JEE Advanced 2024 Paper 1 Q06 Mathematics Matrices & Determinants Systems of Equations Hard

JEE Advanced 2024 Paper 1 · Q06 · Systems of Equations

Let $\mathbb{R}^2$ denote $\mathbb{R}\times\mathbb{R}$. Let $$S = \{(a,b,c) : a, b, c \in \mathbb{R}\ \text{and}\ ax^2 + 2bxy + cy^2 > 0\ \text{for all}\ (x,y) \in \mathbb{R}^2 - \{(0,0)\}\}.$$ Then which of the following statements is (are) TRUE?

  1. A. $\left(2, \dfrac{7}{2}, 6\right) \in S$
  2. B. $\left(3, b, \dfrac{1}{12}\right) \in S$, then $|2b| < 1$
  3. C. For any given $(a, b, c) \in S$, the system of linear equations $ax + by = 1$, $bx + cy = -1$ has a unique solution.
  4. D. For any given $(a, b, c) \in S$, the system of linear equations $(a+1)x + by = 0$, $bx + (c+1)y = 0$ has a unique solution.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:B, C, D

Solution

Quadratic form $ax^2+2bxy+cy^2>0$ on $\mathbb{R}^2\setminus\{0\}$ iff $a>0$ and $ac-b^2>0$ (positive definite). (A) $(2,7/2,6)$: $ac-b^2=12-49/4=-1/4<0$. FALSE. (B) $(3,b,1/12)$: need $\tfrac{3}{12}-b^2>0\Rightarrow b^2<\tfrac{1}{4}\Rightarrow|2b|<1$. TRUE. (C) Determinant $ac-b^2>0\neq 0$, so unique solution exists. TRUE. (D) Determinant of homogeneous system is $(a+1)(c+1)-b^2 = ac+a+c+1-b^2 = (ac-b^2)+(a+c)+1$. With $a>0$ and $c>ac/(... )$ — note $c>0$ since $ac>b^2\geq 0$ and $a>0$. So all three terms positive, determinant $>0$, unique solution (only the trivial one). TRUE. Correct: B, C, D.

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