A) \[{{e}^{a-bx}}\]
B) \[{{e}^{a-bx}}-1\]
C) \[1+a{{\log }_{e}}(a-bx)\]
D) \[{{e}^{-bx}}\]
Correct Answer: A
Solution :
\[1+\frac{(a-bx)}{1\ !}+\frac{{{(a-bx)}^{2}}}{2\ !}+\frac{{{(a-bx)}^{3}}}{3\ !}+......={{e}^{a-bx}}\].You need to login to perform this action.
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