College Algebra Questions With Answers
Sample 2

A set of college algebra multiple choice questions with answers is presented. Each solution is hidden inside a collapsible dropdown so you can practice independently before reviewing the detailed step-by-step explanations.

Multiple Choice Questions

Question 1

How many x-intercepts does the graph of the function \( f(x) = x^3 + 4 \) have?

View Solution

To find the x-intercepts, set \( f(x) = 0 \):

\[ x^3 + 4 = 0 \implies x^3 = -4 \implies x = -\sqrt[3]{4} \]

Since there is only one real solution, the graph has exactly 1 x-intercept.

Correct Answer: B

Question 2

If \( \log_{b}(5) = c \), then \( b^{4c} = \)

View Solution

Using exponent rules and the definition of logarithms:

\[ b^{4c} = (b^c)^4 = 5^4 = 625 \]

Correct Answer: D

Question 3

How many points of intersection do the graphs of the functions \( f(x) = x^2 \) and \( g(x) = 2^x \) have?

View Solution

Testing points for \( x^2 = 2^x \):

  • At \( x = 2 \): \( 2^2 = 4 \) and \( 2^2 = 4 \) (Intersection 1).
  • At \( x = 4 \): \( 4^2 = 16 \) and \( 2^4 = 16 \) (Intersection 2).
  • At \( x = -0.766 \) (approx): negative intersection point.

Analyzing the curves shows there are 2 points of intersection in total for real numbers.

Correct Answer: B

Question 4

The range of the function \( f(x) = -x^2 + 2x - 5 \) is given by

View Solution

Find the vertex of the parabola where \( a = -1, b = 2, c = -5 \):

\[ x_v = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1 \]

Calculate the maximum value at the vertex:

\[ f(1) = -(1)^2 + 2(1) - 5 = -1 + 2 - 5 = -4 \]

Since the parabola opens downward (\( a < 0 \)), the range is \((-\infty, -4]\).

Correct Answer: C

Question 5

Which of these functions does not have an inverse?

View Solution

A function has an inverse if and only if it is one-to-one (passes the horizontal line test). \( f(x) = |x - 5| \) is V-shaped and fails the horizontal line test (e.g., \( f(4) = f(6) = 1 \)), so it is not one-to-one and lacks an inverse.

Correct Answer: A

Question 6

The solution set of the inequality \( (x + 5)^4(x + 3) \geq 0 \) is given by the interval

View Solution

Analyze the factors:

  • \((x + 5)^4 \ge 0\) for all real numbers \( x \) (always non-negative since the exponent is even). At \( x = -5 \), it equals zero.
  • \( x + 3 \ge 0 \implies x \ge -3 \) for the product to be positive or zero.

Thus, the inequality holds when \( x \ge -3 \) as well as at the isolated point \( x = -5 \). The solution set is \(\{-5\} \cup [-3, \infty)\).

Correct Answer: B

Question 7

The range of the function \( h(x) = -|x - 5| - 2 \) is given by

View Solution

Since the absolute value \( |x - 5| \geq 0 \), multiplying by \(-1\) gives \( -|x - 5| \leq 0 \). Subtracting 2 yields:

\[ -|x - 5| - 2 \leq -2 \]

Thus, the range is \((-\infty, -2]\).

Correct Answer: C

Question 8

\( 5 \ln e^{2} - 2 \ln (7e) + \ln 49 = \)

View Solution

Simplify each logarithmic term:

\[ 5 \ln(e^2) = 5(2) = 10 \] \[ -2 \ln(7e) = -2(\ln 7 + \ln e) = -2\ln 7 - 2 \] \[ \ln 49 = \ln(7^2) = 2\ln 7 \]

Summing them together:

\[ 10 - 2\ln 7 - 2 + 2\ln 7 = 8 \]

Correct Answer: A

Question 9

If \( f(x) = \dfrac{x}{-x + 4} \) and \( g(x) = 2x + 2 \), then \( (f \circ g)(1) = \)

View Solution

Evaluate inner function \( g(1) \):

\[ g(1) = 2(1) + 2 = 4 \]

Now evaluate \( f(g(1)) = f(4) \):

\[ f(4) = \frac{4}{-4 + 4} = \frac{4}{0} \quad \text{(division by zero is undefined)} \]

Correct Answer: D

Question 10

\( \sqrt{x^{2} + 6x + 9} = \)

View Solution

Factor the quadratic expression inside the square root:

\[ \sqrt{x^2 + 6x + 9} = \sqrt{(x + 3)^2} = |x + 3| \]

Correct Answer: C

Question 11

If \( f(x) = x^2 + 1 \) for \( x \leq 0 \), then \( f^{-1}(x) = \)

View Solution

Set \( y = x^2 + 1 \) and solve for \( x \):

\[ x^2 = y - 1 \implies x = \pm\sqrt{y - 1} \]

Since the domain of \( f \) is restricted to \( x \leq 0 \), we choose the negative branch: \( x = -\sqrt{y - 1} \). Replacing \( y \) with \( x \) gives \( f^{-1}(x) = -\sqrt{x - 1} \).

Correct Answer: D

Question 12

The domain of \( f(x) = \ln(2 - x) + \ln(3 - x) \) is given by

View Solution

Arguments of logarithms must be strictly positive:

  • \( 2 - x > 0 \implies x < 2 \)
  • \( 3 - x > 0 \implies x < 3 \)

The intersection of these conditions is \( x < 2 \), which corresponds to interval \((-\infty, 2)\).

Correct Answer: A

Question 13

The solution set of the inequality \( (x - 1)^2 \leq 0 \) is given by

View Solution

A squared real number is never negative (\((x - 1)^2 \geq 0\)). Therefore, \((x - 1)^2 \leq 0\) can only be true when \((x - 1)^2 = 0 \implies x = 1\).

Correct Answer: C

Question 14

The solution to the equation \( 2(4^{x+1}) = 4(8^{x+2}) \)

View Solution

Express both sides with base 2:

\[ 2^1 \cdot (2^2)^{x+1} = 2^2 \cdot (2^3)^{x+2} \] \[ 2^{1 + 2x + 2} = 2^{2 + 3x + 6} \implies 2^{2x + 3} = 2^{3x + 8} \]

Equating exponents:

\[ 2x + 3 = 3x + 8 \implies x = -5 \]

Correct Answer: A

Question 15

The number of solutions of the equation \( |-x^2 + 4x + 1| = 5 \) is equal to

View Solution

Split into two equations:

  • \( -x^2 + 4x + 1 = 5 \implies -x^2 + 4x - 4 = 0 \implies (x - 2)^2 = 0 \implies x = 2 \) (1 solution)
  • \( -x^2 + 4x + 1 = -5 \implies -x^2 + 4x + 6 = 0 \implies x^2 - 4x - 6 = 0 \) (Discriminant \( \Delta = 16 - 4(-6) = 40 > 0 \), giving 2 distinct solutions)

Total real solutions = \( 1 + 2 = 3 \).

Correct Answer: C

Question 16

Using the absolute value concept, the statement "The distance from \( x \) to \(-6\) is not less than 7" may be written as follows:

View Solution

The distance between \( x \) and \(-6\) is given by \( |x - (-6)| = |x + 6| \). "Not less than" means greater than or equal to (\(\geq\)). Thus: \( |x + 6| \geq 7 \).

Correct Answer: A

Question 17

If \( f(2x + 3) = 2x - 1 \), then \( f(3x + 4) = \)

View Solution

Let \( u = 2x + 3 \), so \( x = \frac{u - 3}{2} \). Substitute into \( f \):

\[ f(u) = 2\left(\frac{u - 3}{2}\right) - 1 = u - 3 - 1 = u - 4 \]

Thus, \( f(x) = x - 4 \). Evaluating for \( f(3x + 4) \):

\[ f(3x + 4) = (3x + 4) - 4 = 3x \]

Correct Answer: B

Question 18

\( -x^2 + x - 3 = \)

View Solution

Complete the square:

\[ -x^2 + x - 3 = -(x^2 - x) - 3 = -\left(x^2 - x + \frac{1}{4} - \frac{1}{4}\right) - 3 \] \[ = -\left(x - \frac{1}{2}\right)^2 + \frac{1}{4} - 3 = -\left(x - \frac{1}{2}\right)^2 - \frac{11}{4} \]

Correct Answer: C

Question 19

How many solutions does the equation \( \dfrac{1}{x} + \dfrac{1}{x + 1} = \dfrac{1}{3} \) have?

View Solution

Clear denominators by multiplying by \( 3x(x + 1) \):

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

The discriminant is \( \Delta = (-5)^2 - 4(1)(-3) = 25 + 12 = 37 > 0 \), meaning there are 2 distinct real solutions.

Correct Answer: C

Question 20

If the whole graph of function \( f \) lies in quadrant IV, then the graph of the inverse of \( f \) lies in quadrant

View Solution

In quadrant IV, \( x > 0 \) and \( y < 0 \). For an inverse function \( f^{-1} \), the input and output coordinates are swapped \((y, x)\), so the coordinates become \( x < 0 \) and \( y > 0 \), which defines quadrant II.

Correct Answer: B (Note: matching index option D corresponds to IV in the raw list text, but mathematically the inverse of quadrant IV \((+,-)\) is quadrant II \((-,+)\). Let's follow standard coordinate reflection across \(y=x\): \((x, y) \to (y, x)\), so \((+,-) \to (-,+)\) which is Quadrant II).

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