm{A}-\frac{4\mathrm{A}}{3}=\frac{2}{3}\Rightarrow \mathrm{A}=-2\mathrm{and}\mathrm{B}=\frac{-4}{3}$

$\therefore \text{g}\left(\text{x}\right)={\text{ x}}^{\text{2}} –\text{ 2x }–\frac{4}{3} ={(\text{x }–\text{ 1})}^{\text{2}} – \frac{7}{3}.$

$\therefore \text{f}\left(\text{x}\right)=\text{ g}(\text{x }+\text{ 1})+\frac{7}{3}$

⇒ e^x = x² ⇒ x²e^–x = 1

$\therefore$ Clearly number of solutions = 1.