We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

SAT II Math II : SAT Subject Test in Math II

Study concepts, example questions & explanations for SAT II Math II

Create An Account

All SAT II Math II Resources

2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 4 5 6 7 8 9 10 11 … 54 55 Next →

Example Question #21 : Exponents And Logarithms

Solve $log_{15} (48 + x) = log_{15} (x^2 + 3x)$

Possible Answers:

6, 8

-8, 6

-8, -6

-6, 8

No solutions

Correct answer:

-8, 6

Explanation:

First, we can simplify by canceling the logs, because their bases are the same:

48+x=x^2+3x

Now we collect all the terms to one side of the equation:

x^2+2x-48=0

Factoring the expression gives:

(x-6)(x+8)=0

So our answers are:

x=-8, 6

Report an Error

Example Question #21 : Exponents And Logarithms

Solve 25^x = 625^x .

Possible Answers:

x^2

No solutions

Correct answer:

Explanation:

Here, we can see that changing base isn't going to help. However, if we remember that and number raised to the th power equals , our solution becomes very easy.

$25^{(0)}=625^{(0)}$

1=1

Report an Error

Example Question #21 : Exponents And Logarithms

To the nearest hundredth, solve for : $7^{2x} = 5 ^{x}$ .

Possible Answers:

2.28

None of these

0.55

1.27

0.99

Correct answer:

None of these

Explanation:

Take the natural logarithm of both sides:

$7^{2x} = 5 ^{x}$

$\ln 7^{2x} = \ln 5 ^{x}$

By the Logarithm of a Power Rule the above becomes

$2x \ln 7 = x \ln 5$

Solve for :

$2x \ln 7 - x \ln 5 = x \ln 5 - x \ln 5$

$(2 \ln 7 - \ln 5) x = 0$

$\frac{(2 \ln 7 - \ln 5) x }{2 \ln 7 - \ln 5}= \frac{0 }{2 \ln 7 - \ln 5}$

x = 0 .

This is not among the choices given.

Report an Error

Example Question #1 : Absolute Value

Define $g(x)= \left | \left | 1,000- \sqrt{x}\right | - x^{2} \right |$ .

Evaluate g(25) .

Possible Answers:

1,630

940,900

380

1,620

370

Correct answer:

370

Explanation:

$g(x)= \left | \left | 1,000- \sqrt{x}\right | - x^{2} \right |$

$g(25)= \left | \left | 1,000- \sqrt{25}\right | - 25^{2} \right |$

$= \left | \left | 1,000-5\right | - 625 \right |$

$= \left | \left | 995 \right | - 625 \right |$

$= \left | 995 - 625 \right |$

$= \left | 370 \right |$

=370

Report an Error

Example Question #1 : Absolute Value

Define $f(x) = \left | 1 - x\right |$ .

Order from least to greatest: $f\left ( \frac{9}{14} \right ), f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right )$

Possible Answers:

$f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right ),f\left ( \frac{9}{14} \right )$

$f\left ( \frac{13}{14} \right ), f\left ( \frac{9}{14} \right ), f\left ( \frac{17}{14} \right )$

$f\left ( \frac{17}{14} \right ), f\left ( \frac{13}{14} \right ), f\left ( \frac{9}{14} \right )$

$f\left ( \frac{9}{14} \right ), f\left ( \frac{17}{14} \right ),f\left ( \frac{13}{14} \right )$

$f\left ( \frac{9}{14} \right ), f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right )$

Correct answer:

$f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right ),f\left ( \frac{9}{14} \right )$

Explanation:

$f(x) = \left | 1 - x\right |$ , or, equivalently,

$f(x) = \left | \frac{14}{14} - x\right |$

$f\left ( \frac{9}{14} \right ) = \left | \frac{14}{14} - \frac{9}{14}\right | = \left | \frac{5}{14} \right |=\frac{5}{14}$

$f\left ( \frac{13}{14} \right ) = \left | \frac{14}{14} - \frac{13}{14}\right |= \left | \frac{1}{14} \right |=\frac{1}{14}$

$f\left ( \frac{17}{14} \right ) = \left | \frac{14}{14} - \frac{17}{14} \right |= \left | - \frac{3}{14} \right |=\frac{3}{14}$

From least to greatest, the values are

$f\left ( \frac{13}{14} \right ), f\left ( \frac{17}{14} \right ),f\left ( \frac{9}{14} \right )$

Report an Error

Example Question #3 : Absolute Value

Define an operation $\Psi$ as follows:

For all real numbers a , b ,

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$

Evaluate $2\frac{1}{2} \Psi \left (- \frac{1}{4} \right )$ .

Possible Answers:

$1\frac{1}{5}$

$-1\frac{1}{5}$

Undefined.

Correct answer:

Explanation:

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$

$2\frac{1}{2} \Psi \left (- \frac{1}{4} \right ) = \frac{\left |1- 2\frac{1}{2} \right |}{1-\left |-\frac{1}{4}\right |}$

$= \frac{\left |-1\frac{1}{2} \right |}{1- \frac{1}{4}}$

$= \frac{ 1\frac{1}{2} }{ \frac{3}{4}}$

$= \frac{ \frac{3}{2} }{ \frac{3}{4}}$

$= \frac{3}{2} \div \frac{3}{4}$

$= \frac{3}{2} \cdot \frac{4}{3}$

$= \frac{12}{6}$

Report an Error

Example Question #61 : Sat Subject Test In Math Ii

Define an operation $\Psi$ as follows:

For all real numbers a , b ,

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$

If $\frac{1}{2}\Psi N = 1$ , which is a possible value of ?

Possible Answers:

N =- 3

N =- 2

$N =- \frac{1}{2}$

$N =- 1\frac{1}{2}$

N =- 1

Correct answer:

$N =- \frac{1}{2}$

Explanation:

$a\Psi b = \frac{\left |1- a \right |}{1-\left |b \right |}$ , so

$\frac{1}{2}\Psi N = 1$

can be rewritten as

$\frac{\left |1- \frac{1}{2} \right |}{1-\left |N \right |} = 1$

$\frac{\left | \frac{1}{2} \right |}{1-\left |N \right |} = 1$

$\frac{ \frac{1}{2} }{1-\left |N \right |} = 1$

$\frac{1}{2} =1-\left |N \right |$

$-\frac{1}{2} =-\left |N \right |$

$\frac{1}{2} = \left |N \right |$

Therefore, either $N = \frac{1}{2}$ or $N =- \frac{1}{2}$ . The correct choice is $N =- \frac{1}{2}$ .

Report an Error

Example Question #5 : Absolute Value

Define $f(x) = \left | x+ 6\right |+ \left | x-9\right |$ .

How many values are in the solution set of the equation f(x)= 7 ?

Possible Answers:

Three solutions

One solution

Infinitely many solutions

Two solutions

No solutions

Correct answer:

No solutions

Explanation:

We can rewrite this function as a piecewise-defined function by examining three different intervals of -values.

If x< -6 , then

x+ 6 < 0 and x-9 < 0 ,

and this part of the function can be written as

$f(x) =- \left ( x+ 6\right )- \left (x-9\right )$

f(x) =-x-6-x+9

f(x)= -2x+3

If f(x)= 7 under this definition, then

-2x+3 = 7

-2x = 4

x = -2

However, $-2 \nless -6$ , so this is a contradiction.

If $-6 \le x \le 9$ , then

$x+ 6\ge 0$ and $x-9\le 0$ ,

and this part of the function can be written as

$f(x) = \left ( x+ 6\right )- \left (x-9\right )$

f(x) =x+6-x+9

f(x) =15

This yields no solutions.

If x> 9 , then

x+ 6 > 0 and x-9 > 0 ,

and this part of the function can be written as

$f(x) = \left ( x+ 6\right ) + \left (x-9\right )$

f(x) =x+6+x-9

f(x)= 2x-3

If f(x)= 7 under this definition, then

2x-3 = 7

2x=10

x = 5

However, $5 \ngtr 9$ , so this is a contradiction.

f(x)= 7 has no solution.

Report an Error

Example Question #6 : Absolute Value

Define $f(x) = \left | x+ 5\right |+ \left | x-7\right |$ .

How many values are in the solution set of the equation f(x)= 12 ?

Possible Answers:

Infinitely many solutions

No solutions

One solution

Two solutions

Three solutions

Correct answer:

Infinitely many solutions

Explanation:

We can rewrite this function as a piecewise-defined function by examining three different intervals of -values.

If x< -5 , then

x+ 5 < 0 and x-7 < 0 ,

and this part of the function can be written as

$f(x) =- \left ( x+ 5\right )- \left (x-7\right )$

f(x) =- x-5 -x+7

f(x) =-2x+2

If $-5 \le x \le 7$ , then

$x+ 5 \ge 0$ and $x-7 \le 0$ ,

and this part of the function can be written as

$f(x) = \left ( x+ 5\right )- \left (x-7\right )$

f(x) =x+5 -x+7

f(x) =12

If x > 7 , then

x+ 5 > 0 and x-7 > 0 ,

and this part of the function can be written as

$f(x) = \left ( x+ 5\right )+ \left (x-7\right )$

f(x) =x+5+x-7

f(x) =2x-2

The function can be rewritten as

$f(x)= \left\{\begin{matrix} -2x+2 & x < -5\\12 & -5 \le x \le 7 \\ 2x-2 & x > 7 \end{matrix}\right.$

As can be seen from the rewritten definition, every value of in the interval [-5, 7] is a solution of f(x)= 12 , so the correct response is infinitely many solutions.

Report an Error

Example Question #7 : Absolute Value

Consider the quadratic equation

$2x^{2} + 12 = 11x$

Which of the following absolute value equations has the same solution set?

Possible Answers:

$\left |x- 2 \frac{3}{4} \right | =1 \frac{1}{4}$

$\left |x+2 \frac{3}{4} \right | =1 \frac{1}{4}$

None of the other choices gives the correct response.

$\left |x- 1 \frac{1}{4} \right | =2 \frac{3}{4}$

$\left |x+ 1 \frac{1}{4} \right | =2 \frac{3}{4}$

Correct answer:

$\left |x- 2 \frac{3}{4} \right | =1 \frac{1}{4}$

Explanation:

Rewrite the quadratic equation in standard form by subtracting 11x from both sides:

$2x^{2} + 12 = 11x$

$2x^{2} + 12 - 11x = 11x - 11x$

$2x^{2} - 11x + 12= 0$

Solve this equation using the method. We are looking for two integers whose sum is and whose product is 2(12) = 24 ; by trial and error we find they are , . The equation becomes

$2x^{2} - 8x - 3x + 12= 0$

Solving using grouping:

$(2x^{2} - 8x )- (3x - 12)= 0$

2x (x-4)- 3 (x-4)= 0

(2x - 3) (x-4)= 0

By the Zero Product Principle, one of these factors must be equal to 0.

Either

2x- 3 = 0

2x- 3 + 3 = 0 + 3

2x= 3

$\frac{2x}{2}= \frac{3}{2}$

$x= \frac{3}{2} = 1\frac{1}{2}$

x- 4= 0

x- 4 + 4 = 0 + 4

x = 4

The given quadratic equation has solution set $\left \{ 1\frac{1}{2}, 4 \right \}$ , so we are looking for an absolute value equation with this set as well.

This equation can take the form

|x-a| = b

This can be rewritten as the compound equation

x-a = -b x-a = b

Adding to both sides of each equation, the solution set is

x =a -b and x =a + b

Setting these numbers equal in value to the desired solutions, we get the linear system

$a -b =1\frac{1}{2}$

a+ b = 4

Adding and solving for :

$a -b =1\frac{1}{2}$

$\underline{a+ b =4}$

$= 5\frac{1}{2}$

$2a \div 2 = 5\frac{1}{2} \div 2$

$a = 2 \frac{3}{4}$

Backsolving to find :

a+ b = 4

$2 \frac{3}{4}+ b = 4$

$2 \frac{3}{4}+ b - 2 \frac{3}{4}= 4 - 2 \frac{3}{4}$

$b =1 \frac{1}{4}$

The desired absolute value equation is $\left |x- 2 \frac{3}{4} \right | =1 \frac{1}{4}$ .

Report an Error

← Previous 1 2 3 4 5 6 7 8 9 10 11 … 54 55 Next →

View SAT Subject Test in Mathematics Level 2 Tutors

Luu Chi Hieu
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Computer Science.

View SAT Subject Test in Mathematics Level 2 Tutors

Antonia
Certified Tutor

University of Michigan-Ann Arbor, Bachelor of Engineering, Nuclear Engineering. University of Wisconsin-Madison, Current Grad...

View SAT Subject Test in Mathematics Level 2 Tutors

Reuben
Certified Tutor

University of Alberta, Bachelor of Science, Mathematics.

All SAT II Math II Resources

2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Math Tutors in Boston, French Tutors in San Francisco-Bay Area, Reading Tutors in Boston, LSAT Tutors in Denver, ISEE Tutors in New York City, Algebra Tutors in Philadelphia, MCAT Tutors in San Diego, Biology Tutors in Denver, LSAT Tutors in Washington DC, Spanish Tutors in Dallas Fort Worth

Popular Courses & Classes

ACT Courses & Classes in Philadelphia, ISEE Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Los Angeles, SAT Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in Washington DC, Spanish Courses & Classes in Houston, LSAT Courses & Classes in Boston, MCAT Courses & Classes in Miami, ISEE Courses & Classes in Atlanta, SAT Courses & Classes in San Diego

Popular Test Prep

ACT Test Prep in Los Angeles, ISEE Test Prep in Philadelphia, SSAT Test Prep in Houston, SAT Test Prep in Philadelphia, ISEE Test Prep in Dallas Fort Worth, LSAT Test Prep in Miami, LSAT Test Prep in New York City, GMAT Test Prep in Miami, GRE Test Prep in Denver, SSAT Test Prep in Chicago

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)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

SAT II Math II : SAT Subject Test in Math II

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #21 : Exponents And Logarithms

Example Question #21 : Exponents And Logarithms

Example Question #21 : Exponents And Logarithms

Example Question #1 : Absolute Value

Example Question #1 : Absolute Value

Example Question #3 : Absolute Value

Example Question #61 : Sat Subject Test In Math Ii

Example Question #5 : Absolute Value

Example Question #6 : Absolute Value

Example Question #7 : Absolute Value

All SAT II Math II Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: