0

Guest

Courses New

Solve |x-1|+|x-2|+|x-3| > 6

Solve |x-1|+|x-2|+|x-3|>6

Grade:12

4 Answers

Badiuddin askIITians.ismu Expert
148 Points
12 years ago

Dear chilukuri

|x-1|+|x-2|+|x-3|>6

case 1        x>=3

|x-1|+|x-2|+|x-3|>6

or (x-1)+(x-2)+(x-3)>6

x>=4

case 2         2=<x <3

 |x-1|+|x-2|+|x-3|>6

or

(x-1)+(x-2)+(-x+3)>6

x>=6

  so it is not a solution

case 3   1=<x<2

|x-1|+|x-2|+|x-3|>6

or

(x-1)+(-x+2)+(-x+3)>6

x<=-2

 so it is not a solution

case 4  x<1

|x-1|+|x-2|+|x-3|>6

or

(-x+1)+(-x+2)+(-x+3)>6

x<=0  

so final solution

   -∞<x≤0   and   4≤x<∞

Please feel free to post as many doubts on our discussion forum as you can.
 If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

 All the best.
 
Regards,
Askiitians Experts
Badiuddin



Akshay Mr
13 Points
3 years ago
The detailed answer of this question please give me I can't understand the way you Explain the answer
 
Rishi Sharma
askIITians Faculty 646 Points
2 years ago
Dear Student,
Please find below the solution to your problem.

To get rid of the absolute value functions, we need to take 4 cases:
Case 1: x≥3or [3,∞)Here all the bars will open with a positive sign.
x−1+x−2+x−3≥6
3x≥12
x≥4
Case 2: 2≤x<3
or [2,3)
Here, only |x−3|
will open with a negative sign.
x−1+x−2−x+3≥6
x≥6
But this lies out of [2,3)
So no solution
Case 3: 1≤x<2 or [1,2)
Here only |x−1|
will open with a positive sign.
x−1−x+2−x+3≥6
−x≥2
x≤−2
Case 4: x<1(−∞,1)
Here all bars will open with a negative sign.
−x+1−x+2−x+3≥6
x≤0
Eliminating the redundant solutions, we get the final answer as:
x∈(−∞,0]∪[4,∞)
or x∈R−(0,4)

Thanks and Regards
Rishi Sharma
askIITians Faculty 646 Points
2 years ago
Dear Student,
Please find below the solution to your problem.

To get rid of the absolute value functions, we need to take 4 cases:
Case 1: x≥3or [3,∞)Here all the bars will open with a positive sign.
x−1+x−2+x−3≥6
3x≥12
x≥4
Case 2: 2≤x<3
or [2,3)
Here, only |x−3|
will open with a negative sign.
x−1+x−2−x+3≥6
x≥6
But this lies out of [2,3)
So no solution
Case 3: 1≤x<2 or [1,2)
Here only |x−1|
will open with a positive sign.
x−1−x+2−x+3≥6
−x≥2
x≤−2
Case 4: x<1(−∞,1)
Here all bars will open with a negative sign.
−x+1−x+2−x+3≥6
x≤0
Eliminating the redundant solutions, we get the final answer as:
x∈(−∞,0]∪[4,∞)
or x∈R−(0,4)

Thanks and Regards

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

Other Related Questions on Differential Calculus

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More