Linear Inequalities Test -10

We know that:

If \(f(x)=|x|\), then \(f(x)=\left\{\begin{array}{c}x, x \geq 0 \\ -x, x<0\end{array}\right.\)

Given: \(\left|x^{2}-3 x+2\right|>x^{2}-3 x+2\)

Case-1: If \(x \geq 0\)

\(\Rightarrow x^{2}-3 x+2>x^{2}-3 x+2\)

\(\therefore\) No real value of \(x\) satisfies the above equation.

Case-2:- If \(x<0\)

\(\Rightarrow-\left(x^{2}-3 x+2\right)>x^{2}-3 x+2\)

\(\Rightarrow x^{2}-3 x+2<0\)

\(\Rightarrow(x-1)(x-2)<0\)

Given, \(|x+1|+\sqrt{(x-1)}=0\), where each term is non-negative.

So, \(|x+1|=0\) and \(\sqrt{(x-1)}=0\) should be zero simultaneously.

i.e. \(x=-1\) and \(x=1\), which is not possible.

So, there is no value of \(x\) for which each term is zero simultaneously.

Given, 2x + 1 > 3

⇒ 2x > 3 – 1

⇒ 2x > 2

⇒ x > 1

⇒ x ∈ (1, ∞)

Given,

\(\Rightarrow x^{2}-6 x-27>0\)

\(\Rightarrow x^{2}-9 x+3 x-27>0\)

\(\Rightarrow x(x-9)+3(x-9)>0\)

\(\Rightarrow(x-9)(x+3)>0\)

As we know, when \(a b>0\), then there are two cases:

\(\Rightarrow\) Either \(a>0\) and \(b>0\)

\(\Rightarrow\) Or \(a<0\) and \(b<0\)

Considering the first case,

\(\Rightarrow(x-9)>0\) and \((x+3)>0\)

\(\Rightarrow x>9\) and \(x>-3\)

\(\Rightarrow x>9\)

Considering the second case,

\(\Rightarrow(x-9)<0\) and \((x+3)<0\)

\(\Rightarrow x<9\) and \(x<-3\)

\(\Rightarrow x<-3\)

\(\therefore x<-3\) or \(x>9\)

Given,

\(x^{2}<-4\)

\(\Rightarrow {x}^{2}+4<0\)

Which is not possible.

So, there is no solution.

Given,

\(x^{2}<4\)

\(\Rightarrow x^{2}-4<0\)

\(\Rightarrow(x-2) \times(x+2)<0\)

\(\Rightarrow-2

\(\Rightarrow x \in(-2,2)\)

Given, 1 ≤ |x – 1| ≤ 3

⇒ -3 ≤ (x – 1) ≤ -1 or 1 ≤ (x – 1) ≤ 3

i.e. the distance covered is between 1 unit to 3 units

⇒ -2 ≤ x ≤ 0 or 2 ≤ x ≤ 4

Therefore, the solution set of the given inequality is

x ∈ [-2, 0] ∪ [2, 4]

If a is an irrational number which is divisible by b then the number b must be irrational.

Ex: Let the two irrational numbers are \(\sqrt{2}\) and \(\sqrt{3}\)

Now, \(\frac{\sqrt{2}}{\sqrt{3}}\)

\(=\sqrt{(\frac23)}\)

Given, |x| = -5

Since |x| is always positive or zero

So, it can not be negative

Therefore, given inequality has no solution.

Given, f(x) = |x| > 0

We know that modulus is non negative quantity.

So, x ∈ R except that x = 0

⇒ x ∈ R – {0}

This is the required solution.