Q106 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
Express the set $A=\{1,7,17,31,49\}$ in set builder form
- A$\left\{x \mid x=2 n^2+1\right.$, where $n \in \mathcal{N}$ and $\left.n \leq 7\right\}$
- Correct answerB$\left\{x \mid x=2 n^2-1\right.$, where $n \in \mathcal{N}$ and $\left.n \leq 5\right\}$
- C$\left\{x \mid x=2 n^2-3\right.$, where $n \in \mathcal{N}$ and $\left.2 \leq n \leq 8\right\}$
- D$\left\{x \mid x=2 n^2-1\right.$, where $n \in \mathcal{N}$ and $\left.n < 5\right\}$
Solution
The given set is $A = \{1, 7, 17, 31, 49\}$.
We observe the pattern of the elements:
For $n = 1$: $2(1)^2 - 1 = 2 - 1 = 1$
For $n = 2$: $2(2)^2 - 1 = 8 - 1 = 7$
For $n = 3$: $2(3)^2 - 1 = 18 - 1 = 17$
For $n = 4$: $2(4)^2 - 1 = 32 - 1 = 31$
For $n = 5$: $2(5)^2 - 1 = 50 - 1 = 49$
The elements follow the rule $x = 2n^2 - 1$ for $n \in \{1, 2, 3, 4, 5\}$.
This corresponds to the condition $n \in \mathcal{N}$ and $n \leq 5$.
Answer: $\left\{x \mid x=2 n^2-1\right.$, where $n \in \mathcal{N}$ and $\left.n \leq 5\right\}$
We observe the pattern of the elements:
For $n = 1$: $2(1)^2 - 1 = 2 - 1 = 1$
For $n = 2$: $2(2)^2 - 1 = 8 - 1 = 7$
For $n = 3$: $2(3)^2 - 1 = 18 - 1 = 17$
For $n = 4$: $2(4)^2 - 1 = 32 - 1 = 31$
For $n = 5$: $2(5)^2 - 1 = 50 - 1 = 49$
The elements follow the rule $x = 2n^2 - 1$ for $n \in \{1, 2, 3, 4, 5\}$.
This corresponds to the condition $n \in \mathcal{N}$ and $n \leq 5$.
Answer: $\left\{x \mid x=2 n^2-1\right.$, where $n \in \mathcal{N}$ and $\left.n \leq 5\right\}$
Hence, the correct answer is option: 'B'.