Question

Let the function,

$$f(x) = \begin{cases} -3ax^2-2, & x < 1 \\ a^2 + bx, & x \geq 1 \end{cases}$$

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 $\alpha + \beta\sqrt{3}, \alpha, \beta \in \mathbb{Z}$, then the value of $\alpha + \beta$ is

Answer 1

34

Answer 2

The problem involves finding the area enclosed by a piecewise differentiable function and a horizontal line.

Step 1: Determine the values of 'a' and 'b' using differentiability conditions. The function is given as:

$$f(x) = \begin{cases} -3ax^2-2, & x < 1 \\ a^2 + bx, & x \geq 1 \end{cases}$$

For $f(x)$ to be differentiable for all $x \in \mathbb{R}$, it must be continuous at $x=1$ and its left-hand derivative must equal its right-hand derivative at $x=1$.

Condition 1: Continuity at $x=1$ $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$ $-3a(1)^2-2 = a^2+b(1)$ $-3a-2 = a^2+b$ $a^2+b+3a+2 = 0 \quad (\text{Equation 1})$

Condition 2: Differentiability at $x=1$ First, find the derivatives of the two parts: For $x < 1$, $f'(x) = \frac{d}{dx}(-3ax^2-2) = -6ax$. For $x > 1$, $f'(x) = \frac{d}{dx}(a^2+bx) = b$. For differentiability at $x=1$, $f'(1^-) = f'(1^+)$: $-6a(1) = b$ $-6a = b \quad (\text{Equation 2})$

Now, substitute Equation 2 into Equation 1: $a^2 + (-6a) + 3a + 2 = 0$ $a^2 - 3a + 2 = 0$ This is a quadratic equation that can be factored: $(a-1)(a-2) = 0$ So, $a=1$ or $a=2$.

The problem states that $a > 1$, so we choose $a=2$. Substitute $a=2$ into Equation 2 to find $b$: $b = -6(2) = -12$.

Thus, the function $f(x)$ is:

$$f(x) = \begin{cases} -3(2)x^2-2, & x < 1 \\ (2)^2 + (-12)x, & x \geq 1 \end{cases}$$ $$f(x) = \begin{cases} -6x^2-2, & x < 1 \\ 4 - 12x, & x \geq 1 \end{cases}$$

Step 2: Find the intersection points of $y = f(x)$ and $y = -20$. Case 1: For $x < 1$ $-6x^2-2 = -20$ $-6x^2 = -18$ $x^2 = 3$ $x = \pm\sqrt{3}$ Since we are considering $x < 1$, only $x = -\sqrt{3}$ is a valid intersection point.

Case 2: For $x \geq 1$ $4-12x = -20$ $-12x = -24$ $x = 2$ Since we are considering $x \geq 1$, $x=2$ is a valid intersection point.

The region enclosed by $y=f(x)$ and $y=-20$ is between $x=-\sqrt{3}$ and $x=2$.

Step 3: Set up the integral for the area. We need to determine if $f(x)$ is above or below $y=-20$ in the interval $[-\sqrt{3}, 2]$. For $x \in [-\sqrt{3}, 1)$, $f(x) = -6x^2-2$. The maximum value in this interval is at $x=0$, $f(0)=-2$. The minimum value is at $x=-\sqrt{3}$, $f(-\sqrt{3})=-20$. Since $f(x)$ is a downward parabola with vertex at $(0,-2)$, $f(x) \ge -20$ for $x \in [-\sqrt{3}, 1)$. For $x \in [1, 2]$, $f(x) = 4-12x$. This is a decreasing line. At $x=1$, $f(1)=-8$. At $x=2$, $f(2)=-20$. So, $f(x) \ge -20$ for $x \in [1, 2]$. Since $f(x) \ge -20$ over the entire interval $[-\sqrt{3}, 2]$, the area $A$ is given by: $A = \int_{-\sqrt{3}}^{2} (f(x) - (-20)) dx$ We split the integral at $x=1$ due to the piecewise definition: $A = \int_{-\sqrt{3}}^{1} (-6x^2-2 - (-20)) dx + \int_{1}^{2} (4-12x - (-20)) dx$ $A = \int_{-\sqrt{3}}^{1} (-6x^2+18) dx + \int_{1}^{2} (24-12x) dx$

Step 4: Evaluate the integrals. First integral:

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

Second integral:

$$\int_{1}^{2} (24-12x) dx = \left[24x - 12\frac{x^2}{2}\right]_{1}^{2} = \left[24x - 6x^2\right]_{1}^{2} \\ = (24(2) - 6(2)^2) - (24(1) - 6(1)^2) \\ = (48 - 6(4)) - (24 - 6) \\ = (48 - 24) - (18) \\ = 24 - 18 = 6$$

Step 5: Calculate the total area and find $\alpha + \beta$. Total Area $A = (16 + 12\sqrt{3}) + 6 = 22 + 12\sqrt{3}$. The problem states the area is $\alpha + \beta\sqrt{3}$, where $\alpha, \beta \in \mathbb{Z}$. Comparing $22 + 12\sqrt{3}$ with $\alpha + \beta\sqrt{3}$: $\alpha = 22$ $\beta = 12$ The value of $\alpha + \beta = 22 + 12 = 34$.