If$f\left(x\right)=\frac{x-e^x+\cos 2x}{x^2},x\ne 0$ , is continuous at $x=0$ , then

$\underset{x\to 0}{Lt}\frac{x-e^x+1-(1-\cos 2x)}{x^2}=\underset{x\to 0}{Lt}\left[\frac{x-e^x+1}{x^2}-\frac{(1-\cos 2x)}{x^2}\right]$

$=\underset{x\to 0}{Lt}\left[\frac{x+1-\left(1+x+\frac{x^2}{2}\right)}{x^2}-\frac{2{\sin }^{2}x}{x^2}\right]$       (Using expansion of $e^x$ )

$=-\frac{1}{2}-2$

$=-\frac{5}{2}$ ; hence for continuity $f\left(0\right)=-\frac{5}{2}$

Now [f(0)] = -3; {f(0)} $=\left[-\frac{5}{2}\right]=\frac{1}{2}$

Hence, $\left[f\right(0\left)\right]\left\{f\right(0\left)\right\}=-\frac{3}{2}=-1.5$