Question

If f(x) = \(\begin{vmatrix}     cosx & 1 & 0 \\     0 & 2cosx & 3 \\     0 & 1 & 2cosx \\ \end{vmatrix}\) then \(\lim\limits_{x\rightarrow\pi} f(x)=\)

• A. -1
• B. 1
• C. 0
• D. 3

Answer 1

The Correct Option is A

We are given the matrix \( f(x) \): \[ f(x) = \begin{pmatrix} \cos x & 1 & 0 \\ 0 & 2 \cos x & 3 \\ 0 & 1 & 2 \cos x \end{pmatrix} \] We need to evaluate \( \lim_{x \to \pi} f(x) \). The limit of the matrix as \( x \to \pi \) can be found by taking the limit of each element in the matrix. Let's evaluate the limits:
The first element is \( \cos x \), and as \( x \to \pi \), \( \cos \pi = -1 \).
The second element is 1, which is constant.
The third element is 0, which is constant.
The fourth element is 0, which is constant.
The fifth element is \( 2 \cos x \), and as \( x \to \pi \), \( 2 \cos \pi = -2 \).
The sixth element is 3, which is constant.
The seventh element is 0, which is constant.
The eighth element is 1, which is constant.
The ninth element is \( 2 \cos x \), and as \( x \to \pi \), \( 2 \cos \pi = -2 \).
Thus, the limit of the matrix as \( x \to \pi \) is: \[ \lim_{x \to \pi} f(x) = \begin{pmatrix} -1 & 1 & 0 \\ 0 & -2 & 3 \\ 0 & 1 & -2 \end{pmatrix} \]

Therefore, the answer is \({-1} \), corresponding to option (A).

Answer 2

Given the function \(f(x) = \begin{vmatrix} \cos x & 1 & 0 \\ 0 & 2\cos x & 3 \\ 0 & 1 & 2\cos x \end{vmatrix}\).

We need to find \(\lim_{x \to \pi} f(x)\).

First, let's find the determinant of the matrix:

\(f(x) = \cos x \begin{vmatrix} 2\cos x & 3 \\ 1 & 2\cos x \end{vmatrix} - 1 \begin{vmatrix} 0 & 3 \\ 0 & 2\cos x \end{vmatrix} + 0 \begin{vmatrix} 0 & 2\cos x \\ 0 & 1 \end{vmatrix}\)

\(f(x) = \cos x [(2\cos x)(2\cos x) - (3)(1)] - 1 [0 - 0] + 0\)

\(f(x) = \cos x (4\cos^2 x - 3)\)

\(f(x) = 4\cos^3 x - 3\cos x\)

We can recognize that this is the triple angle formula for cosine: \(f(x) = \cos(3x)\).

Now, we need to find the limit as \(x\) approaches \(\pi\):

\(\lim_{x \to \pi} f(x) = \lim_{x \to \pi} \cos(3x) = \cos(3\pi)\)

Since \(\cos(3\pi) = \cos(\pi + 2\pi) = \cos(\pi) = -1\), we have

\(\lim_{x \to \pi} f(x) = -1\).