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Let $f(x)$ be a function continuous on $[1,2]$ and differentiable on $(1,2)$ satisfying
$f(1) = 2, f(2) = 3$ and $f'(x) \geq 1 \forall x \in (1,2)$.Define $g(x)=\int\limits_1^x {f(t)\,dt\,\forall \,x\, \in [1,2]} $ then the greatest value of $g(x)$ on $[1,2]$ is-
$3$
$5$
$\frac{5}{2}$
$\frac{3}{2}$
Solution
Using $LMVT$ on $f(\mathrm{x})$ for intervals $[1, \mathrm{x}]$ and $[\mathrm{x}, 2]$
$\frac{f(\mathrm{x})-f(1)}{\mathrm{x}-1}=f^{\prime}\left(\mathrm{c}_{1}\right) \Rightarrow \frac{f(\mathrm{x})-2}{\mathrm{x}-1} \geq 1$
$\Rightarrow f(\mathrm{x}) \geq \mathrm{x}+1$
and $\frac{f(2)-f(x)}{2-x}=f^{\prime}\left(c_{2}\right) \Rightarrow \frac{3-f(x)}{2-x} \geq 1$
$\Rightarrow f(\mathrm{x}) \leq \mathrm{x}+1$
$\therefore f(\mathrm{x})=\mathrm{x}+1$
$\therefore$ greatest value of $\mathrm{g}(\mathrm{x})=\mathrm{g}(2)$
$ = \int\limits_1^2 {(x + 1)} dx = \frac{5}{2}$
