Question

solve following linear inequality
i) [tex]i) \frac{4}{4x+1} \leq 3\leq \frac{6}{x+1}, (x\ \textgreater \ 0)\\
ii) \frac{|x-2|-1}{|x-2|-2}\leq 0[/tex]

Answer 1

Answer:

hope my answer in the above attachment helps you.........

Now your answer is complete...

Answer 2

Solution :-

1)

$\dfrac{4}{4x+1} \leq 3 \leq \dfrac{6}{x + 1} \: ( x > 0 )$

Now we will split the question into two parts.

1st

$\dfrac{4}{4x+1} \leq 3$

By subtracting 3 both sides multiplication :-

$\dfrac{4}{4x+1} - 3\leq 3- 3$

$\dfrac{4}{4x+1} - 3\leq 0$

$\dfrac{4 - 12x - 3 }{4x+1} \leq 0$

$\dfrac{- 12x + 1 }{4x+1} \leq 0$

multiply by (-)

$\dfrac{12x - 1 }{4x+1} \geq 0$

Now

(1)

$12x - 1 \geq 0 \: and \: 4x + 1 > 0$

$\implies x \geq \dfrac{1}{12} \: and x > \dfrac{-1}{4}$

x belongs to [ 1/12 , + ∞ ]

(2)

$12x - 1 \leq 0 \: and \: 4x + 1 < 0$

$\implies x \leq \dfrac{1}{12} \: and x < \dfrac{-1}{4}$

x belongs to [ -∞, -1/4 ]

From first and second case :-

x = (0 , ∞) , because x > 0

2nd

$\rightarrow \dfrac{6}{x+1}\geq 3$

subtracting 3 from both sides

$\rightarrow \dfrac{6}{x+1} - 3 \geq 3 - 3$

$\rightarrow \dfrac{6 - 3x - 3}{x+1}\geq 0$

$\rightarrow \dfrac{-3x + 3}{x+1}\geq 0$

multiply by (-)

$\rightarrow \dfrac{3x - 3}{x+1}\leq 0$

Now

(1)

$3x - 3\leq 0 \: and \: x + 1 > 0$

$\implies x \leq 1 \: and x > -1$

x belongs to [ 1 ] , x > 0

(2)

$3x - 3 \geq 0 \: and \: x + 1 < 0$

$\implies x \geq 1\: and x < -1$

x belongs to ∅

From first and second case :-

x = [1]

From Both

x = 1

$\line$

2)

$\dfrac{ |x-2| -1}{| x -2| -2} \leq 0$

1......

(i)

$\implies |x -2| - 1\leq 0$

$\implies x - 2 - 1 \leq 0 \: , x \leq 3$

$( x - 2 \geq 0 , x \geq 2)$

$x \: belongs \: to [2 , 3]$

$\implies -(x - 2) - 1 \leq 0 \: , x \geq 1$

$( x - 2 < 0 , x < 2)$

$x \: belongs \: to [1 , 2)$

Union of both

x belongs to [ 1 , 3]

(ii)

$\implies |x -2| -2 > 0 \:$

$\implies x - 2 - 2 > 0 \: , x > 4$

$( x - 2 \geq 0 , x \geq 2)$

$x \: belongs \: to (4, +\infty)$

$\implies -(x - 2) - 2 < 0 \: , x < 0$

$( x - 2 < 0 , x < 2)$

$x \: belongs \: to (-\infty , 0)$

Union of both

x belongs to (- ∞ , 0) U ( 4 , +∞)

Union of (i) and (ii)

x belongs to ∅

Now

2...

(i)

$\implies |x -2| - 1\geq 0$

$\implies x - 2 - 1 \geq 0 \: , x \geq 3$

$( x - 2 \geq 0 , x \geq 2)$

$x \: belongs \: to [ 3 , +\infty]$

$\implies -(x - 2) - 1 \geq 0 \: , x \leq 1$

$( x - 2 < 0 , x < 2)$

$x \: belongs \: to (-\infty , 1]$

Union of both

x belongs to (-∞, 1] U [3 , +∞)

(ii)

$\implies |x -2| -2 < 0 \:$

$\implies x - 2 - 2 < 0 \: , x < 4$

$( x - 2 \geq 0 , x \geq 2)$

$x \: belongs \: to [2 , 4)$

$\implies -(x - 2) - 2 < 0 \: , x > 0$

$( x - 2 < 0 , x < 2)$

$x \: belongs \: to (0 , 2)$

Union of both

x belongs to ( 0 , 4)

Union of (i) and (ii)

x belongs to (0 , 1] U [3 , 4)

Now union of 1 and 2

x belongs ∅

x belongs to (0 , 1] U [3 , 4)

→ x belongs to (0 , 1] U [3 , 4)