Number of integers satisfying x−6−10−x≥1 is

We have x−6−10−x≥1 We must have x≥6 and x≤10 Then x−6≥1+10−x Squaring given inequality, x−6≥1+10−x+210−x⇒2x−17≥210−xClearly LHS must be positive.∴ x≥172so, we have 4x2+289−68x≥4(10−x)⇒4x2−64x+249≥0⇒x2−16x+2494≥0⇒(x−8)2−722≥0⇒ x−8−72x−8+72≥0From the sign scheme, x∈[(16+7)/2,10]