CUET UG 2025 — Mathematics Algebra

The constraints are:

$x \ge 3$x≥3, $x \le 9$x≤9, $y \ge 0$y≥0, $x - y \ge 0$x−y≥0 (i.e. $y \le x$y≤x), $x + y \le 14$x+y≤14

Corner points are formed where two boundary lines intersect inside the feasible region.

---

$x = 3$x=3 meets $y = 0$y=0

Corner point $= (3,\; 0)$=(3,0)

---

$x = 9$x=9 meets $y = 0$y=0

Corner point $= (9,\; 0)$=(9,0)

---

$x = 9$x=9 meets $x + y = 14$x+y=14

$$y = 14 - 9 = 5$$y=14−9=5

Corner point $= (9,\; 5)$=(9,5)

---

$x + y = 14$x+y=14 meets $y = x$y=x

$$x + x = 14$$x+x=14

$$2x = 14$$2x=14

$$x = 7,\; y = 7$$x=7,y=7

Corner point $= (7,\; 7)$=(7,7)

---

$x = 3$x=3 meets $y = x$y=x

$$y = 3$$y=3

Corner point $= (3,\; 3)$=(3,3)

---

Checking option 1: $(0, 0)$(0,0) fails $x \ge 3$x≥3 — not in feasible region. ❌

Checking option 2: $(14, 0)$(14,0) fails $x \le 9$x≤9 — not in feasible region. ❌

Checking option 3: $(7, 7)$(7,7) and $(3, 3)$(3,3) — both are valid corner points. ✅

Checking option 4: At $(3, 6)$(3,6), $x - y = 3 - 6 = -3 < 0$x−y=3−6=−3<0 — fails $x - y \ge 0$x−y≥0. ❌

---

The answer is Option 3 — $(7, 7)$(7,7) and $(3, 3)$(3,3).