# 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 \[\begin{array} &f(a)=f(b)=f(c)=f(d)=e &\text{or} &f(a)=f(b)=f(c)=f(d)=-e. \end{array}\] 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 \(f(a)+f(b)=f(c)+f(d):\) \[\begin{array} &&1. &f(a)=e, &f(b)=-e, &f(c)=e, &f(d)=-e \\ &2. &f(a)=e, &f(b)=-e, &f(c)=-e, &f(d)=e \\ &3. &f(a)=-e, &f(b)=e, &f(c)=e, &f(d)=-e \\ &4. &f(a)=-e, &f(b)=e, &f(c)=-e, &f(d)=e. \end{array}\]

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

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

(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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