I tried by using the following : $ax^2+bx+c<x$ $\implies (b-1)^2<4ac$ But $ax^2+bx+c<x$ also $\implies$ a parabola lying underneath a straight line $y=x$ such that the parabola faces the origin from negative y axis. In that case is $b<1$ the correct choice?
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$ax^2+{(b-1)}x + c < 0$ for all $x$ when $a\neq 0$ implies both
(1) $a < 0$
(2) $(b-1)^2-4ac < 0$
For $a=0$, this becomes a linear inequality $(b-1)x+c<0$ and is true for all $x$ when $b=1$ and $c<0$
cr001
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1(1) is not true. We can have $a=0,b =1,c=-1$ for example. – Kavi Rama Murthy Feb 18 '19 at 08:01
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Thank you for the notice. Have edited to include the linear case. – cr001 Feb 18 '19 at 08:06
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There is a typo at the end. – Kavi Rama Murthy Feb 18 '19 at 08:10
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So, what is the answer ? – Feb 18 '19 at 08:49
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Consider the function
$$ax^2+(b-1)x+c<0.$$
If $a\ne0$, it is a parabola. To stay in the negatives, it needs to be concave down, i.e. $a<0$. Then it may not have real roots, $(b-1)^2<4ac$ (implicitly, $c<0$).
If $a=0$, the function is linear and takes both signs, unless $b=1$. In this case, only $c<0$ remains.
The complete condition is
$$(a<0\land (b-1)^2<4ac)\lor (a=0\land b=1\land c<0).$$
Only $a\le0$ is compatible.