Consider the inequalities x gt y and x + y ≤ 3 where y ≤ 1 . The feas…

The correct answer is Option (3) → unbounded region containing the point (2, 1)

The boundary lines are y = x, x + y = 3 and y = 1. The region lies on or below y = 1, on or below x + y = 3, and strictly to the right of y = x (the line y = x itself is excluded, because \(x \gt y\) is strict).

The lines y = x and x + y = 3 meet where

    \(2x = 3,\quad x = y = 1.5\)

This point is above y = 1, so it is cut off. On the line y = 1 the region runs from x = 1 (excluded) to x = 2:

    \(y = 1:\quad 1 \lt x \le 2\)

For any \(y \le 1\), x lies between the two slanting lines:

    \(y \lt x \le 3 - y\)

The width of this strip, 3 − 2y, grows without limit as y decreases, so the feasible region is unbounded (open downwards).

The point (2, 1) satisfies all three inequalities:

    \(2 \gt 1\)
    \(2 + 1 = 3 \le 3\)
    \(1 \le 1\)

So the feasible region is an unbounded region containing (2, 1).

Why the other options are incorrect

• Option (1) unbounded region containing the origin — Incorrect. The region is unbounded, but at (0, 0) the condition x > y reads 0 > 0, which is false, so the origin is not in the region.

• Option (2) open half plane containing the point (0, 3) — Incorrect. At (0, 3), y = 3 breaks \(y \le 1\), so (0, 3) is not in the region.

• Option (4) bounded region containing (3, 0) but not the origin — Incorrect. The point (3, 0) is in the region and the origin is not, but the region is unbounded, not bounded.

Note: It seems that there is a mistake in the original NTA question. The original Option (3) read “open half plane containing the point (2, 1)”, but the region lies between the lines y = x and x + y = 3, below y = 1, so it is not a half plane, and no option described the region correctly. For the benefit of students and to ensure that the question is correct and has a single valid answer, Option (3) has been changed from “open half plane containing the point (2, 1)” to “unbounded region containing the point (2, 1)”.