Question

If $e^A$ is defined as $e^A=I+A+\frac{A^2}{2!}+\dots \dots =\frac{1}{2}\left[\begin{array}{cc}f(x)& g(x)\\ g(x)& f(x)\end{array}\right]$ where $A=\left[\begin{array}{cc}x& x\\ x& x\end{array}\right]$ and $0<x<1$ I is an identity matrix. Then ${\int }_0^1\frac{g(x)}{f(x)}dx$ is equal to

• A. $ln\left(\frac{e+e^{-1}}{2}\right)$
• B. $ln\left(e+e^{-1}\right)$
• C. $ln\left(e^2+1\right)-ln2$
• D. none of these

Answer 1

Answer: (A)

Put the $A$ in given equation
$\frac{1}{2}\left[\begin{array}{cc}f(x)& g(x)\\ g(x)& f(x)\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}{e}^{2x}+1& e^{2x}-1\\ e^{2x}-1& e^{2x}+1\end{array}\right]$
$\Rightarrow f(x)=e^{2x}+1,g(x)=e^{2x}-1$
${\int }_0^1\left(\frac{e^{2x}-1}{e^{2x}+1}\right)dx={\left(\left[ln\left(e^x+e^{-x}\right)\right]\right)}_0^1$
$=ln\left(e+e^{-1}\right)-ln2$