Question

If $\Delta \left(x\right)=\left|\begin{array}{ccc}1& 1& 1\\ (e^x+e^{-x}{)}^2& ({\pi }^x+{\pi }^{-x}{)}^2& 2\\ (e^x-e^{-x}{)}^2& ({\pi }^x-{\pi }^{-x}{)}^2& -2\end{array}\right|$ then
$\Delta \left(x\right)$ equals

• A. $x^2$
• B. $x^2-1$
• C. $e^{x^2}-{\pi }^{x^2}$
• D. 0

Answer 1

The correct option is D 0

$\Delta \left(x\right)=\left|\begin{array}{ccc}1& 1& 1\\ (e^x+e^{-x}{)}^2& ({\pi }^x+{\pi }^{-x}{)}^2& 2\\ (e^x-e^{-x}{)}^2& ({\pi }^x-{\pi }^{-x}{)}^2& -2\end{array}\right|$
$=\left|\begin{array}{ccc}1& 1& 1\\ e^{2x}+e^{-2x}+2& {\pi }^{2x}+{\pi }^{-2x}+2& 2\\ e^{2x}+e^{-2x}-2& {\pi }^{2x}+{\pi }^{-2x}-2& -2\end{array}\right|$
$R_2\to R_2-R_3$
$=\left|\begin{array}{ccc}1& 1& 1\\ 4& 4& 4\\ e^{2x}+e^{-2x}-2& {\pi }^{2x}+{\pi }^{-2x}-2& -2\end{array}\right|$
$=4\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ e^{2x}+e^{-2x}-2& {\pi }^{2x}+{\pi }^{-2x}+2& -2\end{array}\right|$
$\Rightarrow \Delta \left(x\right)=0$