Given the inequality: \[ -6 < n^2 - 10n + 19 < 6 \]
Split the inequality into two parts:
This simplifies to:
Factorize:
\[ (n - 5)^2 > 0 \]
The solution is \( n \in \mathbb{Z} \setminus \{5\} \) (all integers except 5). Call this result (i).
Find the roots of the equation:
\[ n^2 - 10n + 13 = 0 \]
The roots are:
\[ n = 5 \pm \sqrt{3} \]
Approximating the roots:
\[ 5 - \sqrt{3} \approx 2 \quad \text{and} \quad 5 + \sqrt{3} \approx 8 \]
Thus, \( 2 < n < 8 \). For integer values, \( n \in \{2, 3, 4, 5, 6, 7, 8\} \). Call this result (ii).
From (i) and (ii):
\( n \in \{2, 3, 4, 5, 6, 8\} \).
The number of values of \( n \) is:
\[ \boxed{6} \]
Let \(P(S)\) denote the power set of \(S = \{1, 2, 3, \ldots, 10\}\). Define the relations \(R_1\) and \(R_2\) on \(P(S)\) as \(A R_1 B\) if \[(A \cap B^c) \cup (B \cap A^c) = ,\]and \(A R_2 B\) if\[A \cup B^c = B \cup A^c,\]for all \(A, B \in P(S)\). Then:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)