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
- A \(\log (e-1)\)
- B \(\log (e+1)\)
- C \(\log (e+24)\)
- D \(\log (e-24)\)
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 \(\begin{aligned} & \frac{f(1)-f(0)}{1-0}=f^{\prime}(c) ; \frac{e+24-24-1}{1}=e^c \\ & \Rightarrow e-1=e^c \Rightarrow c=\log (e-1) \end{aligned}\)
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