JEE Main 2025 Apr 2 Shift 2, Mathematics Q25: Roots Lie In An Interval

For both roots to be real and positive, three conditions are needed on the quadratic $(1-a)x^2+2(a-3)x+9=0$ with leading coefficient $(1-a)\ne 0$:

(i) Discriminant $\ge 0$: $4(a-3{)}^2-36(1-a)\ge 0\Rightarrow a^2-6a+9-9+9a\ge 0\Rightarrow a^2+3a\ge 0\Rightarrow a\in (-\infty ,-3]\cup [0,\infty )$.

(ii) Sum of roots positive: $-\frac{2(a-3)}{1-a}>0\Rightarrow \frac{a-3}{a-1}>0\Rightarrow a\in (-\infty ,1)\cup (3,\infty )$.

(iii) Product of roots positive: $\frac{9}{1-a}>0\Rightarrow a<1$.

Intersecting (i), (ii), (iii) with $a\ne 1$: from (iii) we restrict to $a<1$, combined with (i) gives $(-\infty ,-3]\cup [0,1)$, and (ii) is automatically satisfied on this set.

Matching to $(-\infty ,-\alpha ]\cup [\beta ,\gamma )$: $\alpha =3,\beta =0,\gamma =1$.

$\therefore 2\alpha +\beta +\gamma =6+0+1=7$.