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:

Question

Option 1: (A) - (I), (B) - (II), (C) - (III), (D) - (IV)

Option 2: (A) - (I), (B) - (III), (C) - (II), (D) - (IV)

Option 3: (A) - (I), (B) - (II), (C) - (IV), (D) - (III)

Option 4: (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

AfterBoards · Accepted Answer

The answer is '(A) - (I), (B) - (III), (C) - (II), (D) - (IV)'. View full solution:

List-I	List-II
(A) $\vert x\vert < \vert y\vert$	(I) iff $x^2 > y^2$
(B) $\vert x\vert > \vert y\vert$	(II) iff $x^2 \le y^2$
(C) $\vert x\vert \le \vert y\vert$	(III) iff $x^2 < y^2$
(D) $\vert x\vert \ge \vert y\vert$	(IV) iff $x^2 \ge y^2$

List-I	List-II
(A) The solution set of the inequation $3x-6\le 0$ is	(I) $(3,\infty)$
(B) The solution set of the inequation $-5x+20<0$ is	(II) $[3,\infty)$
(C) The solution set of the inequation $3x-9\ge 0$ is	(III) $(-\infty,2]$
(D) The solution set of the inequation $8x+11>35$ is	(IV) $(4,\infty)$

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) $x \in \left[-\frac{3}{2}, 5\right)$

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