Question

Let R² denote R × R. Let
S = {(a, b, c) : a, b, c ∈ R and ax² + 2bxy + cy² > 0 for all (x, y) ∈ R² - {(0, 0}}.
Then which of the following statements is (are) TRUE ?

• A. \((2,\frac{7}{2},6)\in S\)
• B. If \((3,b,\frac{1}{12})\in S\), then |2b| < 1.
• C. For any given (a, b, c) ∈ S, the system of linear equations
  ax + by = 1
  bx + cy = -1
  has a unique solution.
• D. For any given (a, b, c) ∈ S, the system of linear equations
  (a + 1)x + by = 0
  bx + (c + 1)y = 0
  has a unique solution.

Answer 1

The Correct Option is B, C, D

The condition \( ax^2 + 2bxy + cy^2 > 0 \) for all \( (x, y) \in \mathbb{R}^2 \setminus \{(0, 0)\} \) implies the quadratic form is positive definite. The necessary and sufficient conditions for positive definiteness are:

\[ a > 0, \quad c > 0, \quad \text{and} \quad b^2 - ac < 0. \]  

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(A) Consider \( \left( \frac{7}{2}, \frac{7}{2}, 6 \right) \): Here,

\[ a = \frac{7}{2}, \quad b = \frac{7}{2}, \quad c = 6. \]

Check \( b^2 - ac \):

\[ b^2 - ac = \left( \frac{7}{2} \right)^2 - \frac{7}{2} \times 6 = \frac{49}{4} - \frac{42}{4} = \frac{7}{4} > 0. \]

Thus, the quadratic form is not positive definite.

(A is False).

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(B) Consider \( (3, b, \frac{1}{12}) \in S \): Here,

\[ a = 3, \quad c = \frac{1}{12}. \]

For positive definiteness:

\[ b^2 - ac < 0 \Rightarrow b^2 - 3 \times \frac{1}{12} < 0 \Rightarrow b^2 - \frac{1}{4} < 0. \] \[ \Rightarrow |b| < \frac{1}{2}. \]

Thus, \( |2b| < 1 \).

(B is True).

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(C) Consider the system of linear equations:

\[ ax + by = 1, \quad bx + cy = -1. \]

The determinant of the coefficient matrix is:

\[ \Delta = \begin{vmatrix} a & b \\ b & c \end{vmatrix} = ac - b^2. \]

For \( (a, b, c) \in S \), we have \( b^2 - ac < 0 \), which implies \( \Delta \neq 0 \). Thus, the system has a unique solution.

(C is True).

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(D) Consider the system of linear equations:

\[ (a + 1)x + by = 0, \quad bx + (c + 1)y = 0. \]

The determinant of the coefficient matrix is:

\[ \Delta = \begin{vmatrix} a+1 & b \\ b & c+1 \end{vmatrix} = (a + 1)(c + 1) - b^2. \]

For \( (a, b, c) \in S \), since \( b^2 - ac < 0 \) and \( a > 0, c > 0 \), we also have:

\[ (a+1)(c+1) > b^2 \Rightarrow \Delta \neq 0. \]

Thus, the system has a unique solution.

(D is True).

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Final Answer:

\[ (B, C, D) \]