JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023The number of elements in the set \{n \in \mathbb{Z} : |n^2 - 10n + 19| < 6\} is:
Choose the correct answer:
- A.
9
(Correct Answer) - B.
8
- C.
7
- D.
6
9
Explanation
\text{Given that } \{n \in \mathbb{Z} : |n^2 - 10n + 19| < 6\}
\Rightarrow |n^2 - 10n + 19| < 6, \forall n \in \mathbb{Z}
\text{or } -6 < n^2 - 10n + 19 < 6
\Rightarrow n^2 - 10n + 19 > -6 \text{ and } n^2 - 10n + 19 < 6
\Rightarrow n^2 - 10n + 25 > 0 \text{ and } n^2 - 10n + 13 < 0
\Rightarrow (n - 5)^2 > 0 \text{ and } n \in \mathbb{Z}
where α,β are the roots of n2−10n+13=0
⇒n=210±100−52=210±43
⇒α=5−23 and β=5+23
⇒n∈Z−{5} and n∈(5−23,5+23)
⇒n∈Z−{5} and n∈(1.52,8.46)
n∈{2,3,4,…,8}
So, common integers are: } 2, 3, 4, 6, 7, 8
Hence, no. of elements in set are 9
Explanation
\text{Given that } \{n \in \mathbb{Z} : |n^2 - 10n + 19| < 6\}
\Rightarrow |n^2 - 10n + 19| < 6, \forall n \in \mathbb{Z}
\text{or } -6 < n^2 - 10n + 19 < 6
\Rightarrow n^2 - 10n + 19 > -6 \text{ and } n^2 - 10n + 19 < 6
\Rightarrow n^2 - 10n + 25 > 0 \text{ and } n^2 - 10n + 13 < 0
\Rightarrow (n - 5)^2 > 0 \text{ and } n \in \mathbb{Z}
where α,β are the roots of n2−10n+13=0
⇒n=210±100−52=210±43
⇒α=5−23 and β=5+23
⇒n∈Z−{5} and n∈(5−23,5+23)
⇒n∈Z−{5} and n∈(1.52,8.46)
n∈{2,3,4,…,8}
So, common integers are: } 2, 3, 4, 6, 7, 8
Hence, no. of elements in set are 9