Let {a_2},{a_3} in R such that left| {{a_2} - | vedclass

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(D) Given the determinant $f(x) = \left| \begin{array}{ccc} 1 & {a_3} & {a_2} \ 1 & {a_3} & {2{a_2} - x} \ 1 & {2{a_3} - x} & {a_2} \end{array} ig

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$9$

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(D) Given the determinant $f(x) = \left| \begin{array}{ccc} 1 & {a_3} & {a_2} \ 1 & {a_3} & {2{a_2} - x} \ 1 & {2{a_3} - x} & {a_2} \end{array} ight|$.Applying row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$f(x) = \left| \begin{array}{ccc} 1 & {a_3} & {a_2} \ 0 & 0 & {a_2 - x} \ 0 & {a_3 - x} & {a_2 - a_3} \end{array} ight|$.Expanding along the first column:$f(x) = 1 \cdot [0 - (a_2 - x)(a_3 - x)] = -(a_2 - x)(a_3 - x) = -(a_2 a_3 - a_2 x - a_3 x + x^2) = -x^2 + (a_2 + a_3)x - a_2 a_3$.This is a downward-opening parabola $f(x) = ax^2 + bx + c$ where $a = -1$,$b = (a_2 + a_3)$,and $c = -a_2 a_3$.The maximum value of a quadratic $ax^2 + bx + c$ is given by $-\frac{D}{4a}$,where $D = b^2 - 4ac$.$D = (a_2 + a_3)^2 - 4(-1)(-a_2 a_3) = (a_2 + a_3)^2 - 4a_2 a_3 = (a_2 - a_3)^2$.Given $|a_2 - a_3| = 6$,so $D = 6^2 = 36$.The maximum value is $-\frac{36}{4(-1)} = \frac{36}{4} = 9$.

Let ${a_2},{a_3} \in R$ such that $\left| {{a_2} - {a_3}} \right| = 6$ and $f\left( x \right) = \left| \begin{array}{ccc} 1 & {a_3} & {a_2} \\ 1 & {a_3} & {2{a_2} - x} \\ 1 & {2{a_3} - x} & {a_2} \end{array} \right|, x \in R.$ Then the greatest value of $f(x)$ is

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