Let R^2 denote R times R. Let S = {(a, b, c) | vedclass

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(B) The condition $ax^2 + 2bxy + cy^2 > 0$ for all $(x, y) 
eq (0, 0)$ implies that the quadratic form is positive definite. This occurs if and only

Vedclass · Accepted Answer

$B, C, D$

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(B) The condition $ax^2 + 2bxy + cy^2 > 0$ for all $(x, y) 
eq (0, 0)$ implies that the quadratic form is positive definite. This occurs if and only if $a > 0$ and the discriminant $D = (2b)^2 - 4ac $(A)$ For $(2, \frac{7}{2}, 6)$,$a = 2 > 0$ and $b^2 = (\frac{7}{2})^2 = 12.25$,while $ac = 2 	imes 6 = 12$. Since $12.25 > 12$,$(2, \frac{7}{2}, 6) 
otin S$.$(B)$ For $(3, b, \frac{1}{12})$,we need $b^2 $(C)$ The system has a unique solution if the determinant of the coefficient matrix is non-zero. The determinant is $ac - b^2$. Since $(a, b, c) \in S$,$b^2  0$. Thus,the system has a unique solution. This is $TRUE$.$(D)$ The system has a unique solution if the determinant $(a+1)(c+1) - b^2 
eq 0$. Since $ac > b^2$ and $a, c > 0$,we have $ac + a + c + 1 > b^2 + 1 > 0$. Thus,the determinant is always positive,ensuring a unique solution. This is $TRUE$.

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