GMAT Math : Functions/Series

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Example Question #41 : Understanding Functions

Define $f(x)= x ^{3} - 125$ .

What is $f ^{-1}$ ?

Possible Answers:

$f ^{-1} (x)= \sqrt[3]{x+125}$

$f ^{-1} (x)= \sqrt[3]{x}+5$

$f ^{-1} (x)= \sqrt[3]{x}+125$

$f ^{-1} (x)=\frac{1}{x ^{3} - 125}$

$f ^{-1} (x)= \sqrt[3]{x+5}$

Correct answer:

$f ^{-1} (x)= \sqrt[3]{x+125}$

Explanation:

Replace f(x) with .

$y= x ^{3} - 125$

Now exchange and and solve for in the new expression:

$x= y^{3} - 125$

$x + 125= y^{3} - 125 + 125$

$x + 125= y^{3}$

$\sqrt[3]{x + 125 }= y$

$f ^{-1} (x)= \sqrt[3]{x+125}$

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Example Question #41 : Functions/Series

Define an operation $\oplus$ on the set of positive integers as follows:

$a \oplus b = ab$ if and are both odd or both even; otherwise, $a \oplus b = 0$ .

Evaluate:

$\left ( 5 \oplus 5\right ) + \left ( 6 \oplus 6\right )$

Possible Answers:

6 1

Correct answer:

6 1

Explanation:

Both $5 \oplus 5$ and $6 \oplus 6$ can be defined using the first part of the definition of $a \oplus b$ , as in the first case, both numbers are odd, and in the second case, both are even.

$a \oplus b = ab$

$5 \oplus 5 = 5 \times 5 = 25$

$6 \oplus 6 =6 \times 6 = 36$

Add:

$\left ( 5 \oplus 5\right ) + \left ( 6 \oplus 6\right ) = 25 + 36 = 61$

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Example Question #43 : Understanding Functions

Define $f (x) = \frac{5}{4-x}$ .

What is $f ^{-1}$ ?

Possible Answers:

$f^{-1}(x)=\frac{ 4x -5 }{x }$

$f^{-1}(x)=\frac{5-x }{4}$

$f^{-1}(x)=\frac{ 5x -4 }{x }$

$f^{-1}(x)=\frac{ 4-x }{5}$

$f^{-1}(x)=\frac{ 4x -1 }{5x }$

Correct answer:

$f^{-1}(x)=\frac{ 4x -5 }{x }$

Explanation:

Replace f(x) with .

$y = \frac{5}{4-x}$

Now exchange and , and solve for in the new expression:

$x = \frac{5}{4-y}$

$x \left ( 4-y \right ) = \frac{5}{4-y} \left ( 4-y \right )$

$x \left ( 4-y \right ) = 5$

$\frac{x \left ( 4-y \right ) }{x }=\frac{ 5 }{x }$

$4-y=\frac{ 5 }{x }$

$4-y +y -\frac{ 5 }{x } =\frac{ 5 }{x }+y -\frac{ 5 }{x }$

$4-\frac{ 5 }{x } =y$

$f^{-1}(x)= 4-\frac{ 5 }{x } =\frac{ 4x }{x } -\frac{ 5 }{x } = \frac{ 4x -5 }{x }$

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Example Question #41 : Functions/Series

Define $f (x) = \frac{1}{7} x- 13$ .

What is $f ^{-1}$ ?

Possible Answers:

$f ^{-1} (x) = -\frac{1}{7} x + 91$

$f ^{-1} (x) = 7 x + 91$

$f ^{-1} (x) = - 7 x + 91$

$f ^{-1} (x) = 7 x + 13$

$f ^{-1} (x) = -\frac{1}{7} x + 13$

Correct answer:

$f ^{-1} (x) = 7 x + 91$

Explanation:

Replace f(x) with .

$y = \frac{1}{7} x- 13$

Now exchange and and solve for in the new expression:

$x= \frac{1}{7} y- 13$

$x + 13 = \frac{1}{7} y- 13 + 13$

$x + 13 = \frac{1}{7} y$

$\left ( x + 13 \right ) \cdot 7 = \frac{1}{7} y \cdot 7$

7 x + 91 = y

Therefore,

$f ^{-1} (x) = 7 x + 91$ .

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Example Question #45 : Understanding Functions

Define an operation $\bigstar$ on two real numbers as follows:

For all real a,b ,

$a \bigstar b = a+ \sqrt{\left |b \right |}$ .

Evaluate:

$49 \bigstar (-64)$

Possible Answers:

Correct answer:

Explanation:

$a \bigstar b = a+ \sqrt{\left |b \right |}$

$49 \bigstar (-64) = 49 + \sqrt{\left | -64 \right |} = 49 + \sqrt{64 } = 49 + 8 = 57$

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Example Question #46 : Understanding Functions

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

Define $f (x) = \left \lfloor x ^{2} \right \rfloor - 4 x + 7$

Evaluate $f \left ( \frac{7}{5} \right )$

Possible Answers:

$f \left ( \frac{7}{5} \right ) = 3 \frac{2}{5}$

$f \left ( \frac{7}{5} \right ) = 3$

$f \left ( \frac{7}{5} \right ) = 2 \frac{2}{5}$

$f \left ( \frac{7}{5} \right ) = 1 \frac{2}{5}$

$f \left ( \frac{7}{5} \right ) = 2$

Correct answer:

$f \left ( \frac{7}{5} \right ) = 2 \frac{2}{5}$

Explanation:

$f (x) = \left \lfloor x ^{2} \right \rfloor - 4 x + 7$

$f \left ( \frac{7}{5} \right ) = \left \lfloor\left ( \frac{7}{5} \right )^{2} \right \rfloor - 4\cdot \frac{7}{5}+ 7$

$= \left \lfloor \frac{49}{25} \right \rfloor - \frac{28}{5}+ 7$

$= \left \lfloor 1 \frac{24}{25} \right \rfloor -5 \frac{3}{5}+ 7$

$=1 -5 \frac{3}{5}+ 7 = 2 \frac{2}{5}$

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Example Question #47 : Understanding Functions

$f (x) = \left\{\begin{matrix} 2x + 1 & x < 0 \\ 3x - 1 & x \geq 0 \end{matrix}\right.$

$\small g (x) = \left\{\begin{matrix} 3x - 1 & x < 0 \\ 4x +7 & x \geq 0 \end{matrix}\right.$

Evaluate $\small \small (f \circ g) (4) - (g \circ f) (-4)$ .

Possible Answers:

113

162

Correct answer:

Explanation:

$\small \small \small \small (f \circ g) (4) = f (g (4)) = f (4 \cdot 4 + 7) = f (23) = 3 \cdot 23 - 1 = 68$

$\small \small \small \small (g \circ f) (-4) = g (f (-4)) = g (2 \cdot (-4) + 1)$ $\small \small = g (-7) = 3 (-7) -1 = -22$

$\small \small \small \small (f \circ g) (4) - (g \circ f) (-4) = 68 - (-22) = 90$

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Example Question #48 : Understanding Functions

For any real a,b , define $a\odot b =(1-a)(1-b)$ .

For what value or values of would $x\odot x = 4$ ?

Possible Answers:

-2,2

No such value of exists.

$1- \sqrt{2}, 1+ \sqrt{2}$

-1, 3

$-\frac{1}{2}, \frac{1}{2}$

Correct answer:

-1, 3

Explanation:

For such an to exist, it must hold that $x\odot x = (1 -x )(1-x) =(1-x)^{2}=4$ .

Take the square root of both sides:

1-x = -2 or 1-x = 2

Case 1:

1-x = -2

-x = -3

x=3

Case 2:

1-x = 2

-x = 1

x=-1

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Example Question #49 : Understanding Functions

Define an operation $\ddagger$ on the set of real numbers as follows:

$a \ddagger b = 2ab - a - b$

Evaluate $4 \ddagger (3\ddagger 2 )$ .

Possible Answers:

Correct answer:

Explanation:

First, evaluate $3\ddagger 2$ by substituting a = 3,b= 2 :

$a \ddagger b = 2ab - a - b$

$3 \ddagger 2 = 2\cdot 3 \cdot 2 - 3 - 2 = 12 - 3 - 2 = 7$

Second, evaluate $4\ddagger 7$ in the same way.

$a \ddagger b = 2ab - a - b$

$4\ddagger 7 = 2 \cdot 4 \cdot 7 - 4 - 7 = 56 - 4 - 7 = 45$

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Example Question #50 : Understanding Functions

Define an operation $\bigstar$ as follows:

For any real a,b , $a \; \bigstar \; b = a^{2} +b^{2} -8$ .

For what value or values of is it true that $N \; \bigstar \; N = 0$ ?

Possible Answers:

N = 4

No such value of exists.

$N = -2 \textrm{ or } N = 2$

N = 2

$N = -4 \textrm{ or } N =4$

Correct answer:

$N = -2 \textrm{ or } N = 2$

Explanation:

Substitute a = N, b = N into the definition, and then set the expression equal to 0 to solve for :

$a \; \bigstar \; b = a^{2} +b^{2} -8$

$N \; \bigstar \; N = N^{2} +N^{2} -8 = 2N^{2} -8$

$2N^{2} -8 = 0$

$2N^{2} =8$

$N^{2} =4$

$N = -2 \textrm{ or } N = 2$

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GMAT Math : Functions/Series

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #41 : Understanding Functions

Example Question #41 : Functions/Series

Example Question #43 : Understanding Functions

Example Question #41 : Functions/Series

Example Question #45 : Understanding Functions

Example Question #46 : Understanding Functions

Example Question #47 : Understanding Functions

Example Question #48 : Understanding Functions

Example Question #49 : Understanding Functions

Example Question #50 : Understanding Functions

All GMAT Math Resources

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