The value of c for which the Mean Value Theor | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given function is $f(x) = \log_{e} x$ on the interval $[1, 3]$.First,we calculate the values at the endpoints:$f(1) = \log_{e} 1 = 0$$f(3) = \log_

Vedclass · Accepted Answer

$2 \log_{3} e$

Vedclass · Answer

(D) Given function is $f(x) = \log_{e} x$ on the interval $[1, 3]$.First,we calculate the values at the endpoints:$f(1) = \log_{e} 1 = 0$$f(3) = \log_{e} 3$Next,we find the derivative of the function:$f'(x) = \frac{1}{x}$According to Lagrange's Mean Value Theorem,there exists a point $c \in (1, 3)$ such that:$f'(c) = \frac{f(3) - f(1)}{3 - 1}$Substituting the values:$\frac{1}{c} = \frac{\log_{e} 3 - 0}{2}$$\frac{1}{c} = \frac{\log_{e} 3}{2}$$c = \frac{2}{\log_{e} 3}$Using the property of logarithms $\frac{1}{\log_{a} b} = \log_{b} a$,we get:$c = 2 \log_{3} e$Thus,the correct option is $D$.

The value of $c$ for which the Mean Value Theorem holds for the function $f(x) = \log_{e} x$ on the interval $[1, 3]$ is:

Similar Questions

If $f(x) = x^{\alpha} \log x, x > 0, f(0) = 0$ and $f(x)$ satisfies Rolle's theorem on $[0, 1]$,then what is the value of $\alpha$?

Rolle's theorem is not applicable to the function $f(x) = |x|$ defined on $[-1, 1]$ because

Verify the Mean Value Theorem for the function $f(x) = x^{2}$ in the interval $[2, 4]$.

For which interval does the function $f(x) = \frac{x^2 - 3x}{x - 1}$ satisfy all the conditions of Rolle's theorem?

Let $f: [-1, 2] \rightarrow R$ be a differentiable function such that $0 \le f'(t) \le 1$ for $t \in [-1, 0]$ and $-1 \le f'(t) \le 0$ for $t \in [0, 2]$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module