JEE Main Maths: Linear Programming questions with solutions

Q1. What kind of region is described by the inequalities x ≥ 0, y ≥ 0, x + y ≤ 6, and x + y ≤ 3?

1. An unbounded region in the first quadrant
2. A bounded region extending into the first and second quadrants
3. A bounded region in the first quadrant
4. None of the above

Answer: A bounded region in the first quadrant

The inequalities define a region that is limited by the lines x + y = 6 and x + y = 3, both of which intersect the axes in the first quadrant, creating a bounded area where both x and y are non-negative.

Q2. What is the nature of the region defined by the inequalities 2x + 3y - 5 ≤ 0 and 4x - 3y + 2 ≤ 0?

1. It does not lie in the first quadrant
2. It is a bounded region in the first quadrant
3. It is an unbounded region in the first quadrant
4. None of the above

Answer: None of the above

The region defined by the given inequalities does not restrict itself to the first quadrant, and it is not bounded, as the lines intersect outside of that quadrant, leading to an unbounded area that extends into other quadrants.

Q3. For the region determined by the inequalities 3x + 4y ≤ 18, 2x + 3y ≥ 3, and x ≥ 0, y ≥ 0, which of the following is a vertex of the feasible set?

1. (0, 2)
2. (4, 8, 0)
3. (0, 3)
4. None of these

Answer: None of these

The feasible region's corners come from intersecting the boundary lines. They are (0,1) [x=0 & 2x+3y=3], (0,4.5) [x=0 & 3x+4y=18], (1.5,0) and (6,0) [y=0 cases]. Neither (0,2) nor (0,3) is a vertex, and (4,8,0) is malformed, so none of these is correct.

Q4. The area (in sq. units) of the region {x ∈ R: x ≥ 0, y ≥ 0, y ≥ x - 2 and y ≤ √x}, is

1. 13/3
2. 10/3
3. 5/3
4. 8/3

Answer: 10/3

The area is determined by the intersection of the lines and curves defined by the inequalities, specifically the line y = x - 2 and the curve y = √x. By calculating the area between these boundaries in the first quadrant, we find that the total area is 10/3 square units.