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(C) Using the Lagrange Mean Value Theorem $(LMVT)$ on $f(x)$ for the interval $[1, x]$ where $x \in (1, 2)$:$\frac{f(x) - f(1)}{x - 1} = f'(c_1) \geq

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$\frac{5}{2}$

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(C) Using the Lagrange Mean Value Theorem $(LMVT)$ on $f(x)$ for the interval $[1, x]$ where $x \in (1, 2)$:$\frac{f(x) - f(1)}{x - 1} = f'(c_1) \geq 1 \implies f(x) - 2 \geq x - 1 \implies f(x) \geq x + 1$.Using $LMVT$ on $f(x)$ for the interval $[x, 2]$:$\frac{f(2) - f(x)}{2 - x} = f'(c_2) \geq 1 \implies 3 - f(x) \geq 2 - x \implies f(x) \leq x + 1$.Since $f(x) \geq x + 1$ and $f(x) \leq x + 1$,we must have $f(x) = x + 1$.Now,$g(x) = \int_1^x (t + 1) \, dt = \left[ \frac{t^2}{2} + t \right]_1^x = \left( \frac{x^2}{2} + x \right) - \left( \frac{1}{2} + 1 \right) = \frac{x^2}{2} + x - \frac{3}{2}$.Since $g'(x) = f(x) = x + 1 > 0$ for $x \in [1, 2]$,$g(x)$ is an increasing function.Therefore,the greatest value of $g(x)$ on $[1, 2]$ is $g(2) = \int_1^2 (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_1^2 = (2 + 2) - (0.5 + 1) = 4 - 1.5 = \frac{5}{2}$.

Let $f(x)$ be a function continuous on $[1, 2]$ and differentiable on $(1, 2)$ satisfying $f(1) = 2, f(2) = 3$ and $f'(x) \geq 1$ for all $x \in (1, 2)$. Define $g(x) = \int_1^x f(t) \, dt$ for all $x \in [1, 2]$. Then the greatest value of $g(x)$ on $[1, 2]$ is-

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