Let the function $f(x)=\begin{cases}-3ax^{2}-2,&x<1\\ a^{2}+bx,&x\ge1\end{cases}$ be differentiable for all $x\in R,$ where $a>1$, $b\in R$. If the area of the region enclosed by $y=f(x)$ and the line $y = -20$ is $\alpha+\beta\sqrt{3}$, where $\alpha, \beta \in Z$, then the value of $\alpha+\beta$ is:
Solution:
$f(x)$ is continuous at $x=1$
$$ -3a(1)^2 – 2 = a^2 + b(1) $$
$$ a^2 + 3a + b + 2 = 0 \quad \dots(1) $$
$f(x)$ is differentiable at $x=1$ ($LHD = RHD$)
$$ \frac{d}{dx}(-3ax^2 – 2) = \frac{d}{dx}(a^2 + bx) $$
$$ -6ax = b \Rightarrow -6a(1) = b \Rightarrow b = -6a $$
Substituting $b = -6a$ in (1):
$$ a^2 + 3a – 6a + 2 = 0 $$
$$ a^2 – 3a + 2 = 0 \Rightarrow (a-2)(a-1) = 0 $$
Since $a > 1$, we get $a = 2$.
Then $b = -6(2) = -12$.
Function becomes:
$$ f(x)=\begin{cases}-6x^{2}-2,&x<1\\ 4-12x,&x\ge1\end{cases} $$
Area Calculation
Region enclosed by $y=f(x)$ and line $y=-20$.
Intersection points:
1) For $x < 1$: $-6x^2 – 2 = -20 \Rightarrow 6x^2 = 18 \Rightarrow x^2 = 3 \Rightarrow x = -\sqrt{3}$ (since $x<1$).
2) For $x \ge 1$: $4 – 12x = -20 \Rightarrow 12x = 24 \Rightarrow x = 2$.
Area $A = \int_{-\sqrt{3}}^{1} (f(x) – (-20)) dx + \int_{1}^{2} (f(x) – (-20)) dx$
$$ A = \int_{-\sqrt{3}}^{1} (-6x^2 – 2 + 20) dx + \int_{1}^{2} (4 – 12x + 20) dx $$
$$ A = \int_{-\sqrt{3}}^{1} (18 – 6x^2) dx + \int_{1}^{2} (24 – 12x) dx $$
$$ A = [18x – 2x^3]_{-\sqrt{3}}^{1} + [24x – 6x^2]_{1}^{2} $$
$$ A = [(18 – 2) – (-18\sqrt{3} + 2(3\sqrt{3}))] + [(48 – 24) – (24 – 6)] $$
$$ A = [16 – (-12\sqrt{3})] + [24 – 18] $$
$$ A = 16 + 12\sqrt{3} + 6 = 22 + 12\sqrt{3} $$
Comparing with $\alpha + \beta\sqrt{3}$:
$\alpha = 22, \beta = 12$.
$\alpha + \beta = 22 + 12 = 34$.
Ans. (34)