Loading....
Coupon Accepted Successfully!

 

Defined Functions

Defined functions are common on the GMAT, and most students struggle with them. Yet once you get used to them, defined functions can be some of the easiest problems on the test. In these problems, you are given a symbol and a mathematical expression or description that defines the symbol. Some examples will illustrate.
 
Example-1
Define x∇ y by the equation xy = xyy. Then 2∇ 3 =?
  1. 1
  2. 3
  3. 12
  4. 15
  5. 18
Solution
From the above definition, we know that xy = xy – y.
 
So we merely replace x with 2 and y with 3 in the definition: 23 = 2 x 3 – 3 = 3.

Hence, the answer is B.
 
 
Example-2
If a ∆ b is defined to be a2, then what is the value of ?
  1. 2/3
  2. 1
  3. 3/2
  4. 2
  5. 3
Solution
Most students who are unfamiliar with defined functions are unable to solve this problem. Yet it is actually quite easy.
 
By the definition above, ∆ merely squares the first term.
 
So, z ∆ 2 = z2, and z ∆ 3 = z2.
 
Forming the fraction yields .
 
Hence, the answer is (B).
 
 
Example-3
The operation @ is defined for all non-zero x and y by the equation . Then the expression  is equal to
  1. xyz
Solution
.
 
Hence, the answer is (E).
 
Note, though it may appear that choices (A) and (E) are equivalent, they are not: , which is not equal to .
 
 
Example-4
For all real numbers x and y, let x # y = (xy)2 – x + y2. What is the value of y that makes x # y equal to –x for all values of x ?
  1. 0
  2. 2
  3. 5
  4. 7
  5. 10
Solution
Setting x # y equal to –x yields (xy)2 – x + y2 = –x
Canceling –x from both sides of the equation yields (xy)2 + y2 = 0
Expanding the first term yields x2y2 + y2 = 0
Factoring out y2 yields y2(x2 + 1) = 0
Setting each factor equal to zero yields y2 = 0 or x2 + 1 = 0
Now, x2 + 1 is greater than or equal to 1 (why?). Hence, y2 = 0
Taking the square root of both sides of this equation yields y = 0

Hence, the answer is (A).
 
 
Example-5
If [x] denotes the area of a square with sides of length x, then which one of the following is equal to [9] [3] ?
  1. [2]
  2. [3]
  3. [16]
  4. [27]
  5. [81]
Solution
The area of a square with sides of length x is x2.
 
This formula yields [9] [3] = 92 32 = 81 9 = 729
 
Now, [3] = 32 = 9.
 
Hence, the answer is (D).
 
Example-6
If x is a positive integer, define:
(x)* = , if x is even;
(x)* = 4x, if x is odd.
If k is a positive integer, which one of the following equals (2k – 1)* ?
  1. k – 1
  2. 8k – 4
  3. 8k – 1
Solution
First, we must determine whether 2k – 1 is odd or even.
 
(It cannot be both—why?) To this end, let k = 1.
 
Then 2k – 1 = 2 x 1 – 1 = 1, which is an odd number.
 
Therefore, we use the bottom-half of the definition given above.
 
That is, (2k – 1)* = 4(2k – 1) = 8k – 4.
 
The answer is (C).
 

 

You may be wondering how defined functions differ from the functions, f(x), you studied in Intermediate Algebra and more advanced math courses. They don’t differ. They are the same old concept you dealt with in your math classes. The function in Example 6 could just as easily be written  and f(x) = 4x. The purpose of defined functions is to see how well you can adapt unusual structures. Once you realize that defined functions are evaluated and manipulated just as regular functions, they become much less daunting.
 

Example-7
Define x* by the equation x* = π – x. Then ((–π)*)* =
  1. –2π
  2. –1
  3. –π
Solution
Working from the inner parentheses out, we get
((–π)*)* = (π – (–π))* = (π + π)* = (2π)* = π – 2π = –π.
 
Hence, the answer is (C).
 
Method II: We can rewrite this problem using ordinary function notation. Replacing the odd symbol x* with f(x) gives f(x) = π – x. Now, the expression ((–π)*)* becomes the ordinary composite function .
 

 

Example-8
If x is an integer, define:
/x\ = 5, if x is odd;
/x\ = 10, if x is even.
If u and v are integers, and both 3u and 7 – v are odd, then /u\ – /v\ =
  1. –5
  2. 0
  3. 5
  4. 10
  5. 15
Solution
Since 3u is odd, u is odd.
 
(Proof: Suppose u were even, then 3u would be even as well. But we are given that 3u is odd. Hence, u must be odd.)
 
Since 7 – v is odd, v must be even.
 
(Proof: Suppose v were odd, then 7 – v would be even [the difference of two odd numbers is an even number]. But we are given that 7 – v is odd. Hence, v must be even.)
 
Since u is odd, the top part of the definition gives /u\ = 5.
 
Since v is even, the bottom part of the definition gives /v\ = 10.
 
Hence, /u\ – /v\ = 5 – 10 = –5.
 
The answer is (A).
 

Example-9
For all real numbers a and b, where a x b ≠ 0, let ab = ab. Then which of the following must be true?
I.    ab = ba
II.  
III. (ab)◊c = a◊(bc)
  1. I only
  2. II only
  3. III only
  4. I and II only
  5. II and III only
Solution
Statement I is false. For instance, , but This eliminates (A) and (D).
 
Statement II is true: . This eliminates (C).
 
Unfortunately, we have to check Statement III. It is false:  and . This eliminates (E), and the answer is (B).
 
Note: The expression a x b ≠ 0 insures that neither a nor b equals 0: if a x b = 0, then either a = 0 or b = 0, or both. This prevents division by zero from occurring in the problem, otherwise if a = 0 and b = –1, then .
 




Test Your Skills Now!
Take a Quiz now
Reviewer Name