Sample 2022-23 Class 12 Maths Q7 Solution: Solution set of 3x + 5y

The question

The solution set of the inequality $3x + 5y < 4$ is

(a) an open half-plane not containing the origin.

(b) an open half-plane containing the origin.

(c) the whole XY-plane not containing the line $3x + 5y = 4$ ⁠.

(d) a closed half plane containing the origin.

Exam answer

The inequality is strict, so the line $3x + 5y = 4$ is not included: the solution set is an open half-plane.

Why?

The line $3x + 5y = 4$ divides the plane into two half-planes. On one side $3x + 5y < 4$ ⁠, on the other side $3x + 5y > 4$ ⁠.

A half-plane is closed when it includes its boundary line (signs $\le$ or $\ge$ ⁠), and open when it does not (signs $<$ or $>$ ⁠). Here the sign is $<$ ⁠, so option (d), a closed half-plane, is out.

At $(0, 0)$ ⁠: $3(0) + 5(0) = 0 < 4$ is true, so it contains the origin.

Why?

To find which side is the solution, test one point that is not on the line. The origin is the easiest. It satisfies the inequality, so the solution is the side that contains the origin, and option (a) is out.

(c) is false: $(2, 0)$ is not on the line, but $3(2) + 5(0) = 6 > 4$ ⁠, so it is not a solution.

Why?

Option (c) describes both sides of the line together. Points on the far side, such as $(2, 0)$ ⁠, give $3x + 5y > 4$ ⁠, so they do not satisfy the inequality.

Answer: (b) an open half-plane containing the origin.

Where this comes from

Chapter 12, Linear Programming: Solution set of a strict linear inequality.

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