Find the value of `a` for which `a x^2+(a-3)x+1<0` for at least one positive real `x` .

Explanation

Let `f(x)=ax^(2)+ (a-3)x+1`
Case (i)
If `a gt 0`, then `f(x)` will be negative only for those values of x which lie between the roots.
Since `f(0)= 1 gt0`, the favourable graph according to the questions is shown in the following figure.

From the graph, we can see that both the roots are positive.
For this
(i) `D gt 0 rArr (a-3)^(2)-4a gt 0`
`" "rArr a lt 1 or a gt 9 " "` (i)
(ii) Sum `gt 0 ` and product ` gt0`
`" "rArr -(a-3) gt 0 and 1//a gt 0 `
`" "rArr a lt 3 " "`(ii)
From (i) and (ii), ` a lt 1`
But `a gt 0 rArr a in (0, 1)`
Case (ii) : `alt 0`
Since `f(0)=1 gt 0`, the graph is as shown in the following figure :

From the graph `ax^(2)+a(a-3) x+1 lt 0`, for at least one positive x.
Case (iii): `a=0`
If `a=0, f(x) = -3x+1`
`rArr" "f (x) lt 0, AA x gt 1//3`
Thus, from all the cases, the required set of values of `'a'` is `(-oo, 1)`.