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SAT II Math II : SAT Subject Test in Math II

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

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Example Questions

Example Question #33 : Mathematical Relationships

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

Possible Answers:

No solutions

-8, 6

-8, -6

-6, 8

6, 8

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

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Example Question #25 : Exponents And Logarithms

Solve 25^x = 625^x .

Possible Answers:

No solutions

x^2

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

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Example Question #22 : Exponents And Logarithms

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

Possible Answers:

2.28

None of these

0.99

1.27

0.55

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.

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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

370

380

940,900

1,620

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

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Example Question #2 : 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:

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

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

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

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

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

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

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

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Example Question #1 : 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:

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

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

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

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

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

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

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

$= \frac{12}{6}$

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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 =- 1

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

N =- 2

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

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

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Example Question #41 : Mathematical Relationships

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:

One solution

Three solutions

Infinitely many solutions

No solutions

Two 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.

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Example Question #2 : 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:

Three solutions

One solution

No solutions

Infinitely many solutions

Two 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.

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Example Question #3 : Absolute Value

Consider the quadratic equation

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

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

Possible Answers:

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

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

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

None of the other choices gives the correct response.

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

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

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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 #33 : Mathematical Relationships

Example Question #25 : Exponents And Logarithms

Example Question #22 : Exponents And Logarithms

Example Question #1 : Absolute Value

Example Question #2 : Absolute Value

Example Question #1 : Absolute Value

Example Question #61 : Sat Subject Test In Math Ii

Example Question #41 : Mathematical Relationships

Example Question #2 : Absolute Value

Example Question #3 : Absolute Value

All SAT II Math II Resources

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