JEE Advanced 2024 · Paper 1 · Mathematics Q6

MathematicsMultiple correct+4 / -2Question 6 of 51

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

$S = \left\{(a,b,c) : a,b,c \in \mathbb{R} \text{ and } ax^2 + 2bxy + cy^2 > 0 \text{ for all } (x,y) \in \mathbb{R}^2 - \{(0,0)\}\right\}.$

Then which of the following statements is (are) TRUE?

\left(2, \dfrac{7}{2}, 6\right) \in S

\left(3, b, \dfrac{1}{12}\right) \in S

, then

|2b| < 1

For any given

(a,b,c) \in S

, the system of linear equations

\begin{aligned} ax + by &= 1 \\ bx + cy &= -1 \end{aligned}

has a unique solution.

For any given

(a,b,c) \in S

, the system of linear equations

\begin{aligned} (a+1)x + by &= 0 \\ bx + (c+1)y &= 0 \end{aligned}

has a unique solution.

Answer and worked solution

Official answer: B, C, D

$ax^2 + 2bxy + cy^2 > 0$ for all $(x,y)\ne(0,0)$ exactly when $a > 0$ , $c > 0$ and $ac - b^2 > 0$ .

(A) $ac = 12 < b^2 = \frac{49}{4}$ . Not in $S$ . False.

(B) $3\cdot\frac{1}{12} > b^2 \implies |b| < \frac12 \implies |2b| < 1$ . True.

(C) Coefficient determinant $ac - b^2 > 0$ , so the solution is unique. True.

(D) Determinant $(a+1)(c+1) - b^2 = (ac - b^2) + a + c + 1 > 0$ . Unique solution. True.

JEE Advanced 2024 · Paper 1, Mathematics question 6