Let  $f\left(x\right)=\{\begin{array}{cc}\frac{e^{ax}-e^x-x}{x^2}& x\ne 0\\ \frac{3}{2}& x=0\end{array}$ The value of a  so that f is continuous function is 

$f\left(x\right)$is continuous at $x=0$

$\therefore f\left(0\right)=\underset{x\to 0}{Lt}f\left(x\right)$

$\frac{3}{2}=\underset{x\to 0}{Lt}\frac{e^{ax}-e^x-x}{x^2}$

$=\underset{x\to 0}{Lt}\frac{a.e^{ax}-e^x-1}{2x}$(L – Hospital’s Rule)

$\frac{3}{2}=\underset{x\to 0}{Lt}\frac{a^2{e}^{ax}-e^x}{2}$

$\Rightarrow \frac{3}{2}=\frac{a^2{e}^{a\left(0\right)}-e^0}{2}$

$\Rightarrow \frac{3}{2}=\frac{a^2-1}{2}\Rightarrow a^2=4\Rightarrow a=\pm 2$