Key Concepts and Formulas

• Differentiability Implies Continuity: A function is differentiable at a point if and only if it is continuous at that point. This means the left-hand limit (LHL), right-hand limit (RHL), and the function value at that point must be equal.
• Differentiability Condition: For a piecewise function to be differentiable at the point where the definition changes, the left-hand derivative (LHD) must equal the right-hand derivative (RHD).
• Area Under a Curve: The area of the region enclosed by a curve $y=f(x)$ and the x-axis between $x=a$ and $x=b$ is given by $\int_a^b |f(x)| dx$. If the region is enclosed by two curves $y=f(x)$ and $y=g(x)$, the area is $\int_a^b |f(x) - g(x)| dx$.

Step-by-Step Solution

Step 1: Use the differentiability condition at $x=1$ to establish relationships between $a$ and $b$.

Since $f(x)$ is differentiable for all $x \in \mathbf{R}$, it must be differentiable at $x=1$. This implies two conditions: continuity at $x=1$ and equality of the derivatives from the left and right at $x=1$.

• Continuity at $x=1$: LHL = RHL = $f(1)$ $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$ $\lim_{x \to 1^-} (-3ax^2 - 2) = \lim_{x \to 1^+} (a^2 + bx) = a^2 + b(1)$ $-3a(1)^2 - 2 = a^2 + b$ $-3a - 2 = a^2 + b$ (Equation 1)

• Differentiability at $x=1$: LHD = RHD First, find the derivatives of the two pieces: For $x<1$, $f'(x) = \frac{d}{dx}(-3ax^2 - 2) = -6ax$ For $x \geq 1$, $f'(x) = \frac{d}{dx}(a^2 + bx) = b$

  Now, evaluate the limits of these derivatives as $x \to 1$: LHD: $\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} (-6ax) = -6a(1) = -6a$ RHD: $\lim_{x \to 1^+} f'(x) = \lim_{x \to 1^+} b = b$

  Equating LHD and RHD: $-6a = b$ (Equation 2)

Step 2: Solve the system of equations to find the values of $a$ and $b$.

Substitute Equation 2 into Equation 1: $-3a - 2 = a^2 + (-6a)$ $-3a - 2 = a^2 - 6a$ $a^2 - 3a + 2 = 0$

This is a quadratic equation in $a$. We can factor it: $(a-1)(a-2) = 0$ So, $a=1$ or $a=2$.

The problem states that $a > 1$. Therefore, we must choose $a=2$.

Now, use Equation 2 to find $b$: $b = -6a = -6(2) = -12$.

So, $a=2$ and $b=-12$.

Step 3: Write down the explicit form of the function $f(x)$ with the determined values of $a$ and $b$.

Substituting $a=2$ and $b=-12$ into the definition of $f(x)$: $f(x)= \begin{cases}-3(2)x^2-2, & x<1 \\ (2)^2+(-12) x, & x \geqslant 1\end{cases}$ $f(x)= \begin{cases}-6x^2-2, & x<1 \\ 4-12x, & x \geqslant 1\end{cases}$

Step 4: Determine the points of intersection between $y=f(x)$ and the line $y=-20$.

We need to find where $f(x) = -20$.

• Case 1: $x < 1$ $-6x^2 - 2 = -20$ $-6x^2 = -18$ $x^2 = 3$ $x = \pm \sqrt{3}$ Since we are considering $x < 1$, both $x = -\sqrt{3}$ (approximately -1.732) and $x = \sqrt{3}$ (approximately 1.732) need to be checked against the condition $x<1$. $x = -\sqrt{3}$ satisfies $x < 1$. $x = \sqrt{3}$ does NOT satisfy $x < 1$. So, $x = \sqrt{3}$ is not a valid intersection point for this piece.

• Case 2: $x \geq 1$ $4 - 12x = -20$ $-12x = -24$ $x = 2$ This satisfies the condition $x \geq 1$.

So, the intersection points of $y=f(x)$ and $y=-20$ are $x = -\sqrt{3}$ and $x = 2$.

Step 5: Calculate the area of the region enclosed by $y=f(x)$ and $y=-20$.

The region is bounded by $x = -\sqrt{3}$ and $x = 2$. We need to consider the definition of $f(x)$ in different intervals. The critical point is $x=1$.

The area is given by $\int_{-\sqrt{3}}^{2} |f(x) - (-20)| dx = \int_{-\sqrt{3}}^{2} |f(x) + 20| dx$.

Let's analyze the sign of $f(x) + 20$ in the relevant intervals.

• For $x < 1$, $f(x) + 20 = (-6x^2 - 2) + 20 = -6x^2 + 18 = -6(x^2 - 3)$. This expression is positive when $x^2 < 3$, i.e., $-\sqrt{3} < x < \sqrt{3}$. In the interval $[-\sqrt{3}, 1)$, $f(x)+20$ is positive. So, for $-\sqrt{3} \leq x < 1$, $f(x)+20 = -6x^2+18$.

• For $x \geq 1$, $f(x) + 20 = (4 - 12x) + 20 = 24 - 12x = -12(x - 2)$. This expression is positive when $x < 2$. In the interval $[1, 2)$, $f(x)+20$ is positive. So, for $1 \leq x < 2$, $f(x)+20 = 24-12x$. At $x=2$, $f(x)+20=0$.

Therefore, the area can be calculated by splitting the integral at $x=1$: Area $= \int_{-\sqrt{3}}^{1} (f(x) + 20) dx + \int_{1}^{2} (f(x) + 20) dx$ Area $= \int_{-\sqrt{3}}^{1} (-6x^2 - 2 + 20) dx + \int_{1}^{2} (4 - 12x + 20) dx$ Area $= \int_{-\sqrt{3}}^{1} (-6x^2 + 18) dx + \int_{1}^{2} (24 - 12x) dx$

Step 6: Evaluate the integrals.

First integral: $\int_{-\sqrt{3}}^{1} (-6x^2 + 18) dx = [-2x^3 + 18x]_{-\sqrt{3}}^{1}$ $= (-2(1)^3 + 18(1)) - (-2(-\sqrt{3})^3 + 18(-\sqrt{3}))$ $= (-2 + 18) - (-2(-3\sqrt{3}) - 18\sqrt{3})$ $= 16 - (6\sqrt{3} - 18\sqrt{3})$ $= 16 - (-12\sqrt{3})$ $= 16 + 12\sqrt{3}$

Second integral: $\int_{1}^{2} (24 - 12x) dx = [24x - 6x^2]_{1}^{2}$ $= (24(2) - 6(2)^2) - (24(1) - 6(1)^2)$ $= (48 - 6(4)) - (24 - 6)$ $= (48 - 24) - 18$ $= 24 - 18$ $= 6$

Step 7: Combine the results to find the total area.

Total Area = (Result of first integral) + (Result of second integral) Total Area $= (16 + 12\sqrt{3}) + 6$ Total Area $= 22 + 12\sqrt{3}$

Step 8: Determine the values of $\alpha$ and $\beta$ and calculate $\alpha + \beta$.

The area is given in the form $\alpha + \beta\sqrt{3}$, where $\alpha, \beta \in \mathbf{Z}$. Comparing $22 + 12\sqrt{3}$ with $\alpha + \beta\sqrt{3}$, we have: $\alpha = 22$ $\beta = 12$

Both $\alpha$ and $\beta$ are integers, as required.

Finally, calculate $\alpha + \beta$: $\alpha + \beta = 22 + 12 = 34$.

Common Mistakes & Tips

• Forgetting $a>1$ condition: Always check the given constraints on variables. If $a=1$ was allowed, it would lead to different values for $b$ and potentially a different area.
• Incorrectly evaluating limits for derivatives: Ensure you are taking the limit of the derivative function, not the derivative of the limit.
• Sign errors in integration or evaluation: Double-check calculations, especially when dealing with negative numbers and square roots. Pay close attention to the absolute value in the area integral; ensure the integrand is positive in the integration interval.
• Splitting the integral correctly: The integral for the area must be split at $x=1$ because the definition of $f(x)$ changes there. Also, ensure the limits of integration are correct based on the intersection points.

Summary

The problem requires us to first use the conditions of differentiability at the point where the piecewise function definition changes ($x=1$) to find the unknown constants $a$ and $b$. After determining the explicit form of $f(x)$, we find the intersection points of $y=f(x)$ and $y=-20$. The area enclosed by these two curves is then calculated by integrating the absolute difference between the functions over the appropriate interval, splitting the integral at $x=1$ as necessary. Finally, by comparing the calculated area with the given form $\alpha + \beta\sqrt{3}$, we find $\alpha$ and $\beta$ and compute their sum.

The final answer is $\boxed{34}$.