Area Under Curves
Geometric series of bounded areas
MJAT_TS1_P2
Grade 12
Question:
If $A_i$ is the area bounded by $|x - a_i| + |y| = b_i$, $i \in \mathbb{N}$, where $a_{i+1} = a_i + \dfrac{3}{2}b_i$, $b_{i+1} = \dfrac{b_i}{2}$, $a_1 = 0$, $b_1 = 32$, then
A) $A_3 = 128$
B) $A_3 = 256$
C) $\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n} A_i = \dfrac{8}{3}(32)^2$
D) $\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n} A_i = \dfrac{4}{3}(16)^2$
Step-by-Step Solution
Key Concept: The region $|x-a_i|+|y|=b_i$ is a square (rhombus) with diagonals $2b_i$ along $x$ and $2b_i$ along $y$. Area $= \frac{1}{2}(2b_i)(2b_i) = 2b_i^2$. With $b_i = 32/2^{i-1}$: $A_i = 2\cdot(32/2^{i-1})^2 = 2\cdot 32^2/4^{i-1}$.
$b_i = 32/2^{i-1}$, so $A_i = 2b_i^2 = 2\cdot(32)^2/4^{i-1}$. $A_3 = 2\cdot 64 = 128$ ✓ (A). $\sum_{i=1}^\infty A_i = \frac{2\cdot 1024}{1-\frac{1}{4}} = \frac{2048}{3/4} = \frac{8192}{3} = \frac{8}{3}(32)^2$ ✓ (C).
Correct Answer: AC