**21.** Let the function \[ f(x) = \begin{cases} -3a x^2 - 2, & x < 1 \\ a^2 + bx, & x \geq 1 \end{cases} \] 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 \( \alpha + \beta \sqrt{3} \), where \( \alpha, \beta \in \mathbb{Z} \), then the value of \( \alpha + \beta \) is ____.

1. Making the function smooth (differentiable)

For a piece-wise function to be differentiable at the joining point $x=1$ we need continuity and equal slopes.

1. Continuity at $x=1$
   $-3a(1)^2-2 = a^2 + b$
   -3a-2 = a^2 + b \tag{1}

2. Equal derivatives at $x=1$
   Left slope: $\dfrac{d}{dx}(-3ax^2-2) = -6ax$ so at $x=1$ it is $-6a$.
   Right slope: $\dfrac{d}{dx}(a^2+bx) = b$.
   Equate them:
   b=-6a \tag{2}

Put (2) in (1):
$-6a = -3a-2-a^2 \implies a^2-3a+2=0$
$\Rightarrow (a-1)(a-2)=0$
Given $a>1 \;\Rightarrow\; a=2$ and therefore $b=-12$.

So the smooth single curve is

2. Find touch-points with the line $y=-20$

Solve $f(x)=-20$ in each part.

Left part ($x<1$):
$-6x^2-2=-20 \;\Rightarrow\; x^2=3 \;\Rightarrow\; x=\pm\sqrt3$
Only $x=-\sqrt3$ satisfies $x<1$.

Right part ($x\ge1$):
$4-12x=-20 \;\Rightarrow\; x=2$

Hence the region is bounded between $x=-\sqrt3$ and $x=2$.

3. Set up definite integrals for area

Above the line $y=-20$, the height is $f(x)+20$.

1. Quadratic section ($-\sqrt3\;\to\;1$)
   $f(x)+20 = -6x^2-2+20 = 18-6x^2$
   $A_1 = \int_{-\sqrt3}^{1} \left(18-6x^2\right)\,dx$

2. Linear section ($1\;\to\;2$)
   $f(x)+20 = 4-12x+20 = 24-12x$
   $A_2 = \int_{1}^{2} \left(24-12x\right)\,dx$

4. Integrate step by step

Total area $A=A_1+A_2 = (16+12\sqrt3)+6 = 22+12\sqrt3$

Thus $\alpha=22$, $\beta=12$ and $\alpha+\beta=34$.

Think of two different roads for cars

1. First road (left side) is a curvy downhill road: $y=-6x^2-2$ for all cars with $x<1$.
2. Second road (right side) is a straight sloping road: $y=4-12x$ for all cars with $x\ge 1$.

Cars can smoothly shift from the curvy road to the straight road at $x=1$ without any bump because we fixed the height and slope there.

We now draw a horizontal line $y=-20$ (think of water level). Wherever the two roads touch this water, we get the ends of the pool. Between those touch-points the road stays above water, so the space between road and water becomes a nice closed pool of area.

Finally, we add up two easy areas:

• From left touch-point $x=-\sqrt3$ to $x=1$ (over the curvy road)
• From $x=1$ to right touch-point $x=2$ (over the straight road)

When we add those areas, we get $22+12\sqrt3$ square units. Here $\alpha=22$ and $\beta=12$, so $\alpha+\beta=34$.

Complete Step-by-Step Solution

1. Impose differentiability at $x=1$

   Continuity: -3a(1)^2-2 = a^2 + b \;\;\Rightarrow\;\; -3a-2 = a^2+b \tag{1} Equal derivative: -6a = b \tag{2} Substituting (2) in (1): $-6a = -3a-2-a^2 \;\;\Rightarrow\;\; a^2-3a+2=0$ $\Rightarrow (a-1)(a-2)=0 \;\;\Rightarrow\;\; a=2\;\text{(since }a>1)$ $b = -6a = -12$

2. Write the single smooth function

   -6x^2-2 & ,\; x<1 \\ 4-12x & ,\; x\ge1 \end{cases}$$

3. Find intersection with the line $y=-20$

   (i) Left part: $-6x^2-2 = -20 \Rightarrow x^2 = 3 \Rightarrow x = -\sqrt3 \;\text{(valid)}$

   (ii) Right part: $4-12x = -20 \Rightarrow x = 2$

4. Set up the definite integrals for area

   + \int_{1}^{2} \bigl(24-12x\bigr)\,dx$$

5. Compute each integral

   A_1 &= \left[18x-2x^3\right]_{-\sqrt3}^{1} = (18-2) - \bigl(-18\sqrt3+6\sqrt3\bigr) \\[4pt] &= 16 + 12\sqrt3 \\ A_2 &= \left[24x-6x^2\right]_{1}^{2} = (48-24) - (24-6) = 24-18 = 6 \end{aligned}$$ $$A = A_1 + A_2 = 16+12\sqrt3 + 6 = 22 + 12\sqrt3$$

6. Extract $\alpha,\beta$ and answer

   $\alpha = 22, \; \beta = 12 \;\Rightarrow\; \alpha + \beta = 34$

Final Answer: $34$.