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 \( \alpha + \beta\sqrt{3} \), where \( \alpha, \beta \in \mathbb{Z} \), then the value of \( \alpha + \beta \) is:

Hint:

For piecewise functions, ensure both continuity and differentiability at the point where the function changes its form. Then use integration to compute the enclosed area.

Answer 1

Correct Answer: 34

For \( f(x) \) to be continuous and differentiable at \( x = 1 \): \[ LHL = RHL, \quad LHD = RHD \] At \( x = 1 \): \[ -3a - 2 = a^2 + b, \quad -6a = b \] Solving for \( a \) and \( b \): \[ a = 2, \quad b = -12 \] Thus, the function becomes: \[ f(x) = \begin{cases} -6x^2 - 2, & x <1
4 - 12x, & x \geq 1 \end{cases} \] Next, we compute the area enclosed by the curve \( y = f(x) \) and the line \( y = -20 \). The area is calculated by integrating: \[ \text{Area} = \int_{-\sqrt{3}}^1 \left( -6x^2 - 2 + 20 \right)dx + \int_1^2 \left( 4 - 12x + 20 \right)dx \] Evaluating the integrals: \[ 16 + 12\sqrt{3} + 6 = 22 + 12\sqrt{3} \] Thus, \( \alpha + \beta = 34 \).