COMEDK 2024 Shift 1 Paper

To determine which option correctly represents the set A={1,7,17,31,49} in set builder form, we need to evaluate each option by plugging in the values of n and checking if the resulting numbers match the elements of set A.
Let's evaluate each option one by one:
Option A: {x∣x=2n2−1., where n∈mathcal‌N and n<5}
For n=1 :
x=2(1)2−1=2−1=1
For n=2 :
x=2(2)2−1=8−1=7

For n=3 :
x=2(3)2−1=18−1=17
For n=4 :
x=2(4)2−1=32−1=31

For n=5 (Note n<5 so we don't include 5) :
The numbers we get are {1,7,17,31} which matches some elements of set A but only up to n<5.
So this option doesn't list all numbers
Option B: {x∣x=2n2−3., where n∈mathcal‌N and 2≤n≤8}
For n=2 :
x=2(2)2−3=8−3=5
For n=3 :
x=2(3)2−3=18−3=15
For n=4 :
x=2(4)2−3=32−3=29

The numbers we get are not matching the set A. Thus, Option B is incorrect.
Option C: {x∣x=2n2+1., where n∈mathcal‌N and n≤7}
For n=1 :
x=2(1)2+1=2+1=3
For n=2 :
x=2(2)2+1=8+1=9
The numbers we get are not matching the set A. Thus, Option C is incorrect.
Option D: {x∣x=2n2−1., where n∈mathcal‌N and n≤5}
For n=1 :
x=2(1)2−1=2−1=1
For n=2 :
x=2(2)2−1=8−1=7

For n=3 :
x=2(3)2−1=18−1=17
For n=4 :
x=2(4)2−1=32−1=31
For n=5 :
x=2(5)2−1=50−1=49

The numbers we get are {1,7,17,31,49}, which matches the set A exactly. Thus, Option D is correct.
The correct answer is Option D. {x∣x=2n2−1., where n∈mathcal‌N and n≤5}