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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 #3 : Graphing Piecewise And Recusive Functions

Define a function as follows:

$g(x) = \left\{\begin{matrix} \sin \frac{x}{2} -1 &x\le - \pi \\ \cos 2x & -\pi < x \le 0 \\ 1-\tan \frac{x }{2}&0< x < \pi \\ 0 & x\ge \pi \end{matrix}\right.$

At which of the following values of is discontinuous?

I) $x = -\pi$

II) x= 0

III) $x= \pi$

Possible Answers:

I and III only

All of I, II, and III

II and III only

I and II only

None of I, II, and III

Correct answer:

I and III only

Explanation:

To determine whether is continuous at $x = -\pi$ , we examine the definitions of on both sides of $-\pi$ , and evaluate both for $x = -\pi$ :

$\sin \frac{x}{2} -1$ evaluated for $x = -\pi$ :

$\sin \frac{-\pi}{2} -1 = \sin \left (- \frac{ \pi}{2} \right )-1= -1-1= -2$

$\cos 2x$ evaluated for $x = -\pi$ :

$\cos 2(-\pi) = \cos (-2\pi)= 1$

Since the values do not coincide, is discontinuous at $x = -\pi$ .

We do the same thing with the other two boundary values 0 and $\pi$ .

$\cos 2x$ evaluated for x =0 :

$\cos 2(0) = \cos 0= 1$

$1-\tan \frac{x }{2}$ evaluated for x =0 :

$1-\tan \frac{0 }{2} = 1 - \tan 0= 1-0 = 1$

Since the values coincide, is continuous at x =0 .

$1-\tan \frac{\pi }{2}$ turns out to be undefined for $x = \pi$ , (since $\tan \frac{\pi }{2}$ is undefined), so is discontinuous at $x = \pi$ .

The correct response is I and III only.

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Example Question #1 : Graphing Piecewise And Recusive Functions

Define a function as follows:

$g(x) = \left\{\begin{matrix} x^{2}-2x+7 &x\le -1 \\ 9-x & -1 < x \le 0 \\ x+8 &0< x \le 1 \\ x^{2}+3x+ 6& x> 1 \end{matrix}\right.$

At which of the following values of is the graph of discontinuous?

I) x = -1

II) x= 0

III) x= 1

Possible Answers:

None of I, II, and III

I and III only

I and II only

II and III only

All of I, II, and III

Correct answer:

II and III only

Explanation:

To determine whether is continuous at x = -1 , we examine the definitions of on both sides of , and evaluate both for x = -1 :

$x^{2}-2x+7$ evaluated for x = -1 :

$(-1)^{2}-2(-1)+7= 1+2+7 = 10$

9-x evaluated for x = -1 :

9-(-1)= 10

Since the values coincide, the graph of is continuous at x = -1 .

We do the same thing with the other two boundary values 0 and 1:

9-x evaluated for x = 0 :

9-0= 9

x+8 evaluated for x = 0 :

0+8 = 8

Since the values do not coincide, the graph of is discontinuous at x = 0 .

x+8 evaluated for x = 1 :

1+8 = 9

$x^{2}+3x+ 6$ evaluate for x = 1 :

$1^{2}+3\cdot 1+ 6 = 1+ 3 + 6 = 10$

Since the values do not coincide, the graph of is discontinuous at x = 1 .

II and III only is the correct response.

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Example Question #1 : Graphing Piecewise And Recusive Functions

Define a function as follows:

$g(x) = \left\{\begin{matrix} \sin \frac{x}{2} -1 &x\le - \pi \\ \cos 2x & -\pi < x \le 0 \\ 1-\tan \frac{x }{2}&0< x < \pi \\ 0 & x\ge \pi \end{matrix}\right.$

Give the -intercept of the graph of the function.

Possible Answers:

The graph does not have a -intercept.

(0,0)

(0, -2)

(0, -1)

(0, 1)

Correct answer:

(0, 1)

Explanation:

To find the -intercept, evaluate g(0) using the definition of on the interval that includes the value 0. Since

$g(x)= \cos 2x$

on the interval $(-\pi, 0]$ ,

evaluate:

$g(x)= \cos 2 \cdot 0 = \cos 0 = 1$

The -intercept is (0, 1) .

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Example Question #1 : Graphing Parametric Functions

Give the period of the graph of the equation

$f(x) = \frac{4}{7} \cos \left ( \frac{3\pi x}{7} \right )$

Possible Answers:

$\frac{7}{3}$

$\frac{8}{7}$

$\frac{2}{7}$

$\frac{14}{3}$

$\frac{4}{7}$

Correct answer:

$\frac{14}{3}$

Explanation:

The period of the graph of a cosine function $f(x)= A \cos Nx$ is $\frac{2 \pi}{N}$ , or $2 \pi \div N$

Since $N = \frac{3\pi}{7}$ , the period is

$2 \pi \div \frac{3\pi}{7} = \frac{2 \pi }{1}\times \frac{7} {3\pi}= \frac{14}{3}$

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Example Question #1 : Area

Right_triangle_3

Note: Figure NOT drawn to scale.

Refer to the above diagram. AD = 6 , DC = 24 , and $\angle ABC$ and $\angle ADB$ are right angles. What percent of $\Delta ABC$ is colored red?

Possible Answers:

$30 \%$

$25 \%$

$16 \frac{2}{3} \%$

$20 \%$

$33 \frac{1}{3}\%$

Correct answer:

$20 \%$

Explanation:

, as the length of the altitude corresponding to the hypotenuse, is the geometric mean of the lengths of the parts of the hypotenuse it forms; that is, it is the square root of the product of the two:

$BD =\sqrt{ \left ( AD\right ) \left ( DC\right )} = \sqrt{6 \cdot 24} = \sqrt{144} = 12$ .

The area of $\Delta ADB$ , the shaded region, is half the products of its legs:

$A= \frac{1}{2} (AD) \times (DB ) =\frac{1}{2} \times 6 \times 12 = 36$

The area of $\Delta ABC$ is half the product of its hypoteuse, which we can see as the base, and the length of corresponding altitude $\overline{BD}$ :

$A = \frac{1}{2} (AC) (BD) = \frac{1}{2} \cdot 30 \cdot 12 = 180$

$\Delta ADB$ comprises

$\frac{36}{180} \times \100 = 20 \%$

of $\Delta ADC$ .

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Example Question #1 : Geometry

Garden

Note: Figure NOT drawn to scale

Refer to the above figure, which shows a square garden (in green) surrounded by a dirt path (in orange). The dirt path is seven feet wide throughout. What is the area of the dirt path in square feet?

Possible Answers:

$1,372\textrm{ ft}^{2}$

$1,904\textrm{ ft}^{2}$

$2,100\textrm{ ft}^{2}$

$1,050\textrm{ ft}^{2}$

$1,001\textrm{ ft}^{2}$

Correct answer:

$1,904\textrm{ ft}^{2}$

Explanation:

The area of the dirt path is the area of the outer square minus that of the inner square.

The outer square has sidelength 75 feet and therefore has area

$75 \times 75 = 5,625$ square feet.

The inner square has sidelength $75 - 2 \times 7 = 61$ feet and therefore has area

$61 \times 61 = 3,721$ square feet.

Subtract to get the area of the dirt path:

5,625 - 3,721 = 1,904 square feet.

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Example Question #1 : 2 Dimensional Geometry

Garden

Refer to the above figure, which shows a rectangular garden (in green) surrounded by a dirt path (in orange). The dirt path is six feet wide throughout. Which of the following polynomials gives the area of the garden in square feet?

Possible Answers:

$2x^{2} - 36$

$2x ^{2} - 18x + 36$

$2x ^{2} - 36 x + 144$

$2x^{2} - 72$

$2x^{2} - 144$

Correct answer:

$2x ^{2} - 36 x + 144$

Explanation:

The length of the garden is $2 \times 6= 12$ feet less than that of the entire lot, or

L = 2x-12 ;

The width of the garden is $2 \times 6= 12$ less than that of the entire lot, or

W = x-12 ;

The area of the garden is their product:

A = LW = (2x-12)(x - 12)

$=2x \cdot x - 2x \cdot 12 - 12 \cdot x + 12 \cdot 12$

$= 2x ^{2} - 24x - 12 x + 144$

$= 2x ^{2} - 36 x + 144$

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Example Question #1 : Geometry

Decagon

The above figure is a regular decagon. If AB = 100 , then to the nearest whole number, what is ?

Possible Answers:

$AC \approx 185$

$AC \approx 190$

$AC \approx 162$

$AC \approx 173$

$AC \approx 188$

Correct answer:

$AC \approx 190$

Explanation:

As an interior angle of a regular decagon, $\angle B$ measures

$m \angle B = \left [\frac{180(10-2)}{10} \right ] ^{\circ }= 144^{\circ }$ .

AB = BC = 100 .

can be found using the Law of Cosines:

$\left ( AC\right )^{2} = \left ( AB\right )^{2}+ \left ( BC\right )^{2} - 2 \left ( AB\right ) \left ( BC\right ) \cos m\angle B$

$\left ( AC\right )^{2} = 100^{2}+ 100^{2} - 2 \cdot 100 \cdot 100 \cos 144^{\circ }$

$\left ( AC\right )^{2} \approx 10,000 + 10,000 - 20,000 (-0.8090)$

$\left ( AC\right )^{2} \approx 20,000 - (-16,180) \approx 36,180$

$AC \approx \sqrt{36,180} \approx 190$

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Example Question #2 : Geometry

Circle

The circle in the above diagram has its center at the origin. To the nearest tenth, what is the area of the pink region?

Possible Answers:

110.7

124.9

189.3

46.4

32.2

Correct answer:

124.9

Explanation:

First, it is necessary to determine the radius of the circle. This is the distance between (0,0) and (-8,6) , so we apply the distance formula:

$r = \sqrt{(x_{2} - x _{1})^{2}+(y_{2} - y_{1})^{2}}$

$r = \sqrt{\left ( -8 -0 \right )^{2} +\left ( 6-0\right )^{2} }=\sqrt{ 8^{2}+6^{2}} = \sqrt{64+36 }= \sqrt{100} = 10$

Subsequently, the area of the circle is

$A = \pi r^{2} = \pi \cdot 10^{2} = 100 \pi$

Now, we need to find the central angle of the shaded sector. This is found using the relationship

$\tan \theta = \frac{y}{x}$

$\tan \theta = \frac{6}{-8} = -0.75$

Using a calculator, we find that $\tan^{-1}(-0.75) \approx -36.87^{\circ }$ ; since we want a degree measure between $90^{\circ }$ and $180 ^{\circ }$ , we adjust by adding $180 ^{\circ }$ , so

$\theta \approx \left (-36.87+ 180 \right ) ^{\circ } \approx 143.13^{\circ }$

The area of the sector is calculated as follows:

$A \approx \frac{143.13 }{360} \cdot 100 \pi \approx 124.9$

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Example Question #1 : 2 Dimensional Geometry

You own a mug with a circular bottom. If the distance around the outside of the mug's base is $20 \pi cm$ what is the area of the base?

Possible Answers:

100 cm

10 cm

$10 \pi cm^2$

$100 \pi cm^2$

Correct answer:

$100 \pi cm^2$

Explanation:

You own a mug with a circular bottom. If the distance around the outside of the mug's base is $20 \pi cm$ what is the area of the base?

Begin by solving for the radius:

$C=2\pi r$

$20 \pi cm=2\pi r$

r=10cm

Next, plug the radius back into the area formula and solve:

$A_{circle}=\pi r^2$

$A_{circle}=\pi (10cm)^2=100 \pi cm^2$

So our answer is:

$100 \pi cm^2$

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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 #3 : Graphing Piecewise And Recusive Functions

Example Question #1 : Graphing Piecewise And Recusive Functions

Example Question #1 : Graphing Piecewise And Recusive Functions

Example Question #1 : Graphing Parametric Functions

Example Question #1 : Area

Example Question #1 : Geometry

Example Question #1 : 2 Dimensional Geometry

Example Question #1 : Geometry

Example Question #2 : Geometry

Example Question #1 : 2 Dimensional Geometry

All SAT II Math II Resources

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