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KEAM 2018 Math Paper

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Question : 112 of 120

Marks: +1, -0

Let

f(x)=\begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix}.

x=-9

f(x)=0,

then the other root are

2 and 7
3 and 6
7 and 3
6 and 2
6 and 7

Solution:

Given,

f(x)=\begin{vmatrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix}

=\begin{vmatrix} x+9 & x+9 & x+9 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix}

[ applying

R_{1} \rightarrow R_{1}+R_{2}+R_{3}]

=(x+9)\begin{vmatrix} 1 & 1 & 1 \\ 2 & x & 2 \\ 7 & 6 & x \end{vmatrix}

=(x+9)\begin{vmatrix} 0 & 0 & 1 \\ 2-x & x-2 & 2 \\ 1 & 6-x & x \end{vmatrix}

[applying

C_{1} \rightarrow C_{1}-C_{2}

and

C_{2} \rightarrow C_{2}-C_{3}]

=(x+9)[(2-x)(6-x)-(x-2)]

=(x+9)(x-2)[x-6-1]

f(x)=(x+9)(x-2)(x-7)

at

f(x)=0

(x+9)(x-2)(x-7)=0

\Rightarrow \;\; x=-9,2,7

Hence, other roots are 2 and 7 .

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