Limits Continuity And Differentiability Question 67
Question: If $ I _{n}=\frac{d^{n}}{dx^{n}}(x^{n}\log x) $ , then $ I _{n}-n{I _{n-1}}= $
Options:
A) n
B) $ n-1 $
C) $ n! $
D) $ ( n-1 )! $
Show Answer
Answer:
Correct Answer: D
Solution:
$ I _{n}=\frac{{d^{n-1}}}{d{x^{n-1}}}[{x^{n-1}}+n{x^{n-1}}\log x] $
$ I _{n}=(n-1)!+n{I _{n-1}}\Rightarrow I _{n}-n{I _{n-1}}=(n-1)! $
BETA
