Loading....
Coupon Accepted Successfully!

 

Defined Functions
 

Defined functions are very common on the GRE, 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 this type of problem, you will be given a strange symbol and a property that defines the symbol. Some examples will illustrate.

 

Example 1:

Define xy by the equation xy = xy –– y. Then 2∇3 = ?


From the above definition, we know that x∇y = xy –– y. So all we have to do is replace x with 2 and y with 3 in the definition: 2∇3 = 2 • 3−3 = 3. Hence, enter 3.

 

Example 2:

Define a Δ b to be a2 .

 

Column A

zΔ2
 

Column B

zΔ3


(A) Column A is bigger
(B) Column B is bigger
(C) Column A and Column B are equal
(D) There is not enough information to decide
 

Most students who are unfamiliar with defined functions are unable to solve this problem. Yet it is actually quite easy. By the definition given above, Δ merely squares the first term. So z Δ 2 = z2 , and z Δ 3 = z2. In each case, the result is z2 . Hence the two expressions are equal. The answer is C.

 

Example 3:

If x is a positive integer, define:

(x)* = √x, if x is even;
(x)* = 4x, if x is odd.

If k is a positive integer, which of the following equals (2k –– 1)* ?

 

(A) √2k - 1

(B)k - 1

(C) 8k - 4

(D)√8k - 4

(E) 8k - 1

 

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 •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 3 could just as easily be written f (x) = √x and f (x) = 4x . The purpose of defined functions is to see how well you can adapt to unusual structures. Once you realize that defined functions are evaluated and manipulated just as regular functions, they become much less daunting.

 

Example 4:

Define x* by the equation x* = π –– x. Then ((––π)*)* = ?

 

(A) -2π

(B) -1

(C) -π

(D) 2π

(E) 4π

 

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 f ( f (-π) ) = f (π - (-π)) = f (π + π) = f (2π) = π - 2π = -π.

 

Example 5:

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\ =


(A) -5

(B) 0

(C) 5

(D) 10

(E) 15


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 6:

[Select One or More Answer Choices]

For all real numbers a and b, where ab ≠ 0 , let ab = ab . Then which of the following must be true?


A. ab = ba

B. (-a)◊(-a) = (-1)-aaa

C. (a◊b)◊c = a◊(b◊c)


Choice A is false. For instance, 1◊2 = 12 = 1, but 2◊1 = 21 = 2 .
Choice B is true: (-a)◊(-a) = (- a)-a = (-1• a)-a = (-1)-a (a)-a =
(-1)-aaa.
Choice C is false: (2◊2)◊3 = 22 ◊3 = 4◊3 = 43 = 64 and 2◊(2◊3) = 2◊23 = 2◊8 = 28 = 256.
The answer is B.


To confirm that B is true, assume that a = 2 and solve:

(-2)(-2)=(-1)222
-22=(-1)222
 
1-22=1-1241
 
14=14


a × b ≠ 0 insures that neither a nor b equals 0: if a × 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 0◊(-1) = 0-1 =
10

 

Example 7:
The operation @ is defined for all non-zero x and y by the equation x@y = xy . Then the expression (x @ y) @ z is equal to


(A) xyz

(B) xyz

(c) ( xy)z

(D) xyz

(E) ( xy)z


(x@y)@z = ( xy )@z = ( xy)z . Hence, the answer is (E). Note, though it might appear that choices (A) and (E) are equivalent, they are not. ( xy)z = xyz , which is not equal to xyz .

 

Example 8:
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 ?


(A) 0
(B) 2
(C) 5
(D) 7
(E) 10


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. Hence,

y2 = 0

Taking the square root of both sides of this equation yields

y = 0

Hence, the answer is (A).

 

Example 9:
If [x] denotes the area of a square with sides of length x, then which of the following is equal to [9] ÷ [3] ?


(A) [2]
(B) [3]
(C) [16]
(D) [27]
(E) [81]


The area of a square with sides of length x is x2. This formula yields [9] ÷ [3] = 92 ÷ 32 = 81 ÷ 9 = 9 . Now, [3] = 32 = 9. Hence, the answer is (B).





Test Your Skills Now!
Take a Quiz now
Reviewer Name