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Q. If $e^{A}$ is defined as $e^{A}=I+A+\frac{A^{2}}{2 !}+\ldots \ldots =\frac{1}{2}\begin{bmatrix} f\left(\right.x\left.\right) & g\left(\right.x\left.\right) \\ g\left(\right.x\left.\right) & f\left(\right.x\left.\right) \end{bmatrix}$ where $A=\begin{bmatrix} x & x \\ x & x \end{bmatrix}$ and $0 < x < 1$ I is an identity matrix. Then $\displaystyle \int _{0}^{1}\frac{g \left(\right. x \left.\right)}{f \left(\right. x \left.\right)}dx$ is equal to

NTA AbhyasNTA Abhyas 2022

Solution:

Put the $A$ in given equation
$\frac{1}{2}\begin{bmatrix} f\left(\right.x\left.\right) & g\left(\right.x\left.\right) \\ g\left(\right.x\left.\right) & f\left(\right.x\left.\right) \end{bmatrix}=\frac{1}{2}\begin{bmatrix} e^{2 x}+1 & e^{2 x}-1 \\ e^{2 x}-1 & e^{2 x}+1 \end{bmatrix}$
$\Rightarrow f\left(\right.x\left.\right)=e^{2 x}+1,g\left(\right.x\left.\right)=e^{2 x}-1$
$\displaystyle \int _{0}^{1}\left(\frac{e^{2 x} - 1}{e^{2 x} + 1}\right)dx=\left(\left[l n \left(e^{x} + e^{- x}\right)\right]\right)_{0}^{1}$
$=ln\left(e + e^{- 1}\right)-ln2$