Mathematical Relationships - SAT Subject Test in Math II

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Question

Define an operation $\Psi$ as follows:

For all real numbers ,

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

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

Answer

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

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

How many values are in the solution set of the equation ?

Answer

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

If , then

and ,

and this part of the function can be written as

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

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

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

If , then

and ,

and this part of the function can be written as

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

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 is a solution of , so the correct response is infinitely many solutions.

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Question

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

How many values are in the solution set of the equation ?

Answer

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

If , then

and ,

and this part of the function can be written as

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

If under this definition, then

However, , 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

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

This yields no solutions.

If , then

and ,

and this part of the function can be written as

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

If under this definition, then

However, , so this is a contradiction.

has no solution.

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Question

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

Evaluate .

Answer

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

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

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

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

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

$= \left | 370 \right |$

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Question

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

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

Answer

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

Define an operation $\Psi$ as follows:

For all real numbers ,

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

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

Answer

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

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

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

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

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

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

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

$= \frac{12}{6}$

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Question

Consider the quadratic equation

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

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

Answer

Rewrite the quadratic equation in standard form by subtracting 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 ; by trial and error we find they are , . The equation becomes

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

Solving using grouping:

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

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

Either

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

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

Or

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

This can be rewritten as the compound equation

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

and

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

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

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 :

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

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

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

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

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Question

Give the solution set of the inequality:

Answer

To solve an absolute value inequality, first isolate the absolute value expression, which can be done here by subtracting 35 from both sides:

There is no need to go further. The absolute value of any number is always greater than or equal to 0, so, regardless of the value of ,

$|-3y + 12| \ge 0 > -23$ .

Therefore, the solution set is the set of all real numbers.

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Question

Solve .

Answer

First we need to isolate the absolute value term. We do with using some simple algebra:

Now we solve two equations, one with the right side of the equation positive, one negative. Let's start with the positive:

And now the negative:

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

So our answers are:

$x=\frac{4}{5}, 2$

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Question

Solve .

Answer

First, we have to isolate the absolute value:

Let's take a look at our equation right now. It's saying that the absolute value has to be a negative number, which isn't possible. So no solutions exist.

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Question

Solve .

Answer

First, we need to isolate the absolute value:

Because the equation is set equal to , we can drop the absolute value symbols and solve normally:

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Question

Solve: $25\left | 3x+2\right | -3= -5$

Answer

Add 3 on both sides.

$25\left | 3x+2\right | -3+3= -5+3$

$25\left | 3x+2\right | = -2$

Divide by 25 on both sides.

$\frac{25\left | 3x+2\right |}{25} =\frac{ -2}{25}$

$\left | 3x+2\right | = -\frac{2}{25}$

Recall that an absolute value cannot have a negative value. There is no x-value that will equal to the right term.

The answer is: $\textup{No solution.}$

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Question

Solve: .

Answer

Because the absolute value term is "less than" the other side of the equation, we can rewrite the problem like this:

This eliminates the absolute value. Remember, when an operation is performed, it must be performed on all three sets of terms. When we add to each side, we end up with:

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Question

Solve: $|-6x - 8| \geq 12$

Answer

Here, we have to split the problem up into two parts:

$-6x-8\leq-12$ and $-6x-8\geq12$

Let's start with the first equation:

$-6x-8\leq-12$

First, we can add to each side:

$-6x \leq -4$

Now we divide by -6. Remember, when you divide by a negative, you flip the sign of the inequality:

$x \geq \frac{4}{6}$

Which we can reduce:

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

Now let's do the other part of the problem the same way:

$-6x \geq 20$

$x \leq \frac{-20}{6}$

$x \leq \frac{-10}{3}$

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Question

Solve the absolute value: $\left |3x \right |+2 = -12$

Answer

Subtract 2 from both sides.

$\left |3x \right |+2 -2= -12-2$

$\left |3x \right | = -14$

Under this condition, there are no values of that will give a negative 14.

The answer is: $\textup{No solution.}$

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Question

Solve:

$\left | 12-5\right |$

Answer

The lines on the outside of this problem indicate it is an absolute value problem. When solving with absolute value, remember that it is a measure of displacement from 0, meaning the answer will always be positive.

$\left | 12-5\right |=\left | 7\right |=7$

For this problem, that gives us a final answer of 7.

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Question

Solve:

$\left | 5-6\right |$

Answer

The lines on the outside of this problem indicate it is an absolute value problem. When solving with absolute value, remember that it is a measure of displacement from 0, meaning the answer will always be positive.

$\left | 5-6\right |=\left | -1\right |=1$

For this problem, that gives us a final answer of 1.

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Question

Solve:

$\left | 25-39\right |$

Answer

The lines on the outside of this problem indicate it is an absolute value problem. When solving with absolute value, remember that it is a measure of displacement from 0, meaning the answer will always be positive.

$\left | 25-39\right |=\left | -14\right |=14$

For this problem, that gives us a final answer of 14.

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Question

Solve:

$\left | 12-2\right |$

Answer

The lines on the outside of this problem indicate it is an absolute value problem. When solving with absolute value, remember that it is a measure of displacement from 0, meaning the answer will always be positive.

$\left | 12-2\right |=\left | 10\right |=10$

For this problem, that gives us a final answer of 10.

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Question

Solve:

$\left | 3-132\right |$

Answer

The lines on the outside of this problem indicate it is an absolute value problem. When solving with absolute value, remember that it is a measure of displacement from 0, meaning the answer will always be positive.

$\left | 3-132\right |=\left | -129\right |=129$

For this problem, that gives us a final answer of 129.

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