DSQ: Understanding functions - GMAT Quantitative

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Question

True or false: $\left { a_{n} \right }$ . is an arithmetic sequence.

Statement 1: $a_{4} - a_{3} = a_{2} -a_{1}$

Statement 2: $a_{6}- a_{5} = a_{8} - a_{7}$

Answer

An arithmetic sequence is one in which the difference of each term in the sequence and the one preceding is constant.

The two statements together demonstrate that two such differences are equal to each other, and that two other such differences are equal to each other. No information, however, is given about any other differences, of which there are infinitely many. Therefore, the question of whether the sequence is arithmetic or not is unresolved.

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Question

Define and .

Is it true that $f = g^{-1}$ ?

Statement 1:

Statement 2: $B=\frac{1}{2}$

Answer

For $f = g^{-1}$ to be each other's inverse, it must be true that

$(f\circ g)(x) =x$ and $(g\circ f)(x) =x$

We can look at the first condition.

$(f\circ g)(x) =x$

$(f\circ g)(x) =f(g(x)) = f(Bx+5)=4(Bx+5)+A=4Bx+20+A= x$

For this to be true, it must hold that:

$B=\frac{1}{4}$

and

Since both statements violate these conditions, it is impossible for $f = g^{-1}$ , even if you are only given one of them.

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

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Question

The first term of an arithmetic sequence is 100. What is the second term?

Statement 1: The sum of the third and fourth terms is 320.

Statement 2: The tenth term subtracted from the twelfth term yields a difference of 48.

Answer

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.

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Question

Define .

Evaluate $(f \circ g) (9)$ .

Statement 1:

Statement 2:

Answer

If you know that , then you can calculate:

$(f \circ g) (9) = f (; g(9); ) = f(8) = 4\cdot8 -9 = 23$

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

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Question

Define an operation $\diamond$ as follows:

For any real numbers ,

$A \diamond B = A^{2}-B$

Evaluate $5 \diamond 0 + 0 \diamond 5$

Answer

$5 \diamond 0 = 5^{2}-0 = 25$

$0\diamond 5 = 0^{2}-5 = -5$

$5 \diamond 0 + 0 \diamond 5 = 25 + (-5) = 20$

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Question

This relation has the following five ordered pairs: is it a function?

$\begin{matrix} \underline{x}&\underline{y}\ 1& 6\ 2& 4\ 3& 7\ 4&B \ A& 9 \end{matrix}$

Statement 1:

Statement 2:

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 true. Statement 2 is irrelevant; it is possible for more than one -coordinate to be matched with the same -coordinate in a function.

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Question

This relation has five different ordered pairs: is it a function?

$\begin{matrix} \underline{x}&\underline{y}\ 1& 6\ 2& 4\ 3& 7\ 4&B \ A& 9 \end{matrix}$

Statement 1:

Statement 2:

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.

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Question

This relation has five different ordered pairs: is it a function?

$\begin{matrix} \underline{x}&\underline{y}\ 1& A\ 2& A\ 3& A\ B&A + 1 \ 5& A + 2 \end{matrix}$

Statement 1:

Statement 2:

Answer

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.

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Question

$f ^{-1} (A) = 40$

Evaluate .

Statement 1: The graph of includes the point .

Statement 2:

Answer

If $f ^{-1} (A) = 40$ , 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 $f ^{-1} (40) = 19$ .

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Question

$f ^{-1} (A) = 27$

Evaluate .

Statement 1: The graph of includes the point .

Statement 2:

Answer

Both statements are equivalent to the statement $f ^{-1} (27) = 15$ . This is of no help to us.

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Question

What is the first term of the geometric sequence?

Statement 1: The sum of the second and third terms is 90.

Statement 2: The sum of the third and fourth terms is 450.

Answer

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

$A,AR,AR^{2},AR^{3}$ .

The two statements below are equivalent to $AR+ AR^{2} = 90$ and $AR^{2}+AR^{3} = 450$ , respectively. Neither, alone, will help you figure out or . If you know both, you can use algebra to deduce their values:

$AR^{2}(1+R) = 450$

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:

$A \cdot 5 \cdot (1 + 5) = 90$

$A \cdot 30 = 90$

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Question

What is the first term of a geometric sequence?

Statement 1: The product of the second and third terms is 4,096.

Statement 2: The product of the first and fourth terms is 4,096.

Answer

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

$A,AR,AR^{2},AR^{3}$

The two statements are equivalent to the equations $AR \cdot AR^{2} = 4.096$ and $A \cdot AR^{3} = 4,096$ . But they turn out to be equivalent as can be seen here:

$AR \cdot AR^{2} = 4096$

$A \cdot A \cdot R \cdot R^{2} = 4096$

$A^{2}R^{3} = 4.096$

$A \cdot AR^{3} = 4,096$

$A \cdot AR^{3} = 4,096$

$A^{2}R^{3} = 4,096$

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

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Question

$\left \lfloor x\right \rfloor$ is defined to be the greatest integer less than or equal to .

Evaluate $\left \lfloor \frac{A}{B} \right \rfloor$

Statement 1:

Statement 2:

Answer

Statement 1 is not enough to answer this question.

Case 1:

Then $\left \lfloor \frac{A}{B} \right \rfloor = \left \lfloor \frac{1}{5} \right \rfloor = 0$

Case 2:

Then $\left \lfloor \frac{A}{B} \right \rfloor = \left \lfloor \frac{-1}{3} \right \rfloor = \left \lfloor - \frac{1}{3} \right \rfloor = -1$

If Statement 2 is true, however:

$\left \lfloor \frac{A}{B} \right \rfloor = \left \lfloor \frac{A}{2A} \right \rfloor = \left \lfloor \frac{1}{2} \right \rfloor =0$

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Question

$\left \lfloor x\right \rfloor$ is defined to be the greatest integer less than or equal to .

Evaluate $\left \lfloor A^{2} + B \right \rfloor$ .

Statement 1:

Statement 2:

Answer

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

Example 1:

$\left \lfloor A^{2} + B \right \rfloor = \left \lfloor 0.5^{2} + 0.5 \right \rfloor = \left \lfloor 0.25 + 0.5 \right \rfloor= \left \lfloor 0.75\right \rfloor = 0$

Example 2:

$\left \lfloor A^{2} + B \right \rfloor = \left \lfloor 0.8^{2} + 0.8 \right \rfloor = \left \lfloor 0.64 + 0.8 \right \rfloor= \left \lfloor 1.44\right \rfloor = 1$

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Question

Does have an inverse?

Statement 1: There exists only one horizontal line that intersects the graph of more than once.

Statement 2:

Answer

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.

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Question

Does the function have an inverse?

Statement 1:

Statement 2:

Answer

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.

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Question

Is an odd function?

Statement 1: It is a polynomial of degree 3.

Statement 2: Its graph is symmetrical with respect to the origin.

Answer

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: $g(x) = x^{3}$

$g(-x) = (-x)^{3} = (-1)^{3} \cdot x^{3} = -1 x^{3} = -g(x)$

so this is odd.

Case 2: $g(x) = x^{3} -1$

$g(1) = 1^{3}-1 = 1-1 =0$

$g(-1) = (-1)^{3} -1= -1-1 = -2$

, so the function is not odd.

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Question

Is an even function, an odd function, or neither?

Statement 1: The graph of is symmetric with respect to the origin.

Statement 2: The graph of is a line through the origin.

Answer

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.

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Question

$\left \lceil x \right \rceil$ is defined as the least integer greater than or equal to .

Evaluate: $\left \lceil x^2 + y^2 \right \rceil$

Statement 1: $0 < x < \frac{1}{2}$

Statement 2: $0 < y < \frac{1}{2}$

Answer

Statement 1 is insufficient to calculate $\left \lceil x^2 + y^2 \right \rceil$ .

For example, if $x = \frac{1}{4 }$ and ,

$\left \lceil x^2 + y^2 \right \rceil = \left \lceil \left ( \frac{1}{4} \right )^2 + 1^2 \right \rceil = \left \lceil1 \frac{1}{16} \right \rceil = 2$

If $x = \frac{1}{4 }$ and ,

$\left \lceil x^2 + y^2 \right \rceil = \left \lceil \left ( \frac{1}{4} \right )^2 + 2^2 \right \rceil = \left \lceil4 \frac{1}{16} \right \rceil = 5$

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

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

$0 < x < \frac{1}{2} \Rightarrow 0 < x ^{2}< \frac{1}{4}$

$0 < y< \frac{1}{2} \Rightarrow 0 < y ^{2}< \frac{1}{4}$

$0=0+0 <x^{2}+y^{2}< \frac{1}{4} + \frac{1}{4} = \frac{1}{2 }< 1$

Since $0<x^{2}+y^{2}< 1$ ,

$\left \lceil x^2 + y^2 \right \rceil = 1$ .

We have a definitive answer.

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Question

Is an odd function?

Statement 1: For each positive ,

Statement 2:

Answer

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.

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