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Sets and Relations question

2023 · 10 Apr · Shift 1 · Q41

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Sets and Relations
2023 · 10 Apr · Shift 1 · Q41

Sets and Relations question

2023 · 10 Apr · Shift 1 · Q41

JEE MainMathematicsSets and RelationsNumerical+4 / −1

The number of elements in the set

\{ n \in Z:|{n^2} - 10n + 19| \lt 6\}

is

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 6

We need to find the number of integers $n$ satisfying $|n^2-10n+19|<6.$
Rewrite the expression inside the modulus by completing the square: $n^2-10n+19=(n-5)^2-25+19=(n-5)^2-6.$ So the inequality becomes $|(n-5)^2-6|<6.$
Use the property $|A|<6 \iff -6<A<6.$ Thus, $-6<(n-5)^2-6<6.$
Add $6$ throughout: $0<(n-5)^2<12.$
Since $n$ is an integer, $(n-5)^2$ must be a perfect square strictly between $0$ and $12$ . The possible values are $1,4,9.$
Solve for each case:

If $(n-5)^2=1$ , then $n-5=\pm 1$ , so $n=4,6$ .
If $(n-5)^2=4$ , then $n-5=\pm 2$ , so $n=3,7$ .
If $(n-5)^2=9$ , then $n-5=\pm 3$ , so $n=2,8$ .

Hence the set is $\{2,3,4,6,7,8\}$ and the number of elements is $6.$

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