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  • Help me solve this - Limit , continuity and differentiability - JEE Main-14

Let f(x) = x^3 - 12 x then f(x) has 

  • Option 1)

    Local maxima at 2

  • Option 2)

    Local minima at -2

  • Option 3)

    Local mxima at 1

  • Option 4)

    Local maxima at -2

 

Answers (1)

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)

 

 f'(x) = 3x^{2}-12= 3(x^{2}-4)= 3(x-2)(x+2)

 

\Rightarrow f'( x) changes from + to -  at x=-2 from - to + at x=2 so local max at x= -2

and local min at x= 2 

 

 

 


Option 1)

Local maxima at 2

Option 2)

Local minima at -2

Option 3)

Local mxima at 1

Option 4)

Local maxima at -2

Posted by

Himanshu

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