JEE Advanced 2024 Paper 1, Mathematics Section 2 Q2: Linear Equations And Cramer’s Rule

JEE Advanced2024Paper 1Mathematics Section 2

Let $R^{2}$ denote $R \times R$ . Let

$S = {(a, b, c) : a, b, c \in R and a x^{2} + 2 b x y + c y^{2} > 0 for all (x, y) \in R^{2} - {(0, 0)}}$ .

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 $∣2 b ∣ < 1$
C
For any given $(a, b, c) \in S$ , the system of linear equations $a x + b y = 1 b x + cy = - 1$ has a unique solution.
D
For any given $(a, b, c) \in S$ , the system of linear equations $(a + 1) x + b y = 0 b x + (c + 1) y = 0$ has a unique solution.

Solution

The quadratic form $a x^{2} + 2 b x y + c y^{2}$ is positive for all nonzero $(x, y)$ iff $a > 0$ , $c > 0$ , and the discriminant condition $b^{2} < a c$ holds (positive definite form).

(A) Check $(\frac{7}{2})^{2} = \frac{49}{4} > 12 = 2 \times 6$ , so $b^{2} > a c$ . Not in $S$ . False.

(B) $b^{2} < 3 \cdot \frac{1}{12} = \frac{1}{4}$ , hence $4 b^{2} < 1$ , i.e. $∣2 b ∣ < 1$ . True.

(C) Determinant is $a c - b^{2} > 0$ , so the system has a unique solution. True.

(D) Determinant equals $(a + 1) (c + 1) - b^{2} = (a c - b^{2}) + a + c + 1$ . Since $a, c > 0$ and $a c - b^{2} > 0$ , this is strictly positive. Hence a unique solution (the trivial one). True.

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JEE Advanced 2024 Paper 1, Mathematics Section 2 Q2: Linear Equations And Cramer’s Rule

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