Algebra - GMAT Quantitative

by definition. is piecewise defined, with one defintion for negative values of the domain and one for nonnegative values. However, is nonnegative for all real numbers, so the defintion for nonnegative numbers, , is the one that will always be used. Therefore,

for all values of .

by definition.

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

If , then we use the definition . This happens if

or

Therefore, the defintion of for or is

Subsquently, if , we use the defintion , since :

.

The correct choice is

has an inverse on a given domain if and only if there are no two distinct values on the domain such that .

The key to this question is to find the zeroes of the polynomial, which can be done as follows:

The zeroes are , or approximately .

The polynomial being cubic, its graph has two vertices; since all three zeroes are in the interval , so are both vertices. Therefore, this interval must have at least one pair such that . Since a cubic polynomial has two "arms", one going up and one going down, will continuously increase or decrease over the other intevals. The correct choice is .

is the same as .

For solutions sets of integers, we are able to divide it as:

When we multiply this, we notice that the constants must return 7. Since 7 is a prime number, we now know that these constants are 1 and 7. The signs of these can still switch however. If both of the numbers are negative, it will also return a product of positive 7. These are the only ways to make the constant term work. If they are both positive numbers, then we get . If they're both negative numbers then .

For the values of , we simply notice the only way for the overall product to be 0 is for one (or both) of the pieces to be zero. This is done by having equal the additive inverse of the constant piece.

Thus we have 2 integer solution sets: and

Substitute , and, subsequently, , to rewrite this equation as quadratic, then solve by factoring.

We can rewrite the quadratic expression as , where the question marks are replaced with integers whose product is 12 and whose sum is ; these integers are .

Set each factor to zero and solve for ; then substitute back and solve for :

The solution set, which can be confirmed by substitution, is .

First, isolate the absolute value expression on one side:

Rewrite as a compound sentence:

Solve each separately:

or

The solution set is

First, isolate the absolute value expression on one side:

Rewrite as a compound sentence:

Solve each separately:

or

The solution set is .

Subsitute in the defintion, and set it equal to 21:

This sets up a quadratic equation, Move all terms to the left, factor the expression, set each factor to 0, and solve separately.

or

The solution set is

First, rewrite the quadratic equation in standard form by FOILing out the product on the left, then colleciting all of the terms on the left side:

Now factor the quadratic expression to two binomial factors , replacing the question marks with two integers whose product is 36 and whose sum is . These numbers are , so:

or

The solution set is

First, rewrite the quadratic equation in standard form by moving all nonzero terms to the left:

Now factor the quadratic expression into two binomial factors , replacing the question marks with two integers whose product is 16 and whose sum is . These numbers are , so:

or

The solution set is .

The key to answering this question is to note that this equation can be rewritten in piecewise fashion.

If , since both and are nonnegative, we can rewrite as

, or

.

On , this has as its graph a line with positive slope, so it is an increasing function. The range of this part of the function is , or, since

,

.

If , since is negative and is positive, we can rewrite as

, or

is a constant function on this interval and its range is .

If , since both and are nonpositive, we can rewrite as

, or

.

On , this has as its graph a line with negative slope, so it is a decreasing function. The range of this part of the function is , or, since

, .

The union of the ranges is the range of the function - .

The key to answering this question is to note that this equation can be rewritten in piecewise fashion.

If , since both and are positive, we can rewrite as

, or

,

a constant function with range .

If , since is negative and is positive, we can rewrite as

, or

This is decreasing, as its graph is a line with negative slope. The range is ,

or, since

and

,

.

If , since both and are negative, we can rewrite as

, or

,

a constant function with range .

The union of the ranges is the range of the function - - which is not among the choices.

has an inverse on a given domain if and only if there are no two distinct values on the domain such that .

The key to this question is to find the zeroes of the polynomial, which can be done as follows:

The zeroes are , or approximately .

The polynomial being cubic, its graph has two vertices; since all three zeroes are in the interval , so are both vertices. Therefore, this interval must have at least one pair such that . Since a cubic polynomial has two "arms", one going up and one going down, will continuously increase or decrease over the other intevals. The correct choice is .

is the same as .

For solutions sets of integers, we are able to divide it as:

When we multiply this, we notice that the constants must return 7. Since 7 is a prime number, we now know that these constants are 1 and 7. The signs of these can still switch however. If both of the numbers are negative, it will also return a product of positive 7. These are the only ways to make the constant term work. If they are both positive numbers, then we get . If they're both negative numbers then .

For the values of , we simply notice the only way for the overall product to be 0 is for one (or both) of the pieces to be zero. This is done by having equal the additive inverse of the constant piece.

Thus we have 2 integer solution sets: and

A function has an inverse if and only if, if are in the domain of , then implies that , or, contrapositively, if , then . In each of the four choices, no two values of are matched with the same value of , so implies that in all four choices. Since the entire domain is given to be , all of the given functions have inverses.

Substitute , and, subsequently, , to rewrite this equation as quadratic, then solve by factoring.

We can rewrite the quadratic expression as , where the question marks are replaced with integers whose product is 12 and whose sum is ; these integers are .

Set each factor to zero and solve for ; then substitute back and solve for :

The solution set, which can be confirmed by substitution, is .

First, isolate the absolute value expression on one side:

Rewrite as a compound sentence:

Solve each separately:

or

The solution set is

First, isolate the absolute value expression on one side:

Rewrite as a compound sentence:

Solve each separately:

or

The solution set is .