Question

Match the elements of the list I, which contains inequalities, with the elements of list II, which has solution sets.

List-I (Inequality)	List-II (Solution set)
A) $x^2-4x+3>0$	1) $(3,4)$
B) $x^2-5x+6\le 0$	2) $(-1,1)\cup (2,4)$
C) $x^2+6x-27>0,-x^2+3x+4>0$	3) $(-\infty ,1)\cup (3,\infty )$
D) $x^2-3x-4<0,x^2-3x+2>0$	4) $[3,4]$
	5) $[2,3]$

Then the correct order for A B C D, is

• A. 1 2 3 4
• B. 3 5 1 2
• C. 1 3 5 4
• D. 3 2 4 1

Answer 1

Answer: (B)

3 5 1 2

A: $x^2-4x+3>0$

$\Rightarrow (x-1)(x-3)>0$

So, $x\in (-\infty ,1)\cup (3,\infty )$ $\to \left(3\right)$

B: $x^2-5x+6\le 0$

$\Rightarrow (x-2)(x-3)\le 0$

So, $x\in [2,3]$ $\to \left(5\right)$

C: $x^2+6x-27>0$ and $-x^2+3x+4>0$

$\Rightarrow (x+9)(x-3)>0$ and $(x-4)(x+1)<0$

So, $x\in (-\infty ,-9)\cup (3,\infty )$ and $x\in (-1,4)$

Taking intersection, we get

$x\in (3,4)$ $\to \left(1\right)$

D: $x^2-3x-4<0$ and $x^2-3x+2>0$

$\Rightarrow (x-4)(x+1)<0$ and $(x-1)(x-2)>0$

So, $x\in (-1,4)$ and $x\in (-\infty ,1)\cup (2,\infty )$

Taking intersection, we get

$x\in (-1,1)\cup (2,4)$ $\to \left(2\right)$

$\therefore$ The final order for A B C D is 3 5 1 2.

Hence, option B.