InterviewSolution
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InterviewSolution
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If A is a square matrix and `e^a` is defined as `e^A=1+A^2/(2!)+A^3/(3!)...........oo=1/2[f(x) ,g(x) and g(x) ,f(x)],` where `A=[(x,x),(x,x)].` and I being the identity matrix then `int (g(x))/(f(x))dx=`A. `(1)/(2sqrt(e^(x)-1))-"cosec"^(-1)(e^(x))+c`B. `(2)/(sqrt(e^(x)-e^(-x)))-"sec"^(-1)(e^(x))+c`C. `(1)/(2sqrt(e^(2x)-1))+"sec"^(-1)(e^(x))+c`D. none of these |
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Answer» Correct Answer - C `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(e^(2x)+1)/(sqrt(e^(2x)-1))dx=int(e^(2x))/(sqrt(e^(2x)-1))dx+(1)/(sqrt(e^(2x)-1))dx` `=int(e^(2x))/(sqrt(e^(2x)-1))dx+int (e^(x))/(e^(x) sqrt(e^(2x)-1))dx` `=(1)/(2sqrt(e^(2x)-1))+sec^(-1)(e^(x))+C` |
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