Let \( f(x) = 2 - 7 \sin{\left( \frac{2x}{7} \right)} \). Then the maximum value of \( f(x) \) is:
The function \( f(x) \) is given by: \[ f(x) = 2 - 7 \sin{\left( \frac{2x}{7} \right)}. \] The maximum value of \( \sin{\theta} \) is \( 1 \), so the maximum value of \( -7 \sin{\left( \frac{2x}{7} \right)} \) is \( -7 \times (-1) = 7 \).
Thus, the maximum value of \( f(x) \) occurs when \( \sin{\left( \frac{2x}{7} \right)} = -1 \), and is: \[ f(x) = 2 + 7 = 9. \] Thus, the maximum value of \( f(x) \) is \( 9 \), which corresponds to option (D).
If $ | \vec{a} | = 3 $, $ | \vec{b} | = 2 $, then find $ (3\vec{a} - 2\vec{b}) \cdot (3\vec{a} + 2\vec{b}) $.
The value of \( x \) that satisfies the equation:
\[ \begin{vmatrix} x & 1 & 1 \\ 2 & 2 & 0 \\ 1 & 0 & -2 \end{vmatrix} = 6 \]
If \( 2 \) is a solution of the inequality \( \frac{x-a}{a-2x}<-3 \), then \( a \) must lie in the interval:
Let \[ A = \begin{pmatrix} 2 & -1 & 1 \\ -1 & 0 & 2 \\ 1 & -2 & -1 \end{pmatrix} \] and let \( B = \frac{1}{|A|} A \). Then the value of \( |B| \) is equal to:
If $ | \vec{a} | = 3 $, $ | \vec{b} | = 2 $, then find $ (3\vec{a} - 2\vec{b}) \cdot (3\vec{a} + 2\vec{b}) $.