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  • I need help with - Limit , continuity and differentiability - JEE Main-3

Let M and m be respectively the absolute maximum and the absolute minimum
values of the function,
f(x)= 2x^{3}-9x^{2}+12x+5in the interval
[0, 3]. Then M−m is equal to :

  • Option 1)

    5

  • Option 2)

    9

  • Option 3)

    4

  • Option 4)

    1

 

Answers (3)

best_answer

As we have learned

Method for maxima or minima -

First and second derivative method :

Step\:1.\:\:find\:values\:of\:x\:for\:\frac{dy}{dx}=0

Step\:2.\:\:x=x_{\circ }\:\:is\:a\:point\:of\:local\:maximum\:if\:f'(x)>0  and\:local\:minimum\:if\;f'(x)<0.

Step\:\:3.\:\:\:x=x_{\circ }\:\:is\:a\:point\:of\:local\:miximum\:if  f''(x)<0\:\:and\:local\:minimum\:if\:f''(x)>0

- wherein

Where\:\:y=f(x)

\frac{dy}{dx}=f'(x)

 

 

Absolute maximum and Absolute minimum values of a function -

The largest value of a continuous function f(x) in an interval [a, b] and it is maximum (absolute value) and the smallest value of a function is minimum (absolute value)

-

 

 

f(0)= 5 and f(3)=14

f'(x)= 6x^{2}-18x+12

x=1,2

f(1)= 10

f(2)= 16-36+24+5

=9 

M-m = 9-5 = 4


Option 1)

5

This is incorrect 

Option 2)

9

This is incorrect 

Option 3)

4

This is correct 

Option 4)

1

This is incorrect 

Posted by

Himanshu

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