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AP EAMCET · Maths · Application of Derivatives

The value \(c\) of Lagrange's mean value theorem for \(f(x)=e^{\mathrm{x}}+24\) in \([0,1]\) is

  1. A \(\log (e-1)\)
  2. B \(\log (e+1)\)
  3. C \(\log (e+24)\)
  4. D \(\log (e-24)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\log (e-1)\)

Step-by-step Solution

Detailed explanation

\(f(x)=e^x+24 \text { in }[0,1]\) According to Lagrange's mean value theorem \(\begin{aligned} & \frac{f(1)-f(0)}{1-0}=f^{\prime}(c) ; \frac{e+24-24-1}{1}=e^c \\ & \Rightarrow e-1=e^c \Rightarrow c=\log (e-1) \end{aligned}\)