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

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(C) Given,$f(x) = (x - a)(x - b) - (\frac{a + b}{2})$.Expanding this,we get $f(x) = x^2 - (a + b)x + ab - (\frac{a + b}{2})$.The minimum value of a qu

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

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(C) Given,$f(x) = (x - a)(x - b) - (\frac{a + b}{2})$.Expanding this,we get $f(x) = x^2 - (a + b)x + ab - (\frac{a + b}{2})$.The minimum value of a quadratic $Ax^2 + Bx + C$ is given by $-\frac{D}{4A}$,where $D = B^2 - 4AC$.Here,$A = 1$,$B = -(a + b)$,and $C = ab - \frac{a + b}{2}$.$D = (-(a + b))^2 - 4(1)(ab - \frac{a + b}{2}) = (a + b)^2 - 4ab + 2(a + b) = (a - b)^2 + 2(a + b)$.Minimum value $= -\frac{(a - b)^2 + 2(a + b)}{4}$.Since the roots are non-negative,the sum of roots $\alpha + \beta = (a + b) \geq 0$ and the product of roots $\alpha \beta = ab - \frac{a + b}{2} \geq 0$.Also,for real roots,$D \geq 0$,which is satisfied as $(a - b)^2 + 2(a + b) \geq 0$.Given the condition of non-negative roots,the minimum value of the function is attained at $x = \frac{a + b}{2}$.Substituting $x = \frac{a + b}{2}$ into $f(x)$:$f(\frac{a + b}{2}) = (\frac{a + b}{2} - a)(\frac{a + b}{2} - b) - \frac{a + b}{2} = (\frac{b - a}{2})(\frac{a - b}{2}) - \frac{a + b}{2} = -\frac{(a - b)^2}{4} - \frac{a + b}{2}$.Given the constraints,the minimum value is $\geq -\frac{(a + b)^2}{4}$.

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

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