Around the World in 80 Questions (Day 4): How many values of x satisfy

Solve the equation we have 11<x<12. Since the question does not specify the value of x. It can be any number in between 11 and 12.
Option E is the answer.

There are 3 possible scenarios to this question:
1) x>=12:
equation can be written as: |x - 11| + |x - 12| < 1
x-11+x-12<1
x<12; No solution as it contradicts the above condition (x>=12)

2) 11<=x<12
x-11-x+12<1
1<1 No solution, as it contradicts the above condition (11<=x<12)

3) x<11
-2x+11+12<1
22<2x
11<x, No solution, as it contradicts the above condition (x<11)

Hence 0 solution, or option A.

Whatever be the value of x, the minimum value we will get for the above sum will be equal to 1 ( when x is kept close to 11 and 12).

Hence, no condition satisfies the above inequality, and correct choice is A.

How many values of x satisfy the inequality |x - 11| + |x - 12| < 1?

(A) 0
(B) 1
(C) 2
(D) 3
(E) Infinite number

Solution:

Case 1: x<11
-x+11-x+12<1
or x>11. NOT POSSIBLE. CASE REJECTED.

Case 2: 11=<x<12
x-11-x+12<1
or 1<1. NOT POSSIBLE. CASE REJECTED.

Case 3: x>=12
x-11+x-12<1
or x<12. NOT POSSIBLE. CASE REJECTED.

Hence, NO SOLUTION (0). ANSWER A

Case 1: x < 11
In this case, both expressions inside the absolute value signs are negative, so we can drop the absolute value signs. We get:
-(11 - x) - (12 - x) < 1
Simplifying, we get:
-23 + 2x < 1
2x < 24
x < 12

Case 2: 11 ≤ x < 12
In this case, the first expression inside the absolute value signs is negative, and the second expression is positive, so we can drop the absolute value sign around the first expression and keep the second one. We get:
-(11 - x) + (x - 12) < 1
Simplifying, we get:
-23 + 2x < 1
2x < 24
x < 12

Case 3: x ≥ 12
In this case, both expressions inside the absolute value signs are positive, so we can drop the absolute value signs. We get:
(x - 11) + (x - 12) < 1
Simplifying, we get:
2x - 23 < 1
2x < 24
x < 12

We see that all three cases lead to the conclusion that x < 12. Therefore, there is only one value of x that satisfies the inequality, namely x = 11.

The answer is (B)1

|x-11| and |x-12| are the distance from x to 11 and to 12
Draw a line, we can see clearly that any positions of x on the line, including x<=11, 11<x<12, x>=12 will give us a minimum value of 1
Thus, the inequality has 0 values of x that satisfies itself

Case 1: x < 11
When x < 11, both expressions inside the absolute value signs will be negative. Therefore, the inequality becomes:
-(x - 11) - (x - 12) < 1
-2x + 23 < 1
Solving for x:
-2x < 1 - 23
-2x < -22
x > 11
But we had assumed x < 11, so there are no values of x that satisfy the inequality.

Case 2: 11 < x < 12 - the inequality becomes:
(x - 11) - (x - 12) < 1
x - 11 - x + 12 < 1
1 < 1
Since 1 is not less than 1, there are no values of x that satisfy the inequality in this case either.

Case 3: x > 12 - the inequality becomes:
(x - 11) + (x - 12) < 1
2x - 23 < 1
Solving for x:
2x < 1 + 23
2x < 24
x < 12
But in this case, we already assumed x > 12, so there are no values of x that satisfy the inequality in this case as well.

Also for x = 11 and 12, we get 1<1 which is not valid.
Hence, there are no values of x that satisfy the inequality |x - 11| + |x - 12| < 1.
So, the inequality has no solutions.
Answer A

\(|x - 11| + |x - 12| < 1\) cannot yield zero as an answer as the two absolute values will never simultaneously be zero. Nor can the inequality yield a negative answer by virtue of being the sum of two absolute values. As for when \(11<x<12\), \(|x - 11| + |x - 12|\) will always sum to 1 which means that there will never be a fractional value that is less than 1.

Therefore there are no possible values that satisfy the inequality.

Answer A

This question can be solved by two ways:-
(1) Short Way:-

|x-11| represents the magnitude of distance between "11" and "x" on a number line. Similarly, |x-12| represents the magnitude of distance between "12" and "x" on a number line.
So in, |x-11| + |x-12| we are adding the magnitude of distances of "x" from "11" and "12". Now if "x" is anywhere between "11" and "12" inclusive, the expression, |x-11| + |x-12| will always be "1".
And for any other value of "x", this expression will always be necessarily greater than "1".
So in no possible case can the expression, |x-11| + |x-12| be less than "1".

Hence there is no possible value of "x" satisfying the given inequality, therefore correct answer is option A.

(2) Long Way:-

"x" can either be less than "11", between "11" and "12" inclusive or greater than "12".

(i) x < 11

Therefore given inequality can be written as:- -(x - 11) + -(x - 12) < 1
Which gives, 23 - 2x < 1
Therefore, x > 12.
So we have two conditions on "x":- x < 11 & x > 12, which is not possible for any value of "x".

(ii) x > 12

Therefore given inequality can be written as:- (x - 11) + (x - 12) < 1
Which gives, 2x - 23 < 1
Therefore, x < 12.
So we have two conditions on "x":- x > 12 & x < 12, which is not possible for any value of "x".

(iii) x = 11 or 11 < x < 12 or x = 12

Therefore given inequality can be written as:- (x - 11) + -(x - 12) < 1
Which gives, 1 < 1, which is not possible, so "x" cannot be between "11" and "12" inclusive.

As for any value of "x" the given inequality is not possible, the correct answer is option A.

Given inequality : |x - 11| + |x - 12| < 1

To find number of solutions, lets find the zero points by equating the modulus to zero
we get x = 11 and x = 12 .
We will get three regions where the roots can lie as shown in below number line

----------------11----------12----------------
.......I......................II................III..........(regions)

1. Solving for roots in Region I (x<11)

-(x - 11) - (x - 12) < 1
-x+11-x+12-1<0
x>11
No value of x>11 lies in region I
So no roots exist in Region I.

2. Solving for roots in Region II (11<x<12)

(x - 11) - (x - 12) < 1
x-11-x+12-1<0
0<0
Since it is false, no roots exist in Region II.

3. Solving for roots in Region III (x>12)
(x - 11) + (x - 12) < 1
2x - 24 <0
x<12
Since no value of x<12 lies in region III, no roots exist in Region III

Thus, Total values that satisfies given equation = 0

Answer : A

note that abs(x) = x if x>=0 ; -x if x<0 . We will use this definition of absolute value extensively to get to the answer

We are required to deal with abs(x-11) and abs(x-12)

Case 1 : x < 11
so the LHS of the inequality can be rewritten as (11-x)+(12-x) = 23-2x
Now 23 - 2x < 1 => 22 < 2x => x >11. This contradicts the case we are studying : x<11. Therefore no value of x<11 satisfies the inequality.

Case 2 : 11 <= x < 12
so the LHS of the inequality can be rewritten as (x-11)+(12-x) = 1
Now 1 < 1. This is wrong/false. Therefore no value of 11 <= x < 12 satisfies the inequality.

Case 3 : x >= 12
so the LHS of the inequality can be rewritten as (x-11)+(x-12) = 2x-23
Now 2x - 23 < 1 => 2x < 24 => x < 12. This contradicts the case we are studying : x >= 12. Therefore no value of x >=12 satisfies the inequality.

NO VALUE SATISFIES IN ANY OF THE CASES. SO ANSWER IS 0 i.e, A

How many values of x satisfy the inequality |x - 11| + |x - 12| < 1?

For X>12
x-11+x-12<1
2x<24
x<12
Not Possible

For 11<=x<=12
x-11+12-x<1
1<1
Not Possible

For x<11
11-x +12-x<1
22<2x
11<x
Not possible

Option A=0 is correct

How many values of x satisfy the inequality |x - 11| + |x - 12| < 1?
the critical points are 11 and 12
Case 1 - x < 11 - In this region both |x-11| and |x-12| are negative, so the inequality can be written as
-(x-11) - (x-12) <1
-2x +23 <1
-2x<-22
or x > 11
but we are considering the case for x<11 - hence, x>11 is not possible

Case 2 - 11 < x < 12
in this region |x-11| is positive and |x-12| is negative, so the inequality can be written as
x -11 - (x-12) <1
-11+12 < 1
1<1 - Not possible
Case 3 - x>12 - in this region both |x-11| and |x-12| are positive, so the inequality can be written as
x - 11 + x -12 < 1
2x -23 <1
2x< 24
x<12
but we are considering the case of x>12 - hence the solution is not valid
Case 4 - x =11 or x =12
in this case, the sum (LHS of inequality)is 1 and 1<1 is not possible
Hence, in all four possible cases we could not find a value of x for which the inequality holds
Answer A

case1. x<11
-x+11+12-x<1
x>11.. incorrect as taken condition x<11
case 2.. 11<x<12
x-11+12-x<1
1<1..not possible

case 3. x>12
x-11-12+x<1
2x-23<1
x<12...not possible

so ans is A=0

for |x - 11| + |x - 12| < 1 the left hand side is all positive so th minimum value is zero, lets minimize the absolute terms
for |x- 11| to be zero x = 11
then |x - 12| = 1
not satisfying

for |x- 12| to be zero x = 12
then |x - 11| = 1
not satisfying

for all fraction values both the terms will add up and become 1
hence none of the values of x is possible to satisfy the given inequality
correct option A

-|x - 11| is the distance between x and 11
-|x - 12| is the distance between x and 12
-The question basically asks us if the total distance equals 1
-I have attached the image of number line so that everyone can easily understand the concept
-Case 1: x<11 => the total distance will always larger than 1=>no solution
-Case 2: x=11 => the total distance equals 1=>no solution
-Case 3: 11<x<12 => the total distance equals 1=> no solution
-Case 4: x=12 => the total distance equals 1=> no solution
-Case 5: x>12 => the total distance will always larger than 1=>no solution
=> We can't find any x that satisfy the condition
The answer is A

|x-11| + |x-12| <1

The value of x will fall into any 1 of these categories

C1 x<11
In such case, removing the mod
11- x + 12 - x < 1
22<x
x>11
Self Contradictory

C2 x > 12
Removing the mod
x - 11 + x - 12 < 1
2x<24
Self Contradictory

C3 11<x<12
Removing the mod
x - 11 + 12 - x < 1
1<1 Not possible