Comprehensive Math Questions with Answers (Set 2)

This set of 30 multiple-choice math questions covers topics including logarithms, exponentials, quadratic equations, trigonometry, and trigonometric identities. Detailed, step-by-step solutions are provided beneath each question to help you verify your reasoning.

Table of Contents

Multiple-Choice Questions

Question 1 — Inverse of Logarithm

If \( f(x) = \log(x) \), then \( f^{-1}(x) = \)

  1. \( 10^x \)
  2. \( -10^{-x} \)
  3. \( \frac{1}{10^x} \)
  4. \( x^{10} \)
View Step-by-Step Solution

Given \( y = \log(x) \) (common log, base 10), rewrite in exponential form: \( x = 10^y \). Interchanging \( x \) and \( y \) gives \( y = 10^x \). Therefore:

\[ f^{-1}(x) = 10^x \]

Correct Answer: A

Question 2 — Evaluating a Logarithm

\( \log_4 65 = \)

  1. 1.21
  2. 1.81
  3. 3.06
  4. 3.01
View Step-by-Step Solution

Use the change-of-base formula:

\[ \log_4(65) = \frac{\ln(65)}{\ln(4)} \approx \frac{4.1744}{1.3863} \approx 3.011 \]

Correct Answer: D

Question 3 — Exponential Equation

If \( 2^{3x - 1} = 16 \), then \( x = \)

  1. \( 3/5 \)
  2. \( 5/3 \)
  3. \( 1 \)
  4. \( 4 \)
View Step-by-Step Solution

Rewrite 16 as a power of 2: \( 2^{3x-1} = 2^4 \). Since the bases are equal, equate the exponents:

\[ 3x - 1 = 4 \implies 3x = 5 \implies x = \frac{5}{3} \]

Correct Answer: B

Verification: \( 2^{3(5/3)-1} = 2^{5-1} = 2^4 = 16 \) ✓

Question 4 — Logarithmic Equation

If \( \log_x 9 = 2 \), then \( x = \)

  1. \( 2/9 \)
  2. \( 9/2 \)
  3. \( 81 \)
  4. \( 3 \)
View Step-by-Step Solution

Rewrite in exponential form:

\[ x^2 = 9 = 3^2 \implies x = 3 \]

Since a logarithmic base must be positive and not equal to 1, \( x = 3 \) is the only valid solution.

Correct Answer: D

Question 5 — Matching Quadratic Coefficients

If \( x^2 + kx - 6 = (x - 2)(x + 3) \), then \( k = \)

  1. \( 6 \)
  2. \( 3 \)
  3. \( 1 \)
  4. \( 0 \)
View Step-by-Step Solution

Expand the right side: \( (x-2)(x+3) = x^2 + 3x - 2x - 6 = x^2 + x - 6 \). Matching coefficients with \( x^2 + kx - 6 \) gives \( k = 1 \).

Correct Answer: C

Question 6 — Vertex of a Parabola

The vertex of the graph of \( y = 2x^2 + 8x - 3 \) is

  1. \( (8, -3) \)
  2. \( (-2, -11) \)
  3. \( (-4, -11) \)
  4. \( (-2, -3) \)
View Step-by-Step Solution

Complete the square:

\[ y = 2(x^2 + 4x) - 3 = 2(x^2 + 4x + 4) - 8 - 3 = 2(x+2)^2 - 11 \]

The vertex is at \( (-2, -11) \).

Correct Answer: B

Shortcut: Vertex x-coordinate = \( -\frac{b}{2a} = -\frac{8}{4} = -2 \); then \( y = 2(4) + 8(-2) - 3 = 8 - 16 - 3 = -11 \).

Question 7 — Quadratic from Given Roots

The quadratic equation whose roots are \( x = 3 \) and \( x = 5 \) is

  1. \( (x - 3)(x - 5) = 1 \)
  2. \( (x + 3)(x + 5) - 25 = 0 \)
  3. \( (x + 3)(x + 5) = 0 \)
  4. \( x^2 - 8x = -15 \)
View Step-by-Step Solution

A quadratic with roots 3 and 5 is \( (x-3)(x-5) = 0 \), which expands to \( x^2 - 8x + 15 = 0 \). Rearranging:

\[ x^2 - 8x = -15 \]

Correct Answer: D

Check other options: A is false (product ≠ 1); B expands to \( x^2 + 8x - 10 \neq 0 \) with roots ≠ 3, 5; C has roots −3 and −5.

Question 8 — Rational Equation with Extraneous Root

The roots of \( \dfrac{x}{x + 2} + \dfrac{3}{x - 4} = \dfrac{4x + 2}{x^2 - 2x - 8} \) are

  1. \( x = 4 \) and \( x = 1 \)
  2. \( x = 4 \) only
  3. \( x = 1 \) only
  4. \( x = -4 \) only
View Step-by-Step Solution

Note that \( x^2 - 2x - 8 = (x+2)(x-4) \). Multiply the entire equation by \( (x+2)(x-4) \):

\[ x(x-4) + 3(x+2) = 4x + 2 \] \[ x^2 - 4x + 3x + 6 = 4x + 2 \] \[ x^2 - 5x + 4 = 0 \] \[ (x-1)(x-4) = 0 \]

This gives \( x = 1 \) and \( x = 4 \). However, \( x = 4 \) makes the original denominators \( x - 4 \) and \( x^2 - 2x - 8 \) equal to zero, so it is extraneous.

Correct Answer: C (\( x = 1 \) only)

Question 9 — Range of a Quadratic on a Restricted Domain

If the domain of \( f(x) = -x^2 + 6x \) is \( [0,6] \), then the range is

  1. \( [0, 9] \)
  2. \( [0, 6] \)
  3. \( [0, 3] \)
  4. \( [3, 6] \)
View Step-by-Step Solution

Zeros are at \( x=0 \) and \( x=6 \). The maximum occurs at the vertex:

\[ x = \frac{-b}{2a} = \frac{-6}{-2} = 3, \quad f(3) = -9 + 18 = 9 \]

At the endpoints, \( f(0) = 0 \) and \( f(6) = 0 \). Since the parabola opens downward, the range is \( [0, 9] \).

Correct Answer: A

Question 10 — x-Intercepts of a Parabola

The x-intercepts of \( y = -x^2 + 3x + 18 \) are

  1. \( (6, -3) \)
  2. \( (-3, 6) \)
  3. \( (-3, 0) \) and \( (6, 0) \)
  4. \( (3, 0) \) and \( (-6, 0) \)
View Step-by-Step Solution

Set \( y = 0 \):

\[ -x^2 + 3x + 18 = 0 \implies x^2 - 3x - 18 = 0 \implies (x-6)(x+3) = 0 \]

The x-intercepts are \( (6, 0) \) and \( (-3, 0) \).

Correct Answer: C

Question 11 — Domain of a Logarithmic Absolute Value Function

The domain of \( f(x) = 2 \log|x - 2| \) is

  1. \( (-\infty, 2) \cup (2, \infty) \)
  2. \( (-2, \infty) \)
  3. \( (2, \infty) \)
  4. \( (-\infty, \infty) \)
View Step-by-Step Solution

The logarithm is undefined when its argument is zero or negative. Here the argument is \( |x-2| \), which is always non-negative and equals zero only when \( x = 2 \). Thus:

\[ \text{Domain} = (-\infty, 2) \cup (2, \infty) \]

Correct Answer: A

Question 12 — Solving an Exponential Equation

If \( 1.56^x = 2 \), then \( x = \)

  1. \( \frac{\ln 1.56}{\ln 2} \)
  2. \( \frac{\ln 2}{\ln 1.56} \)
  3. \( \frac{2}{\ln 1.56} \)
  4. \( \frac{\ln 2}{1.56} \)
View Step-by-Step Solution

Take the natural log of both sides:

\[ \ln(1.56^x) = \ln(2) \implies x \ln(1.56) = \ln(2) \implies x = \frac{\ln 2}{\ln 1.56} \]

Correct Answer: B

Question 13 — Reference Angle of a Negative Angle

The reference angle of \( \alpha = -1280^\circ \) is

  1. \( 20^\circ \)
  2. \( 30^\circ \)
  3. \( 160^\circ \)
  4. \( 60^\circ \)
View Step-by-Step Solution

Find a positive coterminal angle by adding \( 360^\circ \) repeatedly:

\[ -1280^\circ + 4(360^\circ) = -1280^\circ + 1440^\circ = 160^\circ \]

Since \( 160^\circ \) is in Quadrant II, the reference angle is:

\[ 180^\circ - 160^\circ = 20^\circ \]

Correct Answer: A

Question 14 — Secant from a Point on the Terminal Side

If \( x \) is an angle in standard position with point \( A(-3,4) \) on its terminal side, then \( \sec(x) = \)

  1. \( 5/3 \)
  2. \( -5/3 \)
  3. \( -3/5 \)
  4. \( 3/5 \)
View Step-by-Step Solution

Compute the radius:

\[ r = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = 5 \]

Since \( \sec(x) = \dfrac{r}{x} \):

\[ \sec(x) = \frac{5}{-3} = -\frac{5}{3} \]

Correct Answer: B

Question 15 — Sine on the Line y = −x in Quadrant IV

If the terminal side of \( x \) is in quadrant IV and is given by \( y=-x \), then \( \sin(x) = \)

  1. \( \sqrt{2} \)
  2. \( 1/\sqrt{2} \)
  3. \( -\sqrt{2} \)
  4. \( -1/\sqrt{2} \)
View Step-by-Step Solution

A point on \( y = -x \) in Quadrant IV is \( (3, -3) \). Then:

\[ r = \sqrt{3^2 + (-3)^2} = 3\sqrt{2} \] \[ \sin(x) = \frac{y}{r} = \frac{-3}{3\sqrt{2}} = -\frac{1}{\sqrt{2}} \]

Correct Answer: D

Question 16 — Identifying a False Statement

Which statement is NOT true?

  1. \( \cos(-x) = \cos(x) \)
  2. \( \tan(-x) = \tan(x) \)
  3. The amplitude of \( y=-2\cos(t) \) is 2
  4. The range of \( y=-5\sin(t) \) is \( [-5,5] \)
View Step-by-Step Solution

Tangent is an odd function: \( \tan(-x) = -\tan(x) \), not \( \tan(x) \). Thus B is false.

Correct Answer: B

Question 17 — Identifying a True Statement

Which statement is true?

  1. \( \sec(-x) = -\sec(x) \)
  2. The range of \( y=\tan(x) \) is \( (0, \infty) \)
  3. The period of \( y=-2\cos(3\pi t) \) is \( 2\pi/3 \)
  4. The period of \( y=-5\tan(0.5\pi t) \) is 2
View Step-by-Step Solution

Evaluate each option:

  • A: \( \sec \) is even, so \( \sec(-x) = \sec(x) \) — false.
  • B: The range of \( \tan(x) \) is \( (-\infty, \infty) \) — false.
  • C: The period of \( -2\cos(3\pi t) \) is \( \frac{2\pi}{3\pi} = \frac{2}{3} \) — false.
  • D: The period of \( -5\tan(0.5\pi t) \) is \( \frac{\pi}{0.5\pi} = 2 \) — true.

Correct Answer: D

Question 18 — Degrees to Radians

The radian measure of \( 25^\circ \) is

  1. \( 25\pi \)
  2. \( 5/(36\pi) \)
  3. \( 5\pi/36 \)
  4. \( 36/(5\pi) \)
View Step-by-Step Solution
\[ 25^\circ \cdot \frac{\pi}{180^\circ} = \frac{25\pi}{180} = \frac{5\pi}{36} \]

Correct Answer: C

Question 19 — Period of a Sine Function

The period of \( f(x) = \sin(x \pi/6 + \pi/4) \) is

  1. \( 2\pi \)
  2. \( 12 \)
  3. \( \pi/6 \)
  4. \( 6\pi \)
View Step-by-Step Solution

For \( \sin(bx + \phi) \), the period is \( \dfrac{2\pi}{|b|} \). Here \( b = \pi/6 \):

\[ \text{Period} = \frac{2\pi}{\pi/6} = 12 \]

Correct Answer: B

Question 20 — Identifying an Identity

Which of the following is an identity?

  1. \( \cos(2x) = 2\cos(x) \)
  2. \( \cos(x+y) = \cos(x) + \cos(y) \)
  3. \( \sin(x-y) = \sin(x) - \sin(y) \)
  4. \( \sin(2x) = 2\sin(x)\cos(x) \)
View Step-by-Step Solution

The double-angle identity for sine is \( \sin(2x) = 2\sin(x)\cos(x) \). The other three options are common false identities.

Correct Answer: D

Question 21 — Identifying a Non-Identity

Which of the following is NOT an identity?

  1. \( \cot(x+y) = \dfrac{1 + \cot(x)\cot(y)}{\cot(x) + \cot(y)} \)
  2. \( \tan(x+y) = \dfrac{\tan(x) + \tan(y)}{1 - \tan(x)\tan(y)} \)
  3. \( \sin(x-y) = \sin(x)\cos(y) - \cos(x)\sin(y) \)
  4. \( \cos(2x) = 2\cos^2(x) - 1 \)
View Step-by-Step Solution

The correct cotangent sum identity is:

\[ \cot(x+y) = \frac{\cot(x)\cot(y) - 1}{\cot(x) + \cot(y)} \]

Option A has the numerator reversed (\( 1 + \cot(x)\cot(y) \) instead of \( \cot(x)\cot(y) - 1 \)), so it is not an identity.

Correct Answer: A

Question 22 — Secant in Quadrant III

If \( \tan(x) = 5/12 \) and \( \pi \lt x \lt 3\pi/2 \), then \( \sec(x) = \)

  1. \( 12/13 \)
  2. \( -12/13 \)
  3. \( -13/12 \)
  4. \( 13/12 \)
View Step-by-Step Solution

In Quadrant III, both \( x \) and \( y \) are negative, and \( \tan(x) \gt 0 \). Using \( \tan(x) = y/x = 5/12 \) with both negative: \( (x, y) = (-12, -5) \). Then:

\[ r = \sqrt{(-12)^2 + (-5)^2} = 13, \quad \sec(x) = \frac{r}{x} = -\frac{13}{12} \]

Correct Answer: C

Question 23 — Quadratic Trigonometric Equation

The real solutions to \( \cos^2(x) - 1.5\cos(x) = 1 \) are

  1. \( \cos(x) = 1 \)
  2. \( \cos(x) = 2 \)
  3. \( \cos(x) = 1/2 \)
  4. \( \cos(x) = -1/2 \)
View Step-by-Step Solution

Rearrange as a quadratic in \( \cos(x) \):

\[ \cos^2(x) - 1.5\cos(x) - 1 = 0 \] \[ (2\cos(x) + 1)(\cos(x) - 2) = 0 \]

This gives \( \cos(x) = -1/2 \) or \( \cos(x) = 2 \). Since \( \cos(x) \) must be between \( -1 \) and \( 1 \), the only valid solution is:

\[ \cos(x) = -\frac{1}{2} \]

Correct Answer: D

Question 24 — Point on the Inverse Function

Which point is on the graph of the inverse of \( f(x) = 10^{x+2} \)?

  1. \( (100, 0) \)
  2. \( (0, 100) \)
  3. \( (10, 0) \)
  4. \( (0, 10) \)
View Step-by-Step Solution

Since \( f(0) = 10^{0+2} = 100 \), the point \( (0, 100) \) is on \( f \). The inverse function swaps the coordinates, so \( (100, 0) \) is on \( f^{-1} \).

Correct Answer: A

Question 25 — Solving for a Variable in an Exponential Equation

If \( 10^{x/y} = A/B \), then

  1. \( x/y = \log A / \log B \)
  2. \( y = x / (\log A - \log B) \)
  3. \( x = y / (\log A + \log B) \)
  4. \( y = x \log(A/B) \)
View Step-by-Step Solution

Take log base 10 of both sides:

\[ \frac{x}{y} = \log\left(\frac{A}{B}\right) = \log A - \log B \]

Solve for \( y \):

\[ y = \frac{x}{\log A - \log B} \]

Correct Answer: B

Question 26 — Expressing Sine in Terms of Tangent

If \( \pi \lt x \lt 3\pi/2 \), then \( \sin(x) \) in terms of \( \tan(x) \) is

  1. \( \sin(x) = \dfrac{\tan(x)}{\sqrt{1 + \tan^2(x)}} \)
  2. \( \sin(x) = -\dfrac{\sqrt{1 + \tan^2(x)}}{\tan(x)} \)
  3. \( \sin(x) = -\dfrac{\tan(x)}{\sqrt{1 + \tan^2(x)}} \)
  4. \( \sin(x) = \dfrac{\sqrt{1 + \tan^2(x)}}{\tan(x)} \)
View Step-by-Step Solution

Using the identity \( 1 + \tan^2(x) = \sec^2(x) \), we have \( \sec(x) = \pm\sqrt{1 + \tan^2(x)} \). In Quadrant III, \( \sec(x) \lt 0 \), so:

\[ \sec(x) = -\sqrt{1 + \tan^2(x)} \implies \cos(x) = \frac{-1}{\sqrt{1 + \tan^2(x)}} \]

Since \( \sin(x) = \tan(x)\cos(x) \) and \( \sin(x) \lt 0 \) in Q3:

\[ \sin(x) = -\frac{\tan(x)}{\sqrt{1 + \tan^2(x)}} \]

Correct Answer: C

Question 27 — Coterminal Angles

Angle \( x = 11\pi/3 \) is coterminal to angle \( y \)

  1. \( y = \pi/3 \)
  2. \( y = \pi/6 \)
  3. \( y = 5\pi/3 \)
  4. \( y = 2\pi/3 \)
View Step-by-Step Solution

Subtract \( 2\pi = 6\pi/3 \):

\[ \frac{11\pi}{3} - \frac{6\pi}{3} = \frac{5\pi}{3} \]

Correct Answer: C

Question 28 — Solving sin(x) = −1/2 in Q4

If \( \dfrac{3\pi}{2} \lt x \lt 2\pi \) and \( \sin(x) = -\dfrac{1}{2} \), then \( x = \)

  1. \( \dfrac{13\pi}{6} \)
  2. \( \dfrac{\pi}{3} \)
  3. \( \dfrac{11\pi}{3} \)
  4. \( \dfrac{11\pi}{6} \)
View Step-by-Step Solution

In Quadrant IV, where sine is \( -1/2 \), the reference angle is \( \pi/6 \). Thus:

\[ x = 2\pi - \frac{\pi}{6} = \frac{12\pi}{6} - \frac{\pi}{6} = \frac{11\pi}{6} \]

Correct Answer: D

Question 29 — Exact Value of a Large-Angle Cosine

The exact value of \( \cos\left(\dfrac{127\pi}{3}\right) \) is

  1. \( \dfrac{1}{\sqrt{2}} \)
  2. \( \dfrac{1}{2} \)
  3. \( -\dfrac{1}{2} \)
  4. \( \dfrac{\sqrt{3}}{2} \)
View Step-by-Step Solution

Reduce the angle modulo \( 2\pi = 6\pi/3 \):

\[ \frac{127\pi}{3} = 42\pi + \frac{\pi}{3} \]

Since \( 42\pi \) is a multiple of \( 2\pi \), the angle is coterminal with \( \pi/3 \), so:

\[ \cos\left(\frac{127\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

Correct Answer: B

Question 30 — System of Logarithmic Equations

If \( \log(x-y) = 3 \) and \( \log(x+y) = 4 \), then \( x = \)

  1. \( 3.5 \)
  2. \( 11{,}000 \)
  3. \( 5{,}500 \)
  4. \( 10^{3.5} \)
View Step-by-Step Solution

Convert to exponential form:

\[ x - y = 10^3 = 1000 \] \[ x + y = 10^4 = 10000 \]

Add the two equations:

\[ 2x = 11000 \implies x = 5500 \]

(Then \( y = 4500 \).)

Correct Answer: C

Quick Answer Key

QAnsQAnsQAnsQAnsQAns
1A7D13A19B25B
2D8C14B20D26C
3B9A15D21A27C
4D10C16B22C28D
5C11A17D23D29B
6B12B18C24A30C

Frequently Asked Questions

If \( f(x) = \log(x) \), what is \( f^{-1}(x) \)?

The inverse of the common logarithm function \( f(x) = \log_{10}(x) \) is the exponential function \( f^{-1}(x) = 10^x \).

How do you find the reference angle of a negative angle?

First add \( 360^\circ \) repeatedly until the angle becomes positive (find a coterminal angle). Then compute the reference angle based on which quadrant the angle lies in: subtract from \( 180^\circ \) for Q2, subtract \( 180^\circ \) for Q3, or subtract from \( 360^\circ \) for Q4.

What is the period of \( \tan(bt) \)?

The period of \( \tan(bt) \) is \( \pi/|b| \), not \( 2\pi/|b| \) as for sine and cosine. This is because tangent has period \( \pi \) rather than \( 2\pi \).

How do you solve \( \cos^2(x) - 1.5\cos(x) = 1 \)?

Treat it as a quadratic in \( \cos(x) \): \( \cos^2(x) - 1.5\cos(x) - 1 = 0 \). Factor as \( (2\cos(x) + 1)(\cos(x) - 2) = 0 \). Since \( \cos(x) \) must be between \( -1 \) and \( 1 \), only \( \cos(x) = -0.5 \) is valid.

How do you find the point on the inverse function given a point on the original function?

If \( (a, b) \) is on the original function \( f \), then \( (b, a) \) is on the inverse function \( f^{-1} \), because the inverse swaps the roles of \( x \) and \( y \).

How do you solve \( \log(x-y) = 3 \) and \( \log(x+y) = 4 \) simultaneously?

Convert to exponential form: \( x - y = 10^3 = 1000 \) and \( x + y = 10^4 = 10000 \). Adding the two equations gives \( 2x = 11000 \), so \( x = 5500 \). Then \( y = 4500 \).

More Math Problems with Detailed Solutions