Question: Is |x + 1| + |x - 2| < 4 ?

Let's examine the expression |x + 1| + |x - 2| < 4

i.e. (Distance of x from -1) + (Distance of x from 2) is Less than 4

If x < -1
then, (-1-x)+(2-x) < 4
i.e. 2x > -3
i.e. x > -1.5

If x > 2
then, (x-(-1))+(x-2) < 4
i.e. 2x-1 < 4
i.e. x < 2.5


i.e. Value of x must be between -1.5 and 2.5

Question REPHRASED: Is -1.5 < x < 2.5?

Statement 1: x is between -0.5 and 4

Some values are between -1.5 and 2.5 but some may be outside such as 3.5 as well hence

NOT SUFFICIENT

Statement 2: x is between -2 and 2.5

Some values are between -1.5 and 2.5 but some may be outside such as -1.,75 as well hence

NOT SUFFICIENT

COmbining the statements

-0.5 < x < 2.5

SUFFICIENT

Answer: Option C
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lx+1l + lx-2l <4?
if,x>=2 ,
x+1+x-2<4
or,2x<5
or,x<2.5
so,2≤x<2.5[/b]

if,-1<x<2
x+1+2-x<4
or,3<4 (always true)

if,x<=-1
-x-1+2-x<4
or,-2x<3
or,x>-1.5
so,-1.5<x≤-1

so, question asks if 2<=x<2.5 or -1<x<2 or -1.5<x<=-1
OR simply , -1.5<x<2.5?

1) x is between -0.5 and 4
if, x lies between -0.5 to 2.5 (x=1) ans is YES but if it lies between 2.5 to 4 ans is no (x=3) insufficient

2) x is between -2 and 2.5
if, x=1 ans is YES but x=-1.7 ans is no insufficient

combining 1 and 2
x is between -0.5 and 2.5
ans is always true. sufficient


correct answer C

I have solved this question as follows, kindly let me know if there is any error

The Common area between |x + 1| + |x - 2| is as per the image

Sorry, the image has been posted in side ways. I haven't seen any option to rotate it.

Here we can easily find our answer by subsituting values within the given range in the statements. We just need to see what values to pick.

Lets start with Statement 1 -

Statement 1 -> x is between -0.5 and 4

Lets take \(x = 3\) and substitute in the prompt to check the condition;

\(|x + 1| + |x - 2| < 4\)
\(|3 + 1| + |3 - 2| < 4\)
\(|4| + |1| < 4\)
\(5<4\);
This proves the statement false.

Now lets substitute \(x = 0\)

\(|x + 1| + |x - 2| < 4\)
\(|0 + 1| + |0 - 2| < 4\)
\(|1| + |-2| < 4\)
\(3<4\);

This statement is true.

As two different answers in the same range.

Statement 1 is Insuffucient

Statement 2 -> x is between -2 and 2.5
Lets take \(x = -1.9\) and substitute in the prompt to check the condition;

\(|x + 1| + |x - 2| < 4\)
\(|-1.9 + 1| + |-1.9 - 2| < 4\)
\(|0.9| + |-3.9| < 4\)
\(4.8<4\);

This proves the statement false.

Now lets substitute \(x = 0\)

\(|x + 1| + |x - 2| < 4\)
\(|0 + 1| + |0 - 2| < 4\)
\(|1| + |-2| < 4\)
\(3<4\);

This statement is true.

As two different answers in the same range.

Statement 2 is Insuffucient

Now Combining Statement 1 and Statement 2

The new range becomes : x is between -0.5 and 2.5

Now we can try and substitute directly the end points. If they satisfy, then the value within the range will surely be satisfied.

Lets take \(x = -0.5\) and substitute in the prompt to check the condition;

\(|x + 1| + |x - 2| < 4\)
\(|-0.5 + 1| + |-0.5 - 2| < 4\)
\(|0.5| + |-2.5| < 4\)
\(3<4\);

This satisfies the equation.

Now lets substitute \(x = 2.5\)

\(|x + 1| + |x - 2| < 4\)
\(|2.5 + 1| + |2.5 - 2| < 4\)
\(|3.5| + |0.5| < 4\)
\(4<4\);

Although this is not true but notice that this equal. If we select any value than this, then the equation will be satisfied.
And from the range we surely know that we need to take someting less than 2.5

Hence Combining Statements is Sufficient and answer is therefore, C
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Can somebody please let me know if the graphical method can be applied?

If yes, how?

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