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 If fand g are differentiable functions in [0, 1] satisfying f (0) = 2 = g (1), g (0) = 0 andf(1)=6, then for some c\epsilon]0, 1[ : 

  • Option 1)

    f'\left ( c \right )=g'\left ( c \right )

  • Option 2)

    f'\left ( c \right )=2g'\left ( c \right )

  • Option 3)

    2f'\left ( c \right )=g'\left ( c \right )

  • Option 4)

    2f'\left ( c \right )=3g'\left ( c \right )

 

Answers (1)

best_answer

As we have learned

cauchy's Theorem -

\frac{f'(c)}{g'(c)}= \frac{f(b)-f(a)}{g(b)-g(a)} 

for some C \: \: \epsilon \: \: (a,b)

- wherein

f(x ) \: \: and \: \: g(x ) \: \: diffrentialable\: \: [a,b]

 

 

We have 

\frac{f'(c)}{g'(c)}= \frac{f(1)-f(0)}{g(1)-g(0)}= \frac{6-2}{2-0}= 2

For some 

\epsilon ]0,1[

 


Option 1)

f'\left ( c \right )=g'\left ( c \right )

Option 2)

f'\left ( c \right )=2g'\left ( c \right )

Option 3)

2f'\left ( c \right )=g'\left ( c \right )

Option 4)

2f'\left ( c \right )=3g'\left ( c \right )

Posted by

Himanshu

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