Let S be the set of all functions f:[0,1] rig | vedclass

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(D) By the Lagrange Mean Value Theorem $(LMVT)$ applied to the function $f$ on the interval $[c, 1]$,where $c \in (0, 1)$,there exists at least one po

Vedclass · Accepted Answer

$\frac{f(1) - f(c)}{1 - c} = f'(a)$ for some $a \in (c, 1)$

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(D) By the Lagrange Mean Value Theorem $(LMVT)$ applied to the function $f$ on the interval $[c, 1]$,where $c \in (0, 1)$,there exists at least one point $a \in (c, 1)$ such that $\frac{f(1) - f(c)}{1 - c} = f'(a)$.Options $(A)$,$(B)$,and $(C)$ are not necessarily true for all functions in $S$. For example,if $f(x) = k$ (a constant function),then $f'(x) = 0$. In this case,$|f(c) - f(1)| = 0$ and $(1 - c)|f'(c)| = 0$,so the inequality $|f(c) - f(1)| Option $(D)$ is a direct application of the $LMVT$ on the interval $[c, 1]$,which guarantees the existence of $a \in (c, 1)$ satisfying the equation.

List-$I$	List-$II$
$A. \|x-1\|$	$I. 2 \log (e^3+e^2)$
$B. \log x$	$II. 2$
$C. x^2+x+1$	$III. \log_3 e^2$
$D. e^x$	$IV. \sqrt{2}$
	$V. \log \left(\frac{e^3-e}{2}\right)$

Let $S$ be the set of all functions $f:[0,1] \rightarrow \mathbb{R}$ which are continuous on $[0,1]$ and differentiable on $(0,1)$. Then for every $f \in S$,there exists a $c \in (0,1)$,depending on $f$,such that:

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