GMAT Math : Functions/Series

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Example Questions

Example Question #21 : Understanding Functions

Which of these functions is an example of a function with an inverse?

Possible Answers:

$g(x) = x^{4} - 1$

h(x) = 6x + 7

$C(x) = \cos 2 x$

$R(x) = \frac{1}{x^ {2}}$

$f(x) = 3x^{2} + 5$

Correct answer:

h(x) = 6x + 7

Explanation:

For a function to have an inverse, if f(a) = f(b) , then a = b . We can show this is not the case for four of these functions by providing one counterexample in each case.

$C(x) = \cos 2 x$

$C(0) = \cos 2 \cdot 0 = \cos 0 = 1$

$C( \pi ) = \cos 2 \cdot \pi = \cos 2 \pi = 1$

$f(x) = 3x^{2} + 5$

$f(1) = 3\cdot 1^{2} + 5 = 3\cdot 1 + 5= 3 + 5= 8$

$f(-1) = 3\cdot \left (-1 \right )^{2} + 5 = 3\cdot 1 + 5= 3 + 5= 8$

$g(x) = x^{4} - 1$

$g(1) = 1^{4} - 1 = 1-1 = 0$

$g(-1) = \left (-1 \right )^{4} - 1 = 1-1 = 0$

$R(x) = \frac{1}{x^ {2}}$

$R(1) = \frac{1}{1^ {2}} = \frac{1}{1} = 1$

$R(-1) = \frac{1}{\left (-1 \right )^ {2}} = \frac{1}{1} = 1$

However, we can show that h(x) = 6x + 7 has an inverse function by demonstrating that if h(a) = h(b) , then a = b .

h(a) = h(b)

6a + 7 = 6b + 7

6a + 7 -7 = 6b + 7-7

6a = 6b

$6a \div 6 = 6b \div 6$

a = b

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Example Question #162 : Algebra

Define g(x) = 2x - |x| .

Which of the following would be a valid alternative way of expressing the definition of ?

Possible Answers:

$g (x) = \left\{\begin{matrix} 3x & \textrm{ if } x < 0\\ 2x& \textrm{ if } x \geq 0 \end{matrix}\right.$

$g (x) = \left\{\begin{matrix} 2x & \textrm{ if } x < 0\\ 3x& \textrm{ if } x \geq 0 \end{matrix}\right.$

$g (x) = \left\{\begin{matrix} 3x & \textrm{ if } x < 0\\ x& \textrm{ if } x \geq 0 \end{matrix}\right.$

$g (x) = \left\{\begin{matrix} x & \textrm{ if } x < 0\\ 3x& \textrm{ if } x \geq 0 \end{matrix}\right.$

g (x) = 3 x

Correct answer:

$g (x) = \left\{\begin{matrix} 3x & \textrm{ if } x < 0\\ x& \textrm{ if } x \geq 0 \end{matrix}\right.$

Explanation:

If x < 0 , then |x| = -x ,and subsequently, g(x) = 2x - |x| = 2x - (-x) = 2x + x = 3x

If $x \geq 0$ , then |x| = x ,and subsequently, g(x) = 2x - |x| = 2x - x = x

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Example Question #22 : Understanding Functions

Which of these functions is neither an even function nor an odd function?

Possible Answers:

$F(x) = x^{3} +4x$

$P(x) = x^{4} + 2x^{2} + 3$

$g(x) = x ^{2}\sin x + \sin ^{3} x$

$f(x) = x^{2}\cos x$

$G(x) = x^{3} + x^{2} + x + 1$

Correct answer:

$G(x) = x^{3} + x^{2} + x + 1$

Explanation:

A function is odd if and only if f(-x) = -f(x) for each value of in the domain; it is even if and only if f(-x) = f(x) for each value of in the domain. We can identify each of these functions as even, odd, or neither by evaluating each at and comparing the resulting expression to the definition of the function.

$f(x) = x^{2}\cos x$

$f(-x) = (-x)^{2}\cos (-x) = x^{2} \cos x = f(x)$

making even.

$g(x) = x ^{2}\sin x + \sin ^{3} x$

$g(-x) =\left ( -x \right ) ^{2} \sin \left \( -x \right ) + \sin ^{3} \left ( -x \right )$

$=x ^{2} \left ( -\sin x \right )+\left ( -\sin x \right )^{3}$

$=- x ^{2} \sin x -\sin^{3} x$

$= - \left ( x ^{2}\sin x + \sin ^{3} x \right ) = -g(x)$

making odd.

$F(x) = x^{3} +4x$

$F(-x) = (-x)^{3} +4 (-x)$

$= -x^{3} - 4x$

$= -\left ( x^{3} +4x \right ) = - F(x)$

making odd.

$P(x) = x^{4} + 2x^{2} + 3$

$P(-x) = (-x) ^{4} + 2(-x) ^{2} + 3$

$= x ^{4} + 2x ^{2} + 3 = P(x)$

making even.

$G(x) = x^{3} + x^{2} + x + 1$

$G(-x) = (-x)^{3} + (-x)^{2} +(-x) + 1 = - x^{3} + x^{2} - x + 1$

Since neither G(-x) = G(x) nor G(-x) = -G(x) , is neither even nor odd, making this the correct choice.

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Example Question #1243 : Problem Solving Questions

If $x\ \bigstar\ y = x*y^2$ , then what does $(2\ \bigstar\ 2)\ \bigstar\ 2$ equal?

Possible Answers:

2^6

2^4

2^3

2^5

2^7

Correct answer:

2^5

Explanation:

This problem can be evaluated by determining what the value of the parentheses is and then using that to evaluate the rest of the term.

$(2\bigstar 2)=2*2^2=2^3$

The term then reduces to $2^3\bigstar 2 = 2^3*2^2=2^5$

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Example Question #21 : Functions/Series

Find the next term in the series $1,\frac{-1}{2},\frac{1}{4},\frac{-1}{8}.$

Possible Answers:

$\frac{1}{12}$

$-\frac{1}{16}$

$-\frac{1}{12}$

$\frac{1}{16}$

Correct answer:

$\frac{1}{16}$

Explanation:

We can see that we have a geometric series. The geometric factor can be found by dividing the second term by the first. Doing this, we get $\frac{-\frac{1}{2}}{1}=-\frac{1}{2}.$ To find the next term in the series, we simply multiply the last term by the geometric factor to get $\frac{-1}{8}*\frac{-1}{2}=\frac{1}{16}$ .

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Example Question #22 : Functions/Series

Solve for f(g(4)) .

f(x)=x^2

g(x)=x-7

Possible Answers:

Correct answer:

Explanation:

First, solve for g(4) .

g(4)=4-7=-3

Next, solve for f(-3) .

f(-3)=(-3)^2=9

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Example Question #25 : Understanding Functions

Define an operation $\Cup$ as follows:

$a \Cup b = 0$ if either or , but not both, are negative, and $a \Cup b = a+ b$ otherwise.

Evaluate:

$\left [(-8) \Cup (-8) \right ]+\left ( 8\Cup8 \right )$

Possible Answers:

Correct answer:

Explanation:

Evaluate each of $(-8) \Cup (-8)$ and $8\Cup8 \right$ separately. Since in both cases, the numbers have the same sign, replace the numbers for and in the second expression - that is, simply add them.

$\left [(-8) \Cup (-8) \right ]+\left ( 8\Cup8 \right )$

$= \left [ (-8 )+ (-8) \right ] + (8+8)$

= -16 + 16 = 0

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Example Question #168 : Algebra

Define an operation $\pounds$ as follows:

For all real numbers a,b ,

$a \pounds b = -ab$ if a > b , and $a \pounds b = ab$ if $a \leq b$

Evaluate:

$\left ( 5 \pounds 3\right )\times \left ( 3 \pounds 5\right )$

Possible Answers:

225

Correct answer:

Explanation:

Evaluate $5 \pounds 3$ and $3 \pounds 5$ separately.

Since 5 > 3 , use

$a \pounds b = -ab$ for $5 \pounds 3$

$5\pounds 3 = - 5 \times 3 = -15$

Since 3 <5 , use

$a \pounds b = ab$ for $3 \pounds 5$

$3 \pounds 5 = 3 \times 5 = 15$

$\left ( 5 \pounds 3\right )\times \left ( 3 \pounds 5\right ) = -15 \times 15 = -225$

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Example Question #26 : Understanding Functions

Define the operation $\Cap$ as follows:

For real a,b ,

$a \Cap b = 0$ if both and are negative, and $a \Cap b = ab + 1$ otherwise.

Evaluate $\left [8 \Cap 8 \right ]+ \left [ (-8) \Cap 8 \right ]$

Possible Answers:

128

Correct answer:

Explanation:

Evaluate each of $8 \Cap 8$ and $-8 \Cap 8$ separately. Since in both cases, at least one number is positive, replace the numbers for and in the second expression.

$a \Cap b = ab + 1$

$8 \Cap 8 = 8 \cdot 8 + 1 = 64 + 1 = 65$

$a \Cap b = ab + 1$

$(-8) \Cap 8 = -8 \cdot8 + 1 = -64 + 1 = -63$

$\left [ 8 \Cap 8 \right ]+ \left [ (-8) \Cap 8 \right ] = 65 + (-63) = 2$

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Example Question #171 : Algebra

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

= (x + 1) - (x-1)

= x + 1 - x+ 1

= 2

If -1 < x < 1 , then , then x+ 1 is positive and x - 1 is negative, so

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

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

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

=x+1 +x - 1

= 2x

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

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

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

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

=- x - 1 + x - 1

= -2

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GMAT Math : Functions/Series

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #21 : Understanding Functions

Example Question #162 : Algebra

Example Question #22 : Understanding Functions

Example Question #1243 : Problem Solving Questions

Example Question #21 : Functions/Series

Example Question #22 : Functions/Series

Example Question #25 : Understanding Functions

Example Question #168 : Algebra

Example Question #26 : Understanding Functions

Example Question #171 : Algebra

All GMAT Math Resources

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