GMAT Math : Functions/Series

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Example Question #71 : Understanding Functions

What is the sum of all odd numbers from 0 to 59 inclusive?

Possible Answers:

870

900

360

731

873

Correct answer:

900

Explanation:

We can manipulate summations to make them easier to work with. Here, we are asked for the sum of odd terms from 1 to 59. We can calculate this by subtracting the summation of the even terms from the summation of all numbers from 1 to 59, using the formula $\frac{n^{2}+n}{2}$ to sum all terms from 1 to .

In other words, we have to calculate $\frac{59^{2}+59}{2}-\frac{2\cdot (29^{2}+29)}{2}$ , since the even numbers are given by 2+4+6...+58 or 2(1+2+3...+29) .

We obtain the final answer 900.

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Example Question #72 : Understanding Functions

Define f (x) = x - |x | . Which of the following would be a valid alternative way of expressing the definition of ?

Possible Answers:

$f (x) = \left\{\begin{matrix} 1 & \textrm{ if } x < 1\\ x & \textrm{ if } x \geq 1 \end{matrix}\right.$

$f (x) = \left\{\begin{matrix} 2x & \textrm{ if } x < 0\\ 0 & \textrm{ if } x \geq 0 \end{matrix}\right.$

$f (x) = \left\{\begin{matrix} 1 & \textrm{ if } x < 1\\ 2x & \textrm{ if } x \geq 1 \end{matrix}\right.$

$f (x) = \left\{\begin{matrix} x & \textrm{ if } x < 0\\ 0 & \textrm{ if } x \geq 0 \end{matrix}\right.$

$f (x) = \left\{\begin{matrix} 0 & \textrm{ if } x < 0\\ 2x & \textrm{ if } x \geq 0 \end{matrix}\right.$

Correct answer:

$f (x) = \left\{\begin{matrix} 2x & \textrm{ if } x < 0\\ 0 & \textrm{ if } x \geq 0 \end{matrix}\right.$

Explanation:

By definition:

If $x \geq 0$ , then |x| = x ,and subsequently, f (x) = x - x = 0

If x < 0 , then |x| = -x ,and subsequently, $f (x) = x- \left ( - x \right ) = 2x$

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Example Question #73 : Understanding Functions

Piecewise

Let be the piecewise-defined function graphed above. Define the function $g (x) = x^{2} - 3$ .

Evaluate $\left (g \circ f \right ) (1)$ .

Possible Answers:

Correct answer:

Explanation:

$\left (g \circ f \right ) (1) = g [f(1)]$

As seen in the diagram below, the graph of includes the point (1, -9) .

Piecewise 1

Therefore, f(1) = -9 , and

$\left (g \circ f \right ) (1) = g (-9)$ .

$g (x) = x^{2} - 3$ , so

$g (-9) = (-9)^{2} - 3 = 81 - 3 = 78$ .

Therefore, $\left (g \circ f \right ) (1) = 78$ , the correct choice.

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Example Question #74 : Understanding Functions

Piecewise

Let be the piecewise-defined function graphed above. Define the function $g (x) = x^{2} - 3$ .

Evaluate $\left (f \circ g \right ) (1)$ .

Possible Answers:

Correct answer:

Explanation:

$\left (f \circ g \right ) (1) = f [g(1)]$

$g (x) = x^{2} - 3$ , so

$g (1) = 1^{2} - 3 = 1-3 = -2$ , so

$\left (f \circ g \right ) (1) = f (-2)$

As seen in the diagram below, the graph of includes the point (-2, -1) .

Piecewise 1

Therefore, f(-2) = -1 , and $\left (f \circ g \right ) (1) = -1$ , the correct choice.

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Example Question #75 : Understanding Functions

Piecewise

Let be the piecewise-defined function graphed above. Define function $g(x) = x^{2} - 11$ .

Evaluate $\left (g \circ f^{-1} \right ) (4)$ .

Possible Answers:

$45\frac{1}{2}$

$-10\frac{3}{4}$

4 is not in the domain of $g \circ f^{-1}$

Correct answer:

$-10\frac{3}{4}$

Explanation:

$\left (g \circ f^{-1} \right ) (4) = g [f ^{-1} (4)]$

$f ^{-1} (4) = c$ such that f(c)= 4 .As seen in the diagram below, the graph of includes the point $\left (\frac{1}{2}, 4 \right )$ , so $f ^{-1} (4) = \frac{1}{2}$ .

Piecewise 1

$\left (g \circ f^{-1} \right ) (4) = g \left ( \frac{1}{2} \right )$

$g(x) = x^{2} - 11$ , so

$g\left ( \frac{1}{2} \right )=\left ( \frac{1}{2} \right )^{2} - 11 = \frac{1}{4} -11 = -10 \frac{3}{4}$

$\left (g \circ f^{-1} \right ) (4) =-10 \frac{3}{4}$ , the correct choice.

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Example Question #76 : Understanding Functions

Piecewise

Let be the piecewise-defined function graphed above. Define function $g(x) = x^{2} - 11$ .

Evaluate $\left (f^{-1} \circ g \right ) (4)$ .

Possible Answers:

$-10\frac{3}{4}$

$45\frac{1}{2}$

4 is not in the domain of $f^{-1} \circ g$ .

Correct answer:

4 is not in the domain of $f^{-1} \circ g$ .

Explanation:

$\left (f^{-1} \circ g \right ) (4) = f ^{-1} [g(4)]$

$g(x) = x^{2} - 11$ , so

$g(4) = 4^{2} - 11 = 16 -11 = 5$

Therefore,

$\left (f^{-1} \circ g \right ) (4) = f ^{-1} (5)$ , which is equal to such that f(c) = 5 .

However, the range of the function , as can be seen from the diagram, is $[-9, -6] \cup [-4, 5)$ - 5 lies outside the range of , and, consequently, outside the domain of $f ^{-1}$ . Therefore, the expression $f ^{-1} (5)$ is undefined, and, equivalently, 4 is not in the domain of $f^{-1} \circ g$ .

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Example Question #77 : Understanding Functions

Piecewise

Let be the piecewise-defined function graphed above. Define a function g(x) = x-4 .

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

Possible Answers:

$-10\frac{1}{2}$

6 is not in the domain of $g^{-1} \circ f$ .

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

$-6\frac{1}{2}$

$-8\frac{1}{2}$

Correct answer:

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

Explanation:

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

From the diagram below, it can be seen that the point $\left ( 6, -6 \frac{1}{2}\right )$ is on the graph of .

Therefore, $f(6) = -6\frac{1}{2}$ , and

$\left ( g^{-1} \circ f \right ) (6) = g^{-1}\left ( -6 \frac{1}{2}\right )$

$g^{-1}\left ( -6 \frac{1}{2}\right ) = c$ if $g(c) =- 6 \frac{1}{2}$ , so, since

g(x) = x-4

$x-4 = -6 \frac{1}{2}$

$x = -2\frac{1}{2}$ ,

and $g^{-1}\left ( -6 \frac{1}{2}\right ) = -2\frac{1}{2}$

Therefore, $\left ( g^{-1} \circ f \right ) (6) = -2\frac{1}{2}$ .

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Example Question #78 : Understanding Functions

Piecewise

Let be the piecewise-defined function graphed above. Define a function g(x) = x-4 .

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

Possible Answers:

$\textup{6 is not in the domain of } f \circ g^{-1}$ .

$-8\frac{1}{2}$

$-6\frac{1}{2}$

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

$-10\frac{1}{2}$

Correct answer:

$\textup{6 is not in the domain of } f \circ g^{-1}$ .

Explanation:

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

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

g(x) = x-4

x-4 = 6

x = 10

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 [-8, 7] . 10 is not in the domain of . Therefore, is not in the domain of $f \circ g^{-1}$ .

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Example Question #71 : Understanding Functions

Give the third term of a sequence $a_{n}$ .

Statement 1: The first and second terms are 10 and 20, respectively.

Statement 2: The fourth and fifth terms are 40 and 50, respectively.

Possible Answers:

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

EITHER STATEMENT ALONE provides sufficient information to answer the question.

STATEMENT 2 ALONE provides sufficient information to answer the question, but STATEMENT 1 ALONE does NOT provide sufficient information to answer the question.

BOTH STATEMENTS TOGETHER provide sufficient information to answer the question, but NEITHER STATEMENT ALONE provides sufficient information to answer the question.

STATEMENT 1 ALONE provides sufficient information to answer the question, but STATEMENT 2 ALONE does NOT provide sufficient information to answer the question.

Correct answer:

BOTH STATEMENTS TOGETHER do NOT provide sufficient information to answer the question.

Explanation:

The two statements are insuffcient to determine the third term. Between them, only four of the terms are given. While the sequence seems to be arithmetic with common difference 10, this is not explicitly stated; no specific rule is given for the sequence.

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Example Question #80 : Understanding Functions

Piecewise

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

g(x) = 2x-7 .

Give the domain of the function $f \circ g$ .

Possible Answers:

$\left [-1, \frac{1}{2}\right ] \cup \left [1\frac{1}{2}, 6 \right )$

$(-\infty, \infty)$

$\left [ -\frac{1}{2}, 7 \right ]$

$\left [ -25, -19\right ] \cup \left [-11, 3 \right )$

$\left [ -23, 7\right ]$

Correct answer:

$\left [ -\frac{1}{2}, 7 \right ]$

Explanation:

is a polynomial function, so its own domain is the set of all real numbers. This does not restrict the domain of $f \circ g$ . However, since $(f \circ g)(x) = f[g(x)]$ , it follows that g(x) must fall within the domain of , which is [-8, 7] .

Therefore,

$-8 \le g(x) \le 7$

$-8 \le 2x- 7 \le 7$

$-8 + 7 \le 2x- 7+ 7 \le 7 + 7$

$-1 \le 2x \le 14$

$-1\div 2 \le 2x \div 2 \le 14 \div 2$

$-\frac{1}{2} \le x \le 7$

The domain of $f \circ g$ is the set $\left [ -\frac{1}{2}, 7 \right ]$ .

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GMAT Math : Functions/Series

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #71 : Understanding Functions

Example Question #72 : Understanding Functions

Example Question #73 : Understanding Functions

Example Question #74 : Understanding Functions

Example Question #75 : Understanding Functions

Example Question #76 : Understanding Functions

Example Question #77 : Understanding Functions

Example Question #78 : Understanding Functions

Example Question #71 : Understanding Functions

Example Question #80 : Understanding Functions

All GMAT Math Resources

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