If a2+b2+c2+a b+b c+c a ≤ 0, where a, b, c ∈ R, then domain of the function f(x)=√ operatornamesgn(x) ⋅(a x2+b x+c) will be

Question

If a2+b2+c2+a b+b c+c a ≤ 0, where a, b, c ∈ R, then domain of the function f(x)=√ operatornamesgn(x) ⋅(a x2+b x+c) will be (A) (0, ∞) (B) [0, ∞) (C) (-∞, 0] (D) (-∞, ∞)

Tardigrade Admin · Accepted Answer

Θ a2+b2+c2+a b+b c+c a ≤ 0 ⇒(a+b)2+(b+c)2+(c+a)2 ≤ 0 ⇒ a+b=b+c=c+a=0 ⇒ a=b=c=0 ∴ Df=R

Q. If $a^{2} + b^{2} + c^{2} + ab + b c + c a \leq 0$ , where $a, b, c \in R$ , then domain of the function $f (x) = sgn (x) \cdot (a x^{2} + b x + c)$ will be

$(0, \infty)$

$[0, \infty)$

$(- \infty, 0]$

$(- \infty, \infty)$

$Θ a^{2} + b^{2} + c^{2} + ab + b c + c a \leq 0$
$\Rightarrow (a + b)^{2} + (b + c)^{2} + (c + a)^{2} \leq 0$
$\Rightarrow a + b = b + c = c + a = 0 \Rightarrow a = b = c = 0$
$∴ D_{f} = R$

Q. If a2+b2+c2+ab+bc+ca≤0, where a,b,c∈R, then domain of the function f(x)=sgn(x)⋅(ax2+bx+c)​ will be

(0,∞)

[0,∞)

(−∞,0]

(−∞,∞)