If x2+(a-b) x+(1-a-b)=0, a, b ∈ R then the value of 'a' for which both roots of the equation are real and unequal ∀ b ∈ R is

Question

If x2+(a-b) x+(1-a-b)=0, a, b ∈ R then the value of 'a' for which both roots of the equation are real and unequal ∀ b ∈ R is (A) (2, ∞) (B) (3, ∞) (C) (1, ∞) (D) (-∞, 1)

Tardigrade Admin · Accepted Answer

x2+(a-b) x+(1-a-b)=0 ∵ D >0 ⇒ (a-b)2-4 × 1 ×(1-a-b)>0 a2+b2-2 a b-4+4 a+4 b>0 ∵ b2+2 b(2-a)+(a2+4 a-4)>0 ∵ 4(2-a)2-4 × 1 ×(a2+4 a-4)<0 4+a2-4 a-a2-4 a+4<0 8 a-8 >0 a >1

Q. If $x^{2} + (a - b) x + (1 - a - b) = 0, a, b \in R$ then the value of 'a' for which both roots of the equation are real and unequal $\forall b \in R$ is

$(2, \infty)$

$(3, \infty)$

$(1, \infty)$

$(- \infty, 1)$

Q. If x2+(a−b)x+(1−a−b)=0,a,b∈R then the value of 'a' for which both roots of the equation are real and unequal ∀b∈R is

(2,∞)

(3,∞)

(1,∞)

(−∞,1)

x2+(a−b)x+(1−a−b)=0 ∵D>0 ⇒(a−b)2−4×1×(1−a−b)>0 a2+b2−2ab−4+4a+4b>0 ∵b2+2b(2−a)+(a2+4a−4)>0 ∵4(2−a)2−4×1×(a2+4a−4)<0 4+a2−4a−a2−4a+4<0 8a−8>0 a>1

Q. If $x^{2} + (a - b) x + (1 - a - b) = 0, a, b \in R$ then the value of 'a' for which both roots of the equation are real and unequal $\forall b \in R$ is

$(2, \infty)$

$(3, \infty)$

$(1, \infty)$

$(- \infty, 1)$