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

(B) If $\left(3, \mathrm{~b}, \frac{1}{12}\right) \in \mathrm{S}$, then $|2 \mathrm{~b}|<1$.; (C) For any given $(\mathrm{a}, \mathrm{b}, \mathrm{c}) \in \mathrm{S}$, the system of linear equations \[ \begin{aligned} & a x+b y=1 \\ & b x+c y=-1 \end{aligned} \] has a unique solution.; (D) For any given $(\mathrm{a}, \mathrm{b}, \mathrm{c}) \in \mathrm{S}$, the system of linear equations \[ \begin{aligned} & (a+1) x+b y=0 \\ & b x+(c+1) y=0 \end{aligned} \] has a unique solution

(A) $\left(2, \frac{7}{2}, 6\right) \in S$

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Quadratic Equations and Inequalities

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Let

R^{2}

denote

R \times R

. Let

S = {(a, b, c) : a, b, c \in R

and

ax^{2} + 2 bxy + cy^{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 $∣2 b ∣ < 1$ .✓ Correct
(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.✓ Correct
(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✓ Correct

Answer: (B), (C), (D)

Solution

For

a x^{2} + 2 b x y + c y^{2} > 0

for all

(x, y) \neq = (0, 0)

, the quadratic form must be positive definite. This requires

a > 0

c > 0

, and

b^{2} < a c

.

(A) For

(2, \frac{7}{2}, 6)

b^{2} = \frac{49}{4}

and

a c = 12

. Since

\frac{49}{4} > 12

, we have

b^{2} > a c

, so this is NOT in

S

. FALSE.

(B) For

(3, b, \frac{1}{12}) \in S

b^{2} < 3 \cdot \frac{1}{12} = \frac{1}{4}

, so

∣ b ∣ < \frac{1}{2}

, which means

∣2 b ∣ < 1

. TRUE.

(C) The system has a unique solution if the determinant

a b b c = a c - b^{2} \neq = 0

. Since

(a, b, c) \in S

implies

b^{2} < a c

, we have

a c - b^{2} > 0

, so the system has a unique solution. TRUE.

(D) The system has a unique solution if

(a + 1) (c + 1) - b^{2} \neq = 0

. Since

a c > b^{2}

, we have

a c + a + c + 1 > b^{2} + 1 > b^{2}

, so

(a + 1) (c + 1) > b^{2}

. TRUE.

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JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Quadratic Equations and Inequalities

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