Match each inequality in List-I with the region of the xy-plane it represents in List-II.
| List-I | List-II | ||
|---|---|---|---|
| Inequality | Region | ||
| (A) | \(2x + 3y \le 12\) | (I) | The line through (6, 0) and (0, 4), together with the side of it that does not contain the origin |
| (B) | \(2x + 3y \ge 12\) | (II) | The line \(x = 3\), together with the side of it that contains the origin |
| (C) | \(x \le 3\) | (III) | The line \(y = x\), together with the side of it that contains the point (0, 1) |
| (D) | \(y \ge x\) | (IV) | The line through (6, 0) and (0, 4), together with the side of it that contains the origin |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
In the graphical method, an inequality is drawn as its boundary line (the equation) together with one side of it. A test point, usually the origin, tells which side: if the test point satisfies the inequality, its side is the region.
(A) \(2x + 3y \le 12\)
This matches (IV). The line \(2x + 3y = 12\) meets the axes at (6, 0) and (0, 4). At the origin
\(0 \le 12\), true
so the region is the side containing the origin.
(B) \(2x + 3y \ge 12\)
This matches (I). It has the same boundary line, but at the origin
\(0 \ge 12\), false
so the region is the side that does not contain the origin.
(C) \(x \le 3\)
This matches (II). The boundary is the vertical line \(x = 3\), and the origin gives \(0 \le 3\), true.
(D) \(y \ge x\)
This matches (III). The boundary \(y = x\) passes through the origin, so another test point is needed. At (0, 1)
\(1 \ge 0\), true
so the region is the side containing (0, 1).