MHTCET 17 Apr 2026 Shift 1 Paper

Concept:Graph the given linear inequalities and identify the common feasible region in the first quadrant.Explanation:Rewrite the inequalities as $y \ge x+2$ and $y \le -x+5$.Since $x \ge 0$ and $y \ge 0$, the required region lies in the first quadrant.The line $y = x+2$ cuts the $y$-axis at $(0,2)$.The line $y = -x+5$ cuts the $y$-axis at $(0,5)$ and the $x$-axis at $(5,0)$.The inequality $y \ge x+2$ gives the region above the line $y = x+2$.The inequality $y \le -x+5$ gives the region below the line $y = -x+5$.Find the intersection of $y = x+2$ and $y = -x+5$:$x+2 = -x+5$$2x = 3$$x = 1.5$ and $y = 3.5$Thus, the common region is a triangle with vertices $(0,2)$, $(0,5)$, and $(1.5,3.5)$.Answer:The required region is the triangle with vertices $(0,2)$, $(0,5)$, and $(1.5,3.5)$.Choose the option that shows this triangular region.