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  • If c is a point at which Roll's theorem holds for the function, <img alt="f(x)=\log _{e}\left ( \frac{x^{2}+\alpha }{7x} \right )" src="https://learn.careers360.com/latex-image/?f%28x%29%3D%5Clog%20_%7Be%7D%5Cleft%20%28%20%5Cfrac%7Bx%5E%7B2%7D&

If c is a point at which Roll's theorem holds for the function, f(x)=\log _{e}\left ( \frac{x^{2}+\alpha }{7x} \right ) in the interval \left [ 3,4 \right ], where \alpha \; \epsilon \; R thenf''(c) is equal to :
 
Option: 1 -\frac{1}{24}
Option: 2 -\frac{1}{12}
Option: 3 \frac{\sqrt{3}}{7}
Option: 4 \frac{1}{12}
 

Answers (1)

best_answer

 

 

Rolle’s Theorem -

Rolle’s Theorem

Informally, Rolle’s theorem states that if the outputs of a differentiable function f are equal at the endpoints of an interval, then there must be an interior point c where f ′(c) = 0

Rolle’s Theorem Statement:

Let f be a real-valued function defined on the closed interval [a, b] such that

  1. it is continuous on the closed interval [a, b],

  2. it is differentiable on the open interval (a, b), and

  3. f (a) = f (b).

There then exists at least one c ∈ (a, b) such that f ′(c) = 0.

-

{f(3)=f(4)} \\ {\frac{9+\alpha}{3}=\frac{16+\alpha}{4}}

\alpha=12

\\f'(x)=\frac{x^2-12}{x\left(x^2+12\right)}\\f'(c)=0\\\begin{array}{ll}{\therefore} & {c=\sqrt{12}} \\ {\therefore \quad} & {f^{\prime \prime}(c)=\frac{1}{12}}\end{array}

Correct Option (4)

Posted by

Kuldeep Maurya

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