If the corner points of the bounded feasible region of an LPP are (0,…

The correct answer is Option (4) → every point on the line segment joining (0, 2) and (3, 0)

Note: The given answer is as per the official NTA Answer Key. However, more than one answer is possible: Option (3) and Option (4) are both correct, as explained below.

On a bounded feasible region, the minimum of F occurs at a corner point. The values at the corners:

    \(F(0, 2) = 12\)
\(F(3, 0) = 12\)
\(F(6, 0) = 24\)
\(F(6, 8) = 72\)
\(F(0, 5) = 30\)

The minimum value 12 occurs at two adjacent corners, (0, 2) and (3, 0). The side joining them lies on the line \(2x + 3y = 6\), and on this line

    \(F = 4x + 6y = 2(2x + 3y) = 12\)

So F takes its minimum value 12 at every point of this segment.

Option (1) (0, 2) only — Incorrect. F is also 12 at (3, 0), so the minimum is not at (0, 2) only.

Option (2) (3, 0) only — Incorrect. F is also 12 at (0, 2), so the minimum is not at (3, 0) only.

Option (3) the mid point of the line segment joining (0, 2) and (3, 0) — Should also be correct, though not the answer given by NTA. The mid point is \(\left(\frac{3}{2}, 1\right)\), and \(F = 6 + 6 = 12\) there, the minimum value. The option does not say “only”, so it is true, though it names just one of the minimising points.

Option (4) every point on the line segment joining (0, 2) and (3, 0) — Correct. Every point of the segment has \(2x + 3y = 6\), so \(F = 2(2x + 3y) = 12\) there; this is the complete answer.

Advice for the exam: When more than one option is correct, mark the one that answers the question most completely, here every point on the line segment joining (0, 2) and (3, 0). If NTA’s provisional answer key marks a different option, challenge it during the answer-key challenge window, giving proper reasoning. NTA does revise its key when a challenge is well justified, so a sound challenge can raise your marks.