For the function $f(x) = e^{-2x} (2-x)^2$, the point of local maxima is:

$x=3$

The correct answer is Option (3) → $x=3$

Given: $f(x) = e^{-2x}(2 - x)^2$

To find local maxima, compute $f'(x)$ and solve $f'(x) = 0$

Use product rule:

$f'(x) = \frac{d}{dx}[e^{-2x}] \cdot (2 - x)^2 + e^{-2x} \cdot \frac{d}{dx}[(2 - x)^2]$

$= (-2e^{-2x})(2 - x)^2 + e^{-2x} \cdot 2(2 - x)(-1)$

$= e^{-2x} \left[ -2(2 - x)^2 - 2(2 - x) \right]$

$= -2e^{-2x}(2 - x)[(2 - x) + 1] = -2e^{-2x}(2 - x)(3 - x)$

Set $f'(x) = 0$:

$-2e^{-2x}(2 - x)(3 - x) = 0$

Since $e^{-2x} \ne 0$, we get: $(2 - x)(3 - x) = 0$ ⟹ $x = 2$ or $x = 3$

Check sign of $f'(x)$ around $x = 2$ and $x = 3$:

Pick $x = 1.5$: both $(2 - x)$ and $(3 - x)$ are positive ⟹ $f'(x) < 0$

Pick $x = 2.5$: $(2 - x) < 0$, $(3 - x) > 0$ ⟹ $f'(x) > 0$

Pick $x = 3.5$: both terms negative ⟹ $f'(x) < 0$

So, $f'(x)$ changes from positive to negative at $x = 3$ ⟹ local maximum at $x = 3$