A) \[x\]
B) \[-x\]
C) \[-y\]
D) \[y\]
Correct Answer: D
Solution :
\[y=1-x+\frac{{{x}^{2}}}{(2)!}-\frac{{{x}^{2}}}{(3)!}+............\] Þ \[y={{e}^{-x}}\] Þ \[\frac{dy}{dx}={{e}^{-x}}(-1)\] and \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=(-1)\{{{e}^{-x}}.(-1)\}\]\[={{e}^{-x}}=y\].You need to login to perform this action.
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