x2 + 1 > 0 , x ∊ R ..... (i) (a - 1) x2 - (a + |a - 1| + 2) x + 1 ≥ 0 ..... (2) Solution of equation (2) is x ∊ R a - 1 > 0 and Δ ≤ 0 a > 1 and (a+a−1+2)2 - 4 (a - 1) ≤ 0 4a2 + 5 ≤ 0 ∴ which is not possible for any real value of a ∴ Hence number of integral value of a is zero.