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Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and

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Bunuel
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Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 08:00
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If \(f(x) = |x - 1|\) and \(g(x) = |x + 2|\), then what is the range of all possible values of x such that \(f(g(x)) = 0\) ?

A. 0
B. 1
C. 2
D. 3
E. 4


 


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C
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 08:33
f(g(x))= 0
llx+2l-1ll=0
x+2 must be -1 but it is in absolute term hence it always be positive then the question cannot be zero.
Hence A is answer
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 08:37
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My Answer: C

f(g(x)) = ||x+2|-1|
||x+2|-1|=0,
|x+2|=1

So there are two option, x+2=1, x=-1
and x+2=-1, x=-3

So, range=-1-(-3)=2
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 08:42
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f(g(x)) becomes f(|x+2|).

Now, we know that f(x) = |x-1|. So, f(|x+2|) = ||x+2|-1|.

We are given that f(g(x)) = 0. So, ||x+2|-1| = 0.

The absolute value of a number is always non-negative. It is zero only when the number itself is zero. So, |x+2|-1 = 0.

Solving this equation gives us x+2 = 1 or x+2 = -1.

So, x = -1 or x = -3.

Answer is C: 2
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 08:44
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Bunuel wrote:
If \(f(x) = |x - 1|\) and \(g(x) = |x + 2|\), then what is the range of all possible values of x such that \(f(g(x)) = 0\) ?

A. 0
B. 1
C. 2
D. 3
E. 4


 


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||x+2|-1| = 0

|p| is always non negative.

|x+2| - 1 = 0

|x+2| = 1

1) x + 2 = 1

x = -2 + 1 = -1

2) x + 2 = -1

x = -2 - 1 = -3

Two values are possible x = -1 and x = -3

IMO C
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 08:49
To find the range of all possible values of x such that f(g(x)) = 0, we need to solve the equation f(g(x)) = 0. Let's first evaluate f(g(x)).

f(g(x)) = |g(x) - 1| = |(x + 2) - 1| = |x + 1|

Now, we need to find the values of x that make |x + 1| equal to 0, because that's when f(g(x)) = 0.

|x + 1| = 0 if and only if x + 1 = 0 or x = -1.

So, the values of x that make f(g(x)) = 0 are x = -1.

Thus, the range of all possible values of x such that f(g(x)) = 0 is just the single value x = -1.
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 08:52
Bunuel wrote:
If \(f(x) = |x - 1|\) and \(g(x) = |x + 2|\), then what is the range of all possible values of x such that \(f(g(x)) = 0\) ?

A. 0
B. 1
C. 2
D. 3
E. 4


 


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f(x) = |x - 1|
g(x) = |x + 2|
f(g(x)) = 0
f(x+2) = ||x+2|-1|
therefore g(x) = 1
x+2=1 x=-1
-x-2=1 x=-3
x= -1-(-3)
x=2
(However, range here feels wrong, as x can be only 1 only in these points.. not in between.. does that mean 0?)
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 08:53
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Bunuel wrote:
If \(f(x) = |x - 1|\) and \(g(x) = |x + 2|\), then what is the range of all possible values of x such that \(f(g(x)) = 0\) ?

A. 0
B. 1
C. 2
D. 3
E. 4


 


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for the Around the World in 80 Questions

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||x+2|-1| = 0
|x+2|-1 = 0
|x+2| = 1

|x+2| = 1
if x+2>0, x>-2
x+2 = 1
x = -1 -----right solution as x>-2

if x+2<0, x<-2
-(x+2) = 1
x = -3------right solution as x<-2

Range of solution = -1-(-3) = 2

C is the answer.
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 08:54
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Asked: If \(f(x) = |x - 1|\) and \(g(x) = |x + 2|\), then what is the range of all possible values of x such that \(f(g(x)) = 0\) ?

f(g(x)) = f(|x+2|) = ||x+2|-1| = 0
|x+2| - 1 = 0
|x+2| = 1
x = -1 or -3
Range of all possible values of x = -1 - (-3) = 2

IMO C
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 08:57
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The answer is C=2
Since f(x) = |x - 1| and g(x) = |x + 2|.
f(g(x)) = |g(x) - 1| = | |x + 2| - 1|
| |x + 2| - 1| = 0
Case 1: |x + 2| - 1 = 0
In this case, |x + 2| = 1. This means x + 2 = 1 or x + 2 = -1.
Solving for x in each case:
x + 2 = 1
x = -1
x + 2 = -1
x = -3
Case 2: -|x + 2| + 1 = 0
In this case, -|x + 2| = -1, so |x + 2| = 1.
Solving for x:
x + 2 = 1
x = -1
x + 2 = -1
x = -3
So, the possible values of x that satisfy f(g(x)) = 0 are x = -1 and x = -3.
The range of all possible values of x is {-1, -3}. Hence 2 values
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 08:58
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fog(x)=||x+2|-1|;
graphically, fog(x)=0 gives 2 solutions x=-3,-1
Range =2 C
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 09:19
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Bunuel wrote:
If \(f(x) = |x - 1|\) and \(g(x) = |x + 2|\), then what is the range of all possible values of x such that \(f(g(x)) = 0\) ?

A. 0
B. 1
C. 2
D. 3
E. 4

 


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f(g(x)) = 0
hence, ||x+2|-1|=0
-> |x+2|=1
-> x+2=+1 or -1
-> x=-1 or -3
Hence, range is |-3-(-1)|=2. C is correct.
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 09:20
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f(g(x))= |(|x+2|)-1| = 0

Step (1): In order to satisfy f(g(x))=0, g(x) would have to be +1
Step (2): In order to satisfy g(x)=+1, x can be either -1 or -3

So the answer is 2 possible answers
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 09:21
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given that
\(f(x) = |x - 1|\) and \(g(x) = |x + 2|\)

f(x)= x-1 and -x+1 and g(x) = x+2 and -x-2

range of all possible values of x such that \(f(g(x)) = 0\)

g(x) = x+2 and -x-2
f(g(x))=0
f(x+2)=0
case 1
f(x+2-1)=0
f(x+1)=0
x=-1
case 2
f(-x-2-1)=0
f(-x-3)=0
x=+3
case 2 for g(x)= -x-2
f(-x-2)=0
f(-x-2-1)=0
f(-x-3)=0
x=+3
f(x+2+1)=0
x=-3
values of x -3,-1 are sufficient
range
-1-(-3) ; -1+3 ; 2
OPTION C



Bunuel wrote:
If \(f(x) = |x - 1|\) and \(g(x) = |x + 2|\), then what is the range of all possible values of x such that \(f(g(x)) = 0\) ?

A. 0
B. 1
C. 2
D. 3
E. 4


 


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for the Around the World in 80 Questions

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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 09:31
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g(x) = |x+2|
f(x) = |x-1|

f(g(x)) = | |x+2| - 1|
=> |x+2-1| or |-x-2-1|
=> |x+1| or |-x-3|

1) x+1 = 0 => x = -1
2) -(x+1) = 0 => x = -1

3) -x-3 = 0 => x = -3
4) -(x-3) = 0 => x = -3

Range: -1 - (-3) = -1 + 3 = 2 [C]
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 09:40
IMO B

f(g(x)) = l l x + 2 l - 1 l = 0

On Number line this means distance of X from -2. And whatever that value when substracted with 1, we get a Zero. Only 1 on number line matches this.
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 09:42
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f(g(x)) = ||x+2|-1| = 0

As we have to find value of x only, so, f(g(x)) = 0 only if
|x+2| = 1
x = -3, -1.

Range of values of x = 2.

IMO C.
Bunuel wrote:
If \(f(x) = |x - 1|\) and \(g(x) = |x + 2|\), then what is the range of all possible values of x such that \(f(g(x)) = 0\) ?

A. 0
B. 1
C. 2
D. 3
E. 4


 


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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 09:52
If f(x)=|x−1| and g(x)=|x+2|, then what is the range of all possible values of x such that f(g(x))=0
_____________________________________________________________
We can basically put numbers such as -2,-1,0 and 1. When we do so, we get 1,0,1 and 2. There could be better strategy to solve it. We need to refer to experts reply.
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 10:01
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink] New post   25 Jul 2023, 10:25
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If \(f(x) = |x - 1|\) and \(g(x) = |x + 2|\), then what is the range of all possible values of x such that \(f(g(x)) = 0\) ?

A. 0
B. 1
C. 2
D. 3
E. 4

Answer:
f(x) =|x - 1|

and g(x) = |x +2|
i.e g(x) = x+2 if x>-2
or, -x-2. if x<-2
and 0 if x= -2

f(g(x) = ||x+2| - 1| =0
if x > -2 ,
f(g(x)) = |x+2-1| = |x+1| = 0
so it will become 0 only when x = -1

if x< -2,
f(g(x)) =|-x-2 -1|= 0
i.e |-x-3| = 0
this can become 0 when x = -3

so we get two values of x where this holds
C is the answer
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Re: Around the World in 80 Questions (Day 7): If f(x) = |x - 1| and [#permalink]
25 Jul 2023, 10:25
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