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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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2 Diagnostic Tests 130 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Solving Piecewise And Recusive Functions

Define functions and as follows:

$f(x) = \left\{\begin{matrix} 2x &x< -1 \\ 3x & x > 1 \end{matrix}\right.$

$g(x) = \left\{\begin{matrix} x+ 5 &x< -1 \\ x- 2 & x > 1 \end{matrix}\right.$

Evaluate $(g \circ f)(-3)$ .

Possible Answers:

Undefined

Correct answer:

Explanation:

$(g \circ f)(-3)= g(f(-3))$

First, evaluate f(-3) using the definition of for x < -1 :

f(x)= 2x

f(-3)= 2(-3) = -6

Therefore,

$(g \circ f)(-3)= g(f(-3)) = g(-6)$

Evaluate g(-6) using the definition of for x < -1 :

g(x)= x+ 5

g(-6) = -6+ 5 = -1

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Example Question #3 : Solving Piecewise And Recusive Functions

Define functions and as follows:

$f(x) = \left\{\begin{matrix} 2x+ 1 &x< 0 \\ 3x & x \ge 0 \end{matrix}\right.$

$g(x) = \left\{\begin{matrix} x+ 9 &x< 0 \\ x- 2 & x \ge 0 \end{matrix}\right.$

Evaluate $(f \circ g) \left (-1.7\right )$ .

Possible Answers:

-10.1

15.6

21.9

-2.4

Undefined

Correct answer:

21.9

Explanation:

$(f \circ g) \left ( -1.7\right )=f\left ( g \left ( -1.7\right ) \right )$

First we evaluate g(-1.7) . Since -1.7 < 0 , we use the definition of for the values in the range x < 0 :

g(x)= x + 9

g(-1.7)= -1.7+ 9 = 7.3

Therefore,

$(f \circ g) \left ( -1.7\right )=f\left ( g \left ( -1.7\right ) \right ) = f (7.3)$

Since $7.3 \ge 0$ , we use the definition of for the range $x \ge 0$ :

f(x)= 3x

$f (7.3) = 3 \cdot 7.3 = 21.9$

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Example Question #3 : Solving Piecewise And Recusive Functions

Define two functions as follows:

$f(x) = \begin{Bmatrix} 2x- 7& \textrm{for } x < 0 \\ & \\ 3x+ 3& \textrm{for } x\ge 0 \end{matrix}$

$g(x) = \begin{Bmatrix} 3x+ 3 & \textrm{for } x < 0 \\ & \\ 2x- 7 & \textrm{for } x\ge 0 \end{matrix}$

Evaluate $(f \circ g) (5)$ .

Possible Answers:

Correct answer:

Explanation:

By definition, $(f \circ g) (5) = f \left [ g (5) \right ]$

First, evaluate g(5) , using the definition of g(x) for nonnegative values of . Substituting for 5:

g(x) = 2x-7

$g(5) = 2\cdot 5 -7= 10 -7 = 3$

$(f \circ g) (5) = f \left [ g (5) \right ] = f(3)$ ; evaluate this using the definition of f(x) for nonnegative values of :

f(x) = 3x+ 3

$f(3) = 3 \cdot 3+ 3 = 9 + 3 = 12$

12 is the correct value.

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Example Question #4 : Solving Piecewise And Recusive Functions

Which of the following would be a valid alternative definition for the provided function?

$f(x) = \left | \frac{3}{4} x - \frac{7}{3} \right |$

Possible Answers:

None of these

$f(x) = \left\{\begin{matrix} - \frac{3}{4} x + \frac{7}{3} & \textrm{for }x < \frac{28}{9}\\ \\ \frac{3}{4} x - \frac{7}{3}& \textrm{for }x \ge \frac{28}{9} \end{matrix}\right.$

$f(x) = \left\{\begin{matrix} \frac{3}{4} x - \frac{7}{3} & \textrm{for }x < \frac{28}{9}\\ \\ - \frac{3}{4} x + \frac{7}{3} & \textrm{for }x \ge \frac{28}{9} \end{matrix}\right.$

$f(x) = \left\{\begin{matrix} - \frac{3}{4} x + \frac{7}{3} & \textrm{for }x < -\frac{28}{9}\\ \\ \frac{3}{4} x - \frac{7}{3}& \textrm{for }x \ge - \frac{28}{9} \end{matrix}\right.$

$f(x) = \left\{\begin{matrix} \frac{3}{4} x - \frac{7}{3} & \textrm{for }x < - \frac{28}{9}\\ \\ - \frac{3}{4} x + \frac{7}{3} & \textrm{for }x \ge - \frac{28}{9} \end{matrix}\right.$

Correct answer:

$f(x) = \left\{\begin{matrix} - \frac{3}{4} x + \frac{7}{3} & \textrm{for }x < \frac{28}{9}\\ \\ \frac{3}{4} x - \frac{7}{3}& \textrm{for }x \ge \frac{28}{9} \end{matrix}\right.$

Explanation:

The absolute value of an expression | a | is defined as follows:

| a | = a for $a \ge 0$

| a | = -a for a < 0

Therefore,

$f(x) = \left | \frac{3}{4} x - \frac{7}{3} \right | = \frac{3}{4} x - \frac{7}{3}$

if and only if

$\frac{3}{4} x - \frac{7}{3} \ge 0$ .

Solving this condition for :

$\frac{3}{4} x - \frac{7}{3} + \frac{7}{3} \ge 0 + \frac{7}{3}$

$\frac{3}{4} x \ge \frac{7}{3}$

$\frac{4}{3} \cdot \frac{3}{4} x \ge \frac{4}{3} \cdot \frac{7}{3}$

$x \ge \frac{28}{9}$

Therefore, $f(x) = \frac{3}{4} x - \frac{7}{3}$ for $x \ge \frac{28}{9}$ .

Similarly,

$f(x) = - \left (\frac{3}{4} x - \frac{7}{3} \right ) = - \frac{3}{4} x + \frac{7}{3}$ for $x < \frac{28}{9}$ .

The correct response is therefore

$f(x) = \left\{\begin{matrix} - \frac{3}{4} x + \frac{7}{3} & \textrm{for }x < \frac{28}{9}\\ \\ \frac{3}{4} x - \frac{7}{3}& \textrm{for }x \ge \frac{28}{9} \end{matrix}\right.$

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Example Question #1 : Graphing Linear Functions

Circle

Note: Figure NOT drawn to scale.

Refer to the above figure. The circle has its center at the origin; the line is tangent to the circle at the point indicated. What is the equation of the line in slope-intercept form?

Possible Answers:

$y = \frac{3}{4} x+ \frac{7}{2}$

$y = \frac{4}{3} x$

$y = - \frac{4}{3} x+ 16$

Insufficient information is given to determine the equation of the line.

$y = - \frac{3}{4} x+ \frac{25}{2}$

Correct answer:

$y = - \frac{3}{4} x+ \frac{25}{2}$

Explanation:

A line tangent to a circle at a given point is perpendicular to the radius from the center to that point. That radius, which has endpoints (0,0),(6,8) , has slope

$m = \frac{8-0}{6-0} = \frac{8}{6} = \frac{4}{3}$ .

The line, being perpendicular to this radius, will have slope equal to the opposite of the reciprocal of that of the radius. This slope will be $-\frac{3}{4}$ . Since it includes point (6,8) , we can use the point-slope form of the line to find its equation:

$y - y_{1} = m(x-x_{1})$

$y - 8= - \frac{3}{}4\left ( x-6 \right )$

$y - 8= - \frac{3}{4} x+ \frac{9}{2}$

$y - 8+ 8 = - \frac{3}{4} x+ \frac{9}{2} + 8$

$y = - \frac{3}{4} x+ \frac{25}{2}$

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Example Question #1 : Pre Calculus

What is the center and radius of the circle indicated by the equation?

(x-2)^2+y^2=36

Possible Answers:

$(2,0),\ r=36$

$(-2,0),\ r=36$

$(-2,0),\ r=6$

$(2,0),\ r=6$

Correct answer:

$(2,0),\ r=6$

Explanation:

A circle is defined by an equation in the format (x-h)^2+(y-k)^2=r^2 .

The center is indicated by the point (h,k) and the radius .

In the equation (x-2)^2+(y)^2=36=6^2 , the center is (2,0) and the radius is .

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Example Question #72 : Functions And Graphs

Give the axis of symmetry of the parabola of the equation

$f(x) = 3x^{2} + 9x - 16$

Possible Answers:

$x= \frac{1}{6}$

x = - 16

$x = -\frac{3}{2}$

$x = \frac{3}{2}$

$x= -\frac{1}{6}$

Correct answer:

$x = -\frac{3}{2}$

Explanation:

The line of symmetry of the parabola of the equation

$f(x) = ax^{2}+ bx + c$

is the vertical line

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

Substitute a = 3, b = 9 :

The line of symmetry is

$x = -\frac{9}{2 \cdot 3} = - \frac{9}{6} = -\frac{3}{2}$

That is, the line of the equation $x =- \frac{3}{2}$ .

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Example Question #1 : Graphing Functions

Give the -coordinate of the vertex of the parabola of the function

$f(x) = 4x^{2}- 16x + 1$

Possible Answers:

129

Correct answer:

Explanation:

The -coordinate of the vertex of a parabola of the form

$f(x) = ax^{2}+ bx + c$

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

Substitute a= 4, b = -16 :

$x= -\frac{-16}{2\cdot 4} = -\frac{-16}{8} = 2$

The -coordinate is therefore f(2) :

$f(x) = 4x^{2}- 16x + 1$

$f(2) = 4 \cdot 2 ^{2}- 16 \cdot 2 + 1$

$f(2) = 4 \cdot 4- 16 \cdot 2 + 1$

f(2) = 16 - 32 + 1

f(2) = -15

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Example Question #71 : Functions And Graphs

A baseball is thrown straight up with an initial speed of 60 miles per hour by a man standing on the roof of a 100-foot high building. The height of the baseball in feet is modeled by the function

$h(t) = -16t^{2}+60t + 100$

To the nearest foot, how high is the baseball when it reaches the highest point of its path?

Possible Answers:

$120\textrm{ ft}$

$140\textrm{ ft}$

$172\textrm{ ft}$

$156 \textrm{ ft}$

$132\textrm{ ft}$

Correct answer:

$156 \textrm{ ft}$

Explanation:

We are seeking the value of when the graph of $h\left ( t\right )$ - a parabola - reaches its vertex.

To find this value, we first find the value of . For a parabola of the equation

$f\left ( t\right ) = at^{2}+ bt + c$ ,

the value of the vertex is

$t = -\frac{b}{2a}$ .

Substitute a = -16, b = 60 :

$t = -\frac{60}{2 (-16)} = 1.875$

The height of the baseball after 1.875 seconds will be

$h(1.875) = -16 \cdot 1.875^{2}+60 \cdot 1.875 + 100$

$h(1.875) = -56.25 +112.5 + 100 \approx 156$ feet.

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Example Question #1 : Graphing Quadratic Functions And Conic Sections

Give the -coordinate of the vertex of the parabola of the function

$f(x) = 2x^{2} + 6x + 5$ .

Possible Answers:

$\frac{37}{2}$

$\frac{1}{2}$

$\frac{3}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

The -coordinate of the vertex of a parabola of the form

$f(x) = ax^{2}+ bx + c$

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

Set a = 2, b = 6 :

$x= -\frac{6}{2\cdot 2} = -\frac{6}{4} =- \frac{3}{2}$

The -coordinate is therefore $f\left (- \frac{3}{2} \right )$ :

$f(x) = 2x^{2} + 6x + 5$

$f \left ( -\frac{3}{2} \right )= 2\left ( -\frac{3}{2} \right )^{2}+ 6\left ( -\frac{3}{2} \right ) + 5$

$= 2\left ( \frac{9}{4} \right ) + 6\left (- \frac{3}{2} \right ) + 5$

$= \frac{9}{2} - 9 + 5 = \frac{1}{2}$ , which is the correct choice.

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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 #1 : Solving Piecewise And Recusive Functions

Example Question #3 : Solving Piecewise And Recusive Functions

Example Question #3 : Solving Piecewise And Recusive Functions

Example Question #4 : Solving Piecewise And Recusive Functions

Example Question #1 : Graphing Linear Functions

Example Question #1 : Pre Calculus

Example Question #72 : Functions And Graphs

Example Question #1 : Graphing Functions

Example Question #71 : Functions And Graphs

Example Question #1 : Graphing Quadratic Functions And Conic Sections

All SAT II Math II Resources

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