Lakshman Patel RJIT
asked
in Quantitative Aptitude
Mar 30, 2020
edited
Jun 29, 2020
by soujanyareddy13

553 views
0 votes

We can solve this question, in this way,

- $x^{2}+ 5x+6>0$

$\implies x^{2} + 3x + 2x + 6>0$

$\implies x(x+3) + 2(x+3)>0$

$\implies (x+3)(x+2)>0$

$\implies x>-3 \;\& \;x>-2$

It has infinite range.

- $\left | x+2 \right |>4$

Case$1: x + 2 >4$

$\implies x>2$

Case$2: x + 2 >-4$

$\implies x>-6$

It also has infinite range.

- $9x-7<3x +14$

$\implies 6x<21$

$\implies x < \dfrac{21}{6}$

It also has infinite range.

- $x^{2}- 4x+3<0$

$\implies x^{2}-3x-x+3<0$

$\implies x(x-3)-1(x-3)<0$

$\implies (x-3)(x-1)<0$

Case$1:(x-3)>0\;\&\;(x-1)<0$

$\implies x>3\;\&\;x<1$

It also has infinite range.

Case$1:(x-3)<0\;\&\;(x-1)>0$

$\implies x<3\;\&\;x>1$

$\implies 1<x<3$

It has finite range.

So, the correct answer is $(D).$