|x+1|-|x|+3|x-1|-2|x-2|=|x+2|

Name: Problem 82, Chapter 7: Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry
Uploaded: 2022-02-27T09:56:43-08:00
Duration: 4 min 14 s
Channel: Kavita Kumari
Description: |x+1|-|x|+3|x-1|-2|x-2|=|x+2|

Key Concepts

Solving Absolute Value Equations

Equations involving absolute values require examining each interval separately, rewriting the absolute values as linear expressions based on the sign of the argument, and then solving the resulting linear equations. This methodical approach helps ensure that all possible solutions are considered.

Critical Points and Interval Partitioning

The points where the inputs to the absolute value functions equal zero are known as critical points. These points divide the number line into segments where the expressions inside the absolute values maintain consistent signs, allowing a piecewise approach to solve the equation.

Absolute Value Function

The absolute value function measures the distance of a number from zero, defined as the number itself if it is nonnegative and its opposite if it is negative. This fundamental concept is crucial to understanding equations that involve modulus terms.

Piecewise Analysis

Since the definition of the absolute value function changes based on the sign of its argument, problems involving absolute values are typically solved by breaking the problem into cases or intervals. Within each interval, the absolute value expressions can be replaced with their corresponding linear expressions.

Transcript

00:01 In the following question, we've been given an equation which is in the form of modulus, various components of modulus.

00:09 So the first one is modulus x plus 1 minus modulus x plus 3 into modulus x minus 1, minus 2 into modulus x minus 2, and which will be equal to modulus x plus 2.

00:27 We have to find out the interval of x, the intervals of x wherein this equation will be satisfied.

00:35 So firstly we need to know that we'll be finding the intervals for this complete equation.

00:41 And let's write down those intervals.

00:45 So the intervals here will be x and these will be decided by the zeros of each inside.

00:54 So this will have minus 1, this will have 0, this will have plus 1.

00:58 This will have plus 2 and this will have minus 2.

01:01 So values less than minus 2 is the first interval.

01:04 The second interval will be minus 2 greater than less than equal to x and this will lie between minus 1.

01:16 Then the next interval will be minus 1 to 0.

01:24 The fourth interval will be 0 to 0.

01:29 Plus 1, the fifth interval here will be 1, x greater than equal to 1 and less than 2, and the last interval here will be x is greater than or equal to 2.

01:45 So these are the 6 intervals which we will define for this question.

01:51 And now we need to solve each of the inequality for each of the intervals.

01:57 And we have to find out which interval is actually satisfying this equation and which one isn't.

02:03 The ones which are not satisfying the equation, they will automatically not be considered in our final answer.

02:09 So this interval is satisfying by equation.

02:13 So you can put in any value which is less than minus 2 and check...

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