Functions/Series - GMAT Quantitative Reasoning

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Question

Which of the following would be a valid alternative definition of the function

$f(x) = \left | x + 1 \right | - \left | x - 1\right |$ ?

Answer

If $x \geq 1$ , then and are both positive, so

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

If , then , then is positive and is negative, so

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

$=x+1 - \left [-\left ( x-1\right ) \right ]$

$=x+1 - \left (-x+1\right )$

If $x \leq -1$ , then and are both negative, so

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

$=-\left ( x + 1 \right )- \left [-\left ( x - 1\right ) \right ]$

$=- x - 1- \left (- x + 1\right )$

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Question

Let be the piecewise-defined function graphed above. Define a function .

Evaluate $\left (f \circ g^{-1} \right ) (6)$ .

Answer

$\left (f \circ g^{-1} \right ) (6) = f [g^{-1}(6)]$

$g^{-1}(6) = c$ if , so, since

Therefore, $g^{-1}(6) = 10$ , and

$\left (f \circ g^{-1} \right ) (6) = f (10)$

As can be seen from the diagram, however, the domain of is . 10 is not in the domain of . Therefore, is not in the domain of $f \circ g^{-1}$ .

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Question

Which of the following pairs of statements is sufficient to prove that does not have an inverse?

Answer

For a function to have an inverse, no -coordinate can be paired with more than one -coordinate. Of our choices, only

causes this to happen.

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Question

There is water tank already $\frac{4}{7}$ full. If Jose adds 5 gallons of water to the water tank, the tank will be $\frac{13}{14}$ full. How many gallons of water would the water tank hold if it were full?

Answer

In this case, we need to solve for the volume of the water tank, so we set the full volume of the water tank as $x$ . According to the question, $\frac{4}{7}$ -full can be replaced as $\frac{4}{7}x$ . $\frac{13}{14}$ -full would be $\frac{13}{14}x$ . Therefore, we can write out the equation as:

$\frac{4}{7}x+5=\frac{13}{14}x$ .

Then we can solve the equation and find the answer is 14 gallons.

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Question

There exists a set = {1, 2, 3, 4}. Which of the following defines a function of ?

Answer

Let's look at $f,g,and\ h$ and see if any of them are functions.

1. = {(2, 3), (1, 4), (2, 1), (3, 2), (4, 4)}: This cannot be a function of because two of the ordered pairs, (2, 3) and (2, 1) have the same number (2) as the first coordinate.

2. = {(3, 1), (4, 2), (1, 1)}: This cannot be a function of because it contains no ordered pair with first coordinate 2. Because the set = {1, 2, 3, 4}, we need an ordered pair of the form (2, ) .

3. = {(2, 1), (3, 4), (1, 4), (2, 1), (4, 4)}: This is a function. Even though two of the ordered pairs have the same number (2) as the first coordinate, is still a function of because (2, 1) is simply repeated twice, so the two ordered pairs with first coordinate 2 are equal.

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Question

Let be a function that assigns $x^{2}$ to each real number . Which of the following is NOT an appropriate way to define ?

Answer

This is a definition question. The only choice that does not equal the others is $f(y)=x^{2}$ . This describes a function that assigns $x^{2}$ to some number , instead of assigning $x^{2}$ to its own square root, .

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Question

If $f(x)=x^{2}$ , find $\frac{f(x+h)-f(x)}{h}$ .

Answer

We are given f(x) and h, so the only missing piece is f(x + h).

$f(x+h)=(x+h)^{2}=x^{2}+2xh+h^{2}$

Then $\frac{f(x+h)-f(x)}{h}= \frac{x^{2}+2xh+h^{2}-x^{2}}{h} = \frac{2xh+h^{2}}{h}=2x+h$

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Question

$f (x) = \left{\begin{matrix} 2x + 1 & x < 0 \ 3x - 1 & x \geq 0 \end{matrix}\right.$

$\small g (x) = \left{\begin{matrix} 3x - 1 & x < 0 \ 4x +7 & x \geq 0 \end{matrix}\right.$

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

Answer

$\small \small \small \small (f \circ g) (4) = f (g (4)) = f (4 \cdot 4 + 7) = f (23) = 3 \cdot 23 - 1 = 68$

$\small \small \small \small (g \circ f) (-4) = g (f (-4)) = g (2 \cdot (-4) + 1)$ $\small \small = g (-7) = 3 (-7) -1 = -22$

$\small \small \small \small (f \circ g) (4) - (g \circ f) (-4) = 68 - (-22) = 90$

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Question

Give the range of the function:

$f (x) = \left{\begin{matrix} 2x - 1 & & x < -2\ -3x& & -2\leq x\leq 2\ \5 & & ; x > 2 \end{matrix}\right.$

Answer

We look at the range of the function on each of the three parts of the domain. The overall range is the union of these three intervals.

On $(-\infty , -2)$ , takes the values:

$f\left (x \right )< -5$

or $(-\infty , -5)$

On , takes the values:

$-2 \leq x\leq 2$

$-3(-2) \geq -3x\geq -3 (2)$

$6 \geq -3x\geq -6$

$-6 \leq f(x)\leq 6$ ,

or

On $(2, \infty)$ , takes only value 5.

The range of is therefore $(-\infty , -5) \cup [-6, 6] \cup \left { 5\right }$ , which simplifies to $(-\infty , 6]$ .

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Question

A sequence begins as follows:

It is formed the same way that the Fibonacci sequence is formed. What are the next two numbers in the sequence?

Answer

Each term of the Fibonacci sequence is formed by adding the previous two terms. Therefore, do the same to form this sequence:

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Question

For any values , , define the operation $\Gamma$ as follows:

$a; \Gamma; b = 2ab - a - b$

Which of the following expressions is equal to $2x; \Gamma; 2y$ ?

Answer

Substitute and for and in the expression for $a; \Gamma; b$ :

$2x; \Gamma; 2y = 2(2x)(2y) - 2x - 2y = 8xy -2x-2y$

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Question

For any real , define $a\odot b =(1-a)(1-b)$ .

For what value or values of would $x\odot x = 4$ ?

Answer

For such an to exist, it must hold that $x\odot x = (1 -x )(1-x) =(1-x)^{2}=4$ .

Take the square root of both sides:

or

Case 1:

Case 2:

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Question

Define $f(x) = x^{2}- 4$ and $g(x) = x^{2} + 4$ .

Give the definition of $(f \circ g)(x)$ .

Answer

$(f \circ g)(x) = f(g(x))$

$= f(x^{2}+4)$

$= (x^{2}+4)^{2}-4$

$=x^{4} + 8x^{2}+16-4$

$=x^{4} + 8x^{2}+12$

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Question

Give the inverse of

Answer

The easiest way to find the inverse of is to replace in the definition with , switch with , and solve for in the new equation.

$\frac{1}{5} \cdot\left ( x + 7 \right ) = \frac{1}{5} \cdot 5y$

$\frac{1}{5} x + \frac{7}{5} = y$

$f^{-1}(x)= \frac{1}{5}x + \frac{7}{5}$

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Question

Define $f (x) = x^{3} - 8$ . Give $f^{-1} (x)$

Answer

The easiest way to find the inverse of is to replace in the definition with , switch with , and solve for in the new equation.

$y = x^{3} - 8$

$x= y^{3} - 8$

$x +8= y^{3} - 8 +8$

$x +8= y^{3}$

$\sqrt[3]{x +8}= \sqrt[3]{y^{3}}$

$y = \sqrt[3]{x +8}$

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Question

Define an operation $\odot$ as follows:

For any real numbers ,

$a \odot b = (a-1) (b-1)$

Evaluate $\left (5 \odot 5 \right )\odot 5$ .

Answer

$\left (5 \odot 5 \right )\odot 5$

$= \left [(5 -1) (5 -1) \right ] \odot 5$

$= (4 \cdot 4 ) \odot 5$

$= 16 \odot 5$

$= \left (16-1 \right ) \left (5-1 \right )$

$= 15 \cdot4 = 60$

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Question

Define $f(x) = A (x-B)^{3}$ , where .

Evaluate $f^{-1} (1)$ in terms of and .

Answer

This is equivalent to asking for the value of for which , so we solve for in the following equation:

$A (x-B)^{3} =1$

$\frac{A (x-B)^{3}}{A} =\frac{1}{A}$

$(x-B)^{3} =\frac{1}{A}$

$\sqrt[3]{(x-B)^{3} }=\sqrt[3]{\frac{1}{A}}$

$x-B =\sqrt[3]{\frac{1}{A}}$

$x-B = \frac{1}{\sqrt[3]{A}}$

$x-B + B = B + \frac{1}{\sqrt[3]{A}}$

$x = B + \frac{1}{\sqrt[3]{A}}$

Therefore, $f^{-1} (1)= B + \frac{1}{\sqrt[3]{A}}$ .

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Question

A sequence is formed the same way the Fibonacci sequence is formed. Its third and fourth terms are 16 and 30, respectively. What is its first term?

Answer

A Fibonacci-style sequence starts with two numbers, with each successive number being the sum of the previous two. The second term is therefore the difference of the fourth and third terms, and the first term is the difference of the third and second.

Second term:

First term:

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Question

Define an operation $\bigstar$ as follows:

For any real , $a ; \bigstar ; b = a^{2} +b^{2} -8$ .

For what value or values of is it true that $N ; \bigstar ; N = 0$ ?

Answer

Substitute into the definition, and then set the expression equal to 0 to solve for :

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

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

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

$2N^{2} =8$

$N^{2} =4$

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

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Question

An infinite sequence begins as follows:

Assuming this pattern continues infinitely, what is the sum of the 1000th, 1001st and 1002nd terms?

Answer

This can be seen as a sequence in which the term is equal to if is not divisible by 3, and otherwise. Since 1,000 and 1,001 are not multiples of 3, but 1,002 is, the 1000th, 1001st, and 1002nd terms are, respectively,

and their sum is

$1000+ 1001+ \left (-1002 \right ) = 999$

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