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  • Help me solve this - Limit , continuity and differentiability - JEE Main-11

Let f(x) = x^3 - 3x^2 + 2x in [2,3] when 'c' of L.M.V.T equals ?

  • Option 1)

    \frac{5}{2}

  • Option 2)

    \frac{11}{4}

  • Option 3)

    1 + \frac{\sqrt{21}}{3}

  • Option 4)

    1- \frac{\sqrt{21}}{3}

 

Answers (1)

best_answer

As we have learned

Lagrange's mean value theorem -

If  a  function  f(x) 

1.   is continuous in the closed interval [a,b] and 

2.   is differentiable in the open interval [a, b] then 

f'(c)=\frac{f(b)-f(a)}{b-a}\:\:\:\:where\:\:c\epsilon (a,b)

-

 

 \because f(x) is continous and diffrentiable in given interval to it satisfies condition of L.M.V.T so there exist 'c' such that f'(c)

 

 =\frac{f(3)-f(2)}{3-2} 

\Rightarrow 3c^{2}-6c+2=6-0

\Rightarrow 3c^{2}-6c-4=0

\Rightarrow c = \frac{6\pm \sqrt{36+48}}{6}= \frac{6\pm 2\sqrt21}{6}= 1\pm \sqrt21/3

but c\epsilon (2,3) , so c= 1\pm \sqrt21/3

 

 

 

 

 

 


Option 1)

\frac{5}{2}

Option 2)

\frac{11}{4}

Option 3)

1 + \frac{\sqrt{21}}{3}

Option 4)

1- \frac{\sqrt{21}}{3}

Posted by

Himanshu

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