Q96 / 180 · KCET 2021 · Mathematics

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

  • Correct answer
    A
    96
  • B
    36
  • C
    24
  • D
    120

Solution

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 $96 .$

Hence, the correct answer is option: 'A'.

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