GMAT Math : Understanding functions

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Example Questions

Example Question #11 : Understanding Functions

Define the operation $\Theta$ $\displaystyle \Theta$ as follows:

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

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

Possible Answers:

x = 18 $\displaystyle x = 18$

x = -7 $\displaystyle x = -7$

x = -9 $\displaystyle x = -9$

x = -18 $\displaystyle x = -18$

x = 9 $\displaystyle x = 9$

Correct answer:

x = -7 $\displaystyle x = -7$

Explanation:

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

$\displaystyle 4x + 100 = 72$

$\displaystyle 4x + 100 -100 = 72-100$

$\displaystyle 4x = -28$

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

x = -7 $\displaystyle x = -7$

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Example Question #12 : Understanding Functions

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

Evaluate $f^{-1} (1)$ $\displaystyle f^{-1} (1)$ in terms of $\displaystyle A$ and $\displaystyle B$.

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

$\displaystyle f(x) = 1$

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

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

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

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

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

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

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

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

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

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Example Question #13 : Understanding Functions

Define an operation $\odot$ $\displaystyle \odot$ as follows:

For any real numbers a,b $\displaystyle a,b$ ,

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

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

Possible Answers:

$\displaystyle 64$

$\displaystyle 60$

$\displaystyle 75$

$\displaystyle 96$

100 $\displaystyle 100$

Correct answer:

$\displaystyle 60$

Explanation:

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

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

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

$= 16 \odot 5$ $\displaystyle = 16 \odot 5$

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

$= 15 \cdot4 = 60$ $\displaystyle = 15 \cdot4 = 60$

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Example Question #14 : Understanding Functions

An infinite sequence begins as follows:

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

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

Possible Answers:

999 $\displaystyle 999$

1,001 $\displaystyle 1,001$

1,000 $\displaystyle 1,000$

$\displaystyle 3$

$\displaystyle 0$

Correct answer:

999 $\displaystyle 999$

Explanation:

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

$\displaystyle 1000,1001,-1002$

and their sum is

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

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Example Question #15 : Understanding Functions

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #11 : Understanding Functions

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

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

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

Evaluate $\displaystyle g (3.9 )$.

Possible Answers:

$\displaystyle g (3.9) = 1$

$\displaystyle g (3.9) = 0$

$\displaystyle g (3.9) = -1$

$\displaystyle g (3.9) = 2$

$\displaystyle g (3.9) = -2$

Correct answer:

$\displaystyle g (3.9) = 0$

Explanation:

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

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

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

$\displaystyle g (3.9) = 4 - 4 = 0$

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Example Question #17 : Understanding Functions

Define an operation $\diamond$ $\displaystyle \diamond$ as follows:

For any real numbers A,B $\displaystyle A,B$,

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

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

Possible Answers:

$\displaystyle -25$

$\displaystyle -50$

$\displaystyle 25$

$\displaystyle 50$

$\displaystyle 0$

Correct answer:

$\displaystyle 50$

Explanation:

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

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

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

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Example Question #18 : Understanding Functions

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

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

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

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

Possible Answers:

$\displaystyle 17$

$\displaystyle 25$

$\displaystyle 21$

$\displaystyle 20$

$\displaystyle 16$

Correct answer:

$\displaystyle 21$

Explanation:

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

First, evaluate g(4) $\displaystyle g(4)$:

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

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

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

$\displaystyle g(4) =5- 0.5 = 4.5$

Now, evaluate f (4.5) $\displaystyle f (4.5)$:

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

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

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

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Example Question #19 : Understanding Functions

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

Solve for $\displaystyle y$ as follows:

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

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

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

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

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

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Example Question #20 : Understanding Functions

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

Possible Answers:

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

$\displaystyle f(1) = 10, f (-1) = -10$

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

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

$\displaystyle f(4) = 3,f(-4) = 3$

Correct answer:

$\displaystyle f(4) = 3,f(-4) = 3$

Explanation:

For a function $\displaystyle f$ to have an inverse, no $\displaystyle x$-coordinate can be paired with more than one $\displaystyle y$-coordinate. Of our choices, only

$\displaystyle f(4) = 3,f(-4) = 3$

causes this to happen.

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GMAT Math : Understanding functions

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #11 : Understanding Functions

Example Question #12 : Understanding Functions

Example Question #13 : Understanding Functions

Example Question #14 : Understanding Functions

Example Question #15 : Understanding Functions

Example Question #11 : Understanding Functions

Example Question #17 : Understanding Functions

Example Question #18 : Understanding Functions

Example Question #19 : Understanding Functions

Example Question #20 : Understanding Functions

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