Question

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{a.}& \sqrt{x+2}>\sqrt{8-x^2}& \text{p.}& (-\infty ,-\frac{3}{2})\cup (-\frac{1}{4},\frac{1}{4})\cup (\frac{3}{2},\infty )\\ \text{b.}& ||x|-1|<1-x& \text{q.}& (-\infty ,0)\cup [1,2]\\ \text{c.}& \left|\frac{2-3|x|}{1+|x|}\right|>1& \text{r.}& (2,2\sqrt{2}]\\ \text{d.}& \frac{\sqrt{2-x}+4x-3}{x}\ge 2& \text{s.}& (-\infty ,0)\end{array}$

Which of the following is the correct option ?

• A. $\left(a\right)\to \left(r\right)\left(b\right)\to \left(s\right)\left(c\right)\to \left(p\right)\left(d\right)\to \left(q\right)$
• B. $\left(a\right)\to \left(q\right)\left(b\right)\to \left(s\right)\left(c\right)\to \left(p\right)\left(d\right)\to \left(r\right)$
• C. $\left(a\right)\to \left(r\right)\left(b\right)\to \left(p\right)\left(c\right)\to \left(s\right)\left(d\right)\to \left(q\right)$
• D. $\left(a\right)\to \left(q\right)\left(b\right)\to \left(s\right)\left(c\right)\to \left(r\right)\left(d\right)\to \left(p\right)$

Answer 1

The correct option is A $\left(a\right)\to \left(r\right)\left(b\right)\to \left(s\right)\left(c\right)\to \left(p\right)\left(d\right)\to \left(q\right)$

$a.$
$\sqrt{x+2}>\sqrt{8-x^2}$
Square root will be defined when
$x+2\ge 0\Rightarrow x\ge -28-x^2\ge 0\Rightarrow 8\ge x^2\Rightarrow -2\sqrt{2}\le x\le 2\sqrt{2}\therefore x\in [-2,2\sqrt{2}]\cdots \left(1\right)$
Now, squaring the given inequality, we get
$x+2>8-x^2\Rightarrow x^2+x-6>0\Rightarrow (x+3)(x-2)>0\Rightarrow x\in (-\infty ,-3)\cup (2,\infty )\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, we get
$x\in (2,2\sqrt{2}]$


$\left(a\right)\to \left(r\right)$

$b.$
$||x|-1|<1-x$
When $x\ge 0$
$|x-1|<-(x-1)$
Not possible
When $x<0$
$|-x-1|<1-x\Rightarrow |x+1|<1-x$


Case $1:0>x\ge -1$, we get
$x+1<1-x\Rightarrow 2x<0\Rightarrow x<0$
Case $2:-1>x$, we get
$-x-1<1-x\Rightarrow -1<1$
Always true.
Therefore $x\in (-\infty ,0)$
$\left(b\right)\to \left(s\right)$

$c.$
$\left|\frac{2-3|x|}{1+|x|}\right|>1$
Two possible cases,
$\frac{2-3|x|}{1+|x|}>1\Rightarrow 2-3|x|>1+|x|(\because 1+|x|>0)\Rightarrow -4|x|>-1\Rightarrow |x|<\frac{1}{4}\Rightarrow x\in (-\frac{1}{4},\frac{1}{4})\cdots \left(1\right)$
$\frac{2-3|x|}{1+|x|}<-1\Rightarrow 2-3|x|<-1-|x|\Rightarrow -2|x|<-3\Rightarrow |x|>\frac{3}{2}\Rightarrow x\in (-\infty ,-\frac{3}{2})\cup (\frac{3}{2},\infty )\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, we get
$x\in (-\infty ,-\frac{3}{2})\cup (-\frac{1}{4},\frac{1}{4})\cup (\frac{3}{2},\infty )$


$\left(c\right)\to \left(p\right)$

$d.$
$\frac{\sqrt{2-x}+4x-3}{x}\ge 2$
Here $x\ne 0$
Square root is defined when
$2-x\ge 0\Rightarrow 2\ge x\Rightarrow x\in (-\infty ,2]-\left\{0\right\}$

When $x\in (0,2]$
$\frac{\sqrt{2-x}+4x-3}{x}\ge 2\Rightarrow \sqrt{2-x}+4x-3\ge 2x\Rightarrow \sqrt{2-x}\ge 3-2x$


$\Rightarrow x\in [1,2]\cdots \left(1\right)$

When $x\in (-\infty ,0)$
$\frac{\sqrt{2-x}+4x-3}{x}\ge 2\Rightarrow \sqrt{2-x}+4x-3\le 2x\Rightarrow \sqrt{2-x}\le 3-2x$
Squaring on both sides, we get
$\Rightarrow 2-x\le 9+4x^2-12x\Rightarrow 4x^2-11x+7\ge 0\Rightarrow x\ge \frac{7}{4}\text{or}x\le 1\Rightarrow x\in (-\infty ,0)$
Therefore
$x\in (-\infty ,0)\cup [1,2]$
$\left(d\right)\to \left(q\right)$