Let the function f(x) = begin{cases} -3ax^2 - | vedclass

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(A) For $f(x)$ to be differentiable at $x = 1$,it must be continuous and have equal left-hand and right-hand derivatives.Continuity at $x = 1$: $\lim_

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$34$

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(A) For $f(x)$ to be differentiable at $x = 1$,it must be continuous and have equal left-hand and right-hand derivatives.Continuity at $x = 1$: $\lim_{x 	o 1^-} (-3ax^2 - 2) = \lim_{x 	o 1^+} (a^2 + bx) \implies -3a - 2 = a^2 + b$.Differentiability at $x = 1$: $\frac{d}{dx}(-3ax^2 - 2)|_{x=1} = \frac{d}{dx}(a^2 + bx)|_{x=1} \implies -6a = b$.Substituting $b = -6a$ into the continuity equation: $-3a - 2 = a^2 - 6a \implies a^2 - 3a + 2 = 0 \implies (a - 1)(a - 2) = 0$.Since $a > 1$,we have $a = 2$. Then $b = -6(2) = -12$.Thus,$f(x) = \begin{cases} -6x^2 - 2, & x The region is bounded by $y = f(x)$ and $y = -20$. We find the intersection points:For $x For $x \geq 1$: $4 - 12x = -20 \implies 12x = 24 \implies x = 2$.The area $A = \int_{-\sqrt{3}}^{1} (f(x) - (-20)) dx + \int_{1}^{2} (f(x) - (-20)) dx$.$A = \int_{-\sqrt{3}}^{1} (-6x^2 - 2 + 20) dx + \int_{1}^{2} (4 - 12x + 20) dx$.$A = \int_{-\sqrt{3}}^{1} (-6x^2 + 18) dx + \int_{1}^{2} (24 - 12x) dx$.$A = [-2x^3 + 18x]_{-\sqrt{3}}^{1} + [24x - 6x^2]_{1}^{2}$.$A = (-2 + 18) - (2(3\sqrt{3}) - 18\sqrt{3}) + (48 - 24) - (24 - 6)$.$A = 16 - (6\sqrt{3} - 18\sqrt{3}) + 24 - 18 = 16 + 12\sqrt{3} + 6 = 22 + 12\sqrt{3}$.Comparing with $\alpha + \beta \sqrt{3}$,we get $\alpha = 22, \beta = 12$.Therefore,$\alpha + \beta = 22 + 12 = 34$.

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