Q105 / 160 · AP EAMCET 2024 (22 May Shift 1) · Mathematics
The value $c$ of Lagrange's mean value theorem for $f(x)=e^{\mathrm{x}}+24$ in $[0,1]$ is
- Correct answerA$\log (e-1)$
- B$\log (e+1)$
- C$\log (e+24)$
- D$\log (e-24)$
Solution
$f(x)=e^x+24 \text { in }[0,1]$
According to Lagrange's mean value theorem $\begin{aligned} & \frac{f(1)-f(0)}{1-0}=f^{\prime}(c) ; \frac{e+24-24-1}{1}=e^c \\ & \Rightarrow e-1=e^c \Rightarrow c=\log (e-1) \end{aligned}$
According to Lagrange's mean value theorem $\begin{aligned} & \frac{f(1)-f(0)}{1-0}=f^{\prime}(c) ; \frac{e+24-24-1}{1}=e^c \\ & \Rightarrow e-1=e^c \Rightarrow c=\log (e-1) \end{aligned}$
Hence, the correct answer is option: 'A'.