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  • Given, <img alt="f^{\prime}(1)=1 and \frac{d}{d x}(f(2 x))=f^{\prime}(x), \forall x>0" src="https://entrancecorner.oncodecogs.com/gif.latex?f%5E%7B%5Cprime%7D%281%29%3D1%20and%20%5Cfrac%7Bd

Given, f^{\prime}(1)=1 and \frac{d}{d x}(f(2 x))=f^{\prime}(x), \forall x>0. If f^{\prime}(x) is differentiable, then there exists a number c \in(2,4) such that f^{\prime \prime}(c) equals
  
 

Option: 1

-\frac{1}{4}


Option: 2

-\frac{1}{8}


Option: 3

\frac{1}{4}


Option: 4

\frac{1}{8}


Answers (1)

best_answer

\begin{aligned} & f^{\prime}(1)=1 ; 2 \cdot f^{\prime}(2 x)=f^{\prime}(x)\\ & \text{ Put } x=1, f^{\prime}(2)=\frac{f^{\prime}(1)}{2}=\frac{1}{2}\\ & \text{ and } f^{\prime}(4)=\frac{1}{2} f^{\prime}(2)=\frac{1}{4}\\ & \text{ Applying LMVT for y }=f^{\prime}(x) is [2,4].\\ & f^{\prime \prime}(c)=\frac{f^{\prime}(4)-f^{\prime}(2)}{2}=\frac{\frac{1}{4}-\frac{1}{2}}{2}=-\frac{1}{8} \text {. } \end{aligned}

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