Linear Inequalities Test -9

Given,

|x – 1| ≤ 5, |x| ≥ 2

⇒ -(5 ≤ (x – 1) ≤ 5), (x ≤ -2 or x ≥ 2)

⇒ -(4 ≤ x ≤ 6), (x ≤ -2 or x ≥ 2)

Now, required solution is

x ∈ [-4, -2] ∪ [2, 6]

\(2 x+5>2+3 x\)

\(5-2>3 x-2 x\)

\(3>x\) ..........(1)

\(2 x-3 \leq 4 x-5\)

\(5-3 \leq 4 x-2 x\)

\(1 \leq x\) .........(2)

From (1) and (2)

x=1 or 2

Let \({x}=\frac{\sqrt{22}+\sqrt{10}}{\sqrt{22}-\sqrt{10}} ; {y}=\frac{\sqrt{22}-\sqrt{10}}{\sqrt{22}+\sqrt{10}}\)

We know that, \(x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)\)

\(\Rightarrow({x}+{y})=\frac{\left[(\sqrt{22}+\sqrt{10})^{2}+(\sqrt{22}-\sqrt{10})^{2}\right]}{22-10}\)

\(\Rightarrow(x+y)=\frac{(22+10+22+10)}{12}\)

\(\Rightarrow(x+y)=\frac{64}{12}\)

\(\Rightarrow(x+y)=\frac{16}3\) and \(x y=1\)

\(\Rightarrow x^{2}+y^{2}+2 x y=(\frac{16}{3})^{2}\)

\(\Rightarrow x^{2}+y^{2}+2(1)=\frac{256}9\)

\(\Rightarrow x^{2}+y^{2}=\frac{256}9-2\)

\(\Rightarrow x^{2}+y^{2}=\frac{238}9\)

\(x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)\)

\(\Rightarrow x^{3}+y^{3}=\left(\frac{16}{3}\right) \times\left(\frac{238}{9}-1\right)\)

\(\Rightarrow x^{3}+y^{3}=\left(\frac{16}{3}\right) \times\left(\frac{229}{9}\right)\)

\(\Rightarrow\left(\frac{\sqrt{22}+\sqrt{10}}{\sqrt{22}-\sqrt{10}}\right)^{3}+\left(\frac{\sqrt{22}-\sqrt{10}}{\sqrt{22}+\sqrt{10}}\right)^{3}=\left(\frac{229}{27}\right) \times 16\)

\(\Rightarrow\left(\frac{229}{27}\right) \times 16=\frac{229 a}{27}\)

\(\therefore a=16\)

Given \(-2<2 x-1<2\)

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

\(\Rightarrow-1<2 x<3\)

\(\Rightarrow\frac{-1}2

\(\Rightarrow x \in(\frac{-1}2,\frac32)\)

Given:

\(x=8-2 \sqrt{(15)}\)

\(x=(\sqrt{5})^{2}+(\sqrt{3})^{2}-2 \sqrt{(15)}\)

\(\Rightarrow x=(\sqrt{5}-\sqrt{3})^{2}\)

\(\Rightarrow \sqrt{x}=\sqrt{5}-\sqrt{3}\)

And \(\frac1{\sqrt{x}}=\frac{(\sqrt{5}+\sqrt{3})}2\)

According to question,

\(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}=\left[\sqrt{5}-\sqrt{3}+\left(\frac{\sqrt{5}+\sqrt{3}}{2}\right)\right]^{2}\)

\(\Rightarrow\left[\frac{3 \sqrt{5}-\sqrt{3}}{2}\right]^{2}\)

\(\Rightarrow\left[\frac{45+3-6 \sqrt{15}}{4}\right]\)

\(\Rightarrow \frac{48-6 \sqrt{15}}{4}\)

\(\Rightarrow 12-\frac{3 \sqrt{15}}{2}\)

Given:

\(=a x+9 y=1\) ......(i)

\(=9 y-x-1=0\) ........(ii)

From equation (ii) we get,

\(-x+9 y=1\) ..........(iii)

On comparing eq (iii) and eq (i)

We get, \(a=-1\)

The sum of two rational numbers is a rational number.

Ex: Let two rational numbers are \(\frac12\) and\(\frac13\)

Now, \(\frac12+\frac13=\frac56\) which is a rational number.

Given inequality is:

\(-15<\frac{3(x-2)}{5} \leq 0\)

Multiplying 5 all sides (Eliminating 5)

\(-15 \times 5<5 \times \frac{3(x-2)}{5} \leq 5 \times 0\)

\(-75<3(x-2) \leq 0\)

Dividing 3 all sides (Eliminating 3)

\(\frac{-75}{3}<\frac{3(x-2)}{3} \leq \frac{0}{3}\)

\(-25

Adding 2 all sides (Eliminating 2)

\(-25+2

\(-23

Thus, \(\mathrm{x}\) is a real number which is less than or equal to 2 and greater than \(-23\)

Therefore, \(x \in(-23,2]\) is the solution

Given,

\(|{x}|<-5\)

Now, LHS \(\geq 0\) and RHS \(<0\)

Since LHS is non-negative and RHS is negative So, \(|x|<-5\) does not posses any solution.

Given: \(2(3 x-4)-2<4 x-2 \geq 2 x-4\)

First by solving the inequation: \(2(3 x-4)-2<4 x-2\) we get,

\(\Rightarrow 6 x-10<4 x-2\)

\(\Rightarrow 2 x<8\)

\(\Rightarrow x<4\)\(\quad\).....(1)

Similarly, by solving the inequation \(4 x-2 \geq 2 x-4\) we get,

\(\Rightarrow 2 x \geq-2\)

\(\Rightarrow x \geq-1\)\(\quad\).....(2)

From equation (1) and (2) we can say that \(-1 \leq x<4\)

So, out of the given options the possible value which \(x\) can take is 2.