For all 'x', x2+2ax+10-3a>0, then the interval in which 'a' lies is [2004]
(2)
f(x)=ax2+bx+c has same sign as that of a if D<0.
Since x2+2ax+10-3a>0 ∀x
∴ D<0⇒4a2-4(10-3a)<0
⇒a2+3a-10<0
⇒(a+5)(a-2)<0⇒a∈(-5,2)