Question

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

• A. 15
• B. 20
• C. 25
• D. 30

Answer 1

Answer: (B)

$ \Theta D>0, \forall b \in R$
$\Rightarrow (a-b)^2-4.1(1-a-b)>0, \forall b \in R k $
$\Rightarrow b^2+(4-2 a) b+\left(a^2+4 a-4\right)>0, \forall b \in R $
$\therefore D^{\prime}=(4-2 a)^2-4\left(a^2+4 a-4\right)<0 $
$\Rightarrow 16-16 a+4 a^2-4 a^2-16 a+16<0 $
$\Rightarrow 32 a>32 \Rightarrow a>1 $
$\therefore a=2,3,4,5,6 $
$\therefore \text { sum }=20 $