GMAT Math : Algebra

Study concepts, example questions & explanations for GMAT Math

Create An Account

All GMAT Math Resources

22 Diagnostic Tests 693 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #191 : Algebra

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.

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

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

Report an Error

Example Question #192 : Algebra

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 = 2

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

No such value of exists.

N = 4

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

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 :

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

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

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

$2N^{2} =8$

$N^{2} =4$

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

Report an Error

Example Question #193 : Algebra

Consider the function $f(x) = x^{3} - 2$ .

State whether this function is even, odd, or neither, and give the reason for your answer.

Possible Answers:

is odd because f(-x) = -f(x) for each value of in the domain.

is even because f(-x) = f(x) for each value of in the domain.

is even because it is a polynomial of degree 3.

is not odd, because there exists at least one value of for which $f(-x) \neq -f(x)$ ; is not even, because there exists at least one value of for which $f(-x) \neq f(x)$ .

is odd because it is a polynomial of degree 3.

Correct answer:

is not odd, because there exists at least one value of for which $f(-x) \neq -f(x)$ ; is not even, because there exists at least one value of for which $f(-x) \neq f(x)$ .

Explanation:

A function is odd if and only if f(-x) = -f(x) for each value of in the domain; it is even if and only if f(-x) = f(x) for each value of in the domain. To disprove a function is odd or even, we need only find one value of for which the appropriate statement fails to hold.

Consider x = 1 :

$f(x) = x^{3} - 2$

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

$f(x) = x^{3} - 2$

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

$f(-1 ) \neq - f(1)$ , so is not an odd function; $f(-1 ) \neq f(1)$ , so is not an even function.

Report an Error

Example Question #194 : Algebra

$f (x) = \left\{\begin{matrix} 1\textrm{ if } x< 0\\ x \textrm{ if } x \geq 0 \end{matrix}\right.$

$g (x) = \left\{\begin{matrix} 2x-5 \textrm{ if } x< 0\\ 2\; \; \; \; \; \; \; \; \textrm{ if } x \geq 0 \end{matrix}\right.$ .

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

Possible Answers:

Correct answer:

Explanation:

$(f \circ g) (-4) = f (g(-4))$

First we evaluate g (-4) . Since the parameter is negative, we use the first half of the definition of :

g (x) = 2x-5

$g (-4) = 2 \left ( -4\right )-5 = -8 - 5 = -13$

f (g(-4)) = f(-13) ; since the parameter here is again negative, we use the first half of the definition of :

f(-13) = 1

Therefore, $(f \circ g) (-4) = 1$ .

Report an Error

Example Question #195 : Algebra

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

Define $f (x) = \left \lfloor x ^{2} \right \rfloor - \left \lfloor 4 x \right \rfloor + 8$ .

Evaluate $f \left ( \frac{3}{5} \right )$ .

Possible Answers:

$f \left ( \frac{3}{5} \right ) = 8$

$f \left ( \frac{3}{5} \right ) = 6$

$f \left ( \frac{3}{5} \right ) = 9$

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

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

Correct answer:

$f \left ( \frac{3}{5} \right ) = 6$

Explanation:

$f (x) = \left \lfloor x ^{2} \right \rfloor - \left \lfloor 4 x \right \rfloor + 8$

$f \left ( \frac{3}{5} \right ) = \left \lfloor \left (\frac{3}{5} \right ) ^{2} \right \rfloor - \left \lfloor 4 \cdot \frac{3}{5}\right \rfloor + 8$

$= \left \lfloor \frac{9}{25} \right \rfloor - \left \lfloor \frac{12}{5}\right \rfloor + 8$

$= \left \lfloor \frac{9}{25} \right \rfloor - \left \lfloor 2 \frac{2}{5}\right \rfloor + 8$

=0 - 2 + 8 = 6

Report an Error

Example Question #191 : Algebra

If $f(x)=x^{2}-\sqrt{x}$ and g(x)=3x-2 , what is f(g(2)) ?

Possible Answers:

$10-3\sqrt{2}$

$9-\sqrt{3}$

Correct answer:

Explanation:

We start by finding g(2):

$g(2)=3\cdot2-2 = 6-2 = 4$

Then we find f(g(2)) which is f(4):

$f(4)=4^{2}-\sqrt{4}=16-2=14$

Report an Error

Example Question #55 : Functions/Series

Define two real-valued functions as follows:

$p (x)= x^{2}$

$q(x )= \left\{\begin{matrix} x+1 & x< 0\\ x-1 & x \ge 0 \end{matrix}\right.$

Determine $\left ( q \circ p\right )(x)$ .

Possible Answers:

$\left ( q \circ p\right )(x)= x^{2}- 1$

$\left ( q \circ p\right )(x)= x^{2}-2x + 1$

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} x^{2}+2x + 1 & x < 0\\ x^{2}-2x + 1&x \ge 0 \end{matrix}\right.$

$\left ( q \circ p\right )(x)= x^{2}+1$

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} x^{2}+ 1 & x < 0\\ x^{2}- 1&x \ge 0 \end{matrix}\right.$

Correct answer:

$\left ( q \circ p\right )(x)= x^{2}- 1$

Explanation:

$\left ( q \circ p\right )(x) = q(p(x))= q(x^{2})$ by definition. is piecewise defined, with one defintion for negative values of the domain and one for nonnegative values. However, $p(x)= x^{2}$ is nonnegative for all real numbers, so the defintion for nonnegative numbers, q(x )= x-1 , is the one that will always be used. Therefore,

$\left ( q \circ p\right )(x) = q(p(x))= q(x^{2}) = x^{2}-1$ for all values of .

Report an Error

Example Question #54 : Functions/Series

Define two real-valued functions as follows:

$p (x)= 9 - x^{2}$

$q(x )= \left\{\begin{matrix} 0 & x< 0\\ x & x \ge 0 \end{matrix}\right.$

Determine $\left ( q \circ p\right )(x)$ .

Possible Answers:

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} 9 - x^{2}& x < -3 \\ 0& -3 \le x \le 3 \\ 9 - x^{2}& x > 3 \end{matrix}\right.$

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} 0& x < -3 \\ 9 - x^{2}& -3 \le x \le 3 \\ 0& x > 3 \end{matrix}\right.$

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} 9 & x< 0\\ 9-x^{2} &x \ge 0 \end{matrix}\right.$

$\left ( q \circ p\right )(x) = 9 - x^{2}$

$\left ( q \circ p\right )(x) =0$

Correct answer:

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} 0& x < -3 \\ 9 - x^{2}& -3 \le x \le 3 \\ 0& x > 3 \end{matrix}\right.$

Explanation:

$\left ( q \circ p\right )(x) = q(p(x))= q( 9 - x^{2})$ by definition.

is piecewise defined, with one defintion for negative values of the domain and one for nonnegative values.

If $9 -x^{2} < 0$ , then we use the definition q(x)= 0 . This happens if

$9 -x^{2} < 0$

$9 < x^{2}$

x > 3 or x < -3

Therefore, the defintion of $\left ( q \circ p\right )(x)$ for x > 3 or x < -3 is

$\left ( q \circ p\right )(x) = 0$

Subsquently, if $-3 \le x \le 3$ , we use the defintion q(x) = x , since $9 - x^{2} \ge 0$ :

$\left ( q \circ p\right )(x) =q( 9 - x^{2}) = 9 -x^{2}$ .

The correct choice is

$\left ( q \circ p\right )(x) = \left\{\begin{matrix} 0& x < -3 \\ 9 - x^{2}& -3 \le x \le 3 \\ 0& x > 3 \end{matrix}\right.$

Report an Error

Example Question #53 : Understanding Functions

Define a function on the real numbers as follows:

$f(x) = x +4\sqrt{x}- 6$

Give the range of the function.

Possible Answers:

$\left [ 0, \infty \right )$

$\left [ 2, \infty \right )$

$\left [ -10, \infty \right )$

$\left [ 4, \infty \right )$

$\left [ -6, \infty \right )$

Correct answer:

$\left [ -6, \infty \right )$

Explanation:

This can be understood better by substituting $t = \sqrt{x}$ , and, subsequently, $t^{2} =\left ( \sqrt{x} \right )^{2} = x$ in the function's definition.

$x + 4 \sqrt{x}- 6 = t^{2} +4 t - 6$

which is now in standard quadratic form in terms of .

Write this in vertex form by completing the square:

$t^{2} +4 t - 6 = t^{2}+4 t + \left ( \frac{4}{2} \right ) ^{2} - 6 - \left ( \frac{4}{2} \right ) ^{2}$

$= t^{2}+4 t +4- 6 - 4$

$=\left ( t+2 \right )^{2}- 10$

Substitute $\sqrt{x}$ back for , and the original function can be rewritten as

$f(x)=\left ( \sqrt{x}+2 \right )^{2}- 10$ .

To find the range, note that $\sqrt{x} \ge 0$ . Therefore,

$\sqrt{x} + 2 \ge 2$

$\left ( \sqrt{x}+2 \right )^{2} \ge 4$

and

$f(x)=\left ( \sqrt{x}+2 \right )^{2}- 10 \ge -6$

The range of is the set $\left [ -6, \infty \right )$ .

Report an Error

Example Question #58 : Functions/Series

Define a function on the real numbers as follows:

$f(x) = x^{\frac{2}{3}} + 6x^{\frac{1}{3}}+6$

Give the range of the function.

Possible Answers:

$(-\infty, \infty)$

$[-27, \infty)$

$[6, \infty)$

$[-3, \infty)$

$[ 3, \infty)$

Correct answer:

$[-3, \infty)$

Explanation:

This can be understood better by substituting $t = x^{\frac{1}{3}}$ , and, subsequently, $t^{2} =\left ( x^{\frac{1}{3}} \right )^{2} = x^{\frac{2}{3}}$ in the function's definition.

$x^{\frac{2}{3}} + 6x^{\frac{1}{3}}+6 = t^{2}+ 6t + 6$

which is now in standard quadratic form in terms of .

Write this in vertex form by completing the square:

$t^{2}+ 6t + 6$

$= t^{2}+ 6t +\left ( \frac{6}{2} \right )^{2}+ 6 - \left ( \frac{6}{2} \right )^{2}$

$= t^{2}+ 6t +9+ 6 - 9$

$= (t+3)^{2} - 3$

Substitute $x^{\frac{1}{3}}$ back for . The original function can be rewritten as

$f(x)= (x^{\frac{1}{3}}+3)^{2} - 3$

or, in radical form,

$f(x)= (\sqrt[3]{x }+3)^{2} - 3$

$\sqrt[3]{x}$ can assume any real value; so, subsequently, can $\sqrt[3]{x} + 3$ . But its square must be nonnegative, so

$\left (\sqrt[3]{x} + 3 \right )^{2} \ge 0$

and

$f(x)= (\sqrt[3]{x }+3)^{2} - 3 \ge -3$

The range of is $[-3, \infty)$

Report an Error

All GMAT Math Resources

22 Diagnostic Tests 693 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Spanish Tutors in New York City, Spanish Tutors in Washington DC, ACT Tutors in New York City, Chemistry Tutors in Los Angeles, Spanish Tutors in Dallas Fort Worth, Computer Science Tutors in Phoenix, SSAT Tutors in Miami, Reading Tutors in Phoenix, Statistics Tutors in Dallas Fort Worth, Spanish Tutors in San Diego

Popular Courses & Classes

MCAT Courses & Classes in Miami, SAT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in San Francisco-Bay Area, SAT Courses & Classes in Philadelphia, GMAT Courses & Classes in Miami, SAT Courses & Classes in San Diego, LSAT Courses & Classes in Houston, GMAT Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Washington DC, SSAT Courses & Classes in Boston

Popular Test Prep

ACT Test Prep in Boston, GRE Test Prep in San Diego, GMAT Test Prep in San Francisco-Bay Area, GRE Test Prep in Seattle, ACT Test Prep in Atlanta, LSAT Test Prep in Miami, MCAT Test Prep in New York City, ISEE Test Prep in Phoenix, LSAT Test Prep in Boston, LSAT Test Prep in San Francisco-Bay Area

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

Email address:
Your name:
Feedback:

GMAT Math : Algebra

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #191 : Algebra

Example Question #192 : Algebra

Example Question #193 : Algebra

Example Question #194 : Algebra

Example Question #195 : Algebra

Example Question #191 : Algebra

Example Question #55 : Functions/Series

Example Question #54 : Functions/Series

Example Question #53 : Understanding Functions

Example Question #58 : Functions/Series

All GMAT Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details