Given that |x + 1 | > 2x - 1 and we need to find the correct range of values of x

Let's solve the problem using two methods

Method 1: Substitution

We will values in each option choice and plug in the question and check if it satisfies the question or not. ( Idea is to take such values which can prove the question wrong)

A. x < 0

Lets take x = -1 (which falls in this range of x < 0) and substitute in the equation |x + 1 | > 2x - 1
=> |-1 + 1 | > 2*-1 - 1
=> | 0| > -3
=> 0 > -3 which is TRUE, but let's see if we can find a bigger range

B. x < 2

Lets take x = 1.9 (which falls in this range of x < 2) and substitute in the equation |x + 1 | > 2x - 1
=> |1.9 + 1 | > 2*1.9 - 1
=> | 2.9 | > 2.8
=> 0 > -3 which is TRUE.
Now, this range is bigger than option A, so we can ignore option A now.

C. -2 < x < 0

This is a subset of Option B so we don't need to check this one

D. -1 < x < 2

This is a subset of Option B so we don't need to check this one

E. 0 < x < 2

This is a subset of Option B so we don't need to check this one

So, Answer will be B

Method 2: Algebra

Now, we know that |A| > B can be opened as (Watch this video to know about the Basics of Absolute Value)
A > B for A ≥ 0 and
-A > B for A < 0

=> |x + 1 | > 2x - 1 can be written as

Case 1: x + 1 ≥ 0 or x ≥ -1
=> x + 1 > 2x - 1
=> 2x - x < 1 + 1
=> x < 2
And the condition was x ≥ -1 and we got the answer as x < 2
=> the solution will be the range common in both of them
=> -1 ≤ x < 2



Case 2: x + 1 < 0 or x < -1
=> -(x + 1) > 2x - 1
=> 2x + x < 1 - 1
=> 3x < 0
=> x < 0
And the condition was x < -1 and we got the answer as x < 0
=> the solution will be the range common in both of them
=> x < -1



Combining both the cases we get the answer as -1 ≤ x < 2 and x < -1
=> x < 2

So, Answer will be B
Hope it helps!

Watch the following video to learn How to Solve Absolute Value Problems