MCQ
The nth derivative of $x{e^x}$ vanishes when
- A$x = 0$
- B$x = - 1$
- ✓$x = - n$
- D$x = n$
Now, ${f^n}(x) = 0$
==> $n{e^x} + x{e^x} = 0$ ==>$x = - n$.
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