Basics & LogarithmsmediumJEE Main 2023Free

Basics & Logarithms — JEE Maths practice question (JEE Main 2023)

JEE Maths question with a full step-by-step solution.

Question

The number of elements in the set

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

Solution

Answer: 6

Step 1: The inequality

|n^{2}-10n+19|<6

is equivalent to:

-6<n^{2}-10n+19<6

Step 2: Right inequality:

n^{2}-10n+19<6\Rightarrow n^{2}-10n+13<0

. Roots:

n=\dfrac{10\pm\sqrt{100-52}}{2}=\dfrac{10\pm\sqrt{48}}{2}=5\pm 2\sqrt{3}

. So

5-2\sqrt{3}<n<5+2\sqrt{3}

, i.e., approximately

1.54<n<8.46

. Step 3: Left inequality:

n^{2}-10n+19>-6\Rightarrow n^{2}-10n+25>0\Rightarrow(n-5)^{2}>0

. This holds for all

n\neq 5

. Step 4: Combine:

n\in(5-2\sqrt{3},5+2\sqrt{3})

and

n\neq 5

. Integer values in this interval:

n=2,3,4,5,6,7,8

. Excluding

n=5

n\in\{2,3,4,6,7,8\}

. Step 5: Count:

6

elements. Answer:

6

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