JEE Main On 16 April 2018 Question 20
Question: Let M and m be respectively the absolute maximum and the absolute minimum values of the function, $ f(x)=2x^{3}-9x^{2}+12x+5 $ in the interval $ [0,3] $ . Then $ M-m $ is equal to. [JEE Main 16-4-2018]
Options:
A) 1
B) 5
C) 4
D) 9
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)=2x^{3}-9x^{2}+12x+5 $ $ f’(x)=6x^{2}-18x+12=0 $
$ \Rightarrow $ $ x^{2}-3x+2=0 $ $ x=1 $ or $ x=2 $
$ f’’(x)=12x-18 $ $ f’’(1)=12(1)-18=-6<0 $
Hence, function has maxima at $ x=1 $
$ \Rightarrow $ $ M=f(1)=2-9+12+5=10 $ $ f’’(2)=12(2)-18=6>0 $
Hence, function has minima at $ x=2 $
$ \Rightarrow $ $ m=f(2)=2(8)-9(4)+12(2)+5=9 $
Therefore, $ M-m=10-9=1 $
Answer is option A.
