Mathematics - Application of Derivative Question with Solution | TestHub

Applying LMVT for $\mathrm{f}(\mathrm{x})$ on $[1,3]$ we get\ ${\mathrm{f}}^{\prime }(\mathrm{c})=\frac{\mathrm{f}(3)-\mathrm{f}(1)}{3-1}$ for $\mathrm{c}\in (1,3)\Rightarrow {\mathrm{f}}^{\prime }(\mathrm{c})=2$ $\because f^{\prime }(x)\ge 2\forall x\in (1,3)\Rightarrow f^{\prime }(x)=2$. $\Rightarrow \mathrm{f}(\mathrm{x})=2\mathrm{x}+\mathrm{c}\because \mathrm{f}(1)=5\Rightarrow \mathrm{c}=3\therefore \mathrm{f}(\mathrm{x})=2\mathrm{x}+3$ Now $g(x)={\int }_1^{x}f(t)dt=x^2+3x-4\because g^{\prime }(x)=f(x)>0\forall x\in [1,3]$ ∴ Greatest value $=\mathrm{g}(3)=14$ and least value $=\mathrm{g}(1)=0$ = sum of greatest and least value = 14 .