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High School Math : High School Math

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8 Diagnostic Tests 613 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : Calculus Ii — Integrals

A function is defined by the following piecewise equation:

$f(x)=\left\{\begin{matrix} x^{2}+4x-3, x<3\\10x-12, x\geq 3 \end{matrix}\right.$

At x=3 , the function is:

Possible Answers:

continuous

discontinuous

Correct answer:

continuous

Explanation:

The first step to determine continuity at a point is to determine if the function is defined at that point. When we substitute in 3 for , we get 18 as our -value. x=3 is thus defined for this function.

The next step is determine if the limit of the function is defined at that point. This means that the left-hand limit must be equal to the right-hand limit at x=3 . Substitution reveals the following:

$\lim_{x\rightarrow 3^{-}}f(x)=(3)^{2}+4(3)-3=9+12-3=18$

$\lim_{x\rightarrow 3^{+}}f(x)=10(3)-12=18$

Both sides of the function, therefore, approach a -value of 18.

Finally, we must ensure that the curve is smooth by checking the limit of the derivative of both sides.

$\lim_{x\rightarrow 3^{-}}f^{'}(x)=2(3)+4=10$

$\lim_{x\rightarrow 3^{+}}f^{'}(x)=10$

Since the function passes all three tests, it is continuous.

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Example Question #1 : Limits

Function

The graph depicts a function g(x) . Does $\lim_{x\rightarrow 1} g(x)$ exist?

Possible Answers:

$\lim_{x\rightarrow 1} g(x)$ exists because g(x) is constant on $(-\infty , 1)$ .

$\lim_{x\rightarrow 1} g(x)$ does not exist because g(1) is undefined.

$\lim_{x\rightarrow 1} g(x)$ does not exist because is not continuous at x = 1 .

$\lim_{x\rightarrow 1} g(x)$ does not exist because $\lim_{x\rightarrow 1^{-}} g(x) \neq \lim_{x\rightarrow 1^{+}} g(x)$ .

$\lim_{x\rightarrow 1} g(x)$ exists because $\lim_{x\rightarrow 1^{-}} g(x) = \lim_{x\rightarrow 1^{+}} g(x)$ .

Correct answer:

$\lim_{x\rightarrow 1} g(x)$ exists because $\lim_{x\rightarrow 1^{-}} g(x) = \lim_{x\rightarrow 1^{+}} g(x)$ .

Explanation:

$\lim_{x\rightarrow 1}g(x)$ exists if and only if $\lim_{x\rightarrow 1^{+}} g(x) = \lim_{x\rightarrow 1^{-}} g(x)$ ; the actual value of f(1) is irrelevant.

As can be seen, $\lim_{x\rightarrow 1^{+}} g(x) = 1$ and $\lim_{x\rightarrow 1^{-}}g(x) = 1$ ; therefore, $\lim_{x\rightarrow 1^{+}} g(x) = \lim_{x\rightarrow 1^{-}} g(x)$ , and $\lim_{x\rightarrow 1} g(x)$ exists.

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Example Question #1 : Understanding Polar Coordinates

The polar coordinates of a point are $\left(2.1, \frac{7\pi }{3}\right )$ . Give its -coordinate in the rectangular coordinate system (nearest hundredth).

Possible Answers:

-1.82

1.05

1.82

-1.05

4.20

Correct answer:

1.82

Explanation:

Given the polar coordinates $(r,\theta )$ , the -coordinate is $y= r \sin \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = \frac{7\pi }{3}$ :

$y = r \sin \theta = 2.1 \cdot \sin \frac{7\pi }{3} \approx 2.1 \cdot 0.8660 \approx 1.82$

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Example Question #2 : Understanding Polar Coordinates

The polar coordinates of a point are $\left(2.1, \frac{7\pi }{3}\right )$ . Give its -coordinate in the rectangular coordinate system (nearest hundredth).

Possible Answers:

1.05

-1.82

1.82

-1.05

4.20

Correct answer:

1.05

Explanation:

Given the polar coordinates $(r,\theta )$ , the -coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 2.1, \theta = \frac{7\pi }{3}$ :

$x = r \cos \theta = 2.1 \cdot \cos \frac{7\pi }{3} = 2.1 \cdot 0.5 = 1.05$

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Example Question #1 : Polar

The polar coordinates of a point are $\left(1.2, \frac{2\pi }{5}\right )$ . Give its -coordinate in the rectangular coordinate system (nearest hundredth).

Possible Answers:

1.26

3.88

0.37

1.14

3.69

Correct answer:

1.14

Explanation:

Given the polar coordinates $(r,\theta )$ , the -coordinate is $y= r \sin \theta$ . We can find this coordinate by substituting $r = 1.2, \theta = \frac{2\pi }{5}$ :

$y = r \cos \theta = 1.2 \cdot \sin \frac{2\pi }{5} \approx 1.2 \cdot 0.9511 \approx 1.14$

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Example Question #1 : Calculus Ii — Integrals

The polar coordinates of a point are $\left(1.2, \frac{2\pi }{5}\right )$ . Give its -coordinate in the rectangular coordinate system (nearest hundredth).

Possible Answers:

3.88

1.14

1.26

3.69

0.37

Correct answer:

0.37

Explanation:

Given the polar coordinates $(r,\theta )$ , the -coordinate is $x= r \cos \theta$ . We can find this coordinate by substituting $r = 1.2, \theta = \frac{2\pi }{5}$ :

$x = r \cos \theta = 1.2 \cdot \cos \frac{2\pi }{5} \approx 1.2 \cdot 0.3090 \approx 0.37$

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Example Question #5 : Parametric, Polar, And Vector

Find the vector where its initial point is (-1,4) and its terminal point is (2,2) .

Possible Answers:

<1,6>

<1,2>

<3,-2>

<1,-2>

Correct answer:

<3,-2>

Explanation:

We need to subtract the -coordinate and the -coordinates to solve for a vector when given its initial and terminal coordinates:

Initial pt: (-1,4)

Terminal pt: (2,2)

Vector: <2-(-1),2-4>

Vector:

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Example Question #6 : Parametric, Polar, And Vector

Find the vector where its initial point is (-1,-5) and its terminal point is (2,3) .

Possible Answers:

$\left \langle1,8\right \rangle$

$\left \langle3,2\right \rangle$

$\left \langle3,8\right \rangle$

$\left \langle1,2\right \rangle$

Correct answer:

$\left \langle3,8\right \rangle$

Explanation:

We need to subtract the -coordinate and the -coordinate to solve for a vector when given its initial and terminal coordinates:

Initial pt: (-1,-5)

Terminal pt: (2,3)

Vector: $\left \langle 2-(-1),3-(-5)\right \rangle$

Vector: $\left \langle3,8\right \rangle$

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Example Question #1 : Vector

Let $\vec{a}, \vec{b},\vec{c}$ be vectors. All of the following are defined EXCEPT:

Possible Answers:

$(\vec{a}\times \vec{b})\cdot \vec{c}$

$(\vec{a}\cdot \vec{b})\times\vec{c}$

$(\vec{a}\times \vec{b})\cdot(\vec{a}\times \vec{c})$

$(\vec{a}\times \vec{b})\times\vec{c}$

$(\vec{a}\cdot \vec{b})+(\vec{a}\cdot \vec{c})$

Correct answer:

$(\vec{a}\cdot \vec{b})\times\vec{c}$

Explanation:

The cross product of two vectors (represented by "x") requires two vectors and results in another vector. By contrast, the dot product (represented by " $\cdot$ ") between two vectors requires two vectors and results in a scalar, not a vector.

If we were to evaluate $(\vec{a}\cdot \vec{b})\times\vec{c}$ , we would first have to evaluate $\vec{a}\cdot\vec{b}$ , which would result in a scalar, because it is a dot product.

However, once we have a scalar value, we cannot calculate a cross product with another vector, because a cross product requires two vectors. For example, we cannot find the cross product between 4 and the vector <1, 2, 3>; the cross product is only defined for two vectors, not scalars.

The answer is $(\vec{a}\cdot \vec{b})\times\vec{c}$ .

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Example Question #8 : Parametric, Polar, And Vector

Find the magnitude of vector v=<3,-2> :

Possible Answers:

$\sqrt{5}$

$\sqrt{13}$

$\sqrt{7}$

$\sqrt{11}$

Correct answer:

$\sqrt{13}$

Explanation:

To solve for the magnitude of a vector, we use the following formula:

$\left \| v \right \|=\sqrt{x^2+y^2}$

v=<3,-2>

$\left \| v \right \|=\sqrt{(3)^2+(-2)^2}$

$\left \| v \right \|=\sqrt{9+}4$

$\left \| v \right \|=\sqrt{13}$

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All High School Math Resources

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High School Math : High School Math

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #3 : Calculus Ii — Integrals

Example Question #1 : Limits

Example Question #1 : Understanding Polar Coordinates

Example Question #2 : Understanding Polar Coordinates

Example Question #1 : Polar

Example Question #1 : Calculus Ii — Integrals

Example Question #5 : Parametric, Polar, And Vector

Example Question #6 : Parametric, Polar, And Vector

Example Question #1 : Vector

Example Question #8 : Parametric, Polar, And Vector

All High School Math Resources

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