GMAT Math : Functions/Series

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Example Questions

Example Question #11 : Functions/Series

Define the operation $\Theta$ as follows:

$a \; \Theta \; b = ab + 100$

Solve for : $4\; \Theta \; x = 72$

Possible Answers:

x = -7

x = -18

x = 18

x = -9

x = 9

Correct answer:

x = -7

Explanation:

$4\; \Theta \; x = 72$

4x + 100 = 72

4x + 100 -100 = 72-100

4x = -28

$4x \div 4 = -28 \div 4$

x = -7

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Example Question #12 : Understanding Functions

Define $f(x) = A (x-B)^{3}$ , where $A \neq 0, B \neq 0$ .

Evaluate $f^{-1} (1)$ in terms of and .

Possible Answers:

$f^{-1} (1)= B - \sqrt[3]{A}$

$f^{-1} (1)= B - \frac{1}{\sqrt[3]{A}}$

$f^{-1} (1)= B + \frac{1}{\sqrt[3]{A}}$

$f^{-1} (1)= -B + \frac{1}{\sqrt[3]{A}}$

$f^{-1} (1)= B + \sqrt[3]{A}$

Correct answer:

$f^{-1} (1)= B + \frac{1}{\sqrt[3]{A}}$

Explanation:

This is equivalent to asking for the value of for which f(x) = 1 , so we solve for in the following equation:

f(x) = 1

$A (x-B)^{3} =1$

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

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

$\sqrt[3]{(x-B)^{3} }=\sqrt[3]{\frac{1}{A}}$

$x-B =\sqrt[3]{\frac{1}{A}}$

$x-B = \frac{1}{\sqrt[3]{A}}$

$x-B + B = B + \frac{1}{\sqrt[3]{A}}$

$x = B + \frac{1}{\sqrt[3]{A}}$

Therefore, $f^{-1} (1)= B + \frac{1}{\sqrt[3]{A}}$ .

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Example Question #13 : Understanding Functions

Define an operation $\odot$ as follows:

For any real numbers a,b ,

$a \odot b = (a-1) (b-1)$

Evaluate $\left (5 \odot 5 \right )\odot 5$ .

Possible Answers:

100

Correct answer:

Explanation:

$\left (5 \odot 5 \right )\odot 5$

$= \left [(5 -1) (5 -1) \right ] \odot 5$

$= (4 \cdot 4 ) \odot 5$

$= 16 \odot 5$

$= \left (16-1 \right ) \left (5-1 \right )$

$= 15 \cdot4 = 60$

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Example Question #14 : Understanding Functions

An infinite sequence begins as follows:

1, 2, -3, 4, 5, -6, 7,8,-9 ...

Assuming this pattern continues infinitely, what is the sum of the 1000th, 1001st and 1002nd terms?

Possible Answers:

1,000

1,001

999

Correct answer:

999

Explanation:

This can be seen as a sequence in which the Nth term is equal to if is not divisible by 3, and otherwise. Since 1,000 and 1,001 are not multiples of 3, but 1,002 is, the 1000th, 1001st, and 1002nd terms are, respectively,

1000,1001,-1002

and their sum is

$1000+ 1001+ \left (-1002 \right ) = 999$

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Example Question #15 : Understanding Functions

Define $f(x) = \sqrt{x}$ . What is $\left (f \circ f \right )(x)$ ?

Possible Answers:

$\left (f \circ f \right )(x) = 2\sqrt{x}$

$\left (f \circ f \right )(x) = \sqrt[4]{x}$

$\left (f \circ f \right )(x) = x^{2}$

$\left (f \circ f \right )(x) = \sqrt[3]{x}$

$\left (f \circ f \right )(x) = x$

Correct answer:

$\left (f \circ f \right )(x) = \sqrt[4]{x}$

Explanation:

This can best be solved by rewriting $\sqrt{ x}$ as $x^{\frac{1}{2}}$ and using the power of a power property.

$\left (f \circ f \right )(x) = f (f(x)) = \left [f(x) \right]^{\frac{1}{2}} =\left ( x^{\frac{1}{2}} \right )^{\frac{1}{2}}=x^{\frac{1}{2}\cdot \frac{1}{2} } =x^{\frac{1}{4} } =\sqrt[4]{x}$

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Example Question #12 : Functions/Series

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

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

Define $g (x) = \left \lfloor x+ 0.5 \right \rfloor - \left \lceil x - 0.5 \right \rceil$ .

Evaluate g (3.9 ) .

Possible Answers:

g (3.9) = 1

g (3.9) = 0

g (3.9) = -2

g (3.9) = -1

g (3.9) = 2

Correct answer:

g (3.9) = 0

Explanation:

$g (x) = \left \lfloor x+ 0.5 \right \rfloor - \left \lceil x - 0.5 \right \rceil$

$g (3.9) = \left \lfloor 3.9+ 0.5 \right \rfloor - \left \lceil 3.9 - 0.5 \right \rceil$

$g (3.9) = \left \lfloor 4.4 \right \rfloor - \left \lceil 3.4 \right \rceil$

g (3.9) = 4 - 4 = 0

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Example Question #17 : Understanding Functions

Define an operation $\diamond$ as follows:

For any real numbers A,B ,

$A \diamond B = A^{2}+B^{2}$ .

Evaluate $\left (5 \diamond 0 \right ) + \left (-5 \diamond 0 \right )$ .

Possible Answers:

Correct answer:

Explanation:

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

$-5 \diamond 0 = \left (-5 \right )^{2} - 0 = 25$

$\left (5 \diamond 0 \right ) + \left (-5 \diamond 0 \right ) = 25 + 25 = 50$

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Example Question #18 : Understanding Functions

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

Define $f(x) = \left \lceil x ^{2}\right \rceil$ .

Define $g(x) = \left \lceil x + 0.5\right \rceil - 0.5$ .

Evaluate $\left ( f \circ g \right )(4)$ .

Possible Answers:

Correct answer:

Explanation:

$\left ( f \circ g \right )(4) = f(g(4))$

First, evaluate g(4) :

$g(x) = \left \lceil x + 0.5\right \rceil - 0.5$

$g(4) = \left \lceil 4 + 0.5 \right \rceil - 0.5$

$g(4) = \left \lceil 4.5 \right \rceil - 0.5$

g(4) =5- 0.5 = 4.5

Now, evaluate f (4.5) :

$f(x) = \left \lceil x ^{2}\right \rceil$

$f(4.5) = \left \lceil 4.5 ^{2}\right \rceil$

$f(4.5) = \left \lceil 20.25 \right \rceil = 21$

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Example Question #19 : Understanding Functions

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

Solve for : $\left \lfloor 3y - 1 \right \rfloor = 7$

Possible Answers:

$\left [ \frac{8}{3}, 3 \right )$

$\left (\frac{7}{3}, \frac{8}{3} \right ]$

$\left ( \frac{8}{3}, 3 \right )$

$\left [ \frac{7}{3}, \frac{8}{3} \right )$

$\left (\frac{8}{3}, 3 \right ]$

Correct answer:

$\left [ \frac{8}{3}, 3 \right )$

Explanation:

$\left \lfloor 3y - 1 \right \rfloor = 7$

means that the greatest integer less than or equal to 3y - 1 is 7. The equivalent statement is

$7 \leq 3y - 1 < 8$ .

Solve for as follows:

$7 + 1 \leq 3y - 1 + 1 < 8 + 1$

$8 \leq 3y < 9$

$8 \div 3 \leq 3y \div 3 < 9 \div 3$

$\frac{8}{3} \leq y< 3$

or, in interval form, $\left [ \frac{8}{3}, 3 \right )$

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Example Question #1242 : Problem Solving Questions

Which of the following pairs of statements is sufficient to prove that f(x) does not have an inverse?

Possible Answers:

f (x) is not defined for x = 2 , f (x) is not defined for x = -2 ,

f(4) = 3,f(-4) = 3

f(1) = 10, f (-1) = -10

None of these pairs of statements would be sufficient to prove that does not have an inverse.

$f(5) = 7,f(-5) = \frac{1}{7}$

Correct answer:

f(4) = 3,f(-4) = 3

Explanation:

For a function to have an inverse, no -coordinate can be paired with more than one -coordinate. Of our choices, only

f(4) = 3,f(-4) = 3

causes this to happen.

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GMAT Math : Functions/Series

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #11 : Functions/Series

Example Question #12 : Understanding Functions

Example Question #13 : Understanding Functions

Example Question #14 : Understanding Functions

Example Question #15 : Understanding Functions

Example Question #12 : Functions/Series

Example Question #17 : Understanding Functions

Example Question #18 : Understanding Functions

Example Question #19 : Understanding Functions

Example Question #1242 : Problem Solving Questions

All GMAT Math Resources

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