All points lying inside the triangle formed by the points (5, 0), (-1, 2) and (1, 3) satisfy : (A) 3x + 2y - 18 > 0 (B) 3x + 2y > 0 (C) 2x + y + 13 0 Choose the **correct** answer from the options given below : Option 1: (A) and (B) Only Option 2: (B) and (C) Only Option 3: (D) and (E) Only Option 4: (B), (D) and (E) Only

The answer is '(B), (D) and (E) Only'. View full solution:

CUET Mathematics - All points lying inside the triangle formed by the points (5, 0), (-1, 2) and (1, 3) satisfy : (A) 3x + 2y - 18 > 0 (B) 3x + 2y > 0 (C) 2x + y + 13 < 0 (D) 2x - 3y - 12 < 0 (E) 2x - 3y + 12 > 0 Choose the **correct** answer from the options given below : | PYQs + Solutions

CUET Mathematics 2023 22 May Shift 3

Algebra

Medium

(A) and (B) Only

(B) and (C) Only

(D) and (E) Only

(B), (D) and (E) Only

✅ Correct Option: 4

Related questions:

21 May Shift 2

If a, b, c are positive real numbers, then the least value of $(a+b+c)(ab+bc+ca)$ is:

19 May Shift 1

Match List-I with List-II

List-I	List-II
(Inequality)	(Solution Set)
---	---
(A) $2x - 3 < x + 2 \le 3x + 5, x \in \mathbb{R}$	(I) $x \in (-1, \infty)$
(B) $\vert 2x + 3\vert < 7, x \in \mathbb{R}$	(II) $x \in (-\infty, 120]$
(C) $\frac{1}{2}\left(\frac{3}{5}x + 4\right) \ge \frac{1}{3}(x - 6), x \in \mathbb{R}$	(III) $x \in (-5, 2)$
(D) $\frac{\vert x + 1\vert }{x + 1} > 0, x \in \mathbb{R} - \{-1\}$	(IV) $x \in \left[-\frac{3}{2}, 5\right)$

Choose the correct answer from the options given below:

13 May Shift 1

If $a, b$ and $c$ are positive real numbers, then

Match List-I with List-II

List-I	List-II
(Expression)	(The Least value of the expression)
(A) $(a + b)(b + c)(c + a)$	(I) $8abc$
(B) $(a + b + c)(ab + bc + ca)$	(II) $9a^2b^2c^2$
(C) $(a^2b + b^2c + c^2a)(ab^2 + bc^2 + ca^2)$	(III) $9abc$
(D) $(a + b)^2(b + c)^2(c + a)^2$	(IV) $64a^2b^2c^2$

Choose the correct answer from the options given below: