GMAT Math : Algebra

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Example Question #868 :Data Sufficiency Questions

Letandbe functions, the domain of both of which is the set of real numbers. Let M < 0 .

True or false: $\left ( f \circ g\right )(M) > 0$

Statement 1: The range of我s the set $(0, \infty )$

Statement 2: The range of我s the set $(- \infty, 0 )$

Possible Answers:

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

BOTH statements TOGETHER are insufficient to answer the question.

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

声明单独是充分的回答question.

Correct answer:

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Explanation:

$\left ( f \circ g\right )(M) = f (g (M))$ , so we need to determine whether f (g (M)) > 0 .

From Statement 1 alone, since the range of我s $(0, \infty )$ - that is, the set of all positive numbers, then regardless of the value of g(M) ,

$\left ( f \circ g\right )(M) = f (g (M)) > 0$ .

Therefore, Statement 1 alone yields an affirmative answer to the question.

From Statement 2 alone, regardless of the value or, g(M) < 0 , but we do not know the value or range of values of f(g(M)) . Statement 2 alone is unhelpful.

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Example Question #869 :Data Sufficiency Questions

Is a given relation a function?

Statement 1: The domain of the relation is $\left \{ 1, 2, 3, 4, 5\right \}$ .

Statement 2: The range of the relation is $\left \{ 1, 2, 3, 4, 5\right \}$ .

Possible Answers:

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

BOTH statements TOGETHER are insufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

声明单独是充分的回答question.

Correct answer:

BOTH statements TOGETHER are insufficient to answer the question.

Explanation:

The two statements together do not prove or disprove the relation to be a function.

The relations defined by the sets of points

$\left \{ (1,1), (2,2), (3,3), (4,4), (5,5) \right \}$ .

and

$\left \{ (1,1), (1,2), (2,2), (3,3), (4,4), (5,5) \right \}$

have the domain and range given in the statements, but the former is a function, since each domain element is matched with exaclty one element, and the latter is not a function, since domain element 1 is matched with two different range elements.

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Example Question #870 :Data Sufficiency Questions

A relation comprises ten ordered pairs. Is it a function?

Statement 1: Its domain is $\left \{ 1,3,5, 7, 9, 11\right \}$ .

Statement 2: The line x= 5 passes through its graph twice.

Possible Answers:

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

BOTH statements TOGETHER are insufficient to answer the question.

声明单独是充分的回答question.

Correct answer:

声明单独是充分的回答question.

Explanation:

If Statement 1 alone is assumed, then, since there are only six domain elements and ten points in the relation, at least one of the domain elements must match with more than one range element. This forces the relation to not be a function.

If Statement 2 alone is assumed, then, since x= 5 我s a vertical line that passes through the graph twice, the relation fails the vertical line test and is therefore not a function.

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Example Question #41 :Dsq: Understanding Functions

Defineandto be functions. Does $f \circ g$ have an inverse?

Statement 1:has an inverse.

Statement 2:has an inverse.

Possible Answers:

BOTH statements TOGETHER are insufficient to answer the question.

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

声明单独是充分的回答question.

Correct answer:

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Explanation:

$\left (f \circ g \right )(x) = f \left ( g\left ( x\right )\right )$ . This defintion will come into play here.

A functionhas an inverse if and only if it is "one-to-one" - that is, if

f(a) = f(b) 我f and only if a = b .

Assume statement 1 only.

$f \left ( g\left ( a\right )\right ) = f \left ( g\left ( b\right )\right )$ 我f and only if $g\left ( a\right ) =g(b)$ . However, since it is not known whetherhas an inverse, it is possible for $g\left ( a\right ) =g(b)$ with either a = b or $a \ne b$ . Transitively, it is possible for $f \left ( g\left ( a\right )\right ) = f \left ( g\left ( b\right )\right )$ with either a = b or $a \ne b$ .

Assume Statement 2 only.

$g\left ( a\right ) =g(b)$ 我f and only if a = b . But it is possible for $f \left ( g\left ( a\right )\right ) = f \left ( g\left ( b\right )\right )$ with $g\left ( a\right ) =g(b)$ or $g\left ( a\right ) \ne g(b)$ - and, subsequently, with a = b or $a \ne b$ . Transitively, it is possible for $f \left ( g\left ( a\right )\right ) = f \left ( g\left ( b\right )\right )$ with either a = b or $a \ne b$ .

Assume both statements are true. Then

$f \left ( g\left ( a\right )\right ) = f \left ( g\left ( b\right )\right )$ 我f and only if $g\left ( a\right ) =g(b)$ , and $g\left ( a\right ) =g(b)$ 我f and only if a = b . Transitively,

$f \left ( g\left ( a\right )\right ) = f \left ( g\left ( b\right )\right )$ 我f and only if a = b .

Therefore,

$\left (f \circ g \right )(a) = \left (f \circ g \right )(b)$ 我f and only if a = b , and, subsequently, $f \circ g$ has an inverse.

The two statements together - but neither alone - lead to an answer.

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Example Question #42 :Dsq: Understanding Functions

Defineandto be functions on the real numbers. Does f-g have an inverse?

Statement 1:does not have an inverse.

Statement 2:does not have an inverse.

Possible Answers:

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

BOTH statements TOGETHER are insufficient to answer the question.

声明单独是充分的回答question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

Correct answer:

BOTH statements TOGETHER are insufficient to answer the question.

Explanation:

Examine these two examples.

Example 1: Let f(x) = g(x) = 1

Neither function has an inverse, since both functions pair allvalues with the samevalue, namely, 1.

(f-g)(x) = f(x) - g(x) = 1 - 1= 0

f-g does not have an inverse, since it pairs allvalues with the samevalue, namely, 0.

Therefore,andhave no inverse, and f-g has no inverse.

Example 2: Let $f(x) = x^{2}+x$ and $g(x) = x^{2}$ .

has no inverse, since $f(0) = 0^{2}+0 = 0$ and $f(-1) =(-1) ^{2} +(-1) = 1-1 = 0$ - that is,pairs at least twovalues with the samevalue.

has no inverse, since $g (1) = 1^{2} = 1$ and $g(-1 ) = (-1)^{2} = 1$ ; that is,pairs at least twovalues with the samevalue.

$(f-g)(x) = f(x) - g(x) = (x^{2}+x)-x^{2} = x$

f-g 我s the identity function, which has itself as an inverse.

This demonstrates that, ifanddo not have inverses, it is possible for f-g to have an inverse or to not have an inverse. Therefore, the two statements together are inconclusive.

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Example Question #43 :Dsq: Understanding Functions

Defineandto be functions on the set of real numbers. Does f+g have an inverse?

Statement 1:has an inverse.

Statement 2:has an inverse.

Possible Answers:

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

BOTH statements TOGETHER are insufficient to answer the question.

声明单独是充分的回答question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

Correct answer:

BOTH statements TOGETHER are insufficient to answer the question.

Explanation:

Examine these two examples.

Example 1: f(x) = 3x and g(x) = 2x .

f(x) = 3x can be proved to have an inverse as follows:

y= 3x

Switch the variables:

3y = x

$y = \frac{1}{3}x$

$f^{-1}(x) = \frac{1}{3}x$

Therefore,has an inverse; a smiliar proof shows thathas an inverse $g^{-1} (x)= \frac{1}{2}x$ .

(f+g)(x) = 3x + 2x = 5x , which can be similarly be shown to have inverse

$(f+g)^{-1}(x) = \frac{1}{5}x$ .

Example 2: f(x) = 3x and g(x) = -3x .

Again,andcan be shown to have inverses, $f^{-1}(x) = \frac{1}{3}x$ and $g^{-1}(x) =-\frac{1}{3}x$ .

. However,

(f+g)(x) = 3x + (-3x) = 0 .

This function has no inverse, since this function pairs multiple values ofwith the same value of, 0.

In both cases, bothandhave inverses, but in one case, f+g has an inverse, and in the other case, f+g does not. The two statements together are inconclusive.

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Example Question #44 :Dsq: Understanding Functions

Is a given relation a function?

Statement 1: The line x = 10 passes through its graph infinitely many times.

Statement 2: The line y= 10 passes through its graph infinitely many times.

Possible Answers:

BOTH statements TOGETHER are insufficient to answer the question.

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

声明单独是充分的回答question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

Correct answer:

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Explanation:

If Statement 1 alone is assumed, then, since x= 10 我s a vertical line that passes through the graph of the relation more than once, the relation fails the vertical line test, and the relation can be proved to not be a function.

However, there is no restriction on how many times a horizontal line such as y= 10 can pass through the graph of a relation for it to be or not to be a function. Statement 2 proves nothing either way.

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Example Question #45 :Dsq: Understanding Functions

Is a given relation a function?

Statement 1: Two of its ordered pairs have-coordinate.

Statement 2: Two of its ordered pairs have-coordinate.

Possible Answers:

BOTH statements TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient to answer the question.

Statement 2 ALONE is sufficient to answer the question, but Statement 1 ALONE is NOT sufficient to answer the question.

声明单独是充分的回答question.

BOTH statements TOGETHER are insufficient to answer the question.

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Correct answer:

Statement 1 ALONE is sufficient to answer the question, but Statement 2 ALONE is NOT sufficient to answer the question.

Explanation:

Statement 1 alone disproves that the relation is a function, since in a function, no-coordinate can be paired with two-coordinates. However, statement 2 alone does not prove or disprove the relation to be a function, since it is permissible for two-coordinates to be paired with the same-coordinate.

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Example Question #46 :Dsq: Understanding Functions

The volume of a right circular cylinder is $\pi ^{x}$ ; the radius of its base is $\pi$ .

Give the height of the cylinder.

Possible Answers:

$\pi ^{\frac{1}{2}x- 1}$

$\pi ^{x-3}$

$\pi ^{\sqrt{x-2}}$

$\pi ^{\frac{1}{2}x}$

$\pi ^{\frac{1}{3}x }$

Correct answer:

$\pi ^{x-3}$

Explanation:

The volume of a right circular cylinder, given base with radiusand given height, is

$V = \pi r^{2}h$ .

Setting $V = \pi ^{x}$ and $r = \pi$ :

$\pi ^{x} = \pi \cdot \pi^{2} \cdot h$

$\pi ^{x} = \pi^{2+1} \cdot h$

$\pi ^{x} = \pi^{3} \cdot h$

$\frac{ \pi ^{x}}{ \pi^{3}} =\frac{ \pi^{3} \cdot h }{ \pi^{3}}$

$h = \pi ^{x-3}$

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# 47例问题:Dsq: Understanding Functions

The volume of a right circular cylinder is $\pi ^{x}$ ; its height is $\pi$ .

Give the radius of a base of the cylinder.

Possible Answers:

$\pi ^{x-3}$

$\pi ^{\sqrt{x-2}}$

$\pi ^{\frac{1}{3}x }$

$\pi ^{\frac{1}{2}x- 1}$

$\pi ^{\frac{1}{2}x}$

Correct answer:

$\pi ^{\frac{1}{2}x- 1}$

Explanation:

The volume of a right circular cylinder, given base with radiusand given height, is

$V = \pi r^{2}h$ .

Setting $V = \pi ^{x}$ and $h= \pi$ :

$\pi ^{x} = \pi \cdot r ^{2} \cdot \pi$

$\pi ^{x} = \pi^{2} \cdot r ^{2}$

$\frac{ \pi ^{x}}{\pi^{2} } = \frac{\pi^{2} \cdot r ^{2}}{\pi^{2} }$

$r ^{2} = \pi ^{x-2}$

$\left (r ^{2} \right )^{\frac{1}{2}}= \left (\pi ^{x-2} \right )^{\frac{1}{2}}$

$r ^{2 \cdot \frac{1}{2}}= \pi ^{(x-2)\cdot \frac{1}{2}}$

$r = \pi ^{\frac{1}{2}x- 1}$

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GMAT Math : Algebra

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #868 :Data Sufficiency Questions

Example Question #869 :Data Sufficiency Questions

Example Question #870 :Data Sufficiency Questions

Example Question #41 :Dsq: Understanding Functions

Example Question #42 :Dsq: Understanding Functions

Example Question #43 :Dsq: Understanding Functions

Example Question #44 :Dsq: Understanding Functions

Example Question #45 :Dsq: Understanding Functions

Example Question #46 :Dsq: Understanding Functions

# 47例问题:Dsq: Understanding Functions

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