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(A) Given $f(x) = \left| \begin{array}{ccc} \cos x & 1 & 0 \ 0 & 2 \cos x & 3 \ 0 & 1 & 2 \cos x \end{array} ight|$.Expanding the determinant alon

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$-1$

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(A) Given $f(x) = \left| \begin{array}{ccc} \cos x & 1 & 0 \ 0 & 2 \cos x & 3 \ 0 & 1 & 2 \cos x \end{array} ight|$.Expanding the determinant along the first column $(C_1)$:$f(x) = \cos x \cdot \left| \begin{array}{cc} 2 \cos x & 3 \ 1 & 2 \cos x \end{array} ight| - 0 + 0$$f(x) = \cos x \cdot ((2 \cos x)(2 \cos x) - (3)(1))$$f(x) = \cos x (4 \cos^2 x - 3)$$f(x) = 4 \cos^3 x - 3 \cos x$Using the trigonometric identity $\cos 3x = 4 \cos^3 x - 3 \cos x$,we get:$f(x) = \cos 3x$Now,calculating the limit:$\lim_{x ightarrow \pi} f(x) = \lim_{x ightarrow \pi} \cos 3x = \cos(3\pi)$Since $\cos(3\pi) = -1$,the limit is $-1$.

If $f(x) = \left| \begin{array}{ccc} \cos x & 1 & 0 \\ 0 & 2 \cos x & 3 \\ 0 & 1 & 2 \cos x \end{array} \right|$,then $\lim_{x \rightarrow \pi} f(x)$ is equal to

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