If `e^(A)` is defined as `e^(A)=I+A+A^(2)/(2!)+A^(3)/(3!)+...=1/2 [(f(x),g(x)),(g(x),f(x))]`, where `A=[(x,x),(x,x)], 0 lt x lt 1` and 1 is identity matrix, then find the functions f(x) and g(x).

Options

A. `(e^(x))/(2)(sinx-cosx)`
B. `(e^(2x))/(5)(2sinx-cosx)`
C. `(e^(x))/(5)(sin2x-cos2x)`
D. none of these

Answer

B

Explanation

`A=[(x,x),(x,x)]`
`impliesA^(2)=[(2x^(2),2x^(2)),(2x^(2),2x^(2))],A^(3)=[(2^(2)x^(3),2^(2)x^(3)),(2^(2)x^(3),2^(2)x^(3))]` and so on
Then `e^(A)=I+A+(A^(2))/(2!)+(A^(3))/(3!)+ … +`
`=[(1+x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ... ,x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ...),(x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ..., 1+x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+ ...)]`
`=[((1)/(2)(1+2x+(2^(2)x^(2))/(2!)+(2^(3)x^(3))/(3!)+ ...)+(1)/(2) ,(1)/(2)(1+2x+(2^(2)x^(2))/(2!)+ ...)-(1)/(2)),((1)/(2)(1+2x+(2^(2)x^(2))/(2!)+(2^(3)x^(3))/(3!)+ ...)-(1)/(2) ,(1)/(2)(1+2x+(2^(2)x^(2))/(2!)+ ...)+(1)/(2))]`
`=(1)/(2)[(e^(2x)+1,e^(2x)-1),(e^(2x)-1,e^(2x)+1)]`
` :. f(x)=e^(2x)+1 and g(x)=e^(2x)-1`
`int(g(x)+1)sinx dx=inte^(2x) sinx dx=(e^(2x))/(5)(2sinx-cosx)`