Q104 / 180 · KCET 2018 · Mathematics
If \( x, y, z \in R \), then the value of determinant
\[
\left|\begin{array}{lll}
\left(5^{x}+5^{-x}\right)^{2} & \left(5^{x}-5^{-x}\right)^{2} & 1 \\
\left(6^{x}+6^{-x}\right)^{2} & \left(6^{x}-6^{-x}\right)^{2} & 1 \\
\left(7^{x}+7^{-x}\right)^{2} & \left(7^{x}-7^{-x}\right)^{2} & 1
\end{array}\right| \text { is }
\]
\[ \left|\begin{array}{lll} \left(5^{x}+5^{-x}\right)^{2} & \left(5^{x}-5^{-x}\right)^{2} & 1 \\ \left(6^{x}+6^{-x}\right)^{2} & \left(6^{x}-6^{-x}\right)^{2} & 1 \\ \left(7^{x}+7^{-x}\right)^{2} & \left(7^{x}-7^{-x}\right)^{2} & 1 \end{array}\right| \text { is } \]
- A\( 10 \)
- B\( 12 \)
- C\( 11 \)
- Correct answerD\( 00 \)
Solution
Given that, \( \left|\begin{array}{ccc}\left(5^{x}+5^{-x}\right)^{2} & \left(5^{x}-5^{-x}\right)^{2} & 1 \\ \left(6^{x}+6^{-x}\right)^{2} & \left(6^{x}-6^{-x}\right)^{2} & 1 \\ \left(7^{x}+7^{-x}\right)^{2} & \left(7^{x}-7^{-x}\right)^{2} & 1\end{array}\right| \)
\( C_{1} \rightarrow C_{1}-C_{2} \)
\( \left|\begin{array}{ccc}4 & \left(5^{x}-5^{-x}\right)^{2} & 1 \\ 4 & \left(6^{x}-6^{-x}\right)^{2} & 1 \\ 4 & \left(7^{x}-7^{-x}\right)^{2} & 1\end{array}\right|=4\left|\begin{array}{ccc}1 & \left(5^{x}-5^{-x}\right)^{2} & 1 \\ 1 & \left(6^{x}-6^{-x}\right)^{2} & 1 \\ 1 & \left(7^{x}-7^{-x}\right)^{2} & 1\end{array}\right|=0 \)
\( C_{1} \rightarrow C_{1}-C_{2} \)
\( \left|\begin{array}{ccc}4 & \left(5^{x}-5^{-x}\right)^{2} & 1 \\ 4 & \left(6^{x}-6^{-x}\right)^{2} & 1 \\ 4 & \left(7^{x}-7^{-x}\right)^{2} & 1\end{array}\right|=4\left|\begin{array}{ccc}1 & \left(5^{x}-5^{-x}\right)^{2} & 1 \\ 1 & \left(6^{x}-6^{-x}\right)^{2} & 1 \\ 1 & \left(7^{x}-7^{-x}\right)^{2} & 1\end{array}\right|=0 \)
Hence, the correct answer is option: 'D'.