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SAT II Math II : SAT Subject Test in Math II

Study concepts, example questions & explanations for SAT II Math II

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Example Questions

Example Question #41 : Properties Of Functions And Graphs

What is the vertex of y=-3x^2-5x+3 ? Is it a max or min?

Possible Answers:

$-\frac{5}{6}, \textup{Min}$

$-\frac{5}{3}, \textup{Min}$

$-\frac{5}{6}, \textup{Max}$

$\frac{5}{6}, \textup{Min}$

$-\frac{5}{3}, \textup{Max}$

Correct answer:

$-\frac{5}{6}, \textup{Max}$

Explanation:

The polynomial is in standard form of a parabola.

y=ax^2+bx+c

To determine the vertex, first write the formula.

$x=-\frac{b}{2a}$

Substitute the coefficients.

$x=-\frac{-5}{2(-3)} = -\frac{5}{6}$

Since the is negative is negative, the parabola opens down, and we will have a maximum.

The answer is: $-\frac{5}{6}, \textup{Max}$

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Example Question #1 : Maximum And Minimum

Given the parabola equation y=4x-3x^2 , what is the max or minimum, and where?

Possible Answers:

$\textup{Max}, (-\frac{2}{3},-4)$

$\textup{Min}, (-\frac{2}{3},-4)$

$\textup{Max}, (\frac{2}{3}, \frac{4}{3})$

$\textup{Min}, (\frac{2}{3}, \frac{4}{3})$

$\textup{Max}, (\frac{2}{3},4)$

Correct answer:

$\textup{Max}, (\frac{2}{3}, \frac{4}{3})$

Explanation:

The parabola is in the form: y=ax^2+bx+c

y=-3x^2+4x

The vertex formula will determine the x-value of the max or min. Since the value of is negative, the parabola will open downward, and there will be a maximum.

Write the vertex formula and substitute the correct coefficients.

$x=-\frac{b}{2a} = -\frac{4}{2(-3)} = \frac{2}{3}$

Substitute this value back in the parabolic equation to determine the y-value.

$y=-3( \frac{2}{3})^2+4( \frac{2}{3}) = \frac{4}{3}$

The answer is: $\textup{Max}, (\frac{2}{3}, \frac{4}{3})$

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Example Question #261 : Sat Subject Test In Math Ii

Define the functions and on the set of real numbers as follows:

$f (x) = \sqrt{x- 16}$

$g(x) = \sqrt{x- 4}$

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

Possible Answers:

$[260, \infty)$

$[16, \infty)$

$[4, \infty)$

$[0 , \infty)$

$[32, \infty)$

Correct answer:

$[260, \infty)$

Explanation:

The natural domain of the composite function $f \circ g$ is defined to be the intersection of the two sets.

One set is the natural domain of . Since is defined to be the square root of an expression, the radicand must be nonnegative. Therefore,

$x - 4 \geq 0$

$x \geq 4$

This set is $[4, \infty)$ .

The other set is the set of numbers that the function pairs with a number within the domain of . Since the radicand of the square root in must be nonnegative,

$x - 16 \geq 0$

$x \geq 16$

For g(x) to fall within this set:

$g(x) \geq 16$

$\sqrt{x- 4} \geq 16$

$x - 4 \geq 256$

$x \geq 260$

This set is $[260, \infty)$ .

$[4, \infty) \cap [260, \infty) = [260, \infty)$ , which is the natural domain.

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Example Question #1 : Solving Linear Functions

If f(x) = (3x+5)^2 , what must f(-2) be?

Possible Answers:

121

Correct answer:

Explanation:

Replace the value of negative two with the x-variable.

There is no need to use the FOIL method to expand the binomial.

f(-2) = (3(-2)+5)^2 = (-6+5)^2 = (-1)^2 = 1

The answer is:

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Example Question #1 : Solving Functions

Let f(x) = -3x+9 . What is the value of $f(\frac{7}{8})$ ?

Possible Answers:

$\frac{51}{8}$

$\frac{17}{3}$

$\frac{17}{2}$

$\frac{51}{4}$

Correct answer:

$\frac{51}{8}$

Explanation:

Substitute the fraction as .

$f(\frac{7}{8}) = -3(\frac{7}{8})+9$

Multiply the whole number with the numerator.

$\frac{-21}{8}+9$

Convert the expression so that both terms have similar denominators.

$-\frac{21}{8}+\frac{9(8)}{1(8)} = -\frac{21}{8}+\frac{72}{8}$

The answer is: $\frac{51}{8}$

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Example Question #52 : Functions And Graphs

If f(x) = 5 , what must f(10) be?

Possible Answers:

Correct answer:

Explanation:

A function of x equals five. This can be translated to:

y=5

This means that every point on the x-axis has a y value of five.

Therefore, f(10) =5 .

The answer is:

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Example Question #262 : Sat Subject Test In Math Ii

Rewrite as a single logarithmic expression:

$\ln (y+ 1) - \ln (y + 2) + \ln (y + 3)$

Possible Answers:

$= \ln \frac{y+1}{y^{2}+5y+6}$

$\ln \left (y + 4 \right )$

$= \ln \frac{y^{2}+4y + 3}{y + 2}$

$\ln \frac{y^{2}+3}{y + 2}$

$\ln \left (y + 2 \right )$

Correct answer:

$= \ln \frac{y^{2}+4y + 3}{y + 2}$

Explanation:

Using the properties of logarithms

$\ln a - \ln b = \ln \frac{a}{b}$ and $\ln a + \ln b = \ln ab$ ,

simplify as follows:

$\ln (y+ 1) - \ln (y + 2) + \ln (y + 3)$

$= \ln \frac{y+ 1}{y + 2} + \ln (y + 3)$

$= \ln \left [\frac{y+ 1}{y + 2} \cdot (y + 3) \right ]$

$= \ln \frac{y^{2}+4y + 3}{y + 2}$

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Example Question #263 : Sat Subject Test In Math Ii

Simplify by rationalizing the denominator:

$\frac{11}{6- \sqrt{3}}$

Possible Answers:

$\frac{11\sqrt{3}}{3}$

$2+ \sqrt{3}$

$\frac{6+ \sqrt{3}}{3 }$

$2- \sqrt{3}$

$\frac{6- \sqrt{3}}{3 }$

Correct answer:

$\frac{6+ \sqrt{3}}{3 }$

Explanation:

Multiply the numerator and the denominator by the conjugate of the denominator, which is $6 + \sqrt{3}$ . Then take advantage of the distributive properties and the difference of squares pattern:

$\frac{11}{6- \sqrt{3}}$

$= \frac{11(6+ \sqrt{3})}{\left (6- \sqrt{3} \right )(6+ \sqrt{3})}$

$= \frac{11(6+ \sqrt{3})}{ 6^{2}- \left (\sqrt{3} \right ) ^{2}\right }$

$= \frac{11(6+ \sqrt{3})}{36-3 }$

$= \frac{11(6+ \sqrt{3})}{3 3 }$

$= \frac{6+ \sqrt{3}}{3 }$

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Example Question #264 : Sat Subject Test In Math Ii

Simplify:

$\sqrt[3]{\sqrt[2]{x^{-9}}}$

You may assume that is a nonnegative real number.

Possible Answers:

$x \sqrt{x}$

$\frac{1}{x^{3}}$

$\frac{ \sqrt[3]{x}}{x }$

$\sqrt[3]{x}$

$\frac{ \sqrt{x}}{x^{2}}$

Correct answer:

$\frac{ \sqrt{x}}{x^{2}}$

Explanation:

The best way to simplify a radical within a radical is to rewrite each root as a fractional exponent, then convert back.

First, rewrite the roots as exponents.

$\sqrt[3]{\sqrt[2]{x^{-9}}} = \sqrt[3]{\left ( x^{-9} \right ) ^{\frac{1}{2}} } = \left ( \left ( x^{-9} \right ) ^{\frac{1}{2}} \right )^{\frac{1}{3}} = x ^{-9 \cdot \frac{1}{2} \cdot \frac{1}{3}} } } = x ^{- \frac{3}{2} } }= \frac{1}{x^{\frac{3}{2}}}$

Then convert back to a radical and rationalizing the denominator:

$\frac{1}{x^{\frac{3}{2}}} = \frac{1}{\sqrt {x^{3}}} = \frac{1 \cdot \sqrt{x}}{\sqrt {x^{3} \cdot \sqrt{x}}} = \frac{ \sqrt{x}}{\sqrt {x^{4} }} =\frac{ \sqrt{x}}{x^{2}}$

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Example Question #1 : Solving Exponential, Logarithmic, And Radical Functions

Let $f(x) = 2^{x}+ 3^{2x}$ . What is the value of f(-2) ?

Possible Answers:

$\frac{85}{324}$

$\frac{1}{85}$

$\textup{The answer is not given.}$

Correct answer:

$\frac{85}{324}$

Explanation:

Replace the integer as .

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

Evaluate each negative exponent.

$2^{-2} = \frac{1}{2^2} = \frac{1}{4}$

$3^{-4} = \frac{1}{3^4} = \frac{1}{81}$

Sum the fractions.

$\frac{1}{4}+\frac{1}{81} = \frac{1(81)}{4(81)}+\frac{1(4)}{81(4)} = \frac{85}{324}$

The answer is: $\frac{85}{324}$

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All SAT II Math II Resources

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SAT II Math II : SAT Subject Test in Math II

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #41 : Properties Of Functions And Graphs

Example Question #1 : Maximum And Minimum

Example Question #261 : Sat Subject Test In Math Ii

Example Question #1 : Solving Linear Functions

Example Question #1 : Solving Functions

Example Question #52 : Functions And Graphs

Example Question #262 : Sat Subject Test In Math Ii

Example Question #263 : Sat Subject Test In Math Ii

Example Question #264 : Sat Subject Test In Math Ii

Example Question #1 : Solving Exponential, Logarithmic, And Radical Functions

All SAT II Math II Resources

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