If $t_{1}$ and $t_{2}$ 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 $(t_{1} + t_{2}) - (t_{1}^{2} + t_{2}^{2})$ .

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Solution

Given,

$f_{(x)} = x - x^{2}$

Put $x = t_{1}$ and $x = t_{2}$ we get

$f (t_{1}) = t_{1} - {t_{1}}^{2}$

$f (t_{2}) = t_{2} - {t_{2}}^{2}$

Now,

$f (t_{1}) + f (t_{2}) = t_{1} - {t_{1}}^{2} + t_{2} - {t_{2}}^{2}$

$= t_{1} + t_{2} - ({t_{1}}^{2} + {t_{2}}^{2})$

Taking

$f (x) = x - x^{2}$

Differentiate w.r.to x we get ,

$f^{'} (x) = 1 - 2 x$

For maxima and minima $f^{'} (x) = 0$

$1 - 2 x = 0$

$x = \frac{1}{2}$

Differentiate again $f^{''} (x) = - 2$

Put $x = \frac{1}{2}$

We get, $f^{''} (x) = - 2 < 0$

Now, $f (\frac{1}{2}) = (\frac{1}{2}) - {(\frac{1}{2})}^{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

Hence, $f (t_{1}) + f (t_{2})$ will be maximum when

$f (t_{1}) = f (t_{2}) = \frac{1}{4}$

${[f (t_{1}) + f (t_{2})]}_{max} = {[(t_{1} + t_{2}) - {(t_{1} + t_{2})}^{2}]}_{max} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

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