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If $x^{3}-2 x^{2}-9 x+18=0$ and $A=\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9\end{array}\right|$, then the maximum value of $A$ is
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The correct answer is:
96
Given, $x^{3}-2 x^{2}-9 x+18=0$
and $A=\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9\end{array}\right|$
$x^{3}-2 x^{2}-9 x+18=0$
$\Rightarrow \quad x^{2}(x-2)-9(x-2)=0$
$\Rightarrow \quad\left(x^{2}-9\right)(x-2)=0$
$\Rightarrow \quad(x-3)(x+3)(x-2)=0$
$\therefore \quad x=2,3,-3$
$A=\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9\end{array}\right|$
$=1(9 x-48)-2(36-42)+3(32-7 x)$
$=9 x-48+12+96-21 x$
$=-12 x+60$
$A($ when $x=2)=-12 \times 2+60=36$
$A($ when $x=3)=-12 \times 3+60=24$
$A($ when $x=-3)=-12 \times(-3)+60=96$
More value of $A$ at $x=-3$ is $96 .$
and $A=\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9\end{array}\right|$
$x^{3}-2 x^{2}-9 x+18=0$
$\Rightarrow \quad x^{2}(x-2)-9(x-2)=0$
$\Rightarrow \quad\left(x^{2}-9\right)(x-2)=0$
$\Rightarrow \quad(x-3)(x+3)(x-2)=0$
$\therefore \quad x=2,3,-3$
$A=\left|\begin{array}{lll}1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9\end{array}\right|$
$=1(9 x-48)-2(36-42)+3(32-7 x)$
$=9 x-48+12+96-21 x$
$=-12 x+60$
$A($ when $x=2)=-12 \times 2+60=36$
$A($ when $x=3)=-12 \times 3+60=24$
$A($ when $x=-3)=-12 \times(-3)+60=96$
More value of $A$ at $x=-3$ is $96 .$
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