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  • Help me understand! - Limit , continuity and differentiability - JEE Main-2

 If x= -1 and x = 2 are extreme points of  f\left ( x \right )= \alpha \log \left | x \right |+\beta x^{2} +x then 

  • Option 1)

    \alpha =2,\beta =-\frac{1}{2}

  • Option 2)

    \alpha =2,\beta =\frac{1}{2}

  • Option 3)

    \alpha =-6,\beta =\frac{1}{2}

  • Option 4)

    \alpha =-6,\beta =-\frac{1}{2}

 

Answers (1)

best_answer

As we learnt in

Maxima Minima -

A functions graph follow up and down along x-axis then upper part is known as maxima and lower part is known as minima.

-

 

 \alpha \log x + \beta x^{2}+x

 

\therefore \frac{\alpha}{x}+2\beta x+1

\therefore at\ x=-1,\:\:and\:\:x=2

\therefore -\alpha -2\beta +1=0

\frac{\alpha}{2}+4\beta +1=0

So solve 

\alpha = 2    and    \beta = \frac{-1}{2} 


Option 1)

\alpha =2,\beta =-\frac{1}{2}

This option is correct.

Option 2)

\alpha =2,\beta =\frac{1}{2}

This option is incorrect.

Option 3)

\alpha =-6,\beta =\frac{1}{2}

This option is incorrect.

Option 4)

\alpha =-6,\beta =-\frac{1}{2}

This option is incorrect.

Posted by

Aadil

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