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The value of c prescribed by Largrange's...

The value of c prescribed by Largrange's mean value . Theorem, when `f(x)=sqrt(x^(2)-4),a=2 and b=3` is

A

`2.5`

B

`sqrt(5)`

C

`sqrt(3)`

D

`sqrt(3)+1`

Text Solution

Verified by Experts

The correct Answer is:
B
Clearly, `f(x)=sqrt(x^(2)-4)` is continuous on [2,3] and diferentiable on (2,3)
So, by Largrange's mean value theorem there exists `c in (2,3)` such that
`f'(c)=(f(3)-f(2))/(3-2)`
`rArr (c)/(sqrt(e^(2)-1))=sqrt(5)-0[ :. f(x)=sqrt(x^(2)-4)rArr f'(x)=(x)/(sqrt(x^(2)-4))]`
`rArr c^(2)=5(c^(2)-4)`
`rArr 4c^(2)=20`
`rArr c= sqrt(5)`
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