List of top Mathematics Questions on Mean Value Theorem asked in KEAM

Let $f(x)=x^{2}-10x+16$, $x\in\mathbb{R}.$ If $f^{\prime}(c)$ is equal to slope of the straight line joining the points (2,0) and (8,0), then the value of $c$ is
A value of \( c \) for which the conclusion of mean value theorem holds for the function \( f(x)=\log_e x \) on the interval \([1,3]\) is
Let \( f(x) \) and \( g(x) \) be two differentiable functions for \( 0 \leq x \leq 1 \) such that \( f(0)=2, g(0)=0, f(1)=6 \). If there exists a real number \( c \in (0,1) \) such that \( f'(c)=2g'(c) \), then \( g(1) \) is equal to
If $y = f(x)$ is continuous on $[0, 6]$, differentiable on $(0, 6)$, $f(0) = -2$ and $f(6) = 16$, then at some point between $x = 0$ and $x = 6$, $f'(x)$ must be equal to:

The function \[ f(x)= \begin{cases} 2x^2-1, & \text{if } 1 \leq x \leq 4 \\ 151-30x, & \text{if } 4 < x \leq 5 \end{cases} \] is not suitable to apply Rolle's theorem since:

The function \[ f(x)= \begin{cases} 2x^2-1, & \text{if } 1 \leq x \leq 4 \\ 151-30x, & \text{if } 4 < x \leq 5 \end{cases} \] is not suitable to apply Rolle's theorem since:

The function \[ f(x)= \begin{cases} 2x^2-1, & \text{if } 1 \leq x \leq 4 \\ 151-30x, & \text{if } 4 < x \leq 5 \end{cases} \] is not suitable to apply Rolle's theorem since: