# SAT Functions Perfect Score

To get a perfect score on SAT Math, you need to:

- Get every single problem correct.
- Have complete mastery of all of the SAT skills
- Remember the Tips and use them
- Figure out your common mistakes and avoid them

## SAT Hardest Problems

For two sets $X=\{a, b, c, d\}$ and $Y=\{-e, e\},$ a function $f: X\longrightarrow Y$ satisfies $f(a)+f(b)=f(c)+f(d).$ How many such functions $f$ are there?

(A)$\ \ 4$

(B)$\ \ 5$

(C)$\ \ 6$

(D)$\ \ 7$

(E)$\ \ 8$

Correct Answer: C

Solution:If the range of $f$ is only one element of $Y,$ we have $\begin{array}{c}&f(a)=f(b)=f(c)=f(d)=e &\text{or} &f(a)=f(b)=f(c)=f(d)=-e. \end{array}$ Since both satisfy the given condition $f(a)+f(b)=f(c)+f(d),$ we have found two such functions.

Now, if the range of $f$ is both elements of $Y,$ we have the following four functions $f$ such that $f(a)+f(b)=f(c)+f(d):$ $\begin{array}{c}&&1. &f(a)=e, &f(b)=-e, &f(c)=e, &f(d)=-e \\ &2. &f(a)=e, &f(b)=-e, &f(c)=-e, &f(d)=e \\ &3. &f(a)=-e, &f(b)=e, &f(c)=e, &f(d)=-e \\ &4. &f(a)=-e, &f(b)=e, &f(c)=-e, &f(d)=e. \end{array}$

Therefore, there are a total of $2+4=6$ functions $f$ from $X$ to $Y$ that satisfy $f(a)+f(b)=f(c)+f(d).$ Hence, the correct answer is (C).

Incorrect Choices:

(A),(B),(D), and(E)

The solution explains why these choices are wrong.

What is the area of the triangle bounded by the three lines

$\begin{array}{c}&y=2x, &y=-\frac{x}{2}, &y=x-4? \end{array}$

(A)$\ \ \frac{8\sqrt{5}}{3}$

(B)$\ \ \frac{20}{3}$

(C)$\ \ \frac{40}{3}$

(D)$\ \ \frac{80}{3}$

(E)$\ \ \frac{200}{3}$

Correct Answer: C

Solution:

As shown in the above graph, the triangle of interest $\triangle OAB$ is a right triangle because $y=2x$ and $y=-\frac{x}{2}$ are perpendicular. The area of $\triangle OAB$ is then $\frac{1}{2}\times \lvert \overline{OA}\rvert \times \lvert \overline{OB}\rvert. \qquad (1)$ Let us now calculate $\lvert \overline{OA}\rvert$ and $\lvert \overline{OB}\rvert.$

Equating $y=2x$ and $y=x-4$ gives $x=-4.$ Substituting this into $y=2x$ gives $y=-8,$ which implies $A=(-4, -8).$ Then $\lvert \overline{OA}\rvert=\sqrt{(-4)^2+(-8)^2}=4\sqrt{5}. \qquad (2)$

Similarly, equating $y=-\frac{x}{2}$ and $y=x-4$ gives $x=\frac{8}{3}.$ Substituting this into $y=-\frac{x}{2}$ gives $y=-\frac{4}{3},$ which implies $B=\left(\frac{8}{3}, -\frac{4}{3}\right).$ Then $\lvert \overline{OB}\rvert=\sqrt{\left(\frac{8}{3}\right)^2+\left(-\frac{4}{3}\right)^2}=\frac{4}{3}\sqrt{5}. \qquad (3)$

Substituting $(2)$ and $(3)$ into $(1),$ the area of $\triangle OAB$ is $\frac{1}{2}\times 4\sqrt{5} \times \frac{4}{3}\sqrt{5}=\frac{40}{3}.$

Therefore, the correct answer is (C).

Incorrect Choices:

(A)

If you multiply $\sqrt{5}$ only once, you will get this wrong number.

(B)

If you mistakenly multiply $\frac{1}{2}$ twice when calculating of the area the triangle, you will get this wrong number.

(D)

If you forget about multiplying $\frac{1}{2}$ when calculating of the area the triangle, you will get this wrong number.

(E)

If you miscalculate $\sqrt{5} \times \sqrt{5}=25,$ you will get this wrong number.

The above is the graph of the quadratic function $y=ax^2+bx+c.$ Which of the following statements is true?

$\begin{array}{r r l} & \text{I.} & a+b+c<0\\ & \text{II.} & a-b<0\\ & \text{III.} & 2a-b<0\\ \end{array}$

(A)$\ \$ I only

(B)$\ \$ II only

(C)$\ \$ I and III only

(D)$\ \$ II and III only

(E)$\ \$ I, II and III

Correct Answer: C

Solution:First, since $f(1)=a+b+c<0$ from the above graph, the first statement is true.

Now, observe that the equation of the graph can be rewritten as $y=ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right),$ Then the axis of symmetry is $x=-\frac{b}{2a}. \qquad (1)$

Since the parabola intercepts the $x$-axis at $x=-2$ and also at a point in the interval $0<x<1,$ it must be true that $(1)$ lies to the left of $x=-\frac{1}{2}.$ This is because if the parabola intercepted the $x$-axis at $x=-2$ and $x=1,$ the axis of symmetry would be $x=-\frac{1}{2}.$ Thus, $-\frac{b}{2a}<-\frac{1}{2} \text{ or } \frac{b}{a}>1 \Rightarrow a-b>0,$ which makes the second statement false.Finally, since $f(-2)=4a-2b+c=0,$ we have $2a-b=-\frac{c}{2}<0,$ which makes the third statement true.

Therefore, only statements I and III are true and the correct answer is (C).

Incorrect Choices:

(A),(B),(D), and(E)

The solution explains how to eliminate these choices.

## SAT Tips for Functions

## Functions

- Don't switch the $x-$ and $y-$coordinates of a point.
- The domain of $f(g(x))$ is the set of all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$.
- $\sqrt{x^{2}} = \begin{cases} -x &\mbox{if } x < 0 \\ x & \mbox{if } x \geq 0. \\ \end{cases}$
- For $f(x)=\sqrt{x}, \quad \text{Domain:}\ x\geq 0; \quad \text{Range:}\ f(x) \geq 0.$
## Linear Functions

- The slope of a line is defined as $\frac{\text{change in}\ y}{\text{change in}\ x}.$
- A line with a positive slope rises from left to right.
- A line with a negative slope falls from left to right.
- Slope-intercept form: $y=mx+b,$ where $m$ is the line's slope, and $b$ its $y$-intercept.
- Point-slope form: $y-y_{1}=m(x-x_{1}),$ where $m$ is the line's slope, and $(x_{1}, y_{1})$ is a point on the line.
- The line $x = a$ is a vertical line that crosses the x-axis as $(a,0).$
- The line $y = b$ is a horizontal line that crosses the $y$-axis at $(0,b).$
- If two lines are parallel, their slopes are equal.
- If two lines are perpendicular, their slopes are negative reciprocals of each other.
- If two functions intersect at point $(x,y),$ then $f(x) = y = g(x)$.
- Don't switch the $x-$ and $y-$coordinates of a point.
- When transforming graphs, trace what happens to each point.
## Quadratic Functions

- The parabola $y=a(x-h)^{2}+k$ opens up if $a>0.$
- The parabola $y=a(x-h)^{2}+k$ opens down if $a<0.$
- The parabola $y=ax^{2}+bx+c$ opens up if $a>0.$
- The parabola $y=ax^{2}+bx+c$ opens down if $a<0.$
- The parabola $y=ax^{2}+bx+c$ has a $y-$intercept at $y=c.$
- The parabola $y=a(x-h)^{2}+k$ has a vertex at $(h, k).$
- The parabola $y=a(x-h)^{2}+k$ has an axis of symmetry at $x=h$
- The parabola $y=ax^{2}+bx+c$ has a vertex at $(-\frac{b}{2a}, f(-\frac{b}{2a})).$
- The parabola $y=ax^{2}+bx+c$ has an axis of symmetry at $x=-\frac{b}{2a}$
## Coordinate Geometry

- The line $x = a$ is a vertical line that crosses the x-axis as $(a,0).$
- The line $y = b$ is a horizontal line that crosses the $y$-axis at $(0,b).$
- If two lines are parallel, their slopes are equal.
- If two lines are perpendicular, their slopes are negative reciprocals of each other.
- If two functions intersect at point $(x,y),$ then $f(x) = y = g(x)$.
- Don't switch the $x-$ and $y-$coordinates of a point.
- Distance formula: $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}.$
- Midpoint formula: $M=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right).$
- When transforming graphs, trace what happens to each point.
## Functions as Models

- Exponential growth: $y=ba^{x}$, where $a>1$ and $b > 0.$
- Exponential decay: $y=ba^{x}$, where $0<a<1.$ and $b>0.$
## Newly Defined Functions

- Follow directions exactly.
## Direct and Inverse Variation

- Direct variation: $y=k\cdot x, \quad k\neq0.$
- Inverse variation: $y=k\cdot\frac{1}{x}, \quad k\neq0.$
## Translating Word Problems

## Word Problemes

- Distance = Rate $\times$ Time.
## Student-Produced Response

## SAT General Tips

**Cite as:**SAT Functions Perfect Score.

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