Let the function
\[f(x)={−3ax2−2,a2+bx,x<1x≥1\]
be differentiable for all x∈R, where a>1,b∈R. If the area of the region enclosed by y=f(x) and the line y=−20 is α+β3, α,β∈Z, then the value of α+β is \_\_\_\_\_\_\_\_\_\_\_.
Answer: 34
Solutions · 1
Official solution
Continuity and differentiability at x=1:
\[-3a-2=a^2+b \quad\text{(continuity)}, \qquad -6a=b \quad\text{(equal derivatives)}\]
Substituting b=−6a: −3a−2=a2−6a⟹a2−3a+2=0⟹(a−1)(a−2)=0. Since a>1, a=2, giving b=−12.
\[f(x)={−6x2−2,4−12x,x<1x≥1\]
Finding where f(x)=−20: For x<1: −6x2−2=−20⟹x2=3⟹x=−3. For x≥1: 4−12x=−20⟹x=2.
Area between y=f(x) and y=−20 from x=−3 to x=2:
\[\text{Area}=\int_{-\sqrt{3}}^{1}(18-6x^2)\,dx+\int_{1}^{2}(24-12x)\,dx\]
\[=\Big[18x-2x^3\Big]_{-\sqrt{3}}^{1}+\Big[24x-6x^2\Big]_{1}^{2}=\big(16-(-12\sqrt{3})\big)+(24-18)=16+12\sqrt{3}+6=22+12\sqrt{3}\]
So α=22,β=12, and α+β=34.