Let f(x)=(x-a)(x-b)-left(frac{a+b}{2}right). | vedclass

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(C) Given $f(x) = (x-a)(x-b) - \frac{a+b}{2} = x^2 - (a+b)x + ab - \frac{a+b}{2}$.To find the minimum value,we find the vertex of the parabola. The de

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$\geq \frac{-(a+b)^2}{4}$

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(C) Given $f(x) = (x-a)(x-b) - \frac{a+b}{2} = x^2 - (a+b)x + ab - \frac{a+b}{2}$.To find the minimum value,we find the vertex of the parabola. The derivative is $f'(x) = 2x - (a+b)$.Setting $f'(x) = 0$,we get $x = \frac{a+b}{2}$.The minimum value is $f\left(\frac{a+b}{2}\right) = \left(\frac{a+b}{2} - a\right)\left(\frac{a+b}{2} - b\right) - \frac{a+b}{2}$.$f\left(\frac{a+b}{2}\right) = \left(\frac{b-a}{2}\right)\left(\frac{a-b}{2}\right) - \frac{a+b}{2} = -\frac{(a-b)^2}{4} - \frac{a+b}{2} = -\frac{a^2 - 2ab + b^2 + 2a + 2b}{4}$.We can rewrite this as $f\left(\frac{a+b}{2}\right) = -\frac{(a+b)^2 - 4ab + 2(a+b)}{4} = -\frac{(a+b)^2}{4} + \frac{4ab - 2(a+b)}{4}$.Since the roots of $f(x)=0$ are non-negative,let the roots be $x_1, x_2 \geq 0$. Then $x_1+x_2 = a+b \geq 0$ and $x_1x_2 = ab - \frac{a+b}{2} \geq 0$.From $x_1x_2 \geq 0$,we have $ab \geq \frac{a+b}{2}$,which implies $4ab \geq 2(a+b)$,so $4ab - 2(a+b) \geq 0$.Thus,the minimum value $f\left(\frac{a+b}{2}\right) = -\frac{(a+b)^2}{4} + \frac{4ab - 2(a+b)}{4} \geq -\frac{(a+b)^2}{4}$.

Let $f(x)=(x-a)(x-b)-\left(\frac{a+b}{2}\right)$. If $f(x)=0$ has both non-negative roots,then the minimum value of $f(x)$ is:

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