The number of integral value(s) of x which satisfies x2 – 3x + 2 ≤ 0 and x2 – 3x + 2 > 0, is/are

The given equations are x^2 - 3x + 2 ≤ 0 and x^2 - 3x + 2 > 0.

First, let's solve the quadratic equation x^2 - 3x + 2 = 0.

The roots of the equation can be found using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / 2a.

Here, a = 1, b = -3, and c = 2.

So, the roots are x = [3 ± sqrt((-3)^2 - 412)] / 2*1 = [3 ± sqrt(9 - 8)] / 2 = [3 ± sqrt(1)] / 2 = [3 ± 1] / 2 = 2, 1

So, the roots of the equation are x = 1 and x = 2.

Now, let's consider the inequalities x^2 - 3x + 2 ≤ 0 and x^2 - 3x + 2 > 0.

The solution of the inequality x^2 - 3x + 2 ≤ 0 is the interval [1, 2] including the roots 1 and 2.

The solution of the inequality x^2 - 3x + 2 > 0 is the set of all real numbers except for the interval (1, 2).

However, we are looking for the integral solutions.

The only integral values in the interval [1, 2] are 1 and 2.

But the second inequality x^2 - 3x + 2 > 0 does not include the values 1 and 2.

Therefore, there are no integral values of x that satisfy both inequalities.