Let the function, $f(x) = \begin{cases} -3 a x^2 - 2, & x < 1 \\ a^2 + b x, & x \geqslant 1 \end{cases}$ be differentiable for all $x \in \mathbb{R}$, where $a > 1$, $b \in \mathbb{R}$. If the area of the region enclosed by $y = f(x)$ and the line $y = -20$ is $\alpha + \beta \sqrt{3}$, $\alpha, \beta \in \mathbb{Z}$, then the value of $\alpha + \beta$ is:

Step 1: Apply Differentiability Conditions

For a function to be differentiable at a point, it must first be continuous at that point. This means the left-hand limit and the right-hand limit of the function must be equal to the function's value at that point. Additionally, the left-hand derivative must equal the right-hand derivative at that point.

Step 2: Solve for 'a' and 'b'

By applying the continuity condition, we equate the two parts of the function at $x=1$. Then, we find the derivatives of each part and equate them at $x=1$. Substituting the value of $b$ from the derivative equation into the continuity equation, we get a quadratic equation in terms of $a$. We solve this quadratic equation for $a$.

Step 3: Determine 'a' and 'b' values

Using the quadratic formula, we find two possible values for $a$. Since the problem states that $a > 1$, we select the positive root. Then, we substitute this value of $a$ back into the equation for $b$ to find its value.

Step 4: Find intersection points with $y = -20$

To find the area, we need the intersection points of $y=f(x)$ with $y=-20$. We set each part of $f(x)$ equal to $-20$ and solve for $x$. For the first part, $x < 1$, we find the $x$ -values where $-3ax^2 - 2 = -20$. For the second part, $x \ge 1$, we find the $x$ -value where $a^2 + bx = -20$. We rationalize the denominator for the $x$ values from the first part.

Step 5: Calculate the Area

We identify the relevant intersection points. The area is calculated by integrating the difference between $f(x)$ and $y=-20$ over the appropriate intervals. The first integral is from $x_1$ to $1$ for the parabolic part, and the second integral is from $1$ to $x_2$ for the linear part. After performing the integration and substituting the limits, we find the total area.

Step 6: Determine $\alpha$ and $\beta$

Comparing the calculated area with the given form $\alpha + \beta \sqrt{3}$, we can identify the values of $\alpha$ and $\beta$.

Step 7: Calculate $\alpha + \beta$

Finally, we sum the values of $\alpha$ and $\beta$ to get the required result.

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