Consider all functions given in List-I in the | vedclass

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(D) Lagrange's Mean Value Theorem states that for a function $f(x)$ continuous on $[a,b]$ and differentiable on $(a,b)$, there exists at least one $c

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$A-IV, B-III, C-II, D-V$

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(D) Lagrange's Mean Value Theorem states that for a function $f(x)$ continuous on $[a,b]$ and differentiable on $(a,b)$, there exists at least one $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.$A. f(x) = |x-1|$. This function is not differentiable at $x=1$, which is the endpoint of the interval $[1,3]$. Thus, $LMVT$ is not applicable. However, if we consider the slope of the secant line, $\frac{f(3)-f(1)}{3-1} = \frac{2-0}{2} = 1$. For $x > 1$, $f'(x) = 1$. Any $c \in (1,3)$ satisfies this. Looking at the options, $A-II$ is the intended match.$B. f(x) = \log x$. $f'(c) = \frac{\log 3 - \log 1}{3-1} = \frac{\log 3}{2} = \log 3^{1/2} = \log \sqrt{3}$. Since $f'(x) = 1/x$, $1/c = \log \sqrt{3} \implies c = 1/\log \sqrt{3} = \log_3 e^2$. Thus, $B-III$.$C. f(x) = x^2+x+1$. $f'(c) = \frac{f(3)-f(1)}{3-1} = \frac{(9+3+1)-(1+1+1)}{2} = \frac{13-3}{2} = 5$. Since $f'(x) = 2x+1$, $2c+1 = 5 \implies 2c = 4 \implies c = 2$. Thus, $C-II$.$D. f(x) = e^x$. $f'(c) = \frac{e^3-e^1}{3-1} = \frac{e^3-e}{2}$. Since $f'(x) = e^x$, $e^c = \frac{e^3-e}{2} \implies c = \log \left(\frac{e^3-e}{2}\right)$. Thus, $D-V$.Matching: $A-II, B-III, C-II, D-V$. Note: Option $D$ is the closest match.

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

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