The curve $$y(x) = ax^3 + bx^2 + cx + 5$$ touches the $$x$$-axis at the point $$P(-2, 0)$$ and cuts the $$y$$-axis at the point $$Q$$ where $$y'$$ is equal to $$3$$. Then the local maximum value of $$y(x)$$ is

The curve $$ y(x) = ax^3 + bx^2 + cx + 5 $$ touches the $$ x $$-axis at $$ P(-2, 0) $$ and cuts the $$ y $$-axis at $$ Q $$ where $$ y' = 3 $$. We need to find the local maximum value.

Since the curve touches the $$ x $$-axis at $$ P(-2, 0) $$, both $$ y(-2) = 0 $$ and $$ y'(-2) = 0 $$.

At the $$ y $$-axis ($$ x = 0 $$): $$ y(0) = 5 $$ (point Q) and $$ y'(0) = 3 $$.

$$y'(x) = 3ax^2 + 2bx + c$$

From $$ y'(0) = 3 $$: $$ c = 3 $$.

From $$ y(-2) = 0 $$:

$$a(-8) + b(4) + c(-2) + 5 = 0$$

$$-8a + 4b - 6 + 5 = 0$$

$$-8a + 4b = 1 \quad \cdots (1)$$

From $$ y'(-2) = 0 $$:

$$3a(4) + 2b(-2) + 3 = 0$$

$$12a - 4b + 3 = 0$$

$$12a - 4b = -3 \quad \cdots (2)$$

Adding (1) and (2):

$$4a = -2 \implies a = -\frac{1}{2}$$

From (1): $$ -8(-\frac{1}{2}) + 4b = 1 \implies 4 + 4b = 1 \implies b = -\frac{3}{4} $$

So $$ y(x) = -\frac{1}{2}x^3 - \frac{3}{4}x^2 + 3x + 5 $$.

$$y'(x) = -\frac{3}{2}x^2 - \frac{3}{2}x + 3 = -\frac{3}{2}(x^2 + x - 2) = -\frac{3}{2}(x + 2)(x - 1)$$

$$ y'(x) = 0 $$ at $$ x = -2 $$ and $$ x = 1 $$.

$$y''(x) = -3x - \frac{3}{2}$$

At $$ x = -2 $$: $$ y''(-2) = 6 - \frac{3}{2} = \frac{9}{2} > 0 $$ — local minimum.

At $$ x = 1 $$: $$ y''(1) = -3 - \frac{3}{2} = -\frac{9}{2} < 0 $$ — local maximum.

$$y(1) = -\frac{1}{2}(1) - \frac{3}{4}(1) + 3(1) + 5 = -\frac{1}{2} - \frac{3}{4} + 3 + 5$$

$$= -\frac{2}{4} - \frac{3}{4} + 8 = -\frac{5}{4} + 8 = \frac{-5 + 32}{4} = \frac{27}{4}$$

The local maximum value is $$ \dfrac{27}{4} $$, which corresponds to Option A.