If t1 and t2 are abscissae of two points on the curve f(x)=x - x^2 in the interval (0,1) then find the maximum value of the expression (t1+t2) - (t1^2 +t2^2)

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-{\mathrm{x}}^2 \\ \mathrm{f}\left({\mathrm{t}}_1\right)={\mathrm{t}}_1-{{\mathrm{t}}_1}^2 \\ \mathrm{f}\left({\mathrm{t}}_2\right)={\mathrm{t}}_2-{{\mathrm{t}}_2}^2 \\ \Rightarrow \mathrm{f}\left({\mathrm{t}}_1\right)+\mathrm{f}\left({\mathrm{t}}_2\right)={\mathrm{t}}_1+{\mathrm{t}}_2-{{\mathrm{t}}_1}^2-{{\mathrm{t}}_2}^2 \\ \mathrm{f}\left({\mathrm{t}}_1\right)+\mathrm{f}\left({\mathrm{t}}_2\right)=\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)-\left({{\mathrm{t}}_1}^2+{{\mathrm{t}}_2}^2\right) \\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}-{\mathrm{x}}^2 \\ \mathrm{Differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=1-2\mathrm{x} \\ \mathrm{For} \mathrm{maxima}/\mathrm{minima}, \mathrm{f}\text{'}\left(\mathrm{x}\right)=0 \\ \Rightarrow 1-2\mathrm{x}=0 \\ \Rightarrow \mathrm{x}=\frac{1}{2} \\ \mathrm{f}"\left(\mathrm{x}\right)=-2<0 \\ \mathrm{Hence} \mathrm{x}=\frac{1}{2} \mathrm{is} \mathrm{a} \mathrm{point} \mathrm{of} \mathrm{maxima}. \\ \mathrm{f}\left(\frac{1}{2}\right)=\frac{1}{2}-{\left(\frac{1}{2}\right)}^2=\frac{1}{2}-\frac{1}{4}=\frac{1}{4} \\ \mathrm{Hence} \mathrm{f}\left({\mathrm{t}}_1\right)+\mathrm{f}\left({\mathrm{t}}_2\right) \mathrm{will} \mathrm{be} \mathrm{maximum} \mathrm{when} \mathrm{f}\left({\mathrm{t}}_1\right)=\mathrm{f}\left({\mathrm{t}}_2\right)=\frac{1}{4} \\ {\left[\mathrm{f}\left({\mathrm{t}}_1\right)+\mathrm{f}\left({\mathrm{t}}_2\right)\right]}_{\max }={\left[\left({\mathrm{t}}_1+{\mathrm{t}}_2\right)-\left({{\mathrm{t}}_1}^2+{{\mathrm{t}}_2}^2\right)\right]}_{\max }=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \end{gathered}$$

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