Question

If $a{x}^2+(1-\lambda )x+(a-1-\lambda )=0$ where $a\ne 0,$ has real roots for all $\lambda \in R,$ then

• A. $a=1$
• B. $a>1$
• C. $0<a<1$
• D. $0<a\le 1$

Answer 1

The correct option is D $0<a\le 1$

$a{x}^2+(1-\lambda )x+(a-1-\lambda )=0$
Given equation has real roots, so
$D\ge 0\Rightarrow (1-\lambda {)}^2-4a(a-1-\lambda )\ge 0\Rightarrow {\lambda }^2-2\lambda +1-4a^2+4a+4a\lambda \ge 0\Rightarrow {\lambda }^2+2(2a-1)\lambda +(1-4a^2+4a)\ge 0$
As it is valid for all $\lambda \in R$, so
$D\le 0\Rightarrow 4(2a-1{)}^2-4(1-4a^2+4a)\le 0\Rightarrow 4a^2+1-4a-1+4a^2-4a\le 0\Rightarrow 8a^2-8a\le 0\Rightarrow a(a-1)\le 0\therefore 0<a\le 1$