A  value of \large c for which conclusion of Mean Value Theorem holds for the function

f\left ( x \right )= \log _{e}x  on the interval \left [ 1,3 \right ] is

  • Option 1)

    \log_{3}e

  • Option 2)

    \log_{e}3

  • Option 3)

    2\log_{3}e

  • Option 4)

    \frac{1}{2}\log_{e}3

 

Answers (1)

As we learnt in

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)

-

 

 

 

f(x) =log_ex

f(1) =log'_e =0

f(3) =log^3_e

\therefore f'(x) =\frac{f(b)-f(a)}{b-a} =\frac{log^3_e-0}{2}=\frac{1}{2}log^3_e

\frac{1}{x}=\frac{1}{2}log^3_e

\therefore C= 2log^e_3


Option 1)

\log_{3}e

Incorrect

Option 2)

\log_{e}3

Incorrect

Option 3)

2\log_{3}e

Correct

Option 4)

\frac{1}{2}\log_{e}3

Incorrect

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