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KCET 2021 Math Solved Paper

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Question : 36 of 60

Marks: +1, -0

If

x^{3}-2 x^{2}-9 x+18=0

A=\begin{vmatrix} 1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9 \end{vmatrix},

then the maximum value of A is

96
36
24
120

Solution:

Given

x^{3}-2 x^{2}-9 x+18=0

and

A=\begin{vmatrix} 1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9 \end{vmatrix},

x^{3}-2 x^{2}-9 x+18=0

⇒

x^{2}(x-4-9(x-2)=0

⇒

(x^{2}-9)(x-2)=0

⇒

(x-3)(x+3)(x-2)=0

∴

x=2,3,-3

A=\begin{vmatrix} 1 & 2 & 3 \\ 4 & x & 6 \\ 7 & 8 & 9 \end{vmatrix},

=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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