SAT Advanced Math Domain Test

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Question 1

What is the value of \(\dfrac{5^8}{5^3 \cdot 5^2}\)?




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Combine the denominator: \(5^3 \cdot 5^2 = 5^5\). Then \(\dfrac{5^8}{5^5} = 5^{8-5} = 5^3 = 125\).

    1. 25: Mistakenly treated \(5^3 \cdot 5^2\) as \(5^{3\times 2}=5^6\) instead of \(5^{3+2}=5^5\), giving \(5^{8-6}=5^2\).
    1. 625: Correctly combined the denominator to \(5^5\), but subtracted incorrectly as \(8-5+1=4\) instead of \(8-5=3\).
    1. 3125: Forgot to combine the denominator, subtracting only the first exponent: \(5^{8-3}\).

Answer: B

Question 2

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

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Substitute \(x=-2\): \(f(-2) = 3(-2)^2 - 2(-2) + 1 = 3(4) + 4 + 1 = 12 + 4 + 1 = 17\).

Answer: 17

Question 3

The function \(h\) is defined by \(h(x) = \sqrt{3x - 9}\). What is the domain of \(h\)?




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The radicand must be non-negative: \(3x - 9 \geq 0\), so \(3x \geq 9\), and \(x \geq 3\).

    1. \(x \leq -3\): Made a sign error, solving \(3x + 9 \geq 0\) instead of \(3x - 9 \geq 0\), then flipped the inequality direction.
    1. \(x \geq -3\): Made a sign error, solving \(3x + 9 \geq 0\) instead of \(3x - 9 \geq 0\).
    1. \(x \leq 3\): Solved the inequality correctly for the boundary but flipped the direction.

Answer: D

Question 4

What are the solutions to \(x^2 + 5x + 6 = 0\)?




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Factor: \(x^2+5x+6=(x+2)(x+3)=0\), so \(x=-2\) or \(x=-3\).

    1. \(x = 2\) and \(x = 3\): Flipped the signs of both roots.
    1. \(x = -2\) and \(x = 3\): Correctly found \(-2\) but dropped the negative sign on the second root.
    1. \(x = -1\) and \(x = -6\): Chose a factor pair of 6 (1 and 6) without checking that the pair sums to 5.

Answer: A

Question 5

If \((x - 4)(x + 7) = 0\), what are the solutions for \(x\)?




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By the Zero Product Property: \(x-4=0\) or \(x+7=0\), so \(x=4\) or \(x=-7\).

    1. \(x = -4\) and \(x = 7\): Flipped the signs of both roots.
    1. \(x = 4\) and \(x = 7\): Correctly found \(4\) but dropped the negative sign on the second root.
    1. \(x = -4\) and \(x = -7\): Correctly found \(-7\) but dropped the sign on the first root, using \(-4\) instead of \(4\).

Answer: C

Question 6

If \(p(x) = 3x^2 + 2x - 5\) and \(q(x) = x^2 - 4x + 1\), what is the value of \((p + q)(2)\)?

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\((p+q)(x) = (3x^2+2x-5)+(x^2-4x+1) = 4x^2 - 2x - 4\). Evaluate at \(x=2\): \(4(4) - 2(2) - 4 = 16 - 4 - 4 = 8\).

Answer: 8

Question 7

Which of the following is equivalent to \(\dfrac{x^2 - 16}{x + 4}\) for \(x \neq -4\)?




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Factor: \(\dfrac{(x-4)(x+4)}{x+4}\). Cancel the common factor \((x+4)\): \(x-4\).

    1. \(x + 4\): Cancelled the wrong factor, leaving the surviving term mislabeled as \(x+4\) instead of \(x-4\).
    1. \(4 - x\): Sign flip on the final simplified answer.
    1. \((x-4)(x+4)\): Factored the numerator correctly but forgot to cancel the common factor with the denominator.

Answer: B

Question 8

A population of insects is modeled by \(P(t) = 300 \cdot (1.08)^t\), where \(t\) is the number of weeks and \(P(t)\) is the population size. What does the value \(1.08\) represent in this context?




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In a model \(P(t) = a \cdot b^t\), the base \(b\) is the growth factor. Since \(b = 1.08 = 1 + 0.08\), the population increases by \(8\%\) each week.

    1. The population increases by \(1.08\%\) each week.: Misread the growth factor \(1.08\) directly as the percent increase, rather than computing \(1.08-1=0.08\).
    1. The population increases by 1.08 insects each week.: Treated the growth factor as an additive amount rather than a multiplicative factor.
    1. The initial population was 1.08.: Confused the growth factor with the initial population, which is actually 300.

Answer: D

Question 9

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




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FOIL: \((2x-3)(x+5) = 2x^2 + 10x - 3x - 15 = 2x^2 + 7x - 15\).

    1. \(2x^2 - 7x - 15\): Combined the middle terms with a sign error, computing \(10x-3x\) as \(-7x\) instead of \(7x\).
    1. \(2x^2 + 2x - 15\): Arithmetic slip combining the middle terms, computing \(10x-3x\) as \(2x\) instead of \(7x\).
    1. \(2x^2 + 7x + 15\): Sign error on the constant term, computing \((-3)(5)\) as \(+15\) instead of \(-15\).

Answer: A

Question 10

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




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

    1. \(9x^2 - 16\): Forgot the middle term entirely, treating the expression as a difference of squares pattern instead of a binomial square.
    1. \(9x^2 + 16\): Forgot the middle term and also dropped the sign, squaring each term independently.
    1. \(9x^2 - 12x + 16\): Arithmetic slip computing the middle term, using \(2(3x)(4)=12x\) instead of \(24x\) (forgot the factor of 2).

Answer: C

Question 11

If \(p(x) = x^3 - 4x^2 + x + 9\), what is the remainder when \(p(x)\) is divided by \((x - 3)\)?

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By the Remainder Theorem, the remainder equals \(p(3)\): \(p(3) = 3^3 - 4(3)^2 + 3 + 9 = 27 - 36 + 3 + 9 = 3\).

Answer: 3

Question 12

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




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First find \(g(1) = 2(1)+3 = 5\). Then \(f(g(1)) = f(5) = 5^2 - 1 = 24\).

    1. 3: Reversed the order of composition, computing \(g(f(1))\) instead: \(f(1)=0\), then \(g(0)=3\).
    1. 25: Forgot the \(-1\) in \(f\), computing \(f(5)=5^2=25\) instead of \(5^2-1\).
    1. 26: Sign error in \(f\), treating \(f(x)=x^2+1\) instead of \(x^2-1\), giving \(f(5)=26\).

Answer: B

Question 13

The graph of \(f(x) = x^2\) is shifted left 3 units and down 2 units to form \(g(x)\). Which equation defines \(g(x)\)?




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A horizontal shift left 3 replaces \(x\) with \(x+3\): \((x+3)^2\). A vertical shift down 2 subtracts 2 from the result: \(g(x) = (x+3)^2 - 2\).

    1. \(g(x) = (x-3)^2 + 2\): Sign errors on both shifts — used right instead of left, and up instead of down.
    1. \(g(x) = (x+3)^2 + 2\): Correctly applied the horizontal shift but used up instead of down for the vertical shift.
    1. \(g(x) = (x-3)^2 - 2\): Correctly applied the vertical shift but used right instead of left for the horizontal shift.

Answer: D

Question 14

Which of the following is equivalent to \(x^2 + 8x + 10\)?




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Half of 8 is 4, and \(4^2=16\). \(x^2+8x+10 = (x^2+8x+16-16)+10 = (x+4)^2 - 16 + 10 = (x+4)^2 - 6\).

    1. \((x+4)^2 + 10\): Correctly found the squared term but forgot to update the constant, leaving the original \(+10\).
    1. \((x+8)^2 - 6\): Used the coefficient of \(x\) (8) directly instead of halving it to get 4.
    1. \((x+4)^2 + 6\): Correctly found the squared term but made a sign error on the final constant.

Answer: A

Question 15

The equation \(2x^2 - 8x + k = 0\) has exactly one real solution. What is the value of \(k\)?

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A quadratic has exactly one real solution when its discriminant equals zero: \((-8)^2 - 4(2)(k) = 0\), so \(64 - 8k = 0\), and \(k = 8\).

Answer: 8

Question 16

What is the value of \(x\) that satisfies \(\dfrac{3}{x - 2} = \dfrac{1}{4}\), given that \(x \neq 2\)?




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Cross-multiply: \(3(4) = 1(x-2)\), so \(12 = x - 2\), and \(x = 14\).

    1. 2.75: Inverted the equation, solving \(\dfrac{x-2}{3} = \dfrac{1}{4}\) instead of \(\dfrac{3}{x-2}=\dfrac{1}{4}\).
    1. 10: Sign error, cross-multiplying as \(12 = x + 2\) instead of \(12 = x - 2\).
    1. 18: Arithmetic slip solving \(12 = x - 2\), adding incorrectly to get \(x=18\) instead of \(x=14\).

Answer: C

Question 17

If \(p(x) = 2x^3 + 3x^2 - 5x + 7\), what is the remainder when \(p(x)\) is divided by \((x + 2)\)?

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By the Remainder Theorem, the remainder equals \(p(-2)\): \(p(-2) = 2(-2)^3 + 3(-2)^2 - 5(-2) + 7 = -16 + 12 + 10 + 7 = 13\).

Answer: 13

Question 18

If \(f(x) = 3x - 7\), what is \(f^{-1}(x)\)?




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Set \(y = 3x - 7\) and solve for \(x\): \(y + 7 = 3x\), so \(x = \dfrac{y+7}{3}\). Swapping variables: \(f^{-1}(x) = \dfrac{x+7}{3}\).

    1. \(\dfrac{x-7}{3}\): Sign error, keeping \(-7\) instead of \(+7\) when isolating \(x\).
    1. \(3x + 7\): Did not actually solve for the inverse — just negated the constant term in the original function without dividing.
    1. \(3(x+7)\): Multiplied by 3 instead of dividing by 3 when isolating \(x\).

Answer: B

Question 19

What is the value of \(x\) that satisfies \(\sqrt{2x + 3} = x\)?




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Square both sides: \(2x+3 = x^2\), so \(x^2-2x-3=0\), giving \((x-3)(x+1)=0\), so \(x=3\) or \(x=-1\). Checking \(x=-1\) in the original equation: \(\sqrt{2(-1)+3}=\sqrt{1}=1 \neq -1\), so \(x=-1\) is extraneous. The only valid solution is \(x=3\).

    1. -3: Did not square the equation, incorrectly treating \(\sqrt{2x+3}=x\) as though the radical simply cancels, solving \(2x+3=x\).
    1. -1: Reported the extraneous root without checking it against the original equation.
    1. -1 or 3: Solved the resulting quadratic correctly but reported both roots without eliminating the extraneous one.

Answer: D

Question 20

If \(f(x) = x^2 - 3x\) and \(g(x) = 2x + 1\), what is the positive value of \(x\) for which \(f(g(x)) = 0\)?




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Substitute \(g(x)=2x+1\) into \(f\): \(f(g(x)) = (2x+1)^2 - 3(2x+1)\). Let \(u=2x+1\): this is \(u^2-3u = u(u-3)\), so \(u=0\) or \(u=3\). Since \(u=2x+1\): \(2x+1=0\) gives \(x=-0.5\), and \(2x+1=3\) gives \(x=1\). The positive value is \(x=1\).

    1. -0.5: Reported the non-positive solution instead of the positive one.
    1. 1.78: Arithmetic slip expanding \((2x+1)^2\), forgetting the middle term and using \(4x^2+1\) instead of \(4x^2+4x+1\).
    1. 3: Solved \(f(x)=0\) directly instead of \(f(g(x))=0\), reporting the positive root of the wrong equation.

Answer: B