Let a,b,c be the positive real numbers such that bx2+((a+c)2+4b2)x+a+c≥0 for all x belongs to R, then a,b,c are in
AP
GP
HP
None of these
Explanation for the correct option:
Given that,
bx2+((a+c)2+4b2)x+a+c≥0⇒bx2+((a+c)2+4b2)0.5x+a+c≥0....(i)
Since ∀x is a real number, therefore
Discriminant of (i) ≤0
⇒(a+c)2+4b2–4b(a+c)≤0⇒a2+2ac+c2+4b2-4ab-4bc≤0⇒(a+c-2b)2≤0⇒a+c-2b≤0\
Therefore,
2b=a+c
Thus, a,b,care in AP.
Hence, the correct option is (A)