Limits, Continuity & Differentiability
Differentiability + Area under Curve
nta_pyq_2025_apr
Grade 12
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 ___
Step-by-Step Solution
Key Concept: Continuity at $x=1$: $-3a-2=a^2+b$. Differentiability at $x=1$: $-6a=b$. Solve for $a>1$ and $b$. Then compute the area between $f(x)$ and $y=-20$.
$a=2,b=-12$: $f(x)=\begin{cases}-6x^2-2, & x<1\\4-12x, & x\geq1\end{cases}$. $-6x^2-2=-20\Rightarrow x=-\sqrt{3}$ (for $x<1$). $4-12x=-20\Rightarrow x=2$. Area $=\int_{-\sqrt{3}}^{1}(-6x^2-2+20)dx+\int_1^2(4-12x+20)dx=16+12\sqrt{3}+6=22+12\sqrt{3}$. $\alpha+\beta=22+12=34$.
Correct Answer: 34