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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Determinants
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If $a>0$ and determinant of $ax^2+2bx+c$ is -ve then $\begin{vmatrix}a&b&ax+b\\b&c&bx+c\\ax+b&bx+c&0\end{vmatrix}$ is equal to

$\begin{array}{1 1}(a)\;+ve&(b)\;(ac-b^2)(ax^2+2bx+c)\\(c)\;-ve&(d)\;0\end{array}$

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1 Answer

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We have $\begin{vmatrix}a&b&ax+b\\b&c&bx+c\\ax+b&bx+c&0\end{vmatrix}$
$R_3\rightarrow R_3-(xR_1+R_2)$
$\Rightarrow \begin{vmatrix}a &b&ax+b\\b &c&bx+c\\0&0&-(ax^2+2bx+c)\end{vmatrix}$
$\Rightarrow (ax^2+2bx+c)(b^2-ac)$
$\Rightarrow (+)(-)$
$\Rightarrow -ve$
Hence (c) is the correct answer.
answered Nov 21, 2013 by sreemathi.v
 

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