Functions/Series - GMAT Quantitative

For to be each other's inverse, it must be true that

and

We can look at the first condition.

For this to be true, it must hold that:

and

Since both statements violate these conditions, it is impossible for , even if you are only given one of them.

The answer is that either statement alone is sufficient to answer the question.

Let be the common difference of the sequence. Then the third and fourth terms are, respectively, and . If their sum is 320, then

If the difference between the twelfth term and tenth term is 24, then

From either statement, the common difference can be calculated, then added to 100 to get the second term, 124.

If you know that , then you can calculate:

Knowing the -values of that are paired with -value 9, however, is neither necessary nor useful here.

To prove that a relation is a function, you must prove that no -coordinate is matched with more than one -coordinate. Statement 1 proves that this is true. Statement 2 is irrelevant; it is possible for more than one -coordinate to be matched with the same -coordinate in a function.

Statement 1 is insufficient to calculate .

For example, if and ,

If and ,

A reciprocal argument can be used to show Statement 2 is also insufficient.

From both statements together, however, we know the following:

Since ,

.

We have a definitive answer.

To prove that a relation is a function, you must prove that no -coordinate is matched with more than one -coordinate. Statement 1 proves that this is false, since 3 is now matched with both 7 and 9. Statement 2 is irrelevant, since it does not prove or disprove this condition.

To prove that a relation is a function, you must prove that no -coordinate is matched with more than one -coordinate. Statement 2 proves that this is false, since 5 is now matched with both and , which are different numbers regardless of . Statement 1 is irrelevant, since it does not prove or disprove this condition.

If , then , so this question is equivalent to evaluating .

If the ordered pair is on the graph of , then , so .

Knowing that is of no help, as this just tells us .

Both statements are equivalent to the statement . This is of no help to us.

Let be the first term and be the common ratio. Then the first four terms are:

.

The two statements below are equivalent to and , respectively. Neither, alone, will help you figure out or . If you know both, you can use algebra to deduce their values:

Divide both sides of the first equation by both sides of the second:

Substitute this value into either equation. We'll use the 2nd equation:

Let be the first term and be the common ratio. Then the first four terms are:

The two statements are equivalent to the equations and . But they turn out to be equivalent as can be seen here:

This (common) statement alone is not enough to allow us to calculate .

Statement 1 is not enough to answer this question.

Case 1:

Then

Case 2:

Then

If Statement 2 is true, however:

Even if you know both statements, you cannot answer this question with certainty.

Example 1:

Example 2:

Statement 1 is enough to disprove that has an inverse; fails the horizotal line test, which states that for to have an inverse, no horizontal line can intersect its graph more than once.

Statement 2 is also enough to disprove that has an inverse, since for to have an inverse, no more than one -coordinate can be matched with the same -coordinate.

has an inverse if and only if no two values of map into the same value of . Neither statement is sufficient to prove or disprove this, Both statements together, however, demonstrate that there are two values of , , that map into the same value , so has no inverse.

is odd if and only if, for any value of in its domain, .

It is odd if and only if its graph is symmetrical to the origin, so Statement 2 proves that is an odd function.

Statement 1 does not provide enough information, however; we can give at least one third-degree polynomial that is odd and one that is not:

Case 1:

so this is odd.

Case 2:

, so the function is not odd.

A function is odd if and only if its graph has symmetry with respect to the origin, so Statement 1 proves odd.

A function is odd if and only if, for each in the domain, . A linear function through the origin - that is, one with -intercept 0 - can be written as for some ; since

,

we know is odd.

By definition, for to be odd, then it must hold that for every value of in its domain.

If for every positive , as stated in Statement 1, then for all positive . Equivalently, for all negative . But this does not give us any information about the behavior of at 0, so the picture is incomplete.

But Statement 2 alone proves is not odd. This is because if is odd, the definition forces , which forces . Statement 2 contradicts this.

If , then both statements are true. But both statements can also be true in some cases where .

For example, if , then and .

The two together are inconclusive.