SAT Newly Defined Functions

To solve newly defined functions on the SAT, you need to know how to

• apply the order of operations
• work with fractions
• manipulate algebraic expressions
• make simple substitutions
• apply the rules of exponents
• apply reasoning skills.

Some SAT questions use strange symbols that don't normally appear in a math problem. You will always be told what the symbols mean. For example, let $a \oslash b=a^{b}$. The function $a \oslash b$ is defined as the number to the left of the strange symbol, $a$, raised to the power of $b$, the number to the right of the strange symbol. So, $3 \oslash 2$ will be equal to the number to the left of the strange symbol, $3$, raised to the power of $2$, the number to the right of the strange symbol. That is, $3 \oslash 2=3^{2}$. The key to solving newly defined functions is to follow directions exactly.

For any positive number $x$, let $x \Updownarrow$ be defined as $x \Updownarrow = x^{2} + x$. What is the value of $5\Updownarrow$?

(A) $\ \ 0$
(B) $\ \ 5$
(C) $\ \ 10$
(D) $\ \ 25$
(E) $\ \ 30$

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Correct Answer: E

Solution:

Tip: Follow directions exactly.
Start with the given definition and follow its instructions exactly:

$$\begin{array}{l c l l l} x \Updownarrow &=& x^{2} + x &\qquad \text{given definition} &(1)\\ 5 \Updownarrow &=& 5^{2} + 5 &\qquad \text{plug values in definition} &(2)\\ &=& 25+5 &\qquad 5^{2} = 25 &(3)\\ &=& 30. &\qquad \text{add} &(4)\\ \end{array}$$

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Incorrect Choices:

(A)
Tip: Eliminate obviously wrong answers.
We are told that $x$ is positive. So, $x^{2} + x$ can never equal 0.

(B)
Tip: Read the entire question carefully.
Tip: Follow directions exactly.
If you forget the first term in the definition and solve $x \Updownarrow = x$ instead of the given $x \Updownarrow = x^{2} + x$, you will get this wrong answer.

(C)
Tip: Read the entire question carefully.
Tip: Follow directions exactly.
If you forget to square the first term in the definition and solve $x \Updownarrow = x + x$ instead of the given $x \Updownarrow = x^{2} + x$, you will get this wrong answer.

(D)
Tip: Read the entire question carefully.
Tip: Follow directions exactly.
If you forget the second term in the definition and solve $x \Updownarrow = x^{2}$ instead of the given $x \Updownarrow = x^{2} + x$, you will get this wrong answer.

Let $m$ and $n$ be positive integers. If $m \heartsuit n = (m-n)(m+n)\left(m^{2}+n^{2}\right)$, what is the value of $((0 \heartsuit 1) \heartsuit 0) \heartsuit 1$?

(A) $\ \ -1$
(B) $\ \ 0$
(C) $\ \ 1$
(D) $\ \ 2$
(E) $\ \ 3$

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Correct Answer: B

Solution 1:

Tip: Follow order of operations.
Start with the innermost parentheses $(\underbrace{(0 \heartsuit 1)} \heartsuit 0) \heartsuit 1$.

$$\begin{array}{l c l l l} m \heartsuit n &=& (m-n)(m+n)\left(m^{2}+n^{2}\right) &\qquad \text{given definition} &(1)\\ 0 \heartsuit 1 &=& (0-1)(0+1)\left(0^{2}+1^{2}\right) &\qquad \text{follow directions exactly} &(2)\\ &=& (-1)(1)(1) &\qquad \text{simplify parentheses} &(3)\\ &=& -1. &\qquad \text{multiply} &(4)\\ \end{array}$$

Replacing $0 \heartsuit 1$ in the original expression with $-1$, we again evaluate the parentheses first: $\underbrace{(-1 \heartsuit 0)} \heartsuit 1$.

$$\begin{array}{l c l l l} m \heartsuit n &=& (m-n)(m+n)\left(m^{2}+n^{2}\right)&\qquad \text{given definition} &(5)\\ -1 \heartsuit 0 &=& (-1-0)(-1+0)\left((-1)^{2}+0^{2}\right)&\qquad \text{follow directions exactly} &(6)\\ &=& (-1)(-1)(1) &\qquad \text{simplify parentheses} &(7)\\ &=& 1. &\qquad \text{multiply} &(8) \end{array}$$

Replacing $(-1 \heartsuit 0)$ in the previous expression with $1$, we evaluate $1 \heartsuit 1$.

$$\begin{array}{l c l l l} m \heartsuit n &=& (m-n)(m+n)\left(m^{2}+n^{2}\right) &\qquad \text{given definition} &(9)\\ 1 \heartsuit 1 &=& (1-1)(1+1)\left((1)^{2}+1^{2}\right) &\qquad \text{follow directions exactly} &(10)\\ &=& (0)(2)(1) &\qquad \text{simplify parentheses} &(11)\\ &=& 0. &\qquad \text{multiply} &(12) \end{array}$$

Solution 2:

Tip: $a^{2}-b^{2}=(a-b)(a+b)$
Tip: Follow order of operations.
We simplify the definition of $m \heartsuit n$ first, using the difference of squares formula.

$$\begin{array}{l c l l l} m \heartsuit n &=& (m-n)(m+n)\left(m^{2}+n^{2}\right) &\qquad (1)\\ &=& \left(m^{2}-n^{2}\right)\left(m^{2}+n^{2}\right) &\qquad (2)\\ &=& m^{4}-n^{4}. &\qquad (3)\\ \end{array}$$

Now when we apply $m\heartsuit n$, we can use this simplified version of the definition:

$$\begin{array}{l c l l l} ((0 \heartsuit 1) \heartsuit 0) \heartsuit 1 &=& \left((0^{4}-1^{4}\right)\heartsuit 0) \heartsuit 1 &\qquad \text{apply}\ \heartsuit\ \text{on innermost parentheses} &(4)\\ &=& ((0-1)\heartsuit 0) \heartsuit 1 &\qquad \text{apply}\ 0^{a}=0\ \text{and}\ 1^{a}=1 &(5)\\ &=& (-1\heartsuit 0) \heartsuit 1 &\qquad \text{simplify innermost parentheses} &(6)\\ &=& \left((-1)^{4} - 0^{4}\right) \heartsuit 1 &\qquad \text{apply}\ \heartsuit\ \text{on parentheses} &(7)\\ &=& (1-0) \heartsuit 1 &\qquad (-1)^{4}=1\ \text{and}\ 0^{a}=0&(8)\\ &=& 1\heartsuit 1 &\qquad \text{simplify parentheses} &(9)\\ &=& 1^{4}-1^{4} &\qquad \text{apply}\ \heartsuit &(10)\\ &=& 1-1 &\qquad \text{apply}\ 1^{a}=1 &(11)\\ &=& 0. &\qquad \text{subtract} &(12) \end{array}$$

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Incorrect Choices:

(A)
If in Solution 1 you stop at step $(4)$, only calculating $(\underbrace{(0 \heartsuit 1)} \heartsuit 0) \heartsuit 1$, you will get this wrong answer.

(C)
If in Solution 1 you stop at step $(8)$, only calculating $(\underbrace{(0 \heartsuit 1) \heartsuit 0)} \heartsuit 1$, you will get this wrong answer.

(D)
If you think the operator $\heartsuit$ is a plus sign, you will get this wrong answer.

(E)
This wrong answer choice is just meant to confuse you.

For any positive integer $x$ let $\uplus$ be defined as $y\uplus x = -x(-(-x^{y})-x)-x$. What is the value of $2\uplus 3$?

(A) $\ \ -21$
(B) $\ \ -14$
(C) $\ \ -3$
(D) $\ \ 14$
(E) $\ \ 33$

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Correct Answer: A

Solution:

Tip: Follow directions exactly.
Tip: Follow order of operations.

We have

$$\begin{array}{lclll} y\uplus x &=& -x\left(-\left(-x^{y}\right)-x\right)-x &\qquad \text{begin with the definition} &(1)\\ 2\uplus 3 &=& -3\left(-\left(-3^{2}\right)-3\right)-3 &\qquad \text{plug in}\ x\ \text{and}\ y &(2)\\ &=& -3(-(-9)-3)-3 &\qquad 3^{2}=9 &(3)\\ &=& -3(9-3)-3 &\qquad \text{use distributive property} &(4)\\ &=& -3(6)-3 &\qquad 9-3=6 &(5)\\ &=& -18-3 &\qquad \text{perform multiplication} &(6)\\ &=& -21. &\qquad \text{perform subtraction} &(7)\\ \end{array}$$

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Incorrect Choices:

(B)
Tip: Follow directions exactly.
If you solve $3\uplus 2 =-2\left(-\left(-2^{3}\right)-2\right)-2$, you will get this wrong answer. The question asks for $2\uplus 3$, not for $\boxed{3\uplus 2}$.

(C)
Tip: Follow order of operations.
If you forget to raise the second $x$ to the power of $y$, like this: $-3(-(-\fbox{3})-3)-3=-3(3-3)-3=-3$, you will get this wrong answer.

(D)
This answer is offered to confuse you and make you want to choose between $-14$ and $14$.

(E)
Tip: Be careful with signs!
If in step $(3)$ you forget the negative sign in front of the squared term, you will get this wrong answer: $$\begin{array}{lclll} && -3(-\left(\fbox{-}3^{2}\right)-3)-3 & &(2)\\ &=& -3(-(9)-3)-3. &\qquad \text{mistake: forgot the negative sign} &(3)\\ \end{array}$$

Alternatively, if in step $(4)$ you don't distribute the negative sign as shown below, you will get this wrong answer: $$\begin{array}{lclll} &&-3(\fbox{-}(-9)-3)-3 & &(3)\\ &=&-3(-9-3)-3. &\qquad \text{mistake: didn't distribute the negative sign} &(4)\\ \end{array}$$

If you thought these examples difficult and you need to review the material, these links will help:

• SAT Polynomials
• SAT Fractions and Decimals
• SAT Reasoning Skills
• Multi-Step Equations-Basic
• Applying the Difference of Two Squares Identity
  $a^{2}-b^{2}=(a+b)(a-b)$
• Applying the Perfect Square Identity
  $(a \pm b)^{2}=a^{2} \pm 2ab + b^{2}$
• Rules of Exponents
  $a^m \times a^n = a^{ m + n }$, $a^n / a^m = a^ { n - m }\ldots$

• Follow directions exactly.
• Follow order of operations.
• Be careful with signs!
• SAT General Tips