To solve this problem, we are given a functional equation for \( f(x) \), which is an \( n \)-degree polynomial. We are asked to find the value of \( f(3) \), given that \( f(2) = 33 \).
- Functional Equation: The functional equation given is: \[ f(x) = \frac{1}{2} \left( f(x) \cdot f\left( \frac{1}{x} \right) - f(x) \right) \] This type of equation typically defines the relationship between the values of the function at \( x \) and \( \frac{1}{x} \). The function is a polynomial, and from the given equation, we can derive its degree and possibly its form.
We are given that:
The value of \( f(3) \) is 244 (Option 3).
If \( \vec{u}, \vec{v}, \vec{w} \) are non-coplanar vectors and \( p, q \) are real numbers, then the equality:
\[ [3\vec{u} \quad p\vec{v} \quad p\vec{w}] - [p\vec{v} \quad \vec{w} \quad q\vec{u}] - [2\vec{w} \quad q\vec{v} \quad q\vec{u}] = 0 \]
holds for:
Statement-I: In the interval \( [0, 2\pi] \), the number of common solutions of the equations
\[ 2\sin^2\theta - \cos 2\theta = 0 \]
and
\[ 2\cos^2\theta - 3\sin\theta = 0 \]
is two.
Statement-II: The number of solutions of
\[ 2\cos^2\theta - 3\sin\theta = 0 \]
in \( [0, \pi] \) is two.
If \( A \) and \( B \) are acute angles satisfying
\[ 3\cos^2 A + 2\cos^2 B = 4 \]
and
\[ \frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A}, \]
Then \( A + 2B = \ ? \)
If
\[ \sin \theta + 2 \cos \theta = 1 \]
and
\[ \theta \text{ lies in the 4\textsuperscript{th} quadrant (not on coordinate axes), then } 7 \cos \theta + 6 \sin \theta =\ ? \]