Question

If $e^A$ is defined as $e^A=I+A+\frac{A^2}{2!}+\frac{A^3}{3!}+.....=\frac{1}{2}\left[\begin{array}{cc}f\left(x\right)& g\left(x\right)\\ g\left(x\right)& f\left(x\right)\end{array}\right]$ where $A=\left[\begin{array}{cc}x& x\\ x& x\end{array}\right]$ and $0<x<1$, then I is an identity matrix.

$\int \frac{g\left(x\right)}{f\left(x\right)}dx$ is equal to

• A. $log(e^x+e^{-x})+c$
• B. $log(e^x-e^{-1})+c$
• C. $log(e^{2x}-1)+c$
• D. none of these

Answer 1

The correct option is A $log(e^x+e^{-x})+c$

$A=\left[\begin{array}{cc}x& x\\ x& x\end{array}\right]\Rightarrow A^2=\left[\begin{array}{cc}x& x\\ x& x\end{array}\right]\left[\begin{array}{cc}x& x\\ x& x\end{array}\right]=\left[\begin{array}{cc}2x^2& 2x^2\\ 2x^2& 2x^2\end{array}\right]$
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Similarly, $A^3=\left[\begin{array}{cc}{2}^2{x}^3& 2^2{x}^3\\ 2^2{x}^3& 2^2{x}^3\end{array}\right],A^4=\left[\begin{array}{cc}{2}^3{x}^4& 2^3{x}^4\\ 2^3{x}^4& 2^3{x}^4\end{array}\right],......A^n=\left[\begin{array}{cc}{2}^{n-1}{x}^n& 2^{n-1}{x}^n\\ 2^{n-1}{x}^n& 2^{n-1}{x}^n\end{array}\right]$
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$e^A=I+A+\frac{A^2}{2!}+\frac{A^3}{3!}+.......=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}x& x\\ x& x\end{array}\right]+\frac{\left[\begin{array}{cc}2x^2& 2x^2\\ 2x^2& 2x^2\end{array}\right]}{2!}+\frac{\left[\begin{array}{cc}{2}^2{x}^3& 2^2{x}^3\\ 2^2{x}^3& 2^2{x}^3\end{array}\right]}{3!}+.....=\left[\begin{array}{cc}1+x+\frac{2x^2}{2!}+\frac{2^2{x}^3}{3!}+...& x+\frac{2x^2}{2!}+\frac{2^2{x}^3}{3!}+...\\ x+\frac{2x^2}{2!}+\frac{2^2{x}^3}{3!}+...& 1+x+\frac{2x^2}{2!}+\frac{2^2{x}^3}{3!}+...\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}(1+2x+\frac{2^2{x}^2}{2!}+\frac{2^3{x}^3}{3!}+...)+\frac{1}{2}& \frac{1}{2}(1+2x+\frac{2^2{x}^2}{2!}+\frac{2^3{x}^3}{3!}+...)-\frac{1}{2}\\ \frac{1}{2}(1+2x+\frac{2^2{x}^2}{2!}+\frac{2^3{x}^3}{3!}+...)-\frac{1}{2}& \frac{1}{2}(1+2x+\frac{2^2{x}^2}{2!}+\frac{2^3{x}^3}{3!}+...)+\frac{1}{2}\end{array}\right]$

$=\frac{1}{2}\left[\begin{array}{cc}{e}^{2x}+1& e^{2x}-1\\ e^{2x}-1& e^{2x}+1\end{array}\right]$
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Comparing with the matrix in question, we get $f\left(x\right)=e^{2x}+1$ and $g\left(x\right)=e^{2x}-1$
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$\int \frac{e^{2x}-1}{e^{2x}+1}dx=\int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx$
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Substitute $e^x+e^{-x}=t\Rightarrow e^x-e^{-x}.dx=dt$

$\int \frac{dt}{t}={\log }_{e}t+C={\log }_e(e^x+e^{-x})+C$