Applications Of Derivatives Question 338
Question: The function $ f(x)=2x^{3}-15x^{2}+36x+4 $ is maximum at
[Karnataka CET 2001]
Options:
A) $ x=2 $
B) $ x=4 $
C) $ x=0 $
D) $ x=3 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)=2x^{3}-15x^{2}+36x+4 $
therefore $ {f}’(x)=6x^{2}-30x+36 $ …………..(i) We know that for its maximum value $ {f}’(x)=0. $
$ 6x^{2}-30x+36=0 $
therefore $ (x-2)(x-3)=0 $
therefore $ x=2,,3. $
Again differentiating equation (i), we get $ {f}’’(x)=12x-30 $
therefore $ {f}’’(2)=24-30=-6<0 $ . Therefore $ f(x) $ is maximum at $ x=2. $
BETA
