Q. The value of $\left| \begin{matrix} 1 & 1 & 1 \\ \left( 2^{x} + 2^{- x} \right)^{2} & \left( 3^{x} + 3^{- x} \right)^{2} & \left( 5^{x} + 5^{- x} \right)^{2} \\ \left( 2^{x} - 2^{- x} \right)^{2} & \left( 3^{x} - 3^{- x} \right)^{2} & \left( 5^{x} - 5^{- x} \right)^{2} \end{matrix} \right|$ is

A) 0
B) $30^{x}$
C) $30^{- x}$
D) 1
Explanation

(a)
Applying $R_{2} \rightarrow R_{2} - R_{3},\$we get
$\left| \begin{matrix}
1 & 1 & 1 \\
4 & 4 & 4 \\
\left( 2^{x} - 2^{- x} \right)^{2} & \left( 3^{x} - 3^{- x} \right)^{2} & \left( 5^{x} - 5^{- x} \right)^{2}
\end{matrix} \right|$
$= 4\left| \begin{matrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
\left( 2^{x} - 2^{- x} \right)^{2} & \left( 3^{x} - 3^{- x} \right)^{2} & \left( 5^{x} - 5^{- x} \right)^{2}
\end{matrix} \right|$
$= 4 \times 0 = 0\ \ \ \ \ \ \ \ \ \ \ \lbrack\therefore two\ rows\ are\ identical\rbrack$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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