Applications Of Derivatives Question 303
Question: The function $ x^{5}-5x^{4}+5x^{3}-1 $ is
[MP PET 1993]
Options:
A) Maximum at $ x=3 $ and minimum at $ x=1 $
B) Minimum at $ x=1 $
C) Neither maximum nor minimum at $ x=0 $
D) Maximum at $ x=0 $
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Answer:
Correct Answer: C
Solution:
Let $ f(x)=x^{5}-5x^{4}+5x^{3}-1 $
therefore $ f’(x)=5x^{4}-20x^{3}+15x^{2}=0 $
$ \therefore (x-3)(x-1)=0 $ or $ x=3,1 $
Now $ {f}’’(x)=20x^{3}-60x^{2}+30x $
Put $ x=3 $ and 1, we get $ {f}’’’(3)=+ve $ and $ {f}’’(1)=-ve $ and $ {f}’’(0)=0 $ .
Hence $ f(x) $ neither maximum nor minimum at $ x=0 $ .
BETA
