The range of a function is the set of all possible values of the func

Question

﻿The  range  of a function is the set of all possible values of the function. If the function f is defined by  f(x) = \frac{1}{x^2 + 1}  for all real numbers x, the range of f is the set of

A. all real numbers
B. all real numbers except -1 and 1
C. all positive real numbers
D. all real numbers greater than or equal to 0 and less than or equal to 1
E. all positive real numbers less than or equal to 1

https://gmatclub.com/forum/download/file_private.php?id=76100

gmatophobia · Accepted Answer

x^2   is always non-negative. Hence,  x^2+1  is always greater than or equal to 1.

If  x^2 + 1 = 1  →   \frac{1}{x^2 + 1}  = 1. This is the maximum possible value of f(x).

If  x^2 + 1 > 1  →   \frac{1}{x^2 + 1}  < 1. The value will however be greater than zero.

hence,  0 < f(x) \leq 1

Option E

MartyMurray · Answer

﻿ The  range  of a function is the set of all possible values of the function. If the function f is defined by  f(x) = \frac{1}{x^2 + 1}  for all real numbers x, the range of f is the set of

A. all real numbers
B. all real numbers except -1 and 1
C. all positive real numbers
D. all real numbers greater than or equal to 0 and less than or equal to 1
E. all positive real numbers less than or equal to 1

Since  x^2  is always positive,  x^2 + 1  is always positive. Thus, the numerator and the denominator of  \frac{1}{x^2 + 1}  are always positive.

So, we can eliminate (A) and (B) since they include negative numbers.

Next, we see that the maximum value of  \frac{1}{x^2 + 1}  will exist when the denominator is as small as possible, in other words, when x = 0 and the dennminator 
 is  0^2 + 1 = 1 . When the denominator is 1, the fraction is  \frac{1}{1} = 1 .

So, the maximum value of  \frac{1}{x^2 + 1}  is 1.

To find the minimum value, we consider how small the fraction can be if we increase the denominator. Since the numerator and denominator are both positive, as we increase the denominator, the fraction becomes closer and closer to 0 but never reaches 0.

So, the range of  f(x) = \frac{1}{x^2 + 1}  is all the numbers greater than 0 to 1.

The correct answer is (E).

JeffTargetTestPrep · Answer

x^2 ≥ 0

x^2 + 1 ≥ 1

Since both sides of the inequality have the same sign, taking the reciprocal of both sides will change the direction of the inequality sign.

1/(x^2 + 1) ≤ 1

For any x, both the numerator and the denominator of the fraction 1/(x^2 + 1) are positive, so the fraction itself will always be positive.

0 < f(x) ≤ 1

If x^2 = 0, then the function takes its maximum value. Whereas, if x^2 increases, the value of the function decreases, and it can get as close to zero as we want.

So, the range of the function is all positive real numbers less than or equal to 1.

Answer: E

ayushiii02 · Answer

Where can I read more about the concept of such questions?

Bunuel · Answer

13. Functions 
     Theory   
 How to Interpret Unfamiliar Symbols 
 Functions on GMAT 
 A Closer Look at GMAT Function Questions 
     Questions   
 Operations/functions defining algebraic/arithmetic expressions 
 Symbols Representing Arithmetic Operation 
 Rounding Functions 
 Various Functions 
  DS Questions 
 PS Questions

For more check below:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

KarishmaB · Answer

In GMAT, we deal with only real numbers and they are all available on the number line. So the number line is a great tool to solve domain and range questions. We know that the behaviour of numbers is different in these different sections of the number line. Screenshot 2024-04-15 at 3.30.12 PM.png x > 1 Say x = 2, f(x) = \frac{1}{x^2 + 1} = \frac{1}{(2)^2 + 1} = \frac{1}{5} Value of f(x) is between 0 and 1 x = 1 , f(x) = 1/2 i.e. between 0 and 1. 0 < x < 1 Say x = 1/2, f(x) = \frac{1}{x^2 + 1} =\frac{1}{(1/2)^2 + 1} = \frac{4}{5} Value of f(x) is between 0 and 1 x = 0 , value of f(x) is 1 -1 < x < 0 No different from 0 < x < 1 because x is squared. Value of f(x) is between 0 and 1 x = -1 , value of f(x) is between 0 and 1 x < -1 No different from x > 1 because x is squared. Value of f(x) is between 0 and 1 So for the entire number line, f(x) is between 0 and 1 with 0 excluded but 1 included. (E) all positive real numbers less than or equal to 1 0 < f(x) <= 1 Answer (E)

DanTheGMATMan · Answer

Sit back, breathe, and think about the possible outcomes for   x^2 +1:

https://www.youtube.com/watch?v=emn2mioytMA

Nairobisha · Answer

Can someone help with option D please? How is D different from E? D is also implying the range from 0 to 1, what am I missing here?

Kinshook · Answer

The  range  of a function is the set of all possible values of the function. If the function f is defined by  f(x) = \frac{1}{x^2 + 1}  for all real numbers x, the range of f is the set of

Since x^>=0; 0< f(x) <=1

A. all real numbers
B. all real numbers except -1 and 1
C. all positive real numbers
D. all real numbers greater than or equal to 0 and less than or equal to 1
E. all positive real numbers less than or equal to 1

IMO E

Nairobisha · Answer

Can you help with option D please? How is D different from E? D is also implying the range from 0 to 1, what am I missing here?

Bunuel · Answer

Option D says thar  0 \leq f(x) \leq 1 
Option E says thar  0 < f(x) \leq 1

The key difference is that D allows  f(x)  to be 0, while E does not.

Since  f(x) = \frac{1}{x^2 + 1} ,  f(x)  cannot be 0 for any value of  x . The least value of  x^2 + 1  is 1 when  x = 0 , which gives the greatest value of  f(x)  as 1. The least value of  f(x)  approaches 0 as  x  approaches infinity, but it never reaches 0. Thus,  0 < f(x) \leq 1 .

Hope it's clear.

Hrathen · Answer

What threw me off here was that Choice E makes no mention that the value cant be 0 or less than 0.
It only states less than or equal to 1, no mention of 0 here. It seems to imply a negative is a value range

dhruvie · Answer

I have a doubt, the minimum value of  x^2  is  0 . However, the maximum value is  infinity .

\frac{1}{infinity}=0  so shouldnt the answer be  D .

The only explanation I could think of is that  \frac{1}{infinity}    tends to 0  and isn't really actually 0.

Is that the case?

I got confused because lim (x--> inifnity)  \frac{1}{x}=0

Bunuel · Answer

One over a value that tends to positive infinity is never zero. It only approaches zero but never equals it. It gets closer and closer to zero but never equals zero. That’s why the function never actually reaches zero, only values above it.

Bunuel · Answer

Notice that E mentions positive values:

E. all  positive  real numbers less than or equal to 1

0 is not positive.

Gmat232323 · Answer

f(x) = \frac{1}{x^2 + 1} , all  real numbers  x

Let's look at the denominator first. 
 x^2 + 1  will always be  1  or greater than  1  as  x^2  is non-negative which means that the least value it can take is  0 . When  x^2 = 0 ,  x^2 + 1 = 1

Any number greater than  1  in the denominator will yield a result that is between  0  and  1 . So the range of the function will be all numbers greater than  0  and less than or equal to  1 .

Answer is E.

egmat · Answer

Range problems can be tricky because you need to think about what outputs are actually possible, not just what inputs work. Let me walk you through how to tackle this one systematically.

Understanding What We're Looking For

When we talk about the  range  of a function, we're asking: "What are all the possible output values this function can give us?" So we need to find all possible values of  y = \frac{1}{x^2 + 1}  when  x  can be any real number.

Let's Break This Down Step by Step:

Step 1: Analyze the Denominator

Notice something important about  x^2  – no matter what real number you put in for  x , when you square it, you always get a non-negative result:
 
 If  x = 3 , then  x^2 = 9  
 If  x = -3 , then  x^2 = 9  (still positive!) 
 If  x = 0 , then  x^2 = 0

So  x^2 \geq 0  for any real number  x .

This means  x^2 + 1 \geq 1  for any real number  x .

Here's the key insight:  The denominator is smallest when  x = 0  (giving us 1), and it can grow as large as we want by choosing larger values of  |x| .

Step 2: Find the Maximum Value

Since we're dividing 1 by the denominator, our function will be  largest when the denominator is smallest .

The denominator is smallest when  x^2 + 1 = 1 , which happens when  x = 0 .

When  x = 0 :
 f(0) = \frac{1}{0^2 + 1} = \frac{1}{1} = 1

So the maximum value our function can achieve is  1 .

Step 3: Find the Lower Bound

Now let's think about what happens when  x  gets very large:
 
 If  x = 10 :  f(10) = \frac{1}{100 + 1} = \frac{1}{101} \approx 0.0099  
 If  x = 100 :  f(100) = \frac{1}{10,000 + 1} \approx 0.0001  
 If  x = 1,000 :  f(1,000) \approx 0.000001

As  x  gets larger,  x^2 + 1  gets larger, so  \frac{1}{x^2 + 1}  gets closer and closer to 0.

Critical distinction:  The function  approaches 0 but never equals 0 , because the denominator is never infinite – it's always a finite positive number.

Also notice: since  x^2 + 1  is always positive, and we're dividing the positive number 1 by it, our function is  always positive .

Step 4: Confirm the Range

Let's verify:
 
 Maximum value:  f(x) = 1  when  x = 0  ✓ 
 The function approaches but never reaches 0 ✓ 
 The function is always positive ✓

So the range is:  all positive real numbers less than or equal to 1

In interval notation:  (0, 1]

Watch out for this trap:  Answer choice D says "all real numbers greater than or equal to 0 and less than or equal to 1" – this  includes 0 , which is incorrect because the function never actually equals 0.

Answer choice E says "all positive real numbers less than or equal to 1" – this  excludes 0 , which is correct!

Answer: E

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Want to Master Range Problems Systematically?

While this explanation covers the core approach, you'll find deeper insights on the complete solution. You can check out the  step-by-step solution on Neuron by e-GMAT  to understand the systematic framework that works for all function range problems, including:

The complete strategic approach for analyzing function behavior 
 How to identify all common traps in range vs. domain questions 
 Process skills that help you avoid calculation errors 
 Pattern recognition techniques for similar GMAT function problems

You can also explore other GMAT official questions with detailed solutions on Neuron for structured practice  here .

Happy learning! 🎯

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