(a) ∵
x1,x2,x3 are in AP.
⇒
x1=a−d,x2=a,x3=a+d Where, d is the common difference
Now, since
x1,x2,x3 are the roots of the given equation
x3−x2+βx+γ=0 So, sum of roots =
x1+x2+x3=1⇒ (a − d ) + (a) + (a + d ) =1 ... (i)
3a = 1 ⇒ a =
β=x1x2+x2+x3+x1x3⇒ β = (a − d ) a + a(a + d ) + (a − d )(a + d ) ... (ii)
and product of roots = x
1x
2x
3 = −
γ = (a − d )(a)(a + d ) ... (iii)
Hence from (i), we get a =
and from eq. (ii), we get
β=3a2−d2⇒β=−d2(∵a=)Thus
β=−d2≤ [∵d
2 ≥ 0]
⇒ β < =
∴ β ∈
(−∞,]Again from eq. (iii), we get
a(a2−d2)=−γ⇒()+(−)=−γ⇒
γ=−⇒γ≥ (∵d
2 ≥ 0)
∴ y ∈
[−,∞]Hence option (a) is correct.