Advanced Math Domain Test 1

Calculator is allowed on all questions. Figures are not necessarily drawn to scale.

Question 1

Which of the following is equivalent to $x^3 \cdot x^{-5}$?

A) $x^{-15}$
B) $x^{-2}$
C) $x^{2}$
D) $x^{8}$

Show solution

When multiplying powers with the same base, add the exponents:

\[x^3 \cdot x^{-5} = x^{3 + (-5)} = x^{-2}\]

1. $x^{-15}$: multiplied the exponents instead of adding ($3 \times (-5) = -15$)
1. $x^2$: subtracted in the wrong order ($5 - 3 = 2$) or dropped the negative sign
1. $x^8$: subtracted incorrectly ($3 - (-5) = 8$); this would apply to division, not multiplication

Answer: B

Question 2

Which of the following is equivalent to $(2x + 3)(x - 5)$?

A) $2x^2 - 7x - 15$
B) $2x^2 + 7x - 15$
C) $2x^2 - 7x + 15$
D) $2x^2 - 10x - 15$

Show solution

Using FOIL:

\[\text{First: } 2x \cdot x = 2x^2\] \[\text{Outer: } 2x \cdot (-5) = -10x\] \[\text{Inner: } 3 \cdot x = 3x\] \[\text{Last: } 3 \cdot (-5) = -15\]

\[2x^2 - 10x + 3x - 15 = 2x^2 - 7x - 15\]

1. $2x^2 + 7x - 15$: combined outer and inner as $-10x + 3x = +7x$ (sign error — added absolute values)
1. $2x^2 - 7x + 15$: correct $x$-terms but wrong sign on the last term ($-15$ became $+15$)
1. $2x^2 - 10x - 15$: forgot to include the inner term $3x$ entirely

Answer: A

Question 3

The function $f$ is defined by $f(x) = 3x^2 - 2$. What is the value of $f(-3)$?

A) $-29$
B) $7$
C) $25$
D) $79$

Show solution

\[f(-3) = 3(-3)^2 - 2 = 3(9) - 2 = 27 - 2 = 25\]

1. $-29$: computed $3(-3)^2$ as $3(-9) = -27$ (did not square before applying the negative), then $-27 - 2 = -29$
1. $7$: used $(-3)^2 = 3$ instead of $9$ (square root error), giving $3(3) - 2 = 7$
1. $79$: computed $3(-3)^3 = 3(-27)$… or $3(27) - 2 = 81 - 2 = 79$ (used $(-3)^3$ instead of $(-3)^2$)

Answer: C

Question 4

If $\sqrt{3x + 1} = 7$, what is the value of $x$?

Enter your answer:

Show solution

Square both sides:

\[3x + 1 = 49 \Rightarrow 3x = 48 \Rightarrow x = 16\]

Check: $\sqrt{3(16) + 1} = \sqrt{49} = 7$ ✓

Answer: 16

Question 5

Which of the following is equivalent to $\dfrac{x^2 - 5x + 6}{x^2 - 4}$ for $x \neq 2$ and $x \neq -2$?

A) $\dfrac{x - 3}{x + 2}$
B) $\dfrac{x + 3}{x - 2}$
C) $\dfrac{x - 3}{x - 2}$
D) $\dfrac{x - 2}{x + 2}$

Show solution

Factor numerator and denominator:

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

Cancel the common factor $(x - 2)$:

\[\frac{(x-2)(x-3)}{(x-2)(x+2)} = \frac{x - 3}{x + 2}\]

1. $\dfrac{x+3}{x-2}$: factored the numerator with wrong signs: $(x+3)(x-2)$ gives $x^2 + x - 6 \neq x^2 - 5x + 6$
1. $\dfrac{x-3}{x-2}$: canceled $(x+2)$ from numerator and denominator instead of $(x-2)$
1. $\dfrac{x-2}{x+2}$: canceled $(x-3)$ from numerator and left $(x-2)$ — didn’t factor correctly

Answer: A

Question 6

A population of bacteria doubles every 3 hours. If there are initially 500 bacteria, which of the following expressions gives the number of bacteria after $t$ hours?

A) $500 \cdot 2^t$
B) $500 \cdot 2^{t/3}$
C) $500 \cdot 3^{t/2}$
D) $500 + 2t$

Show solution

The population doubles every 3 hours, so the number of doubling periods in $t$ hours is $\dfrac{t}{3}$.

Each doubling multiplies the population by 2:

\[P(t) = 500 \cdot 2^{t/3}\]

Check: at $t = 3$, $P = 500 \cdot 2^1 = 1000$ (doubled) ✓. At $t = 6$, $P = 500 \cdot 4 = 2000$ ✓.

1. $500 \cdot 2^t$: doubles every hour rather than every 3 hours
1. $500 \cdot 3^{t/2}$: swapped the base (3) and the doubling period (2)
1. $500 + 2t$: models linear growth, not exponential doubling

Answer: B

Question 7

Which of the following is equivalent to $(3x - 4)^2$?

A) $9x^2 - 16$
B) $9x^2 + 16$
C) $9x^2 - 24x + 16$
D) $9x^2 - 12x + 16$

Show solution

Using the perfect square formula $(a - b)^2 = a^2 - 2ab + b^2$ with $a = 3x$ and $b = 4$:

\[(3x - 4)^2 = (3x)^2 - 2(3x)(4) + 4^2 = 9x^2 - 24x + 16\]

1. $9x^2 - 16$: applied the difference of squares pattern $(a-b)(a+b)$ instead of squaring — forgot the middle term
1. $9x^2 + 16$: same error as A but also lost the negative sign
1. $9x^2 - 12x + 16$: computed the middle term as $-2(3)(4)x = -12x$ instead of $-2(3x)(4) = -24x$ (forgot the $x$ in $3x$ when computing $2ab$)

Answer: C

Question 8

If $f(x) = 2x + 1$ and $g(x) = x^2 - 3$, what is the value of $g(f(2))$?

Enter your answer:

Show solution

First evaluate the inner function: $f(2) = 2(2) + 1 = 5$.

Then evaluate $g$ at that result:

\[g(5) = 5^2 - 3 = 25 - 3 = 22\]

Answer: 22

Question 9

Which of the following is equivalent to $\sqrt{50x^3}$, where $x > 0$?

A) $5x\sqrt{2x}$
B) $5x^2\sqrt{2}$
C) $25x\sqrt{2x}$
D) $5\sqrt{2x^3}$

Show solution

Factor out perfect squares:

\[\sqrt{50x^3} = \sqrt{25 \cdot 2 \cdot x^2 \cdot x} = \sqrt{25} \cdot \sqrt{x^2} \cdot \sqrt{2x} = 5 \cdot x \cdot \sqrt{2x} = 5x\sqrt{2x}\]

1. $5x^2\sqrt{2}$: pulled out $x^2$ instead of $x$ from inside the radical (treated $x^3$ as $(x^2)(x^1)$ but then kept both $x$’s outside)
1. $25x\sqrt{2x}$: used $\sqrt{25} = 25$ instead of $5$
1. $5\sqrt{2x^3}$: only extracted $\sqrt{25} = 5$ and left $x^3$ inside, not simplifying the $x^2$ term

Answer: A

Question 10

The function $f$ is defined by $f(x) = x^2 - 4x + 1$. Which of the following is an equivalent form that reveals the minimum value of $f$?

A) $f(x) = (x - 4)^2 - 15$
B) $f(x) = (x - 2)^2 - 3$
C) $f(x) = (x + 2)^2 - 3$
D) $f(x) = (x - 2)^2 + 3$

Show solution

Complete the square:

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

The minimum value is $-3$, occurring at $x = 2$.

Verify by expanding: $(x-2)^2 - 3 = x^2 - 4x + 4 - 3 = x^2 - 4x + 1$ ✓

1. $(x-4)^2 - 15$: used $\left(\frac{-4}{1}\right)^2 = 16$ instead of $\left(\frac{-4}{2}\right)^2 = 4$ for the completing-the-square step
1. $(x+2)^2 - 3$: wrong sign on $h$ — used $+2$ instead of $-2$ in the vertex form
1. $(x-2)^2 + 3$: correct vertex $x$-coordinate but wrong sign on the constant — added $4$ instead of subtracting it

Answer: B

Question 11

The graph of $y = f(x)$ is shown below. Which of the following gives the coordinates of the vertex of $y = f(x - 2) + 1$?

A) $(-1, 1)$
B) $(1, 3)$
C) $(3, 1)$
D) $(3, -1)$

Show solution

The original vertex is at $(1, 0)$.

$f(x - 2)$ shifts the graph 2 units right: $x$-coordinate $= 1 + 2 = 3$.

$+1$ shifts the graph 1 unit up: $y$-coordinate $= 0 + 1 = 1$.

New vertex: $(3, 1)$.

1. $(-1, 1)$: shifted 2 units left instead of right ($1 - 2 = -1$), shifted up correctly
1. $(1, 3)$: did not shift horizontally; shifted $y$ up by $3$ instead of $1$ (or confused the $-2$ as a vertical shift)
1. $(3, -1)$: correct horizontal shift but shifted $y$ down by $1$ instead of up

Answer: C

Question 12

The expression $\dfrac{2}{x+1} + \dfrac{3}{x-1}$ is equivalent to $\dfrac{ax + b}{(x+1)(x-1)}$ for constants $a$ and $b$. What is the value of $a + b$?

Enter your answer:

Show solution

Use the common denominator $(x+1)(x-1)$:

\[\frac{2}{x+1} + \frac{3}{x-1} = \frac{2(x-1)}{(x+1)(x-1)} + \frac{3(x+1)}{(x+1)(x-1)}\]

\[= \frac{2x - 2 + 3x + 3}{(x+1)(x-1)} = \frac{5x + 1}{(x+1)(x-1)}\]

So $a = 5$ and $b = 1$, giving $a + b = 5 + 1 = 6$.

Answer: 6

Question 13

Which of the following are the solutions to $2x^2 + 5x - 3 = 0$?

A) $x = \dfrac{1}{2}$ and $x = -3$
B) $x = -\dfrac{1}{2}$ and $x = 3$
C) $x = 1$ and $x = -\dfrac{3}{2}$
D) $x = \dfrac{3}{2}$ and $x = -1$

Show solution

Factor $2x^2 + 5x - 3$: find two numbers that multiply to $2 \times (-3) = -6$ and sum to $5$. Those are $6$ and $-1$.

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

Setting each factor to zero: $x = \dfrac{1}{2}$ or $x = -3$.

Distractor check: verify all choices. - A) $x = \frac{1}{2}$: $2(\frac{1}{4}) + \frac{5}{2} - 3 = \frac{1}{2} + \frac{5}{2} - 3 = 3 - 3 = 0$ ✓. $x = -3$: $2(9) - 15 - 3 = 18 - 18 = 0$ ✓ - B) $x = -\frac{1}{2}$: $2(\frac{1}{4}) - \frac{5}{2} - 3 = \frac{1}{2} - \frac{5}{2} - 3 = -2 - 3 = -5 \neq 0$ - C) $x = 1$: $2 + 5 - 3 = 4 \neq 0$ - D) $x = \frac{3}{2}$: $2(\frac{9}{4}) + \frac{15}{2} - 3 = \frac{9}{2} + \frac{15}{2} - 3 = 12 - 3 = 9 \neq 0$

1. Negated the fraction root: mixed up signs from $(2x+1)(x-3) = 0$
1. Factored incorrectly as $(x-1)(2x+3) = 0$
1. Factored incorrectly as $(2x-3)(x+1) = 0$

Answer: A

Question 14

A car purchased for $\$24{,}000$ depreciates at a rate of $15\%$ per year. Which of the following expressions gives the value of the car after $t$ years?

A) $24{,}000(0.15)^t$
B) $24{,}000 - 0.15t$
C) $24{,}000(0.85)^t$
D) $24{,}000(1.15)^t$

Show solution

A $15\%$ annual depreciation means the car retains $100\% - 15\% = 85\% = 0.85$ of its value each year.

The exponential decay model is:

\[V(t) = 24{,}000 \cdot (0.85)^t\]

1. $24{,}000(0.15)^t$: used the decay rate itself as the base — the car would lose nearly all its value in year 1
1. $24{,}000 - 0.15t$: linear model, subtracting a tiny fixed amount per year rather than a percentage
1. $24{,}000(1.15)^t$: models $15\%$ annual growth, not depreciation

Answer: C

Question 15

The graph of $y = ax^2 + bx + c$ opens downward and has its vertex in the second quadrant. Which of the following must be true?

A) $a > 0$ and $b > 0$
B) $a < 0$ and $b < 0$
C) $a < 0$ and $b > 0$
D) $a > 0$ and $b < 0$

Show solution

Opens downward requires $a < 0$. This eliminates choices A and D.

Vertex in the second quadrant means the $x$-coordinate of the vertex is negative. The axis of symmetry is $x = -\dfrac{b}{2a}$, so:

\[-\frac{b}{2a} < 0\]

Multiplying both sides by $2a$ (which is negative, so the inequality flips):

\[-b > 0 \Rightarrow b < 0\]

So both $a < 0$ and $b < 0$ must be true.

Verify with an example: vertex at $(-2, 3)$, opens downward gives $y = -(x+2)^2 + 3 = -x^2 - 4x - 1$. Here $a = -1 < 0$, $b = -4 < 0$ ✓, and axis $x = -\frac{-4}{2(-1)} = -2 < 0$ ✓.

1. $a > 0$: line opens upward, not downward
1. $b > 0$: this would place the axis of symmetry at $x = -b/(2a) > 0$ (positive), putting the vertex in Q1 or Q4
1. $a > 0$: same error as A

Answer: B

Question 16

What is the domain of the function $f(x) = \dfrac{x + 3}{x^2 - x - 12}$?

A) All real numbers except $x = -3$
B) All real numbers except $x = 4$
C) All real numbers except $x = -3$ and $x = 4$
D) All real numbers except $x = 3$ and $x = -4$

Show solution

The domain excludes all values where the denominator equals zero. Factor the denominator:

\[x^2 - x - 12 = (x - 4)(x + 3)\]

The denominator equals zero at $x = 4$ and $x = -3$. Both values are excluded from the domain.

Domain: all real numbers except $x = -3$ and $x = 4$.

1. Excludes only $x = -3$: misses $x = 4$
1. Excludes only $x = 4$: $x = -3$ also makes the denominator zero and must be excluded
1. Incorrect factoring — denominator zeros are at $x = 4$ and $x = -3$, not $x = 3$ and $x = -4$

Answer: C

Question 17

The polynomial $p(x) = x^3 - 7x + 6$ has three real roots. Given that $p(1) = 0$, what is the sum of the other two roots?

Enter your answer:

Show solution

Since $p(1) = 0$, $(x - 1)$ is a factor. Perform polynomial division:

\[x^3 - 7x + 6 \div (x - 1)\]

Using synthetic division with root $1$:

\[\begin{array}{r|rrrr} 1 & 1 & 0 & -7 & 6 \\ & & 1 & 1 & -6 \\ \hline & 1 & 1 & -6 & 0 \end{array}\]

Quotient: $x^2 + x - 6 = (x + 3)(x - 2)$.

The other two roots are $x = -3$ and $x = 2$.

Sum $= -3 + 2 = -1$.

By Vieta’s formulas for the full cubic $x^3 + 0x^2 - 7x + 6$: sum of all three roots $= -\dfrac{0}{1} = 0$. So sum of other two roots $= 0 - 1 = -1$ ✓.

Answer: -1

Question 18

Which of the following is equivalent to $\dfrac{1}{x - 2} - \dfrac{1}{x + 2}$?

A) $\dfrac{2}{x^2 - 4}$
B) $\dfrac{4}{x^2 - 4}$
C) $\dfrac{-4}{x^2 - 4}$
D) $\dfrac{2x}{x^2 - 4}$

Show solution

Use common denominator $(x-2)(x+2) = x^2 - 4$:

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

1. $\dfrac{2}{x^2-4}$: correctly found common denominator but computed numerator as $(x+2) - (x+2) = 0$… or added $2 + 0 = 2$ rather than $2 + 2 = 4$
1. $\dfrac{-4}{x^2-4}$: negated the entire numerator — subtracted in the wrong order: $(x-2)-(x+2) = -4$
1. $\dfrac{2x}{x^2-4}$: combined $x + 2 + x - 2 = 2x$ (added instead of subtracted)

Answer: B

Question 19

The function $f$ is defined by $f(x) = \dfrac{2x - 1}{3}$. Which of the following defines $f^{-1}(x)$?

A) $f^{-1}(x) = \dfrac{3x - 1}{2}$
B) $f^{-1}(x) = \dfrac{3x + 1}{2}$
C) $f^{-1}(x) = \dfrac{2x + 1}{3}$
D) $f^{-1}(x) = \dfrac{3}{2x - 1}$

Show solution

Swap $x$ and $y$, then solve for $y$. Starting with $y = \dfrac{2x-1}{3}$:

\[x = \frac{2y - 1}{3} \Rightarrow 3x = 2y - 1 \Rightarrow 2y = 3x + 1 \Rightarrow y = \frac{3x + 1}{2}\]

Verify: $f(f^{-1}(x)) = f\!\left(\dfrac{3x+1}{2}\right) = \dfrac{2 \cdot \frac{3x+1}{2} - 1}{3} = \dfrac{3x+1-1}{3} = \dfrac{3x}{3} = x$ ✓

1. $\dfrac{3x-1}{2}$: correctly multiplied both sides by 3 and divided by 2, but subtracted 1 instead of adding it
1. $\dfrac{2x+1}{3}$: added 1 to the numerator without swapping $x$ and $y$ (just changed the sign on the constant)
1. $\dfrac{3}{2x-1}$: took the reciprocal of $f(x)$ rather than finding the inverse function

Answer: B

Question 20

The equation $x^2 + (k+2)x + 2k = 0$ has two equal real roots. What is the value of $k$?

Enter your answer:

Show solution

For two equal (repeated) real roots, the discriminant must equal zero:

\[\Delta = b^2 - 4ac = 0\]

With $a = 1$, $b = k+2$, $c = 2k$:

\[(k+2)^2 - 4(1)(2k) = 0\] \[k^2 + 4k + 4 - 8k = 0\] \[k^2 - 4k + 4 = 0\] \[(k - 2)^2 = 0\] \[k = 2\]

Verify: with $k = 2$: $x^2 + 4x + 4 = (x+2)^2 = 0$ → repeated root $x = -2$ ✓

Answer: 2