Question

The number of elements in the set $ S = \{x \in \mathbb{Z} : x^2 - 7x + 6 \le 0 \text{ and } x^2 - 3x>0 \} $ is

• A. $ \infty $
• B. 2
• C. 3
• D. 4

Hint:

When solving quadratic inequalities, find the roots and test intervals. For sets with multiple conditions, find the intersection of the solution sets. Remember $ \mathbb{Z} $ is the set of integers.

Answer 1

The Correct Option is C

Step 1: Solve the first inequality $ x^2 - 7x + 6 \le 0 $.
Factor the quadratic expression: $ x^2 - 7x + 6 = (x - 1)(x - 6) $.
The inequality becomes $ (x - 1)(x - 6) \le 0 $.
The solution to this inequality is $ 1 \le x \le 6 $.
Step 2: Solve the second inequality $ x^2 - 3x>0 $.
Factor the quadratic expression: $ x^2 - 3x = x(x - 3) $.
The inequality becomes $ x(x - 3)>0 $.
The solution to this inequality is $ x<0 $ or $ x>3 $.
Step 3: Find the integers $ x $ that satisfy both inequalities.
We need integers $ x $ such that $ (1 \le x \le 6) $ and $ (x<0 $ or $ x>3) $.
The integers satisfying both conditions are $ x = 4, 5, 6 $.
Step 4: Determine the number of elements in the set $ S $.
The set $ S = \{4, 5, 6\} $.
The number of elements in the set $ S $ is 3.
Step 5: Conclusion.
The number of elements in the set $ S $ is 3.