If x2+(a-b) x+(1-a-b)=0, where a ∈ N b ∈ R then sum of the first five smallest values of ' a ' for which equation has unequal real roots for all values of ' b ' is equal to

Question

If x2+(a-b) x+(1-a-b)=0, where a ∈ N b ∈ R then sum of the first five smallest values of ' a ' for which equation has unequal real roots for all values of ' b ' is equal to (A) 15 (B) 20 (C) 25 (D) 30

Tardigrade Admin · Accepted Answer

Θ D>0, ∀ b ∈ R ⇒ (a-b)2-4.1(1-a-b)>0, ∀ b ∈ R k ⇒ b2+(4-2 a) b+(a2+4 a-4)>0, ∀ b ∈ R ∴ D prime=(4-2 a)2-4(a2+4 a-4)<0 ⇒ 16-16 a+4 a2-4 a2-16 a+16<0 ⇒ 32 a>32 ⇒ a>1 ∴ a=2,3,4,5,6 ∴ text sum =20

Q. If x2+(a−b)x+(1−a−b)=0, where a∈N&b∈R then sum of the first five smallest values of ' a ' for which equation has unequal real roots for all values of ' b ' is equal to

15

20

25

30

ΘD>0,∀b∈R ⇒(a−b)2−4.1(1−a−b)>0,∀b∈Rk ⇒b2+(4−2a)b+(a2+4a−4)>0,∀b∈R ∴D′=(4−2a)2−4(a2+4a−4)<0 ⇒16−16a+4a2−4a2−16a+16<0 ⇒32a>32⇒a>1 ∴a=2,3,4,5,6 ∴ sum =20

Q. If $x^{2} + (a - b) x + (1 - a - b) = 0$ , where $a \in N & b \in R$ then sum of the first five smallest values of ' $a$ ' for which equation has unequal real roots for all values of ' $b$ ' is equal to