第11类RD Sharma解决方案–第15章线性不等式–练习15.3

问题1.解决$\left | x + \frac{1}{3} \right | > \frac{8}{3}$

解决方案：

Using the property of modulus operator, we know

|x| > a ⇒ x < -a or x > a

Therefore, $\left ( x + \frac{1}{3} \right ) < \frac{-8}{3}$ or $\left ( x + \frac{1}{3} \right ) > \frac{8}{3}$

⇒ x < $\frac{-8}{3} - \frac{1}{3}$ or x > $\frac{8}{3} - \frac{1}{3}$

⇒ x < -3 or x > $\frac{7}{3}$

Hence, we can conclude x lies in the range ( -∞, -3) ∪ ( $\frac{7}{3}$,∞ )

问题2.解决| 4 – x | +1 <3

解决方案：

We have,  | 4 – x | + 1 < 3

⇒ | 4 – x | < 2

Using the property of modulus operator, we know

|x| < a ⇒ -a < x < a

⇒ -2 < 4 – x <  2 

⇒ -6 < -x < -2

⇒ 6 > x > 2

⇒ 2 < x < 6 

Hence, we can conclude x lies in the range (2, 6)

问题3：解决$\left | \frac{3x - 4}{2} \right |$ ≤ $\frac{5}{12}$

解决方案：

Using the property of modulus operator, we know

|x| ≤ a ⇒ -a ≤ x ≤ a

⇒ – $\frac{5}{12}$ ≤ $\frac{3x - 4}{2}$ ≤ $\frac{5}{12}$ 

⇒ – $\frac{5}{6}$ ≤ 3x – 4 ≤ $\frac{5}{6}$

⇒ – $\frac{5}{6}$ + 4 ≤ 3x ≤ $\frac{5}{6}$ + 4

⇒ $\frac{19}{6}$ ≤ 3x ≤ $\frac{29}{6}$

⇒ $\frac{19}{18}$ ≤ x ≤ $\frac{29}{18}$

Hence, we can conclude x lies in the range [ $\frac{19}{18}$ , $\frac{29}{18}$ ]

问题4：解决$\frac{|x - 2|}{x - 2}$ > 0

解决方案：

Using the property of modulus operator, we have

| x -2 | = x-2 when x ≥ 2 or 2-x when x < 2

since, $\frac{|x-2|}{x-2}$ > 0 for x > 2

Hence, we can conclude x lies in the range ( 2, ∞)

问题5：解决$\frac{1}{|x|-3} < \frac{1}{2}$

解决方案：

We are given, $\frac{1}{|x|-3} < \frac{1}{2}$

⇒ $\frac{1}{|x|-3} - \frac{1}{2}$ < 0

⇒ $\frac{2 - (|x|-3)}{2(|x|-3)}$ < 0

⇒ $\frac{2 - |x|+ 3}{|x|-3}$ < 0

⇒ $\frac{5 - |x|}{|x|-3}$ < 0

Now, we have two cases:

Case 1: When x ≥ 0, then |x| = x

$\frac{5-x}{x-3}$ < 0

⇒ (5 – x < 0 and x – 3 > 0) or ( 5 – x > 0 and x – 3 < 0)

⇒ (x >5 and x > 3) or ( x < 5 and x < 3)

⇒ x > 5 and x < 3

Therefore, we can conclude, from case 1 that x lies in the range [ 0,3) U (5,∞)  

Case 2: When x ≤ 0 then |x| = -x

$\frac{5+x}{-x-3}$ < 0

⇒ $\frac{x+5}{x+3}$ > 0

⇒ (x + 5 > 0 and x + 3 > 0) or ( x + 5 <0 and x + 3 < 0)

⇒ ( x > -5 and x > -3 ) or ( x < -5 and x < -3 )

⇒ x > -3 or x < -5

Therefore, we can conclude, from case 2 that x lies in the range [ -∞, -5) U (-3,0)

Now, taking the union of above two cases we can conclude x lies in the range (-∞, -5) U ( -3,3) U (5, ∞)

问题6：解决$\frac{|x + 2| - x}{x}$ <2

解决方案：

We have $\frac{|x + 2| - x}{x}$ < 2

⇒ $\frac{|x + 2| - x}{x}$ – 2 < 0

⇒ $\frac{|x + 2| - x -2x}{x}$ < 0

⇒ $\frac{|x + 2| - 3x}{x}$ < 0

Now, we have two cases:

Case 1: When x ≥ -2, then |x+2| = x+2,

$\frac{x + 2 - 3x}{x}$ < 0

⇒ $\frac{2-2x}{x}$ < 0

⇒ $\frac{-2(x-1)}{x}$ < 0

⇒ $\frac{x-1}{x}$ > 0

⇒ (x – 1 > 0 and x > 0) or ( x – 1 < 0 and x < 0)

⇒ (x > 1 and x > 0) or ( x < 1 and x < 0)

⇒ x > 1 or x < 0

Therefore, we can conclude, from case 1 that x lies in the range [ -2,0) U (1,∞)

Case 1: When x ≤ -2, then |x+2| = -(x+2),

$\frac{-(x + 2) - 3x}{x}$ < 0

⇒ $\frac{-x-2-3x}{x}$ < 0

⇒ $\frac{-2(2x+1)}{x}$ < 0

⇒ $\frac{2x+1}{x}$ > 0

⇒ (2x -+ 1 > 0 and x > 0) or ( 2x + 1 < 0 and x < 0)

⇒ (x > -1/2 and x > 0) or ( x < -1/2 and x < 0)

⇒ x > 0 or x < -1/2

Therefore, we can conclude, from case 2 that x lies in the range ( -∞,-2 ] U (0,∞)

Now, taking the union of above two cases we can conclude x lies in the range [ -2,0 ) U (1,∞) U (-∞, -2] U ( 0, ∞) i.e., x belongs to (-∞,0) U (1,∞)

问题7：解决$|\frac{2x-1}{x-1}|$ > 2

解决方案：

We have $|\frac{2x-1}{x-1}|$ > 2

⇒ $|\frac{2x-1}{x-1}|$ – 2 > 0

⇒ $\frac{2x-1}{x-1}$ > 0 or $\frac{2x-1}{x-1}$ +2 < 0

⇒ $\frac{1}{x-1}$ > 0 or $\frac{4x-3}{x-1}$ < 0

⇒ x-1 >0 or $\frac{4x-3}{x-1}$ < 0

⇒ x-1 > 0 or [ (4x-3 > 0 and x-1 < 0) or ( 4x-3 < 0 and x-1 > 0) ]

⇒ x > 1 or [ (x > 3/4  and x < 1) or ( x < 3/4 and x > 1) ]

⇒ x > 1 or [ 3/4 < x < 1 or ∅]

⇒ 3/4 < x < 1 or x > 1

Hence, we can conclude x lies in the range ( 3/4, 1) U (1, ∞ )

问题8.解决| x-1 | + | x-2 | + | x-3 | ≥6

解决方案：

We have, |x-1| + |x-2| + |x-3| ≥ 6 ———————let this be equation (1)

As,  | x-1 | = ( x-1, when x ≥ 1 and 1-x when x < 1 )

similarly, | x-2 | = ( x-2, when x ≥ 2 and 2-x when x < 2 )

and | x-3 | = ( x-3, when x ≥ 3 and 3-x when x < 3 )

Now, we have four cases:

Case 1: When x < 1

1 – x + 2 – x + 3 – x ≥ 6

⇒ 6 -3x ≥ 6

⇒ x ≤ 0

So, we see x lies in the range (-∞ ,0]

Case 2: when 1 ≤ x < 2

x – 1 + 2 – x + 3 – x ≥ 6

⇒ 4 – x ≥ 6

⇒ x ≤ -2

using case 2 , we see x has no values so x ∈ ∅

Case 3: when 2 ≤ x < 3

x – 1 + x – 2 + 3 – x ≥ 6

⇒ x  ≥ 6

using case 3 , we see x has no values so x ∈ ∅

Case 4: when x ≥ 3

x – 1 + x – 2 + x – 3 ≥ 6

⇒ 3x – 6 ≥ 6

⇒ x  ≥ 4

using case 4 , we see x has no values so x ∈ [ 4, ∞ )

Combining all the cases, we get to know x lies in the range ( -∞, 0 ] U [ 4, ∞)

问题9：解决$\frac{|x-2|-1}{|x-2|-2}$ ≤0

解决方案：

We have, $\frac{|x-2|-1}{|x-2|-2}$ ≤ 0 

Case 1: when x ≥ 2, then | x -2 | = x – 2

$\frac{x-2-1}{x-2-2}$ ≤ 0

⇒ $\frac{x-3}{x-4}$ ≤ 0

⇒ ( x – 3 ≤ 0 and x – 4 > 0 ) or ( x – 3 ≥ 0 and x – 4 < 0)

⇒ ( x ≤ 3 and x > 4 ) or ( x ≥ 3 and x < 4)

⇒ ∅ or ( 3 ≤ x < 4)

⇒ 3 ≤ x < 4

using case 1 , we see x lies in the range [ 3, 4 )

Case 2: when x ≤ 2, then | x – 2| = 2 – x,

$\frac{2-x-1}{2-x-2}$ ≤ 0

⇒ $\frac{1-x}{-x}$ ≤ 0

⇒ $\frac{x-1}{x}$ ≤ 0

⇒  ( x – 1 ≤ 0 and x > 0 ) or ( x – 1 ≥ 0 and x< 0)

⇒  ( x ≤ 1 and x > 0 ) or ( x ≥ 1 and x< 0)

⇒ ( 0 < x ≤ 1) or ∅

⇒ 0 < x ≤ 1

using case 2 , we see x lies in the range ( 0, 1 ]

Combining all the cases, we get to know x lies in the range ( 0, 1 ] U [ 3, 4)

问题10：解决$\frac{1}{|x|-3} \leq \frac{1}{2}$

解决方案：

We have, $\frac{1}{|x|-3} \leq \frac{1}{2}$

⇒ $\frac{1}{|x|-3} - \frac{1}{2}$ ≤ 0

⇒ $\frac{2 - (|x-3|)}{2(|x-3|)}$ ≤ 0

⇒ $\frac{5 - |x|}{|x|-3}$ ≤ 0

Case 1: when x ≥ 0 then |x| =x

⇒ $\frac{5-x}{x-3}$ ≤ 0

⇒ ( 5 – x ≤ 0 and x – 3 > 0 ) or ( 5 – x ≥ 0 and x – 3 < 0)

⇒  ( x ≥ 5 and x > 3 ) or ( x ≤ 5 and x < 3)

⇒ x ≥ 5 or x < 3

using case 1, we see x lies in the range ( 0, 3 ) U [5, ∞ )

Case 2: when x < 0 then |x| = -x

⇒ $\frac{5+x}{-x-3}$ ≤ 0

⇒ $\frac{x+5}{x+3}$ ≥ 0

⇒ ( x + 5 > 0 and x + 3 > 0 ) or ( x + 5 < 0 and x + 3 < 0)

⇒  ( x > -5 and x > -3 ) or ( x < -5 and x < -3)

⇒ x > -3 or x < -5

using case 2, we see x lies in the range ( -∞, -5 ) U (-3, ∞ )

Combining both the cases, we get to know x lies in the range ( -∞, -5 ) U (-3, ∞ ) U ( 0, 3 ) U [5, ∞ )

问题11.解决| x + 1 | + | x | > 3

解决方案：

We have, |x + 1| + |x| > 3

⇒ |x + 1| = ( x + 1 when x ≥ -1 and -(x + 1) when x < -1 )

similarly, |x| = (x when x ≥ 0 and -x when x < 0)

Case 1: When x < -1

|x + 1| + |x| > 3

⇒ – (x+1) -x > 3

⇒ -2x -1 > 3

⇒ x < -2

using case `1, we see x lies in the range ( -∞, -2 )

Case 2: When -1 ≤ x < 0

|x + 1| + |x| > 3

⇒  (x+1) + x > 3

⇒ 2x > 2

⇒ x > 1

using case `2, we see x lies in the range ( 1, ∞ )

Combining both the cases, we get to know x lies in the range ( -∞, -2 ) U ( 1, ∞ )

问题12.解决1≤| x – 2 | ≤3

解决方案：

We have, 1 ≤ |x – 2| ≤ 3

Case 1: |x – 2| ≥ 1

⇒ ((x – 2) ≤ -1 or (x-2) ≥ 1) 

⇒ ( x ≤ 1 or x ≥ 3)

using case `1, we see x lies in the range ( -∞,1 ] U [3,∞) 

Case 2:  |x – 2| ≤ 3

⇒  ( -3 ≤ (x-2) ≤ 3)

⇒ (-1 ≤ x ≤ 5)

using case `2, we see x lies in the range [-1, 5]

Combining both the cases, we get to know x lies in the range [-1, 1 ] U [ 3, 5 ]

问题13.解决| 3-4x | ≥9

解决方案：

We have, |3-4x| ≥ 9

therefore, by using property of modulus we know, |x| ≥ a ⇒ x ≤ -a or x ≥ a

⇒ (3-4x) ≤ -9 or (3-4x) ≥ 9

⇒ -4x ≤ -12 or  -4x ≥ 6

⇒ x ≥ 3 or  x ≤ -3/2

Hence, we can conclude x lies in the range ( -∞, -3/2] U [ 3, ∞ )