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Question: The value of \(\left| \begin{matrix} 1 & 1 & 1 \\ (2^{x} + 2^{- x})^{2} & (3^{x} + 3^{- x})^{2} & (5...

The value of ∣111(2x+2−x)2(3x+3−x)2(5x+5−x)2(2x−2−x)2(3x−3−x)2(5x−5−x)2∣\left| \begin{matrix} 1 & 1 & 1 \\ (2^{x} + 2^{- x})^{2} & (3^{x} + 3^{- x})^{2} & (5^{x} + 5^{- x})^{2} \\ (2^{x} - 2^{- x})^{2} & (3^{x} - 3^{- x})^{2} & (5^{x} - 5^{- x})^{2} \end{matrix} \right|

A

0

B

30x30^{x}

C

30−x30^{- x}

D

None of these

Answer

0

Explanation

Solution

Applying R2→R2−R3R_{2} \rightarrow R_{2} - R_{3}

1 & 1 & 1 \\ 2.2^{x}.2.2^{- x} & 2.3^{x}.2.3^{- x} & 2.5^{x}.2.5^{- x} \\ (2^{x} - 2^{- x})^{2} & (3^{x} - 3^{- x})^{2} & (5^{x} - 5^{- x})^{2} \end{matrix} \right| = 4\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ (2^{x} - 2^{- x})^{2} & (3^{x} - 3^{- x})^{2} & (5^{x} - 5^{- x})^{2} \end{matrix} \right| = 0\lbrack\because R_{1}\text{ and}R_{2}\text{are identical}\rbrack$$ **Trick :** Putting $x = 0$, we get option (1) is correct