Defined Functions
Defined functions are very common on the SAT, and at first 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 symbol and a property that defines the symbol. Some examples will illustrate.From the above definition, we know that xâˆ‡y = xy â€“ y.
 2/3
 1
 3/2
 2
 3
Most students who are unfamiliar with defined functions are unable to solve this problem.
Yet it is actually quite easy.
By the definition above, âˆ† merely squares the first term.
So, z âˆ† 2 = z^{2}, and z âˆ† 3 = z^{2}.
Forming the fraction yields .
Hence, the answer is (B).
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 â€“ 1)* ?
 k â€“ 1
 8k â€“ 4
 8k â€“ 1
First, we must determine whether 2k â€“ 1 is odd or even.
(It cannot be bothâ€”why?) To this end, let k = 1.
Then 2k â€“ 1 = 2 â€” 1 â€“ 1 = 1, which is an odd number.
Therefore, we use the bottomhalf of the definition given above.
That is, (2k â€“ 1)* = 4(2k â€“ 1) = 8k â€“ 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 donâ€™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 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.
 â€“2Ï€
 â€“1
 â€“Ï€
 2Ï€
 4Ï€
Method II:
We can rewrite this problem using ordinary function notation.
 â€“5
 0
 5
 10
 15
I. aâ—Šb = bâ—Ša
II.
III. (aâ—Šb)â—Šc = aâ—Š(bâ—Šc)
 I only
 II only
 III only
 I and II only
 II and III only
This prevents division by zero from occurring in the problem, otherwise if a = 0 and b = â€“1, then .
 xyz
 0
 2
 5
 7
 10
Setting x # y equal to â€“x yields

(xy)^{2} â€“ x + y^{2} = â€“x

Canceling â€“ x from both sides of the equation yields

(xy)^{2} + y^{2} = 0

Expanding the first term yields

x^{2}y^{2} + y^{2} = 0

Factoring out y^{2} yields

y^{2}(x^{2} + 1) = 0

Setting each factor equal to zero yields

y^{2} = 0 or x2 + 1 = 0

Now, x^{2} + 1 is greater than or equal to 1 (why?). Hence,

y^{2} = 0

Taking the square root of both sides of this equation yields

y = 0

Hence, the answer is (A).

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