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  • Try this! - Limit , continuity and differentiability - JEE Main-12

The value of 'c' of Rolle's Theorem for f(x) = x^3 - 3x^2 + 2x in [1, 2] is 

  • Option 1)

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

  • Option 2)

    1 + \sqrt{3}

  • Option 3)

    1 - \sqrt{3}

  • Option 4)

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

 

Answers (1)

best_answer

As we have learned

Rolle's Theorems -

Let f(x) be a function of x subject to the following conditions.

1.  f(x) is continuous function of    x:x\epsilon [a,b]

2.  f'(x) is exists for every point :  x\epsilon [a,b]

3.  f(a)=f(b)\:\:\:then\:\:f'(c)=0\:\:such \:that\:\:a<c<b.

-

 

 f(x) satisfies all three condition of rolle's theorem , so there exists 'C' such that f'(c)=0 and c \epsilon (1,2) 

\Rightarrow 3c^{2}-6c+2 =0 \Rightarrow c= \frac{6\pm \sqrt 12}{6}= 1\pm 1/\sqrt3

But c\epsilon (1,2) \therefore c= 1+1/\sqrt3

 

 

 

 

 


Option 1)

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

Option 2)

1 + \sqrt{3}

Option 3)

1 - \sqrt{3}

Option 4)

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

Posted by

Himanshu

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