Step 1: Check each option.
- Option (i): \(\exists x \in A\) such that \(x + 3 = 8\).
This is true for \(x = 5\), since \(5 + 3 = 8\).
- Option (ii): \(\exists x \in A\) such that \(x + 2 < 9\).
This is true for \(x = 1, 2, 3, 4\), as \(1 + 2 = 3\), \(2 + 2 = 4\), \(3 + 2 = 5\), \(4 + 2 = 6\).
- Option (iii): \(\forall x \in A, x + 6 \geq 9\).
This is false. For \(x = 1\), \(1 + 6 = 7\), which is less than 9. Hence, this is not true for all \(x\).
- Option (iv): \(\exists x \in A\) such that \(x + 6 < 10\).
This is true for \(x = 1, 2\), since \(1 + 6 = 7\) and \(2 + 6 = 8\).
Step 2: Conclude.
The statement in option (iii) is not true, as there exists at least one \(x\) (i.e., \(x = 1\)) for which \(x + 6 < 9\).
Final Answer: \[ \boxed{(iii) \; \forall x \in A, \; x + 6 \geq 9} \]
The following system of equations: $$ x_1 + x_2 + x_3 = 1, \quad x_1 + 2x_2 + 3x_3 = 2, \quad x_1 + 4x_2 + \alpha x_3 = 4 $$ has a unique solution. Possible value(s) for $\alpha$ is/are:
Solve the following assignment problem for minimization :
Find x if the cost of living index is 150 :