JEE MAIN - Mathematics (2002 - No. 17)
If $$a>0$$ and discriminant of $$\,a{x^2} + 2bx + c$$ is $$-ve$$, then
$$\left| {\matrix{ a & b & {ax + b} \cr b & c & {bx + c} \cr {ax + b} & {bx + c} & 0 \cr } } \right|$$ is equal to
$$\left| {\matrix{ a & b & {ax + b} \cr b & c & {bx + c} \cr {ax + b} & {bx + c} & 0 \cr } } \right|$$ is equal to
$$+ve$$
$$\left( {ac - {b^2}} \right)\left( {a{x^2} + 2bx + c} \right)$$
$$-ve$$
$$0$$
Explanation
We have $$\left| {\matrix{
a & b & {ax + b} \cr
b & c & {bx + c} \cr
{ax + b} & {bx + c} & 0 \cr
} } \right|$$
By $$\,\,\,{R_3} \to {R_3} - \left( {x{R_1} + {R_2}} \right);$$
$$ = \left| {\matrix{ a & b & {ax + b} \cr b & c & {bx + c} \cr 0 & 0 & { - \left( {a{x^2} + 2bx + C} \right)} \cr } } \right|$$
$$ = \left( {a{x^2} + 2bx + c} \right)\left( {{b^2} - ac} \right)$$
$$ = \left( + \right)\left( - \right) = - ve.$$
By $$\,\,\,{R_3} \to {R_3} - \left( {x{R_1} + {R_2}} \right);$$
$$ = \left| {\matrix{ a & b & {ax + b} \cr b & c & {bx + c} \cr 0 & 0 & { - \left( {a{x^2} + 2bx + C} \right)} \cr } } \right|$$
$$ = \left( {a{x^2} + 2bx + c} \right)\left( {{b^2} - ac} \right)$$
$$ = \left( + \right)\left( - \right) = - ve.$$
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